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Chapter 9 · 6 hours

Noise in Communication Systems

IOE past exam questions

Past questions and answers

59 questions set from this chapter, 13 of them more than once. Most asked first.

  • Asked 4 times
  • 2071 Magh (old course) · 4 marks
  • 2068 Jestha (old course) · 5 marks
  • 2080 Bhadra (CS II) · 4 marks
  • 2075 Chaitra (CS II) · 5 marks

Write a short note on threshold effect in demodulation of FM.

Answer

The threshold effect in FM is the sharp drop in output signal-to-noise ratio of an FM receiver when the input carrier-to-noise ratio (CNR) falls below a certain value, called the threshold, typically about 10 dB (about 13 dB in some texts).

Above threshold

When the carrier is much stronger than the noise, noise only causes small phase perturbations. The output SNR grows linearly with input CNR and FM gives a large improvement over AM:

(SNR)o=32β2 (SNR)c(tone modulation)(SNR)_o=\frac{3}{2}\beta^2\,(SNR)_c\quad(\text{tone modulation})

Below threshold

 (SNR)o dB
   |              /  FM above threshold
   |             /
   |            /
   |     ------/  <- knee (threshold, ~10 dB CNR)
   |    /
   |   /   sharp fall below threshold
   +---------------------------> (SNR)c dB

When noise amplitude occasionally exceeds the carrier amplitude, the resultant phasor can encircle the origin, causing a sudden 2π2\pi phase jump. The discriminator output (derivative of phase) then shows sharp impulses, heard as clicks. As CNR falls, clicks become frequent, the noise rises rapidly, and the signal is "captured" by noise. The output SNR then falls much faster than the input CNR and the FM advantage is lost.

Consequences and remedies

  • Large β\beta improves SNR above threshold but needs more bandwidth, so more noise enters and the threshold rises; this sets a practical limit on β\beta.
  • Threshold extension (lowering the threshold by a few dB) uses an FM demodulator with feedback (FMFB) or a phase-locked loop demodulator, which track the signal with a narrower effective bandwidth.
  • Pre-emphasis/de-emphasis improves SNR above threshold but does not remove the threshold.
  • Asked 3 times
  • 2068 Jestha (old course) · 10 marks
  • 2080 Bhadra (CS II) · 8 marks
  • 2074 Asoj (CS II) · 8 marks

Derive the expression for the impulse response of a matched filter for a AWGN channel.

Answer

A matched filter is the linear filter that maximises the peak output signal-to-noise ratio at the sampling instant when a known pulse is received in additive white Gaussian noise (AWGN). It is used as the optimum detector in digital receivers.

Derivation

Consider a known pulse g(t)g(t), 0≤t≤T0\le t\le T, plus white Gaussian noise w(t)w(t) of power spectral density N0/2N_0/2, applied to a linear time-invariant filter with impulse response h(t)h(t) and transfer function H(f)H(f).

 x(t) = g(t) + w(t) --> [ h(t) ] --> y(t) = go(t) + n(t)
                                         |
                                     sample at t = T

Signal output at the sampling instant:

go(T)=∫−∞∞H(f)G(f)ej2πfTdfg_o(T)=\int_{-\infty}^{\infty}H(f)G(f)e^{j2\pi fT}df

Noise output power (white noise through H(f)H(f)):

E[n2(t)]=N02∫−∞∞∣H(f)∣2dfE[n^2(t)]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df

Peak pulse SNR to be maximised:

η=∣go(T)∣2E[n2(t)]=∣∫H(f)G(f)ej2πfTdf∣2N02∫∣H(f)∣2df\eta=\frac{\lvert g_o(T)\rvert^2}{E[n^2(t)]}=\frac{\left|\int H(f)G(f)e^{j2\pi fT}df\right|^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}

Schwarz inequality: for any ϕ1(f),ϕ2(f)\phi_1(f),\phi_2(f),

∣∫ϕ1ϕ2 df∣2≤∫∣ϕ1∣2df∫∣ϕ2∣2df\left|\int\phi_1\phi_2\,df\right|^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df

with equality only when ϕ1(f)=k ϕ2∗(f)\phi_1(f)=k\,\phi_2^*(f). Put ϕ1=H(f)\phi_1=H(f) and ϕ2=G(f)ej2πfT\phi_2=G(f)e^{j2\pi fT}:

∣∫H(f)G(f)ej2πfTdf∣2≤∫∣H(f)∣2df∫∣G(f)∣2df\left|\int H(f)G(f)e^{j2\pi fT}df\right|^2\le\int\lvert H(f)\rvert^2df\int\lvert G(f)\rvert^2df

Substituting into η\eta:

η≤2N0∫−∞∞∣G(f)∣2df=2EN0\eta\le\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert G(f)\rvert^2df=\frac{2E}{N_0}

using Rayleigh's energy theorem E=∫∣G(f)∣2df=∫g2(t)dtE=\int\lvert G(f)\rvert^2df=\int g^2(t)dt.

Optimum filter: the maximum ηmax=2E/N0\eta_{max}=2E/N_0 is reached when equality holds:

Hopt(f)=k G∗(f) e−j2πfTH_{opt}(f)=k\,G^*(f)\,e^{-j2\pi fT}

Impulse response: take the inverse Fourier transform:

hopt(t)=k∫−∞∞G∗(f)e−j2πf(T−t)df=k[∫−∞∞G(f)ej2πf(T−t)df]∗(real g)=k g(T−t)\begin{aligned} h_{opt}(t)&=k\int_{-\infty}^{\infty}G^*(f)e^{-j2\pi f(T-t)}df\\ &=k\left[\int_{-\infty}^{\infty}G(f)e^{j2\pi f(T-t)}df\right]^*\quad(\text{real }g)\\ &=k\,g(T-t) \end{aligned}

So the impulse response of the optimum filter is the time-reversed and delayed (by TT) version of the input signal, scaled by kk. Because it is "matched" to the signal, it is called the matched filter.

  g(t)                     h(t) = g(T - t)
   |\                          /|
   | \                        / |
   |  \                      /  |
   +---+-----> t        +---+---+----> t
   0   T                0       T
  (flip about t = 0, then shift right by T)

Properties

  • Maximum output SNR =2E/N0=2E/N_0 depends only on the signal energy, not on its shape.
  • ∣Hopt(f)∣=k∣G(f)∣\lvert H_{opt}(f)\rvert=k\lvert G(f)\rvert: the filter passes strongly where the signal spectrum is strong.
  • The output go(t)g_o(t) is proportional to the autocorrelation of g(t)g(t) shifted by TT, peaking at t=Tt=T; hence a matched filter is equivalent to a correlator (multiply by g(t)g(t) and integrate over 00 to TT).
  • The delay TT makes h(t)h(t) causal (zero for t<0t<0).

Example

For a rectangular pulse g(t)=Ag(t)=A, 0≤t≤T0\le t\le T, the matched filter is h(t)=kAh(t)=kA for 0≤t≤T0\le t\le T (an integrate-and-dump filter). The output is a triangle peaking at t=Tt=T with value kA2T=kEkA^2T=kE, and ηmax=2A2T/N0\eta_{max}=2A^2T/N_0.

  • Asked 2 times
  • 2079 Chaitra · 2+6 marks
  • 2072 Chaitra (CS II) · 2+5 marks

What do you mean by optimum detector? Show that the impulse response of the matched filter is reverse delayed version of the input signal.

Answer

Optimum detector

An optimum detector is a receiver that makes decisions with the minimum probability of error for given signals and noise. In AWGN this is achieved by first passing the received signal through a filter that maximises the output SNR at the sampling instant (the matched filter or an equivalent correlator), then comparing the sample with a threshold (maximum-likelihood decision).

Matched filter impulse response

Consider a known pulse g(t)g(t), 0≤t≤T0\le t\le T, plus white Gaussian noise w(t)w(t) of power spectral density N0/2N_0/2, applied to a linear time-invariant filter with impulse response h(t)h(t) and transfer function H(f)H(f).

 x(t) = g(t) + w(t) --> [ h(t) ] --> y(t) = go(t) + n(t)
                                         |
                                     sample at t = T

Signal output at the sampling instant:

go(T)=∫−∞∞H(f)G(f)ej2πfTdfg_o(T)=\int_{-\infty}^{\infty}H(f)G(f)e^{j2\pi fT}df

Noise output power (white noise through H(f)H(f)):

E[n2(t)]=N02∫−∞∞∣H(f)∣2dfE[n^2(t)]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df

Peak pulse SNR to be maximised:

η=∣go(T)∣2E[n2(t)]=∣∫H(f)G(f)ej2πfTdf∣2N02∫∣H(f)∣2df\eta=\frac{\lvert g_o(T)\rvert^2}{E[n^2(t)]}=\frac{\left|\int H(f)G(f)e^{j2\pi fT}df\right|^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}

Schwarz inequality: for any ϕ1(f),ϕ2(f)\phi_1(f),\phi_2(f),

∣∫ϕ1ϕ2 df∣2≤∫∣ϕ1∣2df∫∣ϕ2∣2df\left|\int\phi_1\phi_2\,df\right|^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df

with equality only when ϕ1(f)=k ϕ2∗(f)\phi_1(f)=k\,\phi_2^*(f). Put ϕ1=H(f)\phi_1=H(f) and ϕ2=G(f)ej2πfT\phi_2=G(f)e^{j2\pi fT}:

∣∫H(f)G(f)ej2πfTdf∣2≤∫∣H(f)∣2df∫∣G(f)∣2df\left|\int H(f)G(f)e^{j2\pi fT}df\right|^2\le\int\lvert H(f)\rvert^2df\int\lvert G(f)\rvert^2df

Substituting into η\eta:

η≤2N0∫−∞∞∣G(f)∣2df=2EN0\eta\le\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert G(f)\rvert^2df=\frac{2E}{N_0}

using Rayleigh's energy theorem E=∫∣G(f)∣2df=∫g2(t)dtE=\int\lvert G(f)\rvert^2df=\int g^2(t)dt.

Optimum filter: the maximum ηmax=2E/N0\eta_{max}=2E/N_0 is reached when equality holds:

Hopt(f)=k G∗(f) e−j2πfTH_{opt}(f)=k\,G^*(f)\,e^{-j2\pi fT}

Impulse response: take the inverse Fourier transform:

hopt(t)=k∫−∞∞G∗(f)e−j2πf(T−t)df=k[∫−∞∞G(f)ej2πf(T−t)df]∗(real g)=k g(T−t)\begin{aligned} h_{opt}(t)&=k\int_{-\infty}^{\infty}G^*(f)e^{-j2\pi f(T-t)}df\\ &=k\left[\int_{-\infty}^{\infty}G(f)e^{j2\pi f(T-t)}df\right]^*\quad(\text{real }g)\\ &=k\,g(T-t) \end{aligned}

So the impulse response of the optimum filter is the time-reversed and delayed (by TT) version of the input signal, scaled by kk. Because it is "matched" to the signal, it is called the matched filter.

  g(t)                     h(t) = g(T - t)
   |\                          /|
   | \                        / |
   |  \                      /  |
   +---+-----> t        +---+---+----> t
   0   T                0       T
  (flip about t = 0, then shift right by T)

Properties

  • Maximum output SNR =2E/N0=2E/N_0 depends only on the signal energy, not on its shape.
  • ∣Hopt(f)∣=k∣G(f)∣\lvert H_{opt}(f)\rvert=k\lvert G(f)\rvert: the filter passes strongly where the signal spectrum is strong.
  • The output go(t)g_o(t) is proportional to the autocorrelation of g(t)g(t) shifted by TT, peaking at t=Tt=T; hence a matched filter is equivalent to a correlator (multiply by g(t)g(t) and integrate over 00 to TT).
  • The delay TT makes h(t)h(t) causal (zero for t<0t<0).
  • Asked 2 times
  • 2078 Chaitra · 3+5 marks
  • 2070 Bhadra (CS I) · 3+3 marks

List out any three properties of autocorrelation (AC) function. Mention the autocorrelation function of white noise.

Answer

Properties of the autocorrelation function

The autocorrelation (AC) function measures how similar a signal is to a time-shifted copy of itself. For a power signal or a wide-sense stationary (WSS) random process X(t)X(t):

RX(τ)=lim⁡T→∞1T∫−T/2T/2x(t) x(t+τ) dtorRX(τ)=E[X(t)X(t+τ)]R_X(\tau)=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}x(t)\,x(t+\tau)\,dt \quad\text{or}\quad R_X(\tau)=E[X(t)X(t+\tau)]

For an energy signal, Rg(τ)=∫−∞∞g(t)g(t+τ) dtR_g(\tau)=\int_{-\infty}^{\infty}g(t)g(t+\tau)\,dt.

Three (and more) important properties:

  1. Even symmetry: RX(−τ)=RX(τ)R_X(-\tau)=R_X(\tau).
  2. Value at origin = mean-square value (power): RX(0)=E[X2(t)]=PR_X(0)=E[X^2(t)]=P (for an energy signal, Rg(0)=ER_g(0)=E).
  3. Maximum at origin: ∣RX(τ)∣≤RX(0)\lvert R_X(\tau)\rvert\le R_X(0) for all τ\tau.
  4. Fourier pair with PSD (Wiener–Khinchin): SX(f)=∫−∞∞RX(τ)e−j2πfτdτS_X(f)=\int_{-\infty}^{\infty}R_X(\tau)e^{-j2\pi f\tau}d\tau and RX(τ)=∫−∞∞SX(f)ej2πfτdfR_X(\tau)=\int_{-\infty}^{\infty}S_X(f)e^{j2\pi f\tau}df.
  5. Periodicity: if x(t)x(t) is periodic with period T0T_0, RX(τ)R_X(\tau) is also periodic with T0T_0.
  6. DC component: if X(t)X(t) has mean mm and no periodic part, RX(τ)→m2R_X(\tau)\to m^2 as τ→∞\tau\to\infty.

Autocorrelation function of white noise

White noise is an idealised random noise whose power spectral density is flat (constant) over all frequencies, like white light containing all colours. Thermal and shot noise are nearly white up to very high frequencies (∼1012\sim 10^{12} Hz), so the model is accurate for any practical receiver bandwidth. It is usually also assumed Gaussian, zero-mean (AWGN).

PSD (two-sided):

SW(f)=N02for all fS_W(f)=\frac{N_0}{2}\quad\text{for all } f

where N0=kTeN_0=kT_e (W/Hz), kk = Boltzmann constant and TeT_e = equivalent noise temperature.

Autocorrelation (inverse Fourier transform of PSD, using F−1{1}=δ(τ)\mathcal{F}^{-1}\{1\}=\delta(\tau)):

RW(τ)=∫−∞∞N02ej2πfτdf=N02 δ(τ)R_W(\tau)=\int_{-\infty}^{\infty}\frac{N_0}{2}e^{j2\pi f\tau}df=\frac{N_0}{2}\,\delta(\tau)
   S_W(f)                      R_W(tau)
     |                             ^  (N0/2) delta
 N0/2+-----------------            |
     |                             |
 ----+-----------------> f   ------+------> tau
     0                             0

Meaning: RW(τ)=0R_W(\tau)=0 for every τ≠0\tau\ne0, so any two samples of white noise, however close, are uncorrelated (and independent if Gaussian). The total power RW(0)=∫SW(f)dfR_W(0)=\int S_W(f)df is infinite, which is why white noise is only a model; after any filter with finite bandwidth the power becomes finite.

  • Asked 2 times
  • 2079 Chaitra (CS I) · 5 marks
  • 2067 Shrawan (CS I) · 5 marks

Write a short note on autocorrelation function and its properties.

Answer

The autocorrelation (AC) function measures how similar a signal is to a time-shifted copy of itself. For a power signal or a wide-sense stationary (WSS) random process X(t)X(t):

RX(τ)=lim⁡T→∞1T∫−T/2T/2x(t) x(t+τ) dtorRX(τ)=E[X(t)X(t+τ)]R_X(\tau)=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}x(t)\,x(t+\tau)\,dt \quad\text{or}\quad R_X(\tau)=E[X(t)X(t+\tau)]

For an energy signal, Rg(τ)=∫−∞∞g(t)g(t+τ) dtR_g(\tau)=\int_{-\infty}^{\infty}g(t)g(t+\tau)\,dt.

It tells how fast a signal changes: a slowly varying signal stays correlated for large τ\tau, a rapidly varying one loses correlation quickly. It is used to find the PSD, to detect periodic signals buried in noise, and in correlation receivers and radar.

Properties:

  1. Even symmetry: RX(−τ)=RX(τ)R_X(-\tau)=R_X(\tau).
  2. Value at origin = mean-square value (power): RX(0)=E[X2(t)]=PR_X(0)=E[X^2(t)]=P (for an energy signal, Rg(0)=ER_g(0)=E).
  3. Maximum at origin: ∣RX(τ)∣≤RX(0)\lvert R_X(\tau)\rvert\le R_X(0) for all τ\tau.
  4. Fourier pair with PSD (Wiener–Khinchin): SX(f)=∫−∞∞RX(τ)e−j2πfτdτS_X(f)=\int_{-\infty}^{\infty}R_X(\tau)e^{-j2\pi f\tau}d\tau and RX(τ)=∫−∞∞SX(f)ej2πfτdfR_X(\tau)=\int_{-\infty}^{\infty}S_X(f)e^{j2\pi f\tau}df.
  5. Periodicity: if x(t)x(t) is periodic with period T0T_0, RX(τ)R_X(\tau) is also periodic with T0T_0.
  6. DC component: if X(t)X(t) has mean mm and no periodic part, RX(τ)→m2R_X(\tau)\to m^2 as τ→∞\tau\to\infty.

Example: white noise has R(τ)=N02δ(τ)R(\tau)=\frac{N_0}{2}\delta(\tau); samples at different times are uncorrelated.

  • Asked 2 times
  • 2075 Bhadra (CS I) · 4+4 marks
  • 2074 Bhadra (CS I) · 4+4 marks

Define energy spectral density and power spectral density function of a signal and hence derive the auto correlation function of white noise utilizing power spectral density along with necessary diagrams.

Answer

Energy and power spectral density

Energy spectral density (ESD): for an energy signal g(t)↔G(f)g(t)\leftrightarrow G(f), Ψg(f)=∣G(f)∣2\Psi_g(f)=\lvert G(f)\rvert^2 (J/Hz). It shows how the energy is spread over frequency:

E=∫−∞∞∣g(t)∣2dt=∫−∞∞Ψg(f) dfE=\int_{-\infty}^{\infty}\lvert g(t)\rvert^2dt=\int_{-\infty}^{\infty}\Psi_g(f)\,df

and Ψg(f)\Psi_g(f) is the Fourier transform of the energy autocorrelation Rg(τ)R_g(\tau).

Power spectral density (PSD): for a power signal, take a truncated version gT(t)g_T(t) (length TT) with transform GT(f)G_T(f):

Sg(f)=lim⁡T→∞∣GT(f)∣2T  (W/Hz),P=∫−∞∞Sg(f) dfS_g(f)=\lim_{T\to\infty}\frac{\lvert G_T(f)\rvert^2}{T}\ \ \text{(W/Hz)},\qquad P=\int_{-\infty}^{\infty}S_g(f)\,df

For a WSS random process, SX(f)=lim⁡T→∞E[∣XT(f)∣2]/TS_X(f)=\lim_{T\to\infty}E[\lvert X_T(f)\rvert^2]/T. Both ESD and PSD are real, even and non-negative.

Wiener–Khinchin relation (PSD ↔\leftrightarrow AC): for a power signal, Rg(τ)=lim⁡T→∞1T∫−T/2T/2gT(t)gT(t+τ)dtR_g(\tau)=\lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}g_T(t)g_T(t+\tau)dt. The time-domain integral is gT(τ)∗gT(−τ)g_T(\tau)*g_T(-\tau) scaled by 1/T1/T, and gT(−t)↔GT∗(f)g_T(-t)\leftrightarrow G_T^*(f), so

F{Rg(τ)}=lim⁡T→∞1TGT(f)GT∗(f)=lim⁡T→∞∣GT(f)∣2T=Sg(f)\begin{aligned} \mathcal{F}\{R_g(\tau)\} &= \lim_{T\to\infty}\frac{1}{T}G_T(f)G_T^*(f)=\lim_{T\to\infty}\frac{\lvert G_T(f)\rvert^2}{T}=S_g(f) \end{aligned}

Hence

Sg(f)=∫−∞∞Rg(τ)e−j2πfτdτ,Rg(τ)=∫−∞∞Sg(f)ej2πfτdfS_g(f)=\int_{-\infty}^{\infty}R_g(\tau)e^{-j2\pi f\tau}d\tau,\qquad R_g(\tau)=\int_{-\infty}^{\infty}S_g(f)e^{j2\pi f\tau}df

and at τ=0\tau=0: Rg(0)=∫Sg(f)df=PR_g(0)=\int S_g(f)df=P.

Autocorrelation of white noise from its PSD

White noise is an idealised random noise whose power spectral density is flat (constant) over all frequencies, like white light containing all colours. Thermal and shot noise are nearly white up to very high frequencies (∼1012\sim 10^{12} Hz), so the model is accurate for any practical receiver bandwidth. It is usually also assumed Gaussian, zero-mean (AWGN).

PSD (two-sided):

SW(f)=N02for all fS_W(f)=\frac{N_0}{2}\quad\text{for all } f

where N0=kTeN_0=kT_e (W/Hz), kk = Boltzmann constant and TeT_e = equivalent noise temperature.

Autocorrelation (inverse Fourier transform of PSD, using F−1{1}=δ(τ)\mathcal{F}^{-1}\{1\}=\delta(\tau)):

RW(τ)=∫−∞∞N02ej2πfτdf=N02 δ(τ)R_W(\tau)=\int_{-\infty}^{\infty}\frac{N_0}{2}e^{j2\pi f\tau}df=\frac{N_0}{2}\,\delta(\tau)
   S_W(f)                      R_W(tau)
     |                             ^  (N0/2) delta
 N0/2+-----------------            |
     |                             |
 ----+-----------------> f   ------+------> tau
     0                             0

Meaning: RW(τ)=0R_W(\tau)=0 for every τ≠0\tau\ne0, so any two samples of white noise, however close, are uncorrelated (and independent if Gaussian). The total power RW(0)=∫SW(f)dfR_W(0)=\int S_W(f)df is infinite, which is why white noise is only a model; after any filter with finite bandwidth the power becomes finite.

Band-limited white noise (more practical): if S(f)=N02S(f)=\frac{N_0}{2} for ∣f∣≤B\lvert f\rvert\le B and zero elsewhere,

R(τ)=∫−BBN02ej2πfτdf=N0B sin⁡2πBτ2πBτR(\tau)=\int_{-B}^{B}\frac{N_0}{2}e^{j2\pi f\tau}df=N_0B\,\frac{\sin2\pi B\tau}{2\pi B\tau}

Its power is finite, R(0)=N0BR(0)=N_0B, and samples spaced 1/2B1/2B apart are uncorrelated.

  • Asked 2 times
  • 2079 Bhadra (CS II) · 7+3 marks
  • 2070 Chaitra (CS II) · 7+3 marks

Derive the expression for evaluating error probability in binary communication system. What is threshold effect in FM? What are the techniques for the mitigation of threshold effect in FM?

Answer

Error probability in a binary communication system

Model. In a binary system, symbol 1 is sent as s1(t)s_1(t) and 0 as s2(t)s_2(t), each lasting TbT_b, with equal probability. Channel adds white Gaussian noise w(t)w(t) of PSD N0/2N_0/2. The receiver filter output is sampled at t=Tbt=T_b:

 s_i+w  +--------+ y(t)  sample  y  +---------+
 ------>| filter |------>-/ ------->| y > lam?|--> 1/0
        +--------+       t=Tb       +---------+

The sample is y=ai+ny=a_i+n, where a1,a2a_1, a_2 are the noise-free outputs (a1>a2a_1>a_2) and nn is Gaussian with zero mean and variance σ2\sigma^2. Conditional pdfs:

f(y∣1)=12πσe−(y−a1)2/2σ2,f(y∣0)=12πσe−(y−a2)2/2σ2f(y\mid1)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_1)^2/2\sigma^2},\qquad f(y\mid0)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_2)^2/2\sigma^2}
   f(y|0)            f(y|1)
     .--.             .--.
    /    \     |     /    \
   /      \    |    /      \
  /     ...\...|.../...     \
 --------a2---lam---a1----------> y
        Pe0 = area right of lam under f(y|0)
        Pe1 = area left  of lam under f(y|1)

Decision rule: choose 1 if y>λy>\lambda, else 0. For equal priors the optimum threshold is midway, λ=a1+a22\lambda=\frac{a_1+a_2}{2}.

Error when 0 is sent (y>λy>\lambda): with u=(y−a2)/2σu=(y-a_2)/\sqrt2\sigma,

Pe0=∫λ∞12πσe−(y−a2)2/2σ2dy=1π∫(λ−a2)/2σ∞e−u2du=12erfc⁡(a1−a222 σ)\begin{aligned} P_{e0} &= \int_{\lambda}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_2)^2/2\sigma^2}dy=\frac{1}{\sqrt{\pi}}\int_{(\lambda-a_2)/\sqrt2\sigma}^{\infty}e^{-u^2}du \\ &= \frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\,\sigma}\right) \end{aligned}

By symmetry Pe1P_{e1} (1 sent, y<λy<\lambda) is the same. The average error probability is

Pe=12Pe0+12Pe1=12erfc⁡(a1−a222 σ)=Q(a1−a22σ)P_e=\tfrac12P_{e0}+\tfrac12P_{e1}=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\,\sigma}\right)=Q\left(\frac{a_1-a_2}{2\sigma}\right)

where erfc⁡(x)=2π∫x∞e−u2du\operatorname{erfc}(x)=\frac{2}{\sqrt\pi}\int_x^{\infty}e^{-u^2}du and Q(x)=12erfc⁡(x/2)Q(x)=\frac12\operatorname{erfc}(x/\sqrt2).

With a matched filter (matched to s1−s2s_1-s_2), the maximum of (a1−a2)2σ2\frac{(a_1-a_2)^2}{\sigma^2} is 2EdN0\frac{2E_d}{N_0}, where Ed=∫0Tb[s1(t)−s2(t)]2dtE_d=\int_0^{T_b}[s_1(t)-s_2(t)]^2dt. Therefore

Pe=12erfc⁡(Ed4N0)P_e=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_d}{4N_0}}\right)

PeP_e depends only on the energy of the difference signal relative to N0N_0, not on pulse shape. Making the two signals as different as possible (antipodal) gives the lowest PeP_e.

Schemes1(t)s_1(t), s2(t)s_2(t)EdE_dPeP_e
Unipolar / ASK (OOK)Acos⁡ωctA\cos\omega_ct, 002Eb2E_b12erfc⁡Eb/2N0\frac12\operatorname{erfc}\sqrt{E_b/2N_0}
Polar / PSK±Acos⁡ωct\pm A\cos\omega_ct4Eb4E_b12erfc⁡Eb/N0\frac12\operatorname{erfc}\sqrt{E_b/N_0}
Coherent FSKAcos⁡ω1tA\cos\omega_1t, Acos⁡ω2tA\cos\omega_2t2Eb2E_b12erfc⁡Eb/2N0\frac12\operatorname{erfc}\sqrt{E_b/2N_0}

(EbE_b = average energy per bit.)

Threshold effect in FM

Threshold effect in FM. The FM improvement SNRo∝β2\text{SNR}_o\propto\beta^2 holds only when the carrier is much stronger than the noise at the discriminator input. When the carrier-to-noise ratio (CNR) falls below a threshold, about 10 dB (typically 10–13 dB), the noise phasor sometimes becomes larger than the carrier and the resultant phasor encircles the origin, causing a sudden 2π2\pi phase jump.

  high CNR: noise wobbles     low CNR: resultant can swing
  the tip slightly            around origin -> 2pi jump
        .-.                         .---.
  0 -->( * )                   0 --( *   )
        '-'                         '---'

Each 2π2\pi jump produces an impulse (spike) in the discriminator output, heard as clicks. The click rate rises rapidly as CNR drops, so the output SNR falls much faster than linearly and the signal is mutilated. Above threshold, FM is excellent; below it, FM is worse than AM. Wider deviation (larger β\beta) means larger BTB_T, more noise and a higher threshold.

 SNRo(dB)            . FM above threshold
     |             .   (slope 1, offset 3/2 beta^2)
     |           .
     |         .
     |       . <- threshold knee (CNR ~ 10 dB)
     |     .:
     |   . :     AM / baseband
     |  .  :
     +------------------> SNRi (dB)

Mitigation of threshold effect

Techniques to reduce (extend) the threshold:

  1. FM feedback (FMFB) demodulator: the VCO output tracks the signal, so the IF bandwidth needed is reduced (compressed deviation); less noise enters, threshold lowered by about 5–7 dB.
  2. Phase-locked loop (PLL) demodulator: the loop's narrow bandwidth tracks the instantaneous frequency and rejects noise outside it; gives a few dB threshold extension.
  3. Pre-emphasis and de-emphasis: boost high message frequencies before modulation and cut them after detection; reduces the high-frequency (parabolic) noise, improving SNR by about 10–13 dB and helping operation near threshold.
  4. Reducing IF bandwidth / modulation index where possible, or increasing transmitted power (raises CNR above threshold).
  5. Diversity reception and limiters that keep CNR above threshold under fading.
  • Asked 2 times
  • 2076 Asoj (CS II) · 2+6 marks
  • 2071 Chaitra (CS II) · 1+6 marks

What do you mean by optimum detector? Find the impulse response of optimum detector in the presence of additive white noise.

Answer

Optimum detector

An optimum detector is a receiver (filter followed by sampler and threshold device) that makes the decision with the minimum probability of error. For a known pulse in additive white Gaussian noise, this is achieved by the filter that maximises the output signal-to-noise ratio at the sampling instant; this optimum filter is the matched filter.

Impulse response of the optimum detector in additive white noise

Set-up. The received signal is x(t)=s(t)+w(t)x(t)=s(t)+w(t), 0≤t≤T0\le t\le T, where s(t)s(t) is a known pulse and w(t)w(t) is white noise of two-sided PSD N0/2N_0/2. It passes through an LTI filter h(t)↔H(f)h(t)\leftrightarrow H(f) and is sampled at t=Tt=T:

 s(t)+w(t)    +--------+   y(t)    sample     y(T)
 ----------->|  h(t)  |---------->--/ ---->  decision
              +--------+           at t=T

Output y(t)=so(t)+no(t)y(t)=s_o(t)+n_o(t) with

so(T)=∫−∞∞H(f)S(f)ej2πfTdf,E[no2]=N02∫−∞∞∣H(f)∣2dfs_o(T)=\int_{-\infty}^{\infty}H(f)S(f)e^{j2\pi fT}df,\qquad E[n_o^2]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df

We want H(f)H(f) that maximises the output peak signal-to-noise ratio

η=∣so(T)∣2E[no2]=∣∫H(f)S(f)ej2πfTdf∣2N02∫∣H(f)∣2df\eta=\frac{\lvert s_o(T)\rvert^2}{E[n_o^2]}=\frac{\left\lvert\int H(f)S(f)e^{j2\pi fT}df\right\rvert^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}

Schwarz inequality: ∣∫ϕ1ϕ2 df∣2≤∫∣ϕ1∣2df∫∣ϕ2∣2df\left\lvert\int\phi_1\phi_2\,df\right\rvert^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df, with equality only when ϕ1(f)=k ϕ2∗(f)\phi_1(f)=k\,\phi_2^*(f). Put ϕ1=H(f)\phi_1=H(f) and ϕ2=S(f)ej2πfT\phi_2=S(f)e^{j2\pi fT}:

η≤∫∣H∣2df∫∣S∣2dfN02∫∣H∣2df=2N0∫−∞∞∣S(f)∣2df=2EN0\eta\le\frac{\int\lvert H\rvert^2df\int\lvert S\rvert^2df}{\frac{N_0}{2}\int\lvert H\rvert^2df}=\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert S(f)\rvert^2df=\frac{2E}{N_0}

(using Rayleigh's theorem, EE = pulse energy). The maximum is reached when

Hopt(f)=k S∗(f) e−j2πfTH_{opt}(f)=k\,S^*(f)\,e^{-j2\pi fT}

Impulse response. Take the inverse transform. For real s(t)s(t), S∗(f)=S(−f)S^*(f)=S(-f), which is the transform of s(−t)s(-t); the factor e−j2πfTe^{-j2\pi fT} is a delay of TT:

hopt(t)=k∫−∞∞S∗(f)e−j2πfTej2πftdf=k∫−∞∞S(−f)e−j2πf(T−t)df=k s(T−t)\begin{aligned} h_{opt}(t) &= k\int_{-\infty}^{\infty}S^*(f)e^{-j2\pi fT}e^{j2\pi ft}df \\ &= k\int_{-\infty}^{\infty}S(-f)e^{-j2\pi f(T-t)}df \\ &= k\,s(T-t) \end{aligned}

So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by TT. Choosing the sampling instant TT equal to the pulse duration makes h(t)h(t) causal.

   s(t)                      h(t) = s(T - t)
    |\                          /|
    | \                        / |
    |  \                      /  |
 ---+---+---> t          ---+---+---> t
    0   T                   0   T

Results: maximum SNR =2E/N0=2E/N_0 depends only on pulse energy, not on pulse shape; the filter's frequency response is the conjugate of the pulse spectrum, so it passes strongly the frequencies where the signal is strong.

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  • 2075 Chaitra (CS II) · 8 marks
  • 2073 Shrawan (CS II) · 3+3 marks

Find the detection gain for SSB-SC demodulation and compare with DSB-SC.

Answer

Receiver model. The received signal s(t)s(t) plus white noise (N0/2N_0/2) passes through a band-pass (IF) filter of bandwidth BTB_T, then a demodulator and a low-pass filter of bandwidth WW (message bandwidth).

 s(t)  +   +-------+    +-------+   +-----+
 ---->(+)-->|  BPF  |--->| demod |-->| LPF |--> y(t)
       ^    |  B_T  |    |       |   |  W  |
      w(t)  +-------+    +-------+   +-----+

The filtered noise is narrowband and is written in in-phase/quadrature form

n(t)=nc(t)cos⁡ωct−ns(t)sin⁡ωctn(t)=n_c(t)\cos\omega_ct-n_s(t)\sin\omega_ct

where ncn_c and nsn_s are low-pass, each with the same power as n(t)n(t): nc2‾=ns2‾=n2‾=N0BT\overline{n_c^2}=\overline{n_s^2}=\overline{n^2}=N_0B_T.

Definitions. Message power P=m2(t)‾P=\overline{m^2(t)}.

  • Input SNR SNRi=received signal powernoise power in BT\text{SNR}_i=\dfrac{\text{received signal power}}{\text{noise power in } B_T}
  • Output SNR SNRo=message power at outputnoise power at output\text{SNR}_o=\dfrac{\text{message power at output}}{\text{noise power at output}}
  • Detection gain γd=SNRoSNRi\gamma_d=\dfrac{\text{SNR}_o}{\text{SNR}_i}; figure of merit =SNRoSNRref=\dfrac{\text{SNR}_o}{\text{SNR}_{ref}}, where SNRref\text{SNR}_{ref} is the channel SNR measured in the message band WW (baseband reference). Both show how much the demodulator improves or worsens SNR.

SSB-SC (coherent detection). s(t)=Ac2[m(t)cos⁡ωct∓m^(t)sin⁡ωct]s(t)=\frac{A_c}{2}[m(t)\cos\omega_ct\mp\hat m(t)\sin\omega_ct], BT=WB_T=W.

  • Signal power =Ac24⋅P+P2=Ac2P4=\frac{A_c^2}{4}\cdot\frac{P+P}{2}=\frac{A_c^2P}{4} (since m^2‾=P\overline{\hat m^2}=P); noise in WW =N0W=N_0W, so SNRi=Ac2P4N0W\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}.
  • Multiply by cos⁡ωct\cos\omega_ct and low-pass filter:
y(t)=Ac4m(t)+12nc(t)y(t)=\frac{A_c}{4}m(t)+\frac12n_c(t)
  • Signal power Ac2P16\frac{A_c^2P}{16}, noise power N0W4\frac{N_0W}{4}, so SNRo=Ac2P4N0W\text{SNR}_o=\dfrac{A_c^2P}{4N_0W}
γSSB=SNRoSNRi=1\gamma_{SSB}=\frac{\text{SNR}_o}{\text{SNR}_i}=1

With baseband reference (SNRref=Ac2P4N0W\text{SNR}_{ref}=\frac{A_c^2P}{4N_0W}, same transmitted power), figure of merit =1=1.

DSB-SC (coherent detection). s(t)=Acm(t)cos⁡ωcts(t)=A_cm(t)\cos\omega_ct, BT=2WB_T=2W.

  • Signal power =Ac2P2=\frac{A_c^2P}{2}; noise in 2W2W =2N0W=2N_0W, so SNRi=Ac2P4N0W\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}.
  • Multiply x(t)=s(t)+n(t)x(t)=s(t)+n(t) by cos⁡ωct\cos\omega_ct and low-pass filter:
y(t)=Ac2m(t)+12nc(t)y(t)=\frac{A_c}{2}m(t)+\frac12n_c(t)
  • SNRo=Ac2P/42N0W/4=Ac2P2N0W\text{SNR}_o=\dfrac{A_c^2P/4}{2N_0W/4}=\dfrac{A_c^2P}{2N_0W}
γDSB-SC=SNRoSNRi=2\gamma_{DSB\text{-}SC}=\frac{\text{SNR}_o}{\text{SNR}_i}=2

The quadrature noise nsn_s is rejected by coherent detection, giving a 3 dB gain. With baseband reference SNRref=Ac2P2N0W\text{SNR}_{ref}=\frac{A_c^2P}{2N_0W}, figure of merit =1=1.

Comparison:

QuantityDSB-SCSSB-SC
Transmission bandwidth2W2WWW
Noise power at input2N0W2N_0WN0WN_0W
SNRi\text{SNR}_i (same AcA_c)Ac2P4N0W\frac{A_c^2P}{4N_0W}Ac2P4N0W\frac{A_c^2P}{4N_0W}
Detection gain γ\gamma21
Figure of merit (baseband reference)11

The DSB-SC detection gain is twice that of SSB because the two sidebands add coherently (voltage) while the noise in them adds in power. But SSB lets in only half the noise bandwidth, so for the same transmitted power and same noise PSD, both give the same output SNR (figure of merit 1). SSB is preferred because it needs half the bandwidth.

  • Asked 2 times
  • 2074 Asoj (CS II) · 6 marks
  • 2069 Chaitra (CS II) · 6 marks

Derive the expression of error probability for coherent detection of Amplitude Shift Keying (ASK).

Answer

In ASK (on–off keying) a carrier is switched on for bit 1 and off for bit 0. Coherent detection uses a correlator with a locally generated carrier of the same frequency and phase.

ASK (on–off keying) signals over 0≤t≤Tb0\le t\le T_b:

s1(t)=Acos⁡ωct  (bit 1),s2(t)=0  (bit 0)s_1(t)=A\cos\omega_ct\ \ (\text{bit }1),\qquad s_2(t)=0\ \ (\text{bit }0)

Coherent receiver (correlator = matched filter):

 r(t) -->(x)--> integrate --> sample --> y > lam ? 1 : 0
          ^       0..Tb       t=Tb
          |
       cos(wc t)  (local carrier, in phase)

Correlator output at t=Tbt=T_b, with r(t)=si(t)+w(t)r(t)=s_i(t)+w(t):

y=∫0Tbr(t)cos⁡ωct dt=ai+ny=\int_0^{T_b}r(t)\cos\omega_ct\,dt=a_i+n
  • Bit 1: a1=∫0TbAcos⁡2ωct dt=ATb2a_1=\int_0^{T_b}A\cos^2\omega_ct\,dt=\frac{AT_b}{2} (for fcf_c an integer multiple of 1/Tb1/T_b).
  • Bit 0: a2=0a_2=0.
  • Noise: n=∫0Tbw(t)cos⁡ωct dtn=\int_0^{T_b}w(t)\cos\omega_ct\,dt is Gaussian, mean 0, variance
σ2=∫0Tb ⁣ ⁣∫0TbN02δ(t−u)cos⁡ωctcos⁡ωcu dt du=N02⋅Tb2=N0Tb4\sigma^2=\int_0^{T_b}\!\!\int_0^{T_b}\frac{N_0}{2}\delta(t-u)\cos\omega_ct\cos\omega_cu\,dt\,du=\frac{N_0}{2}\cdot\frac{T_b}{2}=\frac{N_0T_b}{4}

Threshold midway: λ=ATb4\lambda=\frac{AT_b}{4}. Using the binary result Pe=12erfc⁡(a1−a222σ)P_e=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\sigma}\right):

a1−a222 σ=ATb/222N0Tb/4=ATb22N0=A2Tb8N0\begin{aligned} \frac{a_1-a_2}{2\sqrt2\,\sigma} &= \frac{AT_b/2}{2\sqrt2\sqrt{N_0T_b/4}}=\frac{A\sqrt{T_b}}{2\sqrt{2N_0}}=\sqrt{\frac{A^2T_b}{8N_0}} \end{aligned}

Energy of a "1" pulse is E1=A2Tb2E_1=\frac{A^2T_b}{2}; the average energy per bit (half the bits are 0) is Eb=E12=A2Tb4E_b=\frac{E_1}{2}=\frac{A^2T_b}{4}. Hence A2Tb8N0=Eb2N0\frac{A^2T_b}{8N_0}=\frac{E_b}{2N_0} and

Pe(ASK)=12erfc⁡(Eb2N0)=Q(EbN0)P_e(\text{ASK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)

(In terms of peak energy E1E_1: Pe=12erfc⁡E1/4N0P_e=\frac12\operatorname{erfc}\sqrt{E_1/4N_0}.)

Remarks: coherent ASK needs 3 dB more average power than BPSK for the same PeP_e, and the threshold depends on received amplitude, so ASK is sensitive to fading. Example: for Eb/N0=10E_b/N_0=10 (10 dB), Pe=Q(10)=Q(3.16)≈7.9×10−4P_e=Q(\sqrt{10})=Q(3.16)\approx7.9\times10^{-4}.

  • Asked 2 times
  • 2073 Shrawan (CS II) · 8 marks
  • 2071 Shrawan (CS II) · 6 marks

Prove that the impulse response of the matched filter is reverse delayed version of the input signal.

Answer

A matched filter is a linear filter whose impulse response is matched to (a time-reversed, delayed copy of) a known input pulse s(t)s(t), h(t)=k s(T−t)h(t)=k\,s(T-t), so that it gives the maximum possible peak signal-to-noise ratio at the sampling instant t=Tt=T when the pulse is received in additive white noise. It is the optimum detector for digital signals in AWGN; in practice it is built as a correlator.

Set-up. The received signal is x(t)=s(t)+w(t)x(t)=s(t)+w(t), 0≤t≤T0\le t\le T, where s(t)s(t) is a known pulse and w(t)w(t) is white noise of two-sided PSD N0/2N_0/2. It passes through an LTI filter h(t)↔H(f)h(t)\leftrightarrow H(f) and is sampled at t=Tt=T:

 s(t)+w(t)    +--------+   y(t)    sample     y(T)
 ----------->|  h(t)  |---------->--/ ---->  decision
              +--------+           at t=T

Output y(t)=so(t)+no(t)y(t)=s_o(t)+n_o(t) with

so(T)=∫−∞∞H(f)S(f)ej2πfTdf,E[no2]=N02∫−∞∞∣H(f)∣2dfs_o(T)=\int_{-\infty}^{\infty}H(f)S(f)e^{j2\pi fT}df,\qquad E[n_o^2]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df

We want H(f)H(f) that maximises the output peak signal-to-noise ratio

η=∣so(T)∣2E[no2]=∣∫H(f)S(f)ej2πfTdf∣2N02∫∣H(f)∣2df\eta=\frac{\lvert s_o(T)\rvert^2}{E[n_o^2]}=\frac{\left\lvert\int H(f)S(f)e^{j2\pi fT}df\right\rvert^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}

Schwarz inequality: ∣∫ϕ1ϕ2 df∣2≤∫∣ϕ1∣2df∫∣ϕ2∣2df\left\lvert\int\phi_1\phi_2\,df\right\rvert^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df, with equality only when ϕ1(f)=k ϕ2∗(f)\phi_1(f)=k\,\phi_2^*(f). Put ϕ1=H(f)\phi_1=H(f) and ϕ2=S(f)ej2πfT\phi_2=S(f)e^{j2\pi fT}:

η≤∫∣H∣2df∫∣S∣2dfN02∫∣H∣2df=2N0∫−∞∞∣S(f)∣2df=2EN0\eta\le\frac{\int\lvert H\rvert^2df\int\lvert S\rvert^2df}{\frac{N_0}{2}\int\lvert H\rvert^2df}=\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert S(f)\rvert^2df=\frac{2E}{N_0}

(using Rayleigh's theorem, EE = pulse energy). The maximum is reached when

Hopt(f)=k S∗(f) e−j2πfTH_{opt}(f)=k\,S^*(f)\,e^{-j2\pi fT}

Impulse response. Take the inverse transform. For real s(t)s(t), S∗(f)=S(−f)S^*(f)=S(-f), which is the transform of s(−t)s(-t); the factor e−j2πfTe^{-j2\pi fT} is a delay of TT:

hopt(t)=k∫−∞∞S∗(f)e−j2πfTej2πftdf=k∫−∞∞S(−f)e−j2πf(T−t)df=k s(T−t)\begin{aligned} h_{opt}(t) &= k\int_{-\infty}^{\infty}S^*(f)e^{-j2\pi fT}e^{j2\pi ft}df \\ &= k\int_{-\infty}^{\infty}S(-f)e^{-j2\pi f(T-t)}df \\ &= k\,s(T-t) \end{aligned}

So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by TT. Choosing the sampling instant TT equal to the pulse duration makes h(t)h(t) causal.

   s(t)                      h(t) = s(T - t)
    |\                          /|
    | \                        / |
    |  \                      /  |
 ---+---+---> t          ---+---+---> t
    0   T                   0   T

Results: maximum SNR =2E/N0=2E/N_0 depends only on pulse energy, not on pulse shape; the filter's frequency response is the conjugate of the pulse spectrum, so it passes strongly the frequencies where the signal is strong.

Example: for a rectangular pulse s(t)=As(t)=A, 0≤t≤T0\le t\le T, h(t)=kAh(t)=kA for 0≤t≤T0\le t\le T, which is the same rectangle; it behaves as an integrate-and-dump circuit. The output is a triangle with peak kA2TkA^2T at t=Tt=T.

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  • 2073 Shrawan (CS II) · 8 marks
  • 2071 Chaitra (CS II) · 8 marks

Derive the expression for evaluation the gain parameter (SNR₀/SNRᵢ) of non-coherent FM detector.

Answer

A non-coherent FM detector (limiter + frequency discriminator) responds to the instantaneous frequency of the received signal; no local carrier is needed. The gain parameter SNRo/SNRi\text{SNR}_o/\text{SNR}_i is found assuming high carrier-to-noise ratio (above threshold).

FM signal and receiver. s(t)=Accos⁡[ωct+2πkf∫m(t)dt]s(t)=A_c\cos\left[\omega_ct+2\pi k_f\int m(t)dt\right], instantaneous frequency deviation kfm(t)k_fm(t). Receiver: BPF (BTB_T) → limiter → discriminator (output =12πdϕdt=\frac{1}{2\pi}\frac{d\phi}{dt}) → LPF (WW).

Noise in phasor form. Write the narrowband noise as n(t)=r(t)cos⁡[ωct+ψ(t)]n(t)=r(t)\cos[\omega_ct+\psi(t)]. At high carrier-to-noise ratio (Ac≫rA_c\gg r), the resultant phase is

θ(t)≈ϕ(t)+r(t)Acsin⁡[ψ(t)−ϕ(t)]\theta(t)\approx\phi(t)+\frac{r(t)}{A_c}\sin[\psi(t)-\phi(t)]
            r(t)
          .------>          resultant = carrier + noise
         /  ^               small noise phasor rotates
 Ac     /   | r sin(psi-phi) the carrier phasor slightly
 ------>----+

The noise term is statistically equivalent to ns(t)Ac\frac{n_s(t)}{A_c} (the phase ϕ\phi can be dropped for noise calculation). Discriminator output:

v(t)=12πdθdt=kfm(t)+12πAcdns(t)dtv(t)=\frac{1}{2\pi}\frac{d\theta}{dt}=k_fm(t)+\frac{1}{2\pi A_c}\frac{dn_s(t)}{dt}

Output signal power: So=kf2PS_o=k_f^2P.

Output noise power. nsn_s has PSD N0N_0 for ∣f∣≤BT/2\lvert f\rvert\le B_T/2. Differentiation multiplies the spectrum by j2πfj2\pi f, so the noise PSD at discriminator output is

Snd(f)=(2πf)2(2πAc)2N0=N0f2Ac2,∣f∣≤BT2S_{n_d}(f)=\frac{(2\pi f)^2}{(2\pi A_c)^2}N_0=\frac{N_0f^2}{A_c^2},\qquad \lvert f\rvert\le\frac{B_T}{2}

This parabolic noise spectrum is the key property of FM. After the LPF (∣f∣≤W\lvert f\rvert\le W):

No=∫−WWN0f2Ac2df=2N0W33Ac2N_o=\int_{-W}^{W}\frac{N_0f^2}{A_c^2}df=\frac{2N_0W^3}{3A_c^2}
 S(f) |\                   /|
      | \                 / |   parabolic noise
      |  \               /  |   after discriminator
      |   \.           ./   |   (only |f|<W kept)
      +----+-----+-----+----+--> f
          -W     0     W

Output SNR:

SNRo=kf2P2N0W3/3Ac2=3Ac2kf2P2N0W3\text{SNR}_o=\frac{k_f^2P}{2N_0W^3/3A_c^2}=\frac{3A_c^2k_f^2P}{2N_0W^3}

Input SNR. Received power Ac22\frac{A_c^2}{2} (constant envelope); noise in BTB_T is N0BTN_0B_T:

SNRi=Ac22N0BT,SNRref=Ac22N0W\text{SNR}_i=\frac{A_c^2}{2N_0B_T},\qquad \text{SNR}_{ref}=\frac{A_c^2}{2N_0W}

Gain and figure of merit:

FOM=SNRoSNRref=3kf2PW2SNRoSNRi=3kf2P BTW3\begin{aligned} \text{FOM} &= \frac{\text{SNR}_o}{\text{SNR}_{ref}}=\frac{3k_f^2P}{W^2} \\ \frac{\text{SNR}_o}{\text{SNR}_i} &= \frac{3k_f^2P\,B_T}{W^3} \end{aligned}

For a single tone m(t)=Amcos⁡ωmtm(t)=A_m\cos\omega_mt: P=Am2/2P=A_m^2/2, Δf=kfAm\Delta f=k_fA_m, β=Δf/W\beta=\Delta f/W, BT=2(β+1)WB_T=2(\beta+1)W (Carson):

FOM=32β2,SNRoSNRi=3β2(β+1)\text{FOM}=\frac{3}{2}\beta^2,\qquad \frac{\text{SNR}_o}{\text{SNR}_i}=3\beta^2(\beta+1)

So FM noise performance improves with the square of the deviation ratio: FM trades bandwidth for SNR. FM beats AM (FOM=1/3\text{FOM}=1/3) when 32β2>13\frac32\beta^2>\frac13, i.e. β>0.471\beta>0.471 (about 0.5).

  • Asked 2 times
  • 2072 Kartik (CS II) · 2+4 marks
  • 2069 Chaitra (CS II) · 4 marks

What do you mean by Random process? Explain white noise with its PSDF and autocorrelation function.

Answer

Random process

A random (stochastic) process X(t,s)X(t,s) is a collection (ensemble) of time functions, one for each outcome ss of a random experiment. At a fixed time t1t_1, X(t1)X(t_1) is a random variable; for a fixed outcome it is an ordinary waveform (a sample function). Example: thermal noise voltages across many identical resistors; the received signal in a communication channel. It is described by its mean mX(t)=E[X(t)]m_X(t)=E[X(t)], autocorrelation RX(t1,t2)R_X(t_1,t_2) and PSD.

White noise

White noise is an idealised random noise whose power spectral density is flat (constant) over all frequencies, like white light containing all colours. Thermal and shot noise are nearly white up to very high frequencies (∼1012\sim 10^{12} Hz), so the model is accurate for any practical receiver bandwidth. It is usually also assumed Gaussian, zero-mean (AWGN).

PSD (two-sided):

SW(f)=N02for all fS_W(f)=\frac{N_0}{2}\quad\text{for all } f

where N0=kTeN_0=kT_e (W/Hz), kk = Boltzmann constant and TeT_e = equivalent noise temperature.

Autocorrelation (inverse Fourier transform of PSD, using F−1{1}=δ(τ)\mathcal{F}^{-1}\{1\}=\delta(\tau)):

RW(τ)=∫−∞∞N02ej2πfτdf=N02 δ(τ)R_W(\tau)=\int_{-\infty}^{\infty}\frac{N_0}{2}e^{j2\pi f\tau}df=\frac{N_0}{2}\,\delta(\tau)
   S_W(f)                      R_W(tau)
     |                             ^  (N0/2) delta
 N0/2+-----------------            |
     |                             |
 ----+-----------------> f   ------+------> tau
     0                             0

Meaning: RW(τ)=0R_W(\tau)=0 for every τ≠0\tau\ne0, so any two samples of white noise, however close, are uncorrelated (and independent if Gaussian). The total power RW(0)=∫SW(f)dfR_W(0)=\int S_W(f)df is infinite, which is why white noise is only a model; after any filter with finite bandwidth the power becomes finite.

  • 2081 Chaitra · 5+3 marks

Derive the expression for figure of merit for the SSB receiver. Define an efficient constellation diagram of a 32-QAM.

Answer

Figure of merit of SSB receiver

Figure of merit =SNRoSNRc=\dfrac{\text{SNR}_o}{\text{SNR}_c}, where SNRc\text{SNR}_c is the channel SNR: average received signal power divided by noise power in the message bandwidth WW.

Coherent SSB receiver: BPF of width WW → product modulator with cos⁡ωct\cos\omega_ct → LPF (WW). Noise PSD N0/2N_0/2.

Signal: s(t)=Ac2[m(t)cos⁡ωct−m^(t)sin⁡ωct]s(t)=\frac{A_c}{2}[m(t)\cos\omega_ct-\hat m(t)\sin\omega_ct] (USB). Since mm and m^\hat m have equal power PP and are orthogonal:

Ps=Ac24⋅P2⋅2=Ac2P4,SNRc=Ac2P4N0WP_s=\frac{A_c^2}{4}\cdot\frac{P}{2}\cdot2=\frac{A_c^2P}{4},\qquad \text{SNR}_c=\frac{A_c^2P}{4N_0W}

Noise after BPF (width WW): n(t)=nccos⁡ωct−nssin⁡ωctn(t)=n_c\cos\omega_ct-n_s\sin\omega_ct with nc2‾=ns2‾=N0W\overline{n_c^2}=\overline{n_s^2}=N_0W.

Detector output (multiply by cos⁡ωct\cos\omega_ct, LPF):

y(t)=Ac4m(t)+12nc(t)y(t)=\frac{A_c}{4}m(t)+\frac12n_c(t) So=Ac2P16,No=N0W4SNRo=Ac2P/16N0W/4=Ac2P4N0W\begin{aligned} S_o &= \frac{A_c^2P}{16},\qquad N_o=\frac{N_0W}{4} \\ \text{SNR}_o &= \frac{A_c^2P/16}{N_0W/4}=\frac{A_c^2P}{4N_0W} \end{aligned} FOMSSB=SNRoSNRc=1\text{FOM}_{SSB}=\frac{\text{SNR}_o}{\text{SNR}_c}=1

SSB has the same noise performance as DSB-SC and baseband, while using only half the bandwidth.

Efficient constellation of 32-QAM

In 32-QAM each symbol carries log⁡232=5\log_232=5 bits. 32 is not a perfect square, so a square grid cannot be used. The efficient arrangement is the cross constellation: a 6×66\times6 square grid of points (36) with the 4 corner points removed, giving 32 points. Removing the corners (which have the highest energy) lowers the average and peak power for the same minimum distance dd.

              Q
              |
        .  *  *  *  *  .
        *  *  *  *  *  *
        *  *  *  *  *  *
   -----------+------------ I
        *  *  *  *  *  *
        *  *  *  *  *  *
        .  *  *  *  *  .
              |
  * = point (32), . = removed corner

Points lie at I,Q∈{±1,±3,±5}⋅d2I,Q\in\{\pm1,\pm3,\pm5\}\cdot\frac d2 except the four corners (±5,±5)(\pm5,\pm5).

  • 2080 Chaitra · 2+6+2 marks

Define optimum detector and find the impulse response of optimum detector in the presence of additive white noise. Briefly explain Hilbert Transform.

Answer

Optimum detector

An optimum detector is a receiver (filter followed by sampler and threshold device) that makes the decision with the minimum probability of error. For a known pulse in additive white Gaussian noise, this is achieved by the filter that maximises the output signal-to-noise ratio at the sampling instant; this optimum filter is the matched filter.

Impulse response of the optimum detector

Set-up. The received signal is x(t)=s(t)+w(t)x(t)=s(t)+w(t), 0≤t≤T0\le t\le T, where s(t)s(t) is a known pulse and w(t)w(t) is white noise of two-sided PSD N0/2N_0/2. It passes through an LTI filter h(t)↔H(f)h(t)\leftrightarrow H(f) and is sampled at t=Tt=T:

 s(t)+w(t)    +--------+   y(t)    sample     y(T)
 ----------->|  h(t)  |---------->--/ ---->  decision
              +--------+           at t=T

Output y(t)=so(t)+no(t)y(t)=s_o(t)+n_o(t) with

so(T)=∫−∞∞H(f)S(f)ej2πfTdf,E[no2]=N02∫−∞∞∣H(f)∣2dfs_o(T)=\int_{-\infty}^{\infty}H(f)S(f)e^{j2\pi fT}df,\qquad E[n_o^2]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df

We want H(f)H(f) that maximises the output peak signal-to-noise ratio

η=∣so(T)∣2E[no2]=∣∫H(f)S(f)ej2πfTdf∣2N02∫∣H(f)∣2df\eta=\frac{\lvert s_o(T)\rvert^2}{E[n_o^2]}=\frac{\left\lvert\int H(f)S(f)e^{j2\pi fT}df\right\rvert^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}

Schwarz inequality: ∣∫ϕ1ϕ2 df∣2≤∫∣ϕ1∣2df∫∣ϕ2∣2df\left\lvert\int\phi_1\phi_2\,df\right\rvert^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df, with equality only when ϕ1(f)=k ϕ2∗(f)\phi_1(f)=k\,\phi_2^*(f). Put ϕ1=H(f)\phi_1=H(f) and ϕ2=S(f)ej2πfT\phi_2=S(f)e^{j2\pi fT}:

η≤∫∣H∣2df∫∣S∣2dfN02∫∣H∣2df=2N0∫−∞∞∣S(f)∣2df=2EN0\eta\le\frac{\int\lvert H\rvert^2df\int\lvert S\rvert^2df}{\frac{N_0}{2}\int\lvert H\rvert^2df}=\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert S(f)\rvert^2df=\frac{2E}{N_0}

(using Rayleigh's theorem, EE = pulse energy). The maximum is reached when

Hopt(f)=k S∗(f) e−j2πfTH_{opt}(f)=k\,S^*(f)\,e^{-j2\pi fT}

Impulse response. Take the inverse transform. For real s(t)s(t), S∗(f)=S(−f)S^*(f)=S(-f), which is the transform of s(−t)s(-t); the factor e−j2πfTe^{-j2\pi fT} is a delay of TT:

hopt(t)=k∫−∞∞S∗(f)e−j2πfTej2πftdf=k∫−∞∞S(−f)e−j2πf(T−t)df=k s(T−t)\begin{aligned} h_{opt}(t) &= k\int_{-\infty}^{\infty}S^*(f)e^{-j2\pi fT}e^{j2\pi ft}df \\ &= k\int_{-\infty}^{\infty}S(-f)e^{-j2\pi f(T-t)}df \\ &= k\,s(T-t) \end{aligned}

So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by TT. Choosing the sampling instant TT equal to the pulse duration makes h(t)h(t) causal.

   s(t)                      h(t) = s(T - t)
    |\                          /|
    | \                        / |
    |  \                      /  |
 ---+---+---> t          ---+---+---> t
    0   T                   0   T

Results: maximum SNR =2E/N0=2E/N_0 depends only on pulse energy, not on pulse shape; the filter's frequency response is the conjugate of the pulse spectrum, so it passes strongly the frequencies where the signal is strong.

Hilbert transform

Hilbert transform. The Hilbert transform g^(t)\hat g(t) of g(t)g(t) is obtained by shifting the phase of every frequency component by −90∘-90^\circ (positive frequencies) and +90∘+90^\circ (negative frequencies), without changing amplitudes:

g^(t)=g(t)∗1πt=1π∫−∞∞g(τ)t−τdτ,G^(f)=−j sgn⁡(f) G(f)\hat g(t)=g(t)*\frac{1}{\pi t}=\frac1\pi\int_{-\infty}^{\infty}\frac{g(\tau)}{t-\tau}d\tau,\qquad \hat G(f)=-j\,\operatorname{sgn}(f)\,G(f)

Example: the Hilbert transform of cos⁡ω0t\cos\omega_0t is sin⁡ω0t\sin\omega_0t. Properties: gg and g^\hat g have the same amplitude spectrum and energy, and are orthogonal. Uses: SSB generation by the phase-shift method s(t)=m(t)cos⁡ωct∓m^(t)sin⁡ωcts(t)=m(t)\cos\omega_ct\mp\hat m(t)\sin\omega_ct, analytic signals g+(t)=g(t)+jg^(t)g_+(t)=g(t)+j\hat g(t), and band-pass signal representation.

  • 2080 Chaitra (CS I) · 2+6 marks

State Parseval's Theorem. Explain the relation between power spectral density function and autocorrelation function with the example of white noise.

Answer

Parseval's theorem

Parseval's theorem states that the power (or energy) of a signal is the same whether computed in the time domain or in the frequency domain.

  • Periodic power signal with Fourier coefficients cnc_n: P=1T0∫T0∣g(t)∣2dt=∑n=−∞∞∣cn∣2P=\frac{1}{T_0}\int_{T_0}\lvert g(t)\rvert^2dt=\sum_{n=-\infty}^{\infty}\lvert c_n\rvert^2
  • Energy signal (Rayleigh form): E=∫∣g(t)∣2dt=∫∣G(f)∣2dfE=\int\lvert g(t)\rvert^2dt=\int\lvert G(f)\rvert^2df

Relation between PSD and autocorrelation

Wiener–Khinchin relation (PSD ↔\leftrightarrow AC): for a power signal, Rg(τ)=lim⁡T→∞1T∫−T/2T/2gT(t)gT(t+τ)dtR_g(\tau)=\lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}g_T(t)g_T(t+\tau)dt. The time-domain integral is gT(τ)∗gT(−τ)g_T(\tau)*g_T(-\tau) scaled by 1/T1/T, and gT(−t)↔GT∗(f)g_T(-t)\leftrightarrow G_T^*(f), so

F{Rg(τ)}=lim⁡T→∞1TGT(f)GT∗(f)=lim⁡T→∞∣GT(f)∣2T=Sg(f)\begin{aligned} \mathcal{F}\{R_g(\tau)\} &= \lim_{T\to\infty}\frac{1}{T}G_T(f)G_T^*(f)=\lim_{T\to\infty}\frac{\lvert G_T(f)\rvert^2}{T}=S_g(f) \end{aligned}

Hence

Sg(f)=∫−∞∞Rg(τ)e−j2πfτdτ,Rg(τ)=∫−∞∞Sg(f)ej2πfτdfS_g(f)=\int_{-\infty}^{\infty}R_g(\tau)e^{-j2\pi f\tau}d\tau,\qquad R_g(\tau)=\int_{-\infty}^{\infty}S_g(f)e^{j2\pi f\tau}df

and at τ=0\tau=0: Rg(0)=∫Sg(f)df=PR_g(0)=\int S_g(f)df=P.

Example: white noise. The PSD is flat, SW(f)=N0/2S_W(f)=N_0/2. Then by the relation above:

RW(τ)=∫−∞∞N02ej2πfτdf=N02δ(τ)R_W(\tau)=\int_{-\infty}^{\infty}\frac{N_0}{2}e^{j2\pi f\tau}df=\frac{N_0}{2}\delta(\tau)
   S_W(f)                      R_W(tau)
 N0/2+-----------------            ^ (N0/2) delta(tau)
     |                             |
 ----+-----------------> f   ------+------> tau

A flat (infinitely wide) spectrum corresponds to an impulse autocorrelation: white noise is uncorrelated with itself at any non-zero shift. Conversely, if white noise is band-limited to BB, R(τ)=N0B sinc(2Bτ)R(\tau)=N_0B\,\text{sinc}(2B\tau) spreads out (with sinc(x)=sin⁡πx/πx\text{sinc}(x)=\sin\pi x/\pi x); the narrower the spectrum, the wider the autocorrelation.

Link with Parseval: putting τ=0\tau=0 in R(τ)=∫S(f)ej2πfτdfR(\tau)=\int S(f)e^{j2\pi f\tau}df gives R(0)=∫S(f)df=PR(0)=\int S(f)df=P, which is exactly Parseval's theorem for power: the time-average power equals the area under the PSD.

  • 2071 Magh (CS I) · 3+3 marks

Define white noise. Establish relation between psdf and the AC function of a white noise.

Answer

White noise

White noise is an idealised random noise whose power spectral density is flat (constant) over all frequencies, like white light containing all colours. Thermal and shot noise are nearly white up to very high frequencies (∼1012\sim 10^{12} Hz), so the model is accurate for any practical receiver bandwidth. It is usually also assumed Gaussian, zero-mean (AWGN).

Relation between PSDF and AC function

By the Wiener–Khinchin theorem, PSD and autocorrelation are a Fourier transform pair:

S(f)=∫−∞∞R(τ)e−j2πfτdτ,R(τ)=∫−∞∞S(f)ej2πfτdfS(f)=\int_{-\infty}^{\infty}R(\tau)e^{-j2\pi f\tau}d\tau,\qquad R(\tau)=\int_{-\infty}^{\infty}S(f)e^{j2\pi f\tau}df

PSD (two-sided):

SW(f)=N02for all fS_W(f)=\frac{N_0}{2}\quad\text{for all } f

where N0=kTeN_0=kT_e (W/Hz), kk = Boltzmann constant and TeT_e = equivalent noise temperature.

Autocorrelation (inverse Fourier transform of PSD, using F−1{1}=δ(τ)\mathcal{F}^{-1}\{1\}=\delta(\tau)):

RW(τ)=∫−∞∞N02ej2πfτdf=N02 δ(τ)R_W(\tau)=\int_{-\infty}^{\infty}\frac{N_0}{2}e^{j2\pi f\tau}df=\frac{N_0}{2}\,\delta(\tau)
   S_W(f)                      R_W(tau)
     |                             ^  (N0/2) delta
 N0/2+-----------------            |
     |                             |
 ----+-----------------> f   ------+------> tau
     0                             0

Meaning: RW(τ)=0R_W(\tau)=0 for every τ≠0\tau\ne0, so any two samples of white noise, however close, are uncorrelated (and independent if Gaussian). The total power RW(0)=∫SW(f)dfR_W(0)=\int S_W(f)df is infinite, which is why white noise is only a model; after any filter with finite bandwidth the power becomes finite.

  • 2068 Bhadra (CS I) · 4+4 marks

Define white noise with PSDF and auto correlation function. State the properties of auto correlation function.

Answer

White noise, its PSDF and AC function

White noise is an idealised random noise whose power spectral density is flat (constant) over all frequencies, like white light containing all colours. Thermal and shot noise are nearly white up to very high frequencies (∼1012\sim 10^{12} Hz), so the model is accurate for any practical receiver bandwidth. It is usually also assumed Gaussian, zero-mean (AWGN).

PSD (two-sided):

SW(f)=N02for all fS_W(f)=\frac{N_0}{2}\quad\text{for all } f

where N0=kTeN_0=kT_e (W/Hz), kk = Boltzmann constant and TeT_e = equivalent noise temperature.

Autocorrelation (inverse Fourier transform of PSD, using F−1{1}=δ(τ)\mathcal{F}^{-1}\{1\}=\delta(\tau)):

RW(τ)=∫−∞∞N02ej2πfτdf=N02 δ(τ)R_W(\tau)=\int_{-\infty}^{\infty}\frac{N_0}{2}e^{j2\pi f\tau}df=\frac{N_0}{2}\,\delta(\tau)
   S_W(f)                      R_W(tau)
     |                             ^  (N0/2) delta
 N0/2+-----------------            |
     |                             |
 ----+-----------------> f   ------+------> tau
     0                             0

Meaning: RW(τ)=0R_W(\tau)=0 for every τ≠0\tau\ne0, so any two samples of white noise, however close, are uncorrelated (and independent if Gaussian). The total power RW(0)=∫SW(f)dfR_W(0)=\int S_W(f)df is infinite, which is why white noise is only a model; after any filter with finite bandwidth the power becomes finite.

Properties of the autocorrelation function

  1. Even symmetry: RX(−τ)=RX(τ)R_X(-\tau)=R_X(\tau).
  2. Value at origin = mean-square value (power): RX(0)=E[X2(t)]=PR_X(0)=E[X^2(t)]=P (for an energy signal, Rg(0)=ER_g(0)=E).
  3. Maximum at origin: ∣RX(τ)∣≤RX(0)\lvert R_X(\tau)\rvert\le R_X(0) for all τ\tau.
  4. Fourier pair with PSD (Wiener–Khinchin): SX(f)=∫−∞∞RX(τ)e−j2πfτdτS_X(f)=\int_{-\infty}^{\infty}R_X(\tau)e^{-j2\pi f\tau}d\tau and RX(τ)=∫−∞∞SX(f)ej2πfτdfR_X(\tau)=\int_{-\infty}^{\infty}S_X(f)e^{j2\pi f\tau}df.
  5. Periodicity: if x(t)x(t) is periodic with period T0T_0, RX(τ)R_X(\tau) is also periodic with T0T_0.
  6. DC component: if X(t)X(t) has mean mm and no periodic part, RX(τ)→m2R_X(\tau)\to m^2 as τ→∞\tau\to\infty.

White noise satisfies these: N02δ(τ)\frac{N_0}{2}\delta(\tau) is even, maximum at τ=0\tau=0, and its Fourier transform is the flat PSD.

  • 2081 Bhadra (CS II) · 4+6 marks

Explain what Noise Equivalent Bandwidth represents of a filter. Derive the impulse response of the optimum detector in the presence of additive white noise.

Answer

Noise equivalent bandwidth

The noise equivalent bandwidth (NEB) BNB_N of a filter is the bandwidth of an ideal rectangular filter, with the same peak (usually zero-frequency or centre) gain ∣H(0)∣\lvert H(0)\rvert, that passes the same total noise power as the actual filter when both are fed with white noise.

White noise N0/2N_0/2 into H(f)H(f) gives output power

Pactual=N02∫−∞∞∣H(f)∣2df=N0∫0∞∣H(f)∣2dfP_{actual}=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df=N_0\int_0^{\infty}\lvert H(f)\rvert^2df

The ideal filter of gain ∣H(0)∣\lvert H(0)\rvert and bandwidth BNB_N gives Pideal=N0BN∣H(0)∣2P_{ideal}=N_0B_N\lvert H(0)\rvert^2. Equating:

BN=∫0∞∣H(f)∣2df∣H(0)∣2B_N=\frac{\int_0^{\infty}\lvert H(f)\rvert^2df}{\lvert H(0)\rvert^2}
 |H(f)|^2
 |H(0)|^2 +---------+     equal areas
          |  ...    |
          |      .. |
          |        .|..
          |         |  ......
        --+---------+----------> f
          0        B_N

NEB lets us write the output noise power simply as N=N0BN∣H(0)∣2N=N_0B_N\lvert H(0)\rvert^2 (or kTBNkTB_N per unit gain), which is used in noise figure and SNR calculations. For a band-pass filter ∣H(0)∣\lvert H(0)\rvert is replaced by the centre-frequency gain ∣H(f0)∣\lvert H(f_0)\rvert.

Example: an RC low-pass filter has BN=14RC=π2f3dBB_N=\frac{1}{4RC}=\frac\pi2f_{3dB}.

Impulse response of the optimum detector

An optimum detector is a receiver (filter followed by sampler and threshold device) that makes the decision with the minimum probability of error. For a known pulse in additive white Gaussian noise, this is achieved by the filter that maximises the output signal-to-noise ratio at the sampling instant; this optimum filter is the matched filter.

Set-up. The received signal is x(t)=s(t)+w(t)x(t)=s(t)+w(t), 0≤t≤T0\le t\le T, where s(t)s(t) is a known pulse and w(t)w(t) is white noise of two-sided PSD N0/2N_0/2. It passes through an LTI filter h(t)↔H(f)h(t)\leftrightarrow H(f) and is sampled at t=Tt=T:

 s(t)+w(t)    +--------+   y(t)    sample     y(T)
 ----------->|  h(t)  |---------->--/ ---->  decision
              +--------+           at t=T

Output y(t)=so(t)+no(t)y(t)=s_o(t)+n_o(t) with

so(T)=∫−∞∞H(f)S(f)ej2πfTdf,E[no2]=N02∫−∞∞∣H(f)∣2dfs_o(T)=\int_{-\infty}^{\infty}H(f)S(f)e^{j2\pi fT}df,\qquad E[n_o^2]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df

We want H(f)H(f) that maximises the output peak signal-to-noise ratio

η=∣so(T)∣2E[no2]=∣∫H(f)S(f)ej2πfTdf∣2N02∫∣H(f)∣2df\eta=\frac{\lvert s_o(T)\rvert^2}{E[n_o^2]}=\frac{\left\lvert\int H(f)S(f)e^{j2\pi fT}df\right\rvert^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}

Schwarz inequality: ∣∫ϕ1ϕ2 df∣2≤∫∣ϕ1∣2df∫∣ϕ2∣2df\left\lvert\int\phi_1\phi_2\,df\right\rvert^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df, with equality only when ϕ1(f)=k ϕ2∗(f)\phi_1(f)=k\,\phi_2^*(f). Put ϕ1=H(f)\phi_1=H(f) and ϕ2=S(f)ej2πfT\phi_2=S(f)e^{j2\pi fT}:

η≤∫∣H∣2df∫∣S∣2dfN02∫∣H∣2df=2N0∫−∞∞∣S(f)∣2df=2EN0\eta\le\frac{\int\lvert H\rvert^2df\int\lvert S\rvert^2df}{\frac{N_0}{2}\int\lvert H\rvert^2df}=\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert S(f)\rvert^2df=\frac{2E}{N_0}

(using Rayleigh's theorem, EE = pulse energy). The maximum is reached when

Hopt(f)=k S∗(f) e−j2πfTH_{opt}(f)=k\,S^*(f)\,e^{-j2\pi fT}

Impulse response. Take the inverse transform. For real s(t)s(t), S∗(f)=S(−f)S^*(f)=S(-f), which is the transform of s(−t)s(-t); the factor e−j2πfTe^{-j2\pi fT} is a delay of TT:

hopt(t)=k∫−∞∞S∗(f)e−j2πfTej2πftdf=k∫−∞∞S(−f)e−j2πf(T−t)df=k s(T−t)\begin{aligned} h_{opt}(t) &= k\int_{-\infty}^{\infty}S^*(f)e^{-j2\pi fT}e^{j2\pi ft}df \\ &= k\int_{-\infty}^{\infty}S(-f)e^{-j2\pi f(T-t)}df \\ &= k\,s(T-t) \end{aligned}

So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by TT. Choosing the sampling instant TT equal to the pulse duration makes h(t)h(t) causal.

   s(t)                      h(t) = s(T - t)
    |\                          /|
    | \                        / |
    |  \                      /  |
 ---+---+---> t          ---+---+---> t
    0   T                   0   T

Results: maximum SNR =2E/N0=2E/N_0 depends only on pulse energy, not on pulse shape; the filter's frequency response is the conjugate of the pulse spectrum, so it passes strongly the frequencies where the signal is strong.

  • 2081 Bhadra (CS II) · 7 marks

Evaluate the error probability in a binary communication system with appropriate expression.

Answer

Model. In a binary system, symbol 1 is sent as s1(t)s_1(t) and 0 as s2(t)s_2(t), each lasting TbT_b, with equal probability. Channel adds white Gaussian noise w(t)w(t) of PSD N0/2N_0/2. The receiver filter output is sampled at t=Tbt=T_b:

 s_i+w  +--------+ y(t)  sample  y  +---------+
 ------>| filter |------>-/ ------->| y > lam?|--> 1/0
        +--------+       t=Tb       +---------+

The sample is y=ai+ny=a_i+n, where a1,a2a_1, a_2 are the noise-free outputs (a1>a2a_1>a_2) and nn is Gaussian with zero mean and variance σ2\sigma^2. Conditional pdfs:

f(y∣1)=12πσe−(y−a1)2/2σ2,f(y∣0)=12πσe−(y−a2)2/2σ2f(y\mid1)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_1)^2/2\sigma^2},\qquad f(y\mid0)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_2)^2/2\sigma^2}
   f(y|0)            f(y|1)
     .--.             .--.
    /    \     |     /    \
   /      \    |    /      \
  /     ...\...|.../...     \
 --------a2---lam---a1----------> y
        Pe0 = area right of lam under f(y|0)
        Pe1 = area left  of lam under f(y|1)

Decision rule: choose 1 if y>λy>\lambda, else 0. For equal priors the optimum threshold is midway, λ=a1+a22\lambda=\frac{a_1+a_2}{2}.

Error when 0 is sent (y>λy>\lambda): with u=(y−a2)/2σu=(y-a_2)/\sqrt2\sigma,

Pe0=∫λ∞12πσe−(y−a2)2/2σ2dy=1π∫(λ−a2)/2σ∞e−u2du=12erfc⁡(a1−a222 σ)\begin{aligned} P_{e0} &= \int_{\lambda}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_2)^2/2\sigma^2}dy=\frac{1}{\sqrt{\pi}}\int_{(\lambda-a_2)/\sqrt2\sigma}^{\infty}e^{-u^2}du \\ &= \frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\,\sigma}\right) \end{aligned}

By symmetry Pe1P_{e1} (1 sent, y<λy<\lambda) is the same. The average error probability is

Pe=12Pe0+12Pe1=12erfc⁡(a1−a222 σ)=Q(a1−a22σ)P_e=\tfrac12P_{e0}+\tfrac12P_{e1}=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\,\sigma}\right)=Q\left(\frac{a_1-a_2}{2\sigma}\right)

where erfc⁡(x)=2π∫x∞e−u2du\operatorname{erfc}(x)=\frac{2}{\sqrt\pi}\int_x^{\infty}e^{-u^2}du and Q(x)=12erfc⁡(x/2)Q(x)=\frac12\operatorname{erfc}(x/\sqrt2).

With a matched filter (matched to s1−s2s_1-s_2), the maximum of (a1−a2)2σ2\frac{(a_1-a_2)^2}{\sigma^2} is 2EdN0\frac{2E_d}{N_0}, where Ed=∫0Tb[s1(t)−s2(t)]2dtE_d=\int_0^{T_b}[s_1(t)-s_2(t)]^2dt. Therefore

Pe=12erfc⁡(Ed4N0)P_e=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_d}{4N_0}}\right)

PeP_e depends only on the energy of the difference signal relative to N0N_0, not on pulse shape. Making the two signals as different as possible (antipodal) gives the lowest PeP_e.

Special cases:

Schemes1(t)s_1(t), s2(t)s_2(t)EdE_dPeP_e
Unipolar / ASK (OOK)Acos⁡ωctA\cos\omega_ct, 002Eb2E_b12erfc⁡Eb/2N0\frac12\operatorname{erfc}\sqrt{E_b/2N_0}
Polar / PSK±Acos⁡ωct\pm A\cos\omega_ct4Eb4E_b12erfc⁡Eb/N0\frac12\operatorname{erfc}\sqrt{E_b/N_0}
Coherent FSKAcos⁡ω1tA\cos\omega_1t, Acos⁡ω2tA\cos\omega_2t2Eb2E_b12erfc⁡Eb/2N0\frac12\operatorname{erfc}\sqrt{E_b/2N_0}

(EbE_b = average energy per bit.)

  • 2081 Baisakh (CS II) · 2+6 marks

What is Ergodic Stochastic Process? Derive the expression of impulse response of the matched filter.

Answer

Ergodic stochastic process

A stationary random process is ergodic if its time averages (computed from a single sample function over a long time) are equal to its ensemble averages (computed across all sample functions at one time):

⟨x(t)⟩=lim⁡T→∞1T∫−T/2T/2x(t)dt=E[X(t)]=mX\langle x(t)\rangle=\lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}x(t)dt=E[X(t)]=m_X

and similarly ⟨x(t)x(t+τ)⟩=RX(τ)\langle x(t)x(t+\tau)\rangle=R_X(\tau). Ergodicity lets us measure mean, power and autocorrelation of noise from one long record (e.g. with a DC meter or power meter). Every ergodic process is stationary, but not every stationary process is ergodic.

Impulse response of the matched filter

Set-up. The received signal is x(t)=s(t)+w(t)x(t)=s(t)+w(t), 0≤t≤T0\le t\le T, where s(t)s(t) is a known pulse and w(t)w(t) is white noise of two-sided PSD N0/2N_0/2. It passes through an LTI filter h(t)↔H(f)h(t)\leftrightarrow H(f) and is sampled at t=Tt=T:

 s(t)+w(t)    +--------+   y(t)    sample     y(T)
 ----------->|  h(t)  |---------->--/ ---->  decision
              +--------+           at t=T

Output y(t)=so(t)+no(t)y(t)=s_o(t)+n_o(t) with

so(T)=∫−∞∞H(f)S(f)ej2πfTdf,E[no2]=N02∫−∞∞∣H(f)∣2dfs_o(T)=\int_{-\infty}^{\infty}H(f)S(f)e^{j2\pi fT}df,\qquad E[n_o^2]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df

We want H(f)H(f) that maximises the output peak signal-to-noise ratio

η=∣so(T)∣2E[no2]=∣∫H(f)S(f)ej2πfTdf∣2N02∫∣H(f)∣2df\eta=\frac{\lvert s_o(T)\rvert^2}{E[n_o^2]}=\frac{\left\lvert\int H(f)S(f)e^{j2\pi fT}df\right\rvert^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}

Schwarz inequality: ∣∫ϕ1ϕ2 df∣2≤∫∣ϕ1∣2df∫∣ϕ2∣2df\left\lvert\int\phi_1\phi_2\,df\right\rvert^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df, with equality only when ϕ1(f)=k ϕ2∗(f)\phi_1(f)=k\,\phi_2^*(f). Put ϕ1=H(f)\phi_1=H(f) and ϕ2=S(f)ej2πfT\phi_2=S(f)e^{j2\pi fT}:

η≤∫∣H∣2df∫∣S∣2dfN02∫∣H∣2df=2N0∫−∞∞∣S(f)∣2df=2EN0\eta\le\frac{\int\lvert H\rvert^2df\int\lvert S\rvert^2df}{\frac{N_0}{2}\int\lvert H\rvert^2df}=\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert S(f)\rvert^2df=\frac{2E}{N_0}

(using Rayleigh's theorem, EE = pulse energy). The maximum is reached when

Hopt(f)=k S∗(f) e−j2πfTH_{opt}(f)=k\,S^*(f)\,e^{-j2\pi fT}

Impulse response. Take the inverse transform. For real s(t)s(t), S∗(f)=S(−f)S^*(f)=S(-f), which is the transform of s(−t)s(-t); the factor e−j2πfTe^{-j2\pi fT} is a delay of TT:

hopt(t)=k∫−∞∞S∗(f)e−j2πfTej2πftdf=k∫−∞∞S(−f)e−j2πf(T−t)df=k s(T−t)\begin{aligned} h_{opt}(t) &= k\int_{-\infty}^{\infty}S^*(f)e^{-j2\pi fT}e^{j2\pi ft}df \\ &= k\int_{-\infty}^{\infty}S(-f)e^{-j2\pi f(T-t)}df \\ &= k\,s(T-t) \end{aligned}

So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by TT. Choosing the sampling instant TT equal to the pulse duration makes h(t)h(t) causal.

   s(t)                      h(t) = s(T - t)
    |\                          /|
    | \                        / |
    |  \                      /  |
 ---+---+---> t          ---+---+---> t
    0   T                   0   T

Results: maximum SNR =2E/N0=2E/N_0 depends only on pulse energy, not on pulse shape; the filter's frequency response is the conjugate of the pulse spectrum, so it passes strongly the frequencies where the signal is strong.

  • 2081 Baisakh (CS II) · 6 marks

Show that the output signal of LTI (linear time invariant) system is also a wide sense stationary process if the input to it is also a WSSP.

Answer

Let a WSS process X(t)X(t) (mean mXm_X, autocorrelation RX(τ)R_X(\tau), PSD SX(f)S_X(f)) be applied to a stable LTI system with impulse response h(t)h(t) and H(f)H(f):

Y(t)=∫−∞∞h(α)X(t−α) dαY(t)=\int_{-\infty}^{\infty}h(\alpha)X(t-\alpha)\,d\alpha
  X(t) (WSS)   +--------------+   Y(t)
 ------------->| h(t) , H(f)  |------------->
  R_X, S_X     +--------------+  R_Y, S_Y

1. Mean of output. Expectation and integration can be interchanged:

E[Y(t)]=∫−∞∞h(α)E[X(t−α)] dα=mX∫−∞∞h(α) dα=mX H(0)(a constant, independent of t)\begin{aligned} E[Y(t)] &= \int_{-\infty}^{\infty}h(\alpha)E[X(t-\alpha)]\,d\alpha = m_X\int_{-\infty}^{\infty}h(\alpha)\,d\alpha \\ &= m_X\,H(0) \quad(\text{a constant, independent of } t) \end{aligned}

2. Autocorrelation of output.

RY(t,t+τ)=E[Y(t)Y(t+τ)]=E[∫h(α)X(t−α)dα∫h(β)X(t+τ−β)dβ]=∫ ⁣ ⁣∫h(α)h(β) E[X(t−α)X(t+τ−β)] dα dβ=∫ ⁣ ⁣∫h(α)h(β) RX(τ+α−β) dα dβ\begin{aligned} R_Y(t,t+\tau) &= E[Y(t)Y(t+\tau)] \\ &= E\left[\int h(\alpha)X(t-\alpha)d\alpha\int h(\beta)X(t+\tau-\beta)d\beta\right] \\ &= \int\!\!\int h(\alpha)h(\beta)\,E[X(t-\alpha)X(t+\tau-\beta)]\,d\alpha\,d\beta \\ &= \int\!\!\int h(\alpha)h(\beta)\,R_X(\tau+\alpha-\beta)\,d\alpha\,d\beta \end{aligned}

because XX is WSS, its autocorrelation depends only on the time difference (t+τ−β)−(t−α)=τ+α−β(t+\tau-\beta)-(t-\alpha)=\tau+\alpha-\beta. The result depends only on τ\tau, not on tt:

RY(τ)=h(τ)∗h(−τ)∗RX(τ)R_Y(\tau)=h(\tau)*h(-\tau)*R_X(\tau)

3. Mean-square value E[Y2]=RY(0)=∫ ⁣ ⁣∫h(α)h(β)RX(α−β)dαdβE[Y^2]=R_Y(0)=\int\!\!\int h(\alpha)h(\beta)R_X(\alpha-\beta)d\alpha d\beta is finite and constant.

Since the mean is constant and the autocorrelation depends only on τ\tau, Y(t)Y(t) is also wide-sense stationary.

4. Output PSD. Taking the Fourier transform of RY(τ)R_Y(\tau) (h(τ)→H(f)h(\tau)\to H(f), h(−τ)→H∗(f)h(-\tau)\to H^*(f)):

SY(f)=H(f)H∗(f)SX(f)=∣H(f)∣2SX(f)S_Y(f)=H(f)H^*(f)S_X(f)=\lvert H(f)\rvert^2S_X(f)

and the output power is PY=RY(0)=∫−∞∞∣H(f)∣2SX(f) dfP_Y=R_Y(0)=\int_{-\infty}^{\infty}\lvert H(f)\rvert^2S_X(f)\,df. The phase of H(f)H(f) has no effect on the output PSD.

Example. White noise (SX=N0/2S_X=N_0/2, WSS) through an RC low-pass filter, ∣H(f)∣2=11+(2πfRC)2\lvert H(f)\rvert^2=\frac{1}{1+(2\pi fRC)^2}:

SY(f)=N0/21+(2πfRC)2,RY(τ)=N04RCe−∣τ∣/RCS_Y(f)=\frac{N_0/2}{1+(2\pi fRC)^2},\qquad R_Y(\tau)=\frac{N_0}{4RC}e^{-\lvert\tau\rvert/RC}

RYR_Y depends only on τ\tau, confirming that the output is WSS; its power RY(0)=N0/4RCR_Y(0)=N_0/4RC is now finite.

  • 2081 Baisakh (CS II) · 5+2 marks

Evaluate the noise performance of the coherent detection of DSB-FC using necessary derivations. Use the result to evaluate the noise performance of the single-tone DSB-FC modulation with 100% modulation.

Answer

Noise performance of coherent DSB-FC detection

Noise performance is measured by the output SNR and the detection gain γ=SNRo/SNRi\gamma=\text{SNR}_o/\text{SNR}_i (or figure of merit, with SNR referred to the message band WW).

 s(t)+w(t) +-----+       +-----+   +-------+
 --------->| BPF |-->(x)-->| LPF |-->|  DC   |--> y(t)
           | 2W  |    ^    |  W  |   | block |
           +-----+    |    +-----+   +-------+
                  cos(wc t)

Receiver: BPF (BT=2WB_T=2W) → product detector with cos⁡ωct\cos\omega_ct → LPF (WW) → DC block. Noise PSD N0/2N_0/2, narrowband noise n(t)=nccos⁡ωct−nssin⁡ωctn(t)=n_c\cos\omega_ct-n_s\sin\omega_ct with nc2‾=2N0W\overline{n_c^2}=2N_0W.

  • Signal power =Ac22(1+ka2P)=\frac{A_c^2}{2}(1+k_a^2P); noise in 2W2W =2N0W=2N_0W:
SNRi=Ac2(1+ka2P)4N0W\text{SNR}_i=\frac{A_c^2(1+k_a^2P)}{4N_0W}
  • Coherent detection (or envelope detection at high SNR) and DC blocking:
y(t)=Ac2kam(t)+12nc(t) ⇒ SNRo=Ac2ka2P/42N0W/4=Ac2ka2P2N0Wy(t)=\frac{A_c}{2}k_am(t)+\frac12n_c(t)\ \Rightarrow\ \text{SNR}_o=\frac{A_c^2k_a^2P/4}{2N_0W/4}=\frac{A_c^2k_a^2P}{2N_0W} γDSB-FC=SNRoSNRi=2ka2P1+ka2P,FOM=ka2P1+ka2P\gamma_{DSB\text{-}FC}=\frac{\text{SNR}_o}{\text{SNR}_i}=\frac{2k_a^2P}{1+k_a^2P},\qquad \text{FOM}=\frac{k_a^2P}{1+k_a^2P}

The carrier uses power but carries no information, so the gain is reduced.

Single-tone DSB-FC with 100% modulation

Single tone, 100% modulation. m(t)=Amcos⁡ωmtm(t)=A_m\cos\omega_mt, μ=kaAm=1\mu=k_aA_m=1, so ka2P=μ22=12k_a^2P=\frac{\mu^2}{2}=\frac12:

γDSB-FC=2(μ2/2)1+μ2/2=2μ22+μ2=23≈0.667FOM=μ22+μ2=13\begin{aligned} \gamma_{DSB\text{-}FC} &= \frac{2(\mu^2/2)}{1+\mu^2/2}=\frac{2\mu^2}{2+\mu^2}=\frac{2}{3}\approx0.667 \\ \text{FOM} &= \frac{\mu^2}{2+\mu^2}=\frac13 \end{aligned}

So with 100% single-tone modulation the output SNR is only one-third of that of DSB-SC or SSB for the same transmitted power (10log⁡103=4.7710\log_{10}3=4.77 dB worse), and the detection gain is less than 1, because two-thirds of the power is in the carrier.

  • 2081 Baisakh (CS II) · 6 marks

Derive the bit error probability for the coherent binary FSK system.

Answer

In binary FSK, bit 1 and bit 0 are sent as two different carrier frequencies f1f_1 and f2f_2 with the same amplitude.

Coherent BFSK signals over 0≤t≤Tb0\le t\le T_b:

s1(t)=Acos⁡ω1t  (1),s2(t)=Acos⁡ω2t  (0),Eb=A2Tb2s_1(t)=A\cos\omega_1t\ \ (1),\qquad s_2(t)=A\cos\omega_2t\ \ (0),\qquad E_b=\frac{A^2T_b}{2}

The tones are chosen orthogonal over TbT_b: ∫0Tbcos⁡ω1tcos⁡ω2t dt=0\int_0^{T_b}\cos\omega_1t\cos\omega_2t\,dt=0 (e.g. ∣f1−f2∣=n/2Tb\lvert f_1-f_2\rvert=n/2T_b).

Coherent receiver: two correlators, one per tone; their outputs are subtracted and compared with zero.

        +->(x)--> integ 0..Tb --> y1 --+
        |   ^ cos w1t                  |(+)
 r(t) --+                            (sum)--> l = y1 - y2
        |   v cos w2t                  |(-)   l > 0 ? 1 : 0
        +->(x)--> integ 0..Tb --> y2 --+

Signal part of ll.

  • 1 sent: y1=ATb2y_1=\frac{AT_b}{2}, y2=0y_2=0 (orthogonality) so l=+ATb2l=+\frac{AT_b}{2}.
  • 0 sent: y1=0y_1=0, y2=ATb2y_2=\frac{AT_b}{2} so l=−ATb2l=-\frac{AT_b}{2}.

Noise part. n1=∫wcos⁡ω1t dtn_1=\int w\cos\omega_1t\,dt and n2=∫wcos⁡ω2t dtn_2=\int w\cos\omega_2t\,dt each have variance N0Tb4\frac{N_0T_b}{4} and are uncorrelated (orthogonal tones), so n=n1−n2n=n_1-n_2 has

σ2=N0Tb4+N0Tb4=N0Tb2\sigma^2=\frac{N_0T_b}{4}+\frac{N_0T_b}{4}=\frac{N_0T_b}{2}

Error probability. ll is Gaussian with mean ±ATb/2\pm AT_b/2 and variance σ2\sigma^2; threshold 0. For 0 sent, error if l>0l>0:

Pe=12erfc⁡(ATb/22 σ)=12erfc⁡(ATb/2N0Tb)=12erfc⁡(A2Tb4N0)\begin{aligned} P_e &= \frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt2\,\sigma}\right)=\frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt{N_0T_b}}\right) \\ &= \frac12\operatorname{erfc}\left(\sqrt{\frac{A^2T_b}{4N_0}}\right) \end{aligned}

With Eb=A2Tb/2E_b=A^2T_b/2:

Pe(coherent BFSK)=12erfc⁡(Eb2N0)=Q(EbN0)P_e(\text{coherent BFSK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)

Check with the general formula: Ed=∫(s1−s2)2dt=Eb+Eb=2EbE_d=\int(s_1-s_2)^2dt=E_b+E_b=2E_b, so 12erfc⁡Ed/4N0=12erfc⁡Eb/2N0\frac12\operatorname{erfc}\sqrt{E_d/4N_0}=\frac12\operatorname{erfc}\sqrt{E_b/2N_0}.

Comparison: BFSK has the same PeP_e as ASK (for equal average energy) and needs 3 dB more Eb/N0E_b/N_0 than BPSK, because orthogonal signals are 2Eb\sqrt{2E_b} apart while antipodal ones are 2Eb2\sqrt{E_b} apart. Non-coherent FSK gives Pe=12e−Eb/2N0P_e=\frac12e^{-E_b/2N_0}, which is slightly worse but needs no carrier phase.

  • 2080 Bhadra (CS II) · 8 marks

Derive the expression for evaluating error probability in Binary baseband system.

Answer

In a binary baseband system, bits are sent as pulses (e.g. polar NRZ ±A\pm A) over a channel adding white Gaussian noise. The receiver samples the filtered signal once per bit and compares it with a threshold; an error occurs when noise pushes the sample across the threshold.

Model. In a binary system, symbol 1 is sent as s1(t)s_1(t) and 0 as s2(t)s_2(t), each lasting TbT_b, with equal probability. Channel adds white Gaussian noise w(t)w(t) of PSD N0/2N_0/2. The receiver filter output is sampled at t=Tbt=T_b:

 s_i+w  +--------+ y(t)  sample  y  +---------+
 ------>| filter |------>-/ ------->| y > lam?|--> 1/0
        +--------+       t=Tb       +---------+

The sample is y=ai+ny=a_i+n, where a1,a2a_1, a_2 are the noise-free outputs (a1>a2a_1>a_2) and nn is Gaussian with zero mean and variance σ2\sigma^2. Conditional pdfs:

f(y∣1)=12πσe−(y−a1)2/2σ2,f(y∣0)=12πσe−(y−a2)2/2σ2f(y\mid1)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_1)^2/2\sigma^2},\qquad f(y\mid0)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_2)^2/2\sigma^2}
   f(y|0)            f(y|1)
     .--.             .--.
    /    \     |     /    \
   /      \    |    /      \
  /     ...\...|.../...     \
 --------a2---lam---a1----------> y
        Pe0 = area right of lam under f(y|0)
        Pe1 = area left  of lam under f(y|1)

Decision rule: choose 1 if y>λy>\lambda, else 0. For equal priors the optimum threshold is midway, λ=a1+a22\lambda=\frac{a_1+a_2}{2}.

Error when 0 is sent (y>λy>\lambda): with u=(y−a2)/2σu=(y-a_2)/\sqrt2\sigma,

Pe0=∫λ∞12πσe−(y−a2)2/2σ2dy=1π∫(λ−a2)/2σ∞e−u2du=12erfc⁡(a1−a222 σ)\begin{aligned} P_{e0} &= \int_{\lambda}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_2)^2/2\sigma^2}dy=\frac{1}{\sqrt{\pi}}\int_{(\lambda-a_2)/\sqrt2\sigma}^{\infty}e^{-u^2}du \\ &= \frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\,\sigma}\right) \end{aligned}

By symmetry Pe1P_{e1} (1 sent, y<λy<\lambda) is the same. The average error probability is

Pe=12Pe0+12Pe1=12erfc⁡(a1−a222 σ)=Q(a1−a22σ)P_e=\tfrac12P_{e0}+\tfrac12P_{e1}=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\,\sigma}\right)=Q\left(\frac{a_1-a_2}{2\sigma}\right)

where erfc⁡(x)=2π∫x∞e−u2du\operatorname{erfc}(x)=\frac{2}{\sqrt\pi}\int_x^{\infty}e^{-u^2}du and Q(x)=12erfc⁡(x/2)Q(x)=\frac12\operatorname{erfc}(x/\sqrt2).

With a matched filter (matched to s1−s2s_1-s_2), the maximum of (a1−a2)2σ2\frac{(a_1-a_2)^2}{\sigma^2} is 2EdN0\frac{2E_d}{N_0}, where Ed=∫0Tb[s1(t)−s2(t)]2dtE_d=\int_0^{T_b}[s_1(t)-s_2(t)]^2dt. Therefore

Pe=12erfc⁡(Ed4N0)P_e=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_d}{4N_0}}\right)

PeP_e depends only on the energy of the difference signal relative to N0N_0, not on pulse shape. Making the two signals as different as possible (antipodal) gives the lowest PeP_e.

Polar NRZ example. Levels a1=+Aa_1=+A, a2=−Aa_2=-A, threshold 0: Pe=12erfc⁡(A2σ)P_e=\frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\sigma}\right); with matched filter, Pe=12erfc⁡Eb/N0P_e=\frac12\operatorname{erfc}\sqrt{E_b/N_0}. Unipolar (AA, 0) needs twice the average power for the same PeP_e.

  • 2080 Baisakh (CS II) · 4+6 marks

Define Noise Equivalent bandwidth. Find noise equivalent bandwidth of the first order RC low pass filter.

Answer

Noise equivalent bandwidth

The noise equivalent bandwidth (NEB) BNB_N of a filter is the bandwidth of an ideal rectangular filter, with the same peak (usually zero-frequency or centre) gain ∣H(0)∣\lvert H(0)\rvert, that passes the same total noise power as the actual filter when both are fed with white noise.

White noise N0/2N_0/2 into H(f)H(f) gives output power

Pactual=N02∫−∞∞∣H(f)∣2df=N0∫0∞∣H(f)∣2dfP_{actual}=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df=N_0\int_0^{\infty}\lvert H(f)\rvert^2df

The ideal filter of gain ∣H(0)∣\lvert H(0)\rvert and bandwidth BNB_N gives Pideal=N0BN∣H(0)∣2P_{ideal}=N_0B_N\lvert H(0)\rvert^2. Equating:

BN=∫0∞∣H(f)∣2df∣H(0)∣2B_N=\frac{\int_0^{\infty}\lvert H(f)\rvert^2df}{\lvert H(0)\rvert^2}
 |H(f)|^2
 |H(0)|^2 +---------+     equal areas
          |  ...    |
          |      .. |
          |        .|..
          |         |  ......
        --+---------+----------> f
          0        B_N

NEB lets us write the output noise power simply as N=N0BN∣H(0)∣2N=N_0B_N\lvert H(0)\rvert^2 (or kTBNkTB_N per unit gain), which is used in noise figure and SNR calculations. For a band-pass filter ∣H(0)∣\lvert H(0)\rvert is replaced by the centre-frequency gain ∣H(f0)∣\lvert H(f_0)\rvert.

NEB of first-order RC low-pass filter

  vin o---/\/\/\---+---o vout
           R       |
                  === C
                   |
  gnd o------------+---o

For the RC low-pass filter (series RR, shunt CC):

H(f)=11+j2πfRC,∣H(f)∣2=11+(2πfRC)2,∣H(0)∣=1H(f)=\frac{1}{1+j2\pi fRC},\qquad \lvert H(f)\rvert^2=\frac{1}{1+(2\pi fRC)^2},\qquad \lvert H(0)\rvert=1 BN=∫0∞df1+(2πfRC)2(let u=2πfRC, df=du2πRC)=12πRC∫0∞du1+u2=12πRC⋅π2=14RC\begin{aligned} B_N &= \int_0^{\infty}\frac{df}{1+(2\pi fRC)^2}\qquad (\text{let } u=2\pi fRC,\ df=\tfrac{du}{2\pi RC}) \\ &= \frac{1}{2\pi RC}\int_0^{\infty}\frac{du}{1+u^2}=\frac{1}{2\pi RC}\cdot\frac{\pi}{2} \\ &= \frac{1}{4RC} \end{aligned}

Since the 3 dB bandwidth is f3dB=12πRCf_{3dB}=\frac{1}{2\pi RC},

BN=π2f3dB≈1.57 f3dBB_N=\frac{\pi}{2}f_{3dB}\approx1.57\,f_{3dB}

The NEB is larger than the 3 dB bandwidth because the RC response rolls off slowly (only −20-20 dB/decade) and passes noise well beyond f3dBf_{3dB}. Output noise power =N0BN=N04RC=N_0B_N=\frac{N_0}{4RC}.

Example: R=1 kΩR=1\ \text{k}\Omega, C=1 μFC=1\ \mu\text{F}: f3dB=159.2f_{3dB}=159.2 Hz, BN=14×10−3=250B_N=\frac{1}{4\times10^{-3}}=250 Hz.

  • 2080 Baisakh (CS II) · 5+5 marks

Compare A.M and F.M in terms of power efficiency and system complexity. Calculate the error probability of coherent ASK.

Answer

AM vs FM: power efficiency and complexity

PointAM (DSB-FC)FM
Transmitted powerVaries with modulation, Pc(1+μ2/2)P_c(1+\mu^2/2)Constant, Ac2/2A_c^2/2, independent of modulation
Useful (sideband) powerAt most 1/3 of total (μ=1\mu=1)All power is useful; carrier power is redistributed to sidebands
Power amplifierMust be linear (class A/B), lower efficiencyClass C (non-linear), high efficiency, since envelope is constant
Noise performanceFOM ≤1/3\le 1/3FOM =32β2=\frac32\beta^2; much better for β>0.5\beta>0.5
Bandwidth2W2W2(β+1)W2(\beta+1)W (Carson), much wider
Transmitter complexitySimple (high-level collector/plate modulation)More complex (VCO, AFC/frequency stabilisation, multipliers)
Receiver complexityVery simple (diode envelope detector)Needs limiter, discriminator/PLL, de-emphasis
Noise immunityAmplitude noise directly affects outputLimiter removes amplitude noise; capture effect

So FM is more power efficient (better SNR per watt and efficient class C amplifiers) but uses more bandwidth and more complex circuits; AM is simpler and cheaper but wastes power in the carrier.

Error probability of coherent ASK

ASK (on–off keying) signals over 0≤t≤Tb0\le t\le T_b:

s1(t)=Acos⁡ωct  (bit 1),s2(t)=0  (bit 0)s_1(t)=A\cos\omega_ct\ \ (\text{bit }1),\qquad s_2(t)=0\ \ (\text{bit }0)

Coherent receiver (correlator = matched filter):

 r(t) -->(x)--> integrate --> sample --> y > lam ? 1 : 0
          ^       0..Tb       t=Tb
          |
       cos(wc t)  (local carrier, in phase)

Correlator output at t=Tbt=T_b, with r(t)=si(t)+w(t)r(t)=s_i(t)+w(t):

y=∫0Tbr(t)cos⁡ωct dt=ai+ny=\int_0^{T_b}r(t)\cos\omega_ct\,dt=a_i+n
  • Bit 1: a1=∫0TbAcos⁡2ωct dt=ATb2a_1=\int_0^{T_b}A\cos^2\omega_ct\,dt=\frac{AT_b}{2} (for fcf_c an integer multiple of 1/Tb1/T_b).
  • Bit 0: a2=0a_2=0.
  • Noise: n=∫0Tbw(t)cos⁡ωct dtn=\int_0^{T_b}w(t)\cos\omega_ct\,dt is Gaussian, mean 0, variance
σ2=∫0Tb ⁣ ⁣∫0TbN02δ(t−u)cos⁡ωctcos⁡ωcu dt du=N02⋅Tb2=N0Tb4\sigma^2=\int_0^{T_b}\!\!\int_0^{T_b}\frac{N_0}{2}\delta(t-u)\cos\omega_ct\cos\omega_cu\,dt\,du=\frac{N_0}{2}\cdot\frac{T_b}{2}=\frac{N_0T_b}{4}

Threshold midway: λ=ATb4\lambda=\frac{AT_b}{4}. Using the binary result Pe=12erfc⁡(a1−a222σ)P_e=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\sigma}\right):

a1−a222 σ=ATb/222N0Tb/4=ATb22N0=A2Tb8N0\begin{aligned} \frac{a_1-a_2}{2\sqrt2\,\sigma} &= \frac{AT_b/2}{2\sqrt2\sqrt{N_0T_b/4}}=\frac{A\sqrt{T_b}}{2\sqrt{2N_0}}=\sqrt{\frac{A^2T_b}{8N_0}} \end{aligned}

Energy of a "1" pulse is E1=A2Tb2E_1=\frac{A^2T_b}{2}; the average energy per bit (half the bits are 0) is Eb=E12=A2Tb4E_b=\frac{E_1}{2}=\frac{A^2T_b}{4}. Hence A2Tb8N0=Eb2N0\frac{A^2T_b}{8N_0}=\frac{E_b}{2N_0} and

Pe(ASK)=12erfc⁡(Eb2N0)=Q(EbN0)P_e(\text{ASK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)

(In terms of peak energy E1E_1: Pe=12erfc⁡E1/4N0P_e=\frac12\operatorname{erfc}\sqrt{E_1/4N_0}.)

  • 2079 Bhadra (CS II) · 5+5 marks

Define matched filter. Find the impulse response of optimum detector in the presence of additive white noise.

Answer

Matched filter

A matched filter is a linear filter whose impulse response is matched to (a time-reversed, delayed copy of) a known input pulse s(t)s(t), h(t)=k s(T−t)h(t)=k\,s(T-t), so that it gives the maximum possible peak signal-to-noise ratio at the sampling instant t=Tt=T when the pulse is received in additive white noise. It is the optimum detector for digital signals in AWGN; in practice it is built as a correlator.

Key features: maximum output SNR =2E/N0=2E/N_0; H(f)=kS∗(f)e−j2πfTH(f)=kS^*(f)e^{-j2\pi fT}; it maximises SNR but does not preserve pulse shape (output is the autocorrelation of the pulse).

Impulse response of optimum detector in additive white noise

An optimum detector is a receiver (filter followed by sampler and threshold device) that makes the decision with the minimum probability of error. For a known pulse in additive white Gaussian noise, this is achieved by the filter that maximises the output signal-to-noise ratio at the sampling instant; this optimum filter is the matched filter.

Set-up. The received signal is x(t)=s(t)+w(t)x(t)=s(t)+w(t), 0≤t≤T0\le t\le T, where s(t)s(t) is a known pulse and w(t)w(t) is white noise of two-sided PSD N0/2N_0/2. It passes through an LTI filter h(t)↔H(f)h(t)\leftrightarrow H(f) and is sampled at t=Tt=T:

 s(t)+w(t)    +--------+   y(t)    sample     y(T)
 ----------->|  h(t)  |---------->--/ ---->  decision
              +--------+           at t=T

Output y(t)=so(t)+no(t)y(t)=s_o(t)+n_o(t) with

so(T)=∫−∞∞H(f)S(f)ej2πfTdf,E[no2]=N02∫−∞∞∣H(f)∣2dfs_o(T)=\int_{-\infty}^{\infty}H(f)S(f)e^{j2\pi fT}df,\qquad E[n_o^2]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df

We want H(f)H(f) that maximises the output peak signal-to-noise ratio

η=∣so(T)∣2E[no2]=∣∫H(f)S(f)ej2πfTdf∣2N02∫∣H(f)∣2df\eta=\frac{\lvert s_o(T)\rvert^2}{E[n_o^2]}=\frac{\left\lvert\int H(f)S(f)e^{j2\pi fT}df\right\rvert^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}

Schwarz inequality: ∣∫ϕ1ϕ2 df∣2≤∫∣ϕ1∣2df∫∣ϕ2∣2df\left\lvert\int\phi_1\phi_2\,df\right\rvert^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df, with equality only when ϕ1(f)=k ϕ2∗(f)\phi_1(f)=k\,\phi_2^*(f). Put ϕ1=H(f)\phi_1=H(f) and ϕ2=S(f)ej2πfT\phi_2=S(f)e^{j2\pi fT}:

η≤∫∣H∣2df∫∣S∣2dfN02∫∣H∣2df=2N0∫−∞∞∣S(f)∣2df=2EN0\eta\le\frac{\int\lvert H\rvert^2df\int\lvert S\rvert^2df}{\frac{N_0}{2}\int\lvert H\rvert^2df}=\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert S(f)\rvert^2df=\frac{2E}{N_0}

(using Rayleigh's theorem, EE = pulse energy). The maximum is reached when

Hopt(f)=k S∗(f) e−j2πfTH_{opt}(f)=k\,S^*(f)\,e^{-j2\pi fT}

Impulse response. Take the inverse transform. For real s(t)s(t), S∗(f)=S(−f)S^*(f)=S(-f), which is the transform of s(−t)s(-t); the factor e−j2πfTe^{-j2\pi fT} is a delay of TT:

hopt(t)=k∫−∞∞S∗(f)e−j2πfTej2πftdf=k∫−∞∞S(−f)e−j2πf(T−t)df=k s(T−t)\begin{aligned} h_{opt}(t) &= k\int_{-\infty}^{\infty}S^*(f)e^{-j2\pi fT}e^{j2\pi ft}df \\ &= k\int_{-\infty}^{\infty}S(-f)e^{-j2\pi f(T-t)}df \\ &= k\,s(T-t) \end{aligned}

So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by TT. Choosing the sampling instant TT equal to the pulse duration makes h(t)h(t) causal.

   s(t)                      h(t) = s(T - t)
    |\                          /|
    | \                        / |
    |  \                      /  |
 ---+---+---> t          ---+---+---> t
    0   T                   0   T

Results: maximum SNR =2E/N0=2E/N_0 depends only on pulse energy, not on pulse shape; the filter's frequency response is the conjugate of the pulse spectrum, so it passes strongly the frequencies where the signal is strong.

  • 2076 Chaitra (CS II) · 4+6 marks

Explain threshold effect in detection of FM signal. Derive the expression of error probability for coherent detection of Phase Shift Keying (PSK).

Answer

Threshold effect in FM detection

Threshold effect in FM. The FM improvement SNRo∝β2\text{SNR}_o\propto\beta^2 holds only when the carrier is much stronger than the noise at the discriminator input. When the carrier-to-noise ratio (CNR) falls below a threshold, about 10 dB (typically 10–13 dB), the noise phasor sometimes becomes larger than the carrier and the resultant phasor encircles the origin, causing a sudden 2π2\pi phase jump.

  high CNR: noise wobbles     low CNR: resultant can swing
  the tip slightly            around origin -> 2pi jump
        .-.                         .---.
  0 -->( * )                   0 --( *   )
        '-'                         '---'

Each 2π2\pi jump produces an impulse (spike) in the discriminator output, heard as clicks. The click rate rises rapidly as CNR drops, so the output SNR falls much faster than linearly and the signal is mutilated. Above threshold, FM is excellent; below it, FM is worse than AM. Wider deviation (larger β\beta) means larger BTB_T, more noise and a higher threshold.

 SNRo(dB)            . FM above threshold
     |             .   (slope 1, offset 3/2 beta^2)
     |           .
     |         .
     |       . <- threshold knee (CNR ~ 10 dB)
     |     .:
     |   . :     AM / baseband
     |  .  :
     +------------------> SNRi (dB)

It is reduced by FM feedback demodulators, PLL demodulators and pre-emphasis/de-emphasis.

Error probability of coherent PSK

BPSK signals over 0≤t≤Tb0\le t\le T_b (antipodal):

s1(t)=Acos⁡ωct  (1),s2(t)=−Acos⁡ωct  (0),Eb=A2Tb2s_1(t)=A\cos\omega_ct\ \ (1),\qquad s_2(t)=-A\cos\omega_ct\ \ (0),\qquad E_b=\frac{A^2T_b}{2}

Coherent receiver: multiply by the locally generated cos⁡ωct\cos\omega_ct, integrate over TbT_b, sample, compare with threshold λ=0\lambda=0.

 r(t) -->(x)--> integrate --> sample --> y > 0 ? 1 : 0
          ^       0..Tb       t=Tb
          |
       cos(wc t)  (from carrier recovery)

Correlator output y=ai+ny=a_i+n with

a1=+ATb2,a2=−ATb2,σ2=N0Tb4a_1=+\frac{AT_b}{2},\qquad a_2=-\frac{AT_b}{2},\qquad \sigma^2=\frac{N_0T_b}{4}

The pdfs of yy are Gaussian centred at ±ATb/2\pm AT_b/2. Error when 0 is sent (y>0y>0):

Pe=∫0∞12πσe−(y+ATb/2)2/2σ2dy=12erfc⁡(ATb/22 σ)=12erfc⁡(ATb/2N0Tb/2)=12erfc⁡(A2Tb2N0)\begin{aligned} P_e &= \int_0^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-(y+AT_b/2)^2/2\sigma^2}dy=\frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt2\,\sigma}\right) \\ &= \frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt{N_0T_b/2}}\right)=\frac12\operatorname{erfc}\left(\sqrt{\frac{A^2T_b}{2N_0}}\right) \end{aligned}

By symmetry the error for a sent 1 is the same, so

Pe(BPSK)=12erfc⁡(EbN0)=Q(2EbN0)P_e(\text{BPSK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{N_0}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)

BPSK is the best binary scheme: it needs 3 dB less Eb/N0E_b/N_0 than coherent ASK or FSK for the same PeP_e. Example: Eb/N0=10E_b/N_0=10 dB gives Pe=Q(20)≈3.9×10−6P_e=Q(\sqrt{20})\approx3.9\times10^{-6}.

  • 2076 Asoj (CS II) · 2+10 marks

What do you mean by Stochastic Process? Explain with necessary derivation passage of wide-sense random signals through a LTI.

Answer

Stochastic process

A stochastic (random) process X(t)X(t) is an ensemble (collection) of time functions, one sample function for each outcome of a random experiment. At any fixed time t1t_1, X(t1)X(t_1) is a random variable. Example: the noise voltage at the output of many identical receivers. It is described by its mean mX(t)=E[X(t)]m_X(t)=E[X(t)] and autocorrelation RX(t1,t2)=E[X(t1)X(t2)]R_X(t_1,t_2)=E[X(t_1)X(t_2)].

A process is wide-sense stationary (WSS) if (i) its mean is constant and (ii) its autocorrelation depends only on the time difference τ=t2−t1\tau=t_2-t_1: RX(t,t+τ)=RX(τ)R_X(t,t+\tau)=R_X(\tau).

Passage of WSS random signals through an LTI system

Let a WSS process X(t)X(t) (mean mXm_X, autocorrelation RX(τ)R_X(\tau), PSD SX(f)S_X(f)) be applied to a stable LTI system with impulse response h(t)h(t) and H(f)H(f):

Y(t)=∫−∞∞h(α)X(t−α) dαY(t)=\int_{-\infty}^{\infty}h(\alpha)X(t-\alpha)\,d\alpha
  X(t) (WSS)   +--------------+   Y(t)
 ------------->| h(t) , H(f)  |------------->
  R_X, S_X     +--------------+  R_Y, S_Y

1. Mean of output. Expectation and integration can be interchanged:

E[Y(t)]=∫−∞∞h(α)E[X(t−α)] dα=mX∫−∞∞h(α) dα=mX H(0)(a constant, independent of t)\begin{aligned} E[Y(t)] &= \int_{-\infty}^{\infty}h(\alpha)E[X(t-\alpha)]\,d\alpha = m_X\int_{-\infty}^{\infty}h(\alpha)\,d\alpha \\ &= m_X\,H(0) \quad(\text{a constant, independent of } t) \end{aligned}

2. Autocorrelation of output.

RY(t,t+τ)=E[Y(t)Y(t+τ)]=E[∫h(α)X(t−α)dα∫h(β)X(t+τ−β)dβ]=∫ ⁣ ⁣∫h(α)h(β) E[X(t−α)X(t+τ−β)] dα dβ=∫ ⁣ ⁣∫h(α)h(β) RX(τ+α−β) dα dβ\begin{aligned} R_Y(t,t+\tau) &= E[Y(t)Y(t+\tau)] \\ &= E\left[\int h(\alpha)X(t-\alpha)d\alpha\int h(\beta)X(t+\tau-\beta)d\beta\right] \\ &= \int\!\!\int h(\alpha)h(\beta)\,E[X(t-\alpha)X(t+\tau-\beta)]\,d\alpha\,d\beta \\ &= \int\!\!\int h(\alpha)h(\beta)\,R_X(\tau+\alpha-\beta)\,d\alpha\,d\beta \end{aligned}

because XX is WSS, its autocorrelation depends only on the time difference (t+τ−β)−(t−α)=τ+α−β(t+\tau-\beta)-(t-\alpha)=\tau+\alpha-\beta. The result depends only on τ\tau, not on tt:

RY(τ)=h(τ)∗h(−τ)∗RX(τ)R_Y(\tau)=h(\tau)*h(-\tau)*R_X(\tau)

3. Mean-square value E[Y2]=RY(0)=∫ ⁣ ⁣∫h(α)h(β)RX(α−β)dαdβE[Y^2]=R_Y(0)=\int\!\!\int h(\alpha)h(\beta)R_X(\alpha-\beta)d\alpha d\beta is finite and constant.

Since the mean is constant and the autocorrelation depends only on τ\tau, Y(t)Y(t) is also wide-sense stationary.

4. Output PSD. Taking the Fourier transform of RY(τ)R_Y(\tau) (h(τ)→H(f)h(\tau)\to H(f), h(−τ)→H∗(f)h(-\tau)\to H^*(f)):

SY(f)=H(f)H∗(f)SX(f)=∣H(f)∣2SX(f)S_Y(f)=H(f)H^*(f)S_X(f)=\lvert H(f)\rvert^2S_X(f)

and the output power is PY=RY(0)=∫−∞∞∣H(f)∣2SX(f) dfP_Y=R_Y(0)=\int_{-\infty}^{\infty}\lvert H(f)\rvert^2S_X(f)\,df. The phase of H(f)H(f) has no effect on the output PSD.

Example (white noise through RC low-pass filter). Input SX(f)=N02S_X(f)=\frac{N_0}{2}, ∣H(f)∣2=11+(2πfRC)2\lvert H(f)\rvert^2=\frac{1}{1+(2\pi fRC)^2}:

SY(f)=N0/21+(2πfRC)2,RY(τ)=N04RCe−∣τ∣/RC,PY=N04RCS_Y(f)=\frac{N_0/2}{1+(2\pi fRC)^2},\qquad R_Y(\tau)=\frac{N_0}{4RC}e^{-\lvert\tau\rvert/RC},\qquad P_Y=\frac{N_0}{4RC}

The filter "colours" the noise: the output is correlated over about one time constant RCRC, and its power becomes finite.

Cross-correlation between input and output (useful for system identification): RXY(τ)=h(τ)∗RX(τ)R_{XY}(\tau)=h(\tau)*R_X(\tau); with white-noise input RXY(τ)=N02h(τ)R_{XY}(\tau)=\frac{N_0}{2}h(\tau), so the impulse response can be measured by correlating output with input noise.

  • 2076 Asoj (CS II) · 8 marks

Derive the expression for evaluating error probability in binary baseband system and compare it with M-ary system.

Answer

Error probability in binary baseband system

Model. In a binary system, symbol 1 is sent as s1(t)s_1(t) and 0 as s2(t)s_2(t), each lasting TbT_b, with equal probability. Channel adds white Gaussian noise w(t)w(t) of PSD N0/2N_0/2. The receiver filter output is sampled at t=Tbt=T_b:

 s_i+w  +--------+ y(t)  sample  y  +---------+
 ------>| filter |------>-/ ------->| y > lam?|--> 1/0
        +--------+       t=Tb       +---------+

The sample is y=ai+ny=a_i+n, where a1,a2a_1, a_2 are the noise-free outputs (a1>a2a_1>a_2) and nn is Gaussian with zero mean and variance σ2\sigma^2. Conditional pdfs:

f(y∣1)=12πσe−(y−a1)2/2σ2,f(y∣0)=12πσe−(y−a2)2/2σ2f(y\mid1)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_1)^2/2\sigma^2},\qquad f(y\mid0)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_2)^2/2\sigma^2}
   f(y|0)            f(y|1)
     .--.             .--.
    /    \     |     /    \
   /      \    |    /      \
  /     ...\...|.../...     \
 --------a2---lam---a1----------> y
        Pe0 = area right of lam under f(y|0)
        Pe1 = area left  of lam under f(y|1)

Decision rule: choose 1 if y>λy>\lambda, else 0. For equal priors the optimum threshold is midway, λ=a1+a22\lambda=\frac{a_1+a_2}{2}.

Error when 0 is sent (y>λy>\lambda): with u=(y−a2)/2σu=(y-a_2)/\sqrt2\sigma,

Pe0=∫λ∞12πσe−(y−a2)2/2σ2dy=1π∫(λ−a2)/2σ∞e−u2du=12erfc⁡(a1−a222 σ)\begin{aligned} P_{e0} &= \int_{\lambda}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_2)^2/2\sigma^2}dy=\frac{1}{\sqrt{\pi}}\int_{(\lambda-a_2)/\sqrt2\sigma}^{\infty}e^{-u^2}du \\ &= \frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\,\sigma}\right) \end{aligned}

By symmetry Pe1P_{e1} (1 sent, y<λy<\lambda) is the same. The average error probability is

Pe=12Pe0+12Pe1=12erfc⁡(a1−a222 σ)=Q(a1−a22σ)P_e=\tfrac12P_{e0}+\tfrac12P_{e1}=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\,\sigma}\right)=Q\left(\frac{a_1-a_2}{2\sigma}\right)

where erfc⁡(x)=2π∫x∞e−u2du\operatorname{erfc}(x)=\frac{2}{\sqrt\pi}\int_x^{\infty}e^{-u^2}du and Q(x)=12erfc⁡(x/2)Q(x)=\frac12\operatorname{erfc}(x/\sqrt2).

With a matched filter (matched to s1−s2s_1-s_2), the maximum of (a1−a2)2σ2\frac{(a_1-a_2)^2}{\sigma^2} is 2EdN0\frac{2E_d}{N_0}, where Ed=∫0Tb[s1(t)−s2(t)]2dtE_d=\int_0^{T_b}[s_1(t)-s_2(t)]^2dt. Therefore

Pe=12erfc⁡(Ed4N0)P_e=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_d}{4N_0}}\right)

PeP_e depends only on the energy of the difference signal relative to N0N_0, not on pulse shape. Making the two signals as different as possible (antipodal) gives the lowest PeP_e.

For polar NRZ (±A\pm A at the sampler): Pe=12erfc⁡(A2σ)P_e=\frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\sigma}\right).

Comparison with M-ary system

For M-ary PAM with levels ±A,±3A,…,±(M−1)A\pm A,\pm3A,\dots,\pm(M-1)A (spacing 2A2A), inner levels can err on both sides and outer levels on one side, giving

Pe=(1−1M)erfc⁡(A2 σ),Pav=(M2−1)A23P_e=\left(1-\frac1M\right)\operatorname{erfc}\left(\frac{A}{\sqrt2\,\sigma}\right),\qquad P_{av}=\frac{(M^2-1)A^2}{3}

which reduces to the binary result for M=2M=2.

PointBinary (M=2M=2)M-ary (M=2kM=2^k)
Bits per symbol1k=log⁡2Mk=\log_2M
Bandwidth for same bit rateRb/2R_b/2 (Nyquist)Rb/(2log⁡2M)R_b/(2\log_2M), i.e. kk times less
Level spacing for same peak powerLargeSmaller, by factor M−1M-1
PeP_e for same average powerLowestHigher; needs about (M2−1)/3(M^2-1)/3 times more power
ThresholdsOneM−1M-1
Receiver complexitySimpleMore complex
Typical useNoisy, power-limited linksBand-limited links (e.g. telephone modems)

So M-ary signalling trades power for bandwidth: it saves bandwidth but needs more signal power for the same error probability.

  • 2075 Chaitra (CS II) · 8 marks

Explain and compare ideal and practical RC filtering of white noise with respect to change in autocorrelation function.

Answer

White noise (SW(f)=N0/2S_W(f)=N_0/2, RW(τ)=N02δ(τ)R_W(\tau)=\frac{N_0}{2}\delta(\tau)) has infinite power and zero correlation time. Passing it through a filter H(f)H(f) gives SY(f)=∣H(f)∣2N02S_Y(f)=\lvert H(f)\rvert^2\frac{N_0}{2} and RY(τ)=F−1{SY(f)}R_Y(\tau)=\mathcal{F}^{-1}\{S_Y(f)\}: the noise becomes band-limited (coloured) and its samples become correlated.

Ideal low-pass filtering

Ideal LPF, gain 1 for ∣f∣≤B\lvert f\rvert\le B, zero elsewhere:

SY(f)={N02,∣f∣≤B0,otherwiseS_Y(f)=\begin{cases}\frac{N_0}{2}, & \lvert f\rvert\le B\\ 0, & \text{otherwise}\end{cases} RY(τ)=∫−BBN02ej2πfτdf=N02⋅sin⁡2πBτπτ=N0B sin⁡2πBτ2πBτ\begin{aligned} R_Y(\tau) &= \int_{-B}^{B}\frac{N_0}{2}e^{j2\pi f\tau}df=\frac{N_0}{2}\cdot\frac{\sin2\pi B\tau}{\pi\tau} \\ &= N_0B\,\frac{\sin2\pi B\tau}{2\pi B\tau} \end{aligned}
  • Power RY(0)=N0BR_Y(0)=N_0B.
  • RY(τ)R_Y(\tau) is a sinc with zeros at τ=±k2B\tau=\pm\frac{k}{2B}: samples taken at the Nyquist rate 2B2B are uncorrelated (independent if Gaussian).
  • It oscillates (takes negative values) and decays slowly as 1/τ1/\tau.

Practical RC low-pass filtering

H(f)=11+j2πfRCH(f)=\frac{1}{1+j2\pi fRC}, so

SY(f)=N0/21+(2πfRC)2S_Y(f)=\frac{N_0/2}{1+(2\pi fRC)^2}

Using the pair e−a∣τ∣↔2aa2+(2πf)2e^{-a\lvert\tau\rvert}\leftrightarrow\frac{2a}{a^2+(2\pi f)^2} with a=1/RCa=1/RC:

RY(τ)=N04RC e−∣τ∣/RCR_Y(\tau)=\frac{N_0}{4RC}\,e^{-\lvert\tau\rvert/RC}
  • Power RY(0)=N04RCR_Y(0)=\frac{N_0}{4RC}, i.e. NEB =14RC=π2f3dB=\frac{1}{4RC}=\frac\pi2f_{3dB}.
  • RY(τ)R_Y(\tau) is a decaying exponential: always positive, never exactly zero, falls to 1/e1/e at τ=RC\tau=RC. Correlation time ≈RC\approx RC.
 R(tau)  ideal LPF: sinc         RC LPF: exponential
   N0B .                     N0/4RC .
      / \                          /|\
  .  /   \  .                     / | \
 -.-/-----\-.-- tau          ----'--+--'---- tau
   -1/2B  1/2B                 -RC  0  RC

Comparison

PointIdeal LPF (bandwidth BB)RC LPF (f3dB=1/2πRCf_{3dB}=1/2\pi RC)
Output PSDFlat up to BB, sharp cut-offLorentzian, gradual roll-off
AutocorrelationN0B sinc(2Bτ)N_0B\,\text{sinc}(2B\tau)N04RCe−∣τ∣/RC\frac{N_0}{4RC}e^{-\lvert\tau\rvert/RC}
Output noise powerN0BN_0BN0/4RC=N0⋅π2f3dBN_0/4RC=N_0\cdot\frac\pi2f_{3dB}
Zero crossings of R(τ)R(\tau)At τ=k/2B\tau=k/2BNone
Sign of R(τ)R(\tau)Positive and negativeAlways positive
Decay of R(τ)R(\tau)Slow, ∝1/τ\propto1/\tauFast, exponential
RealisableNo (non-causal)Yes

In both cases a narrower bandwidth gives a wider autocorrelation (longer correlation time) and less noise power.

  • 2075 Asoj (CS II) · 8 marks

Derive expression for evaluating error probability of M-ary system.

Answer

In an M-ary system each symbol represents k=log⁡2Mk=\log_2M bits. For baseband M-ary PAM the receiver samples the filtered signal and decides which of the MM levels was sent using M−1M-1 thresholds.

M-ary baseband PAM. Each symbol carries k=log⁡2Mk=\log_2M bits and takes one of MM equally likely levels spaced 2A2A apart, symmetric about zero:

ai=±A, ±3A, …, ±(M−1)Aa_i=\pm A,\ \pm3A,\ \dots,\ \pm(M-1)A

The received sample is y=ai+ny=a_i+n, n∼N(0,σ2)n\sim\mathcal{N}(0,\sigma^2). Thresholds are placed midway between adjacent levels (at 0,±2A,±4A,…0,\pm2A,\pm4A,\dots).

  levels:  -3A    -A     +A    +3A        (M = 4)
    -------*--|---*--|---*--|---*-------> y
             -2A     0     +2A       thresholds
  outer levels: error on one side only
  inner levels: error on both sides
  • Inner level (there are M−2M-2): error if ∣n∣>A\lvert n\rvert>A: Pinner=2⋅12erfc⁡(A2σ)=erfc⁡(A2σ)P_{inner}=2\cdot\frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\sigma}\right)=\operatorname{erfc}\left(\frac{A}{\sqrt2\sigma}\right)
  • Outer level (there are 2): error only if noise pushes it inward by more than AA: Pouter=12erfc⁡(A2σ)P_{outer}=\frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\sigma}\right)

Average symbol error probability:

Pe=1M[(M−2)erfc⁡(A2σ)+2⋅12erfc⁡(A2σ)]=(1−1M)erfc⁡(A2 σ)=2(1−1M)Q(Aσ)\begin{aligned} P_e &= \frac{1}{M}\left[(M-2)\operatorname{erfc}\left(\frac{A}{\sqrt2\sigma}\right)+2\cdot\frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\sigma}\right)\right] \\ &= \left(1-\frac1M\right)\operatorname{erfc}\left(\frac{A}{\sqrt2\,\sigma}\right)=2\left(1-\frac1M\right)Q\left(\frac{A}{\sigma}\right) \end{aligned}

In terms of signal power. Average symbol power

Pav=2M∑i=1M/2(2i−1)2A2=(M2−1)A23 ⇒ A2=3PavM2−1P_{av}=\frac{2}{M}\sum_{i=1}^{M/2}(2i-1)^2A^2=\frac{(M^2-1)A^2}{3}\ \Rightarrow\ A^2=\frac{3P_{av}}{M^2-1}

so

Pe=(1−1M)erfc⁡(32(M2−1)⋅Pavσ2)P_e=\left(1-\frac1M\right)\operatorname{erfc}\left(\sqrt{\frac{3}{2(M^2-1)}\cdot\frac{P_{av}}{\sigma^2}}\right)

For M=2M=2 this reduces to the binary polar result Pe=12erfc⁡(A2σ)P_e=\frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\sigma}\right).

Observations:

  • For fixed A/σA/\sigma, PeP_e rises slowly with MM (factor 1−1/M1-1/M), but for fixed average power the spacing shrinks as 3/(M2−1)\sqrt{3/(M^2-1)}, so PeP_e rises sharply.
  • With Gray coding, bit error probability ≈Pe/log⁡2M\approx P_e/\log_2M.
  • M-ary signalling reduces bandwidth by log⁡2M\log_2M but needs more power: a bandwidth–power trade-off.
  • 2074 Asoj (CS II) · 6+2 marks

Compute the figure of merit of non coherent FM System and explain the threshold effects.

Answer

Figure of merit of non-coherent FM

FM signal and receiver. s(t)=Accos⁡[ωct+2πkf∫m(t)dt]s(t)=A_c\cos\left[\omega_ct+2\pi k_f\int m(t)dt\right], instantaneous frequency deviation kfm(t)k_fm(t). Receiver: BPF (BTB_T) → limiter → discriminator (output =12πdϕdt=\frac{1}{2\pi}\frac{d\phi}{dt}) → LPF (WW).

Noise in phasor form. Write the narrowband noise as n(t)=r(t)cos⁡[ωct+ψ(t)]n(t)=r(t)\cos[\omega_ct+\psi(t)]. At high carrier-to-noise ratio (Ac≫rA_c\gg r), the resultant phase is

θ(t)≈ϕ(t)+r(t)Acsin⁡[ψ(t)−ϕ(t)]\theta(t)\approx\phi(t)+\frac{r(t)}{A_c}\sin[\psi(t)-\phi(t)]
            r(t)
          .------>          resultant = carrier + noise
         /  ^               small noise phasor rotates
 Ac     /   | r sin(psi-phi) the carrier phasor slightly
 ------>----+

The noise term is statistically equivalent to ns(t)Ac\frac{n_s(t)}{A_c} (the phase ϕ\phi can be dropped for noise calculation). Discriminator output:

v(t)=12πdθdt=kfm(t)+12πAcdns(t)dtv(t)=\frac{1}{2\pi}\frac{d\theta}{dt}=k_fm(t)+\frac{1}{2\pi A_c}\frac{dn_s(t)}{dt}

Output signal power: So=kf2PS_o=k_f^2P.

Output noise power. nsn_s has PSD N0N_0 for ∣f∣≤BT/2\lvert f\rvert\le B_T/2. Differentiation multiplies the spectrum by j2πfj2\pi f, so the noise PSD at discriminator output is

Snd(f)=(2πf)2(2πAc)2N0=N0f2Ac2,∣f∣≤BT2S_{n_d}(f)=\frac{(2\pi f)^2}{(2\pi A_c)^2}N_0=\frac{N_0f^2}{A_c^2},\qquad \lvert f\rvert\le\frac{B_T}{2}

This parabolic noise spectrum is the key property of FM. After the LPF (∣f∣≤W\lvert f\rvert\le W):

No=∫−WWN0f2Ac2df=2N0W33Ac2N_o=\int_{-W}^{W}\frac{N_0f^2}{A_c^2}df=\frac{2N_0W^3}{3A_c^2}
 S(f) |\                   /|
      | \                 / |   parabolic noise
      |  \               /  |   after discriminator
      |   \.           ./   |   (only |f|<W kept)
      +----+-----+-----+----+--> f
          -W     0     W

Output SNR:

SNRo=kf2P2N0W3/3Ac2=3Ac2kf2P2N0W3\text{SNR}_o=\frac{k_f^2P}{2N_0W^3/3A_c^2}=\frac{3A_c^2k_f^2P}{2N_0W^3}

Input SNR. Received power Ac22\frac{A_c^2}{2} (constant envelope); noise in BTB_T is N0BTN_0B_T:

SNRi=Ac22N0BT,SNRref=Ac22N0W\text{SNR}_i=\frac{A_c^2}{2N_0B_T},\qquad \text{SNR}_{ref}=\frac{A_c^2}{2N_0W}

Gain and figure of merit:

FOM=SNRoSNRref=3kf2PW2SNRoSNRi=3kf2P BTW3\begin{aligned} \text{FOM} &= \frac{\text{SNR}_o}{\text{SNR}_{ref}}=\frac{3k_f^2P}{W^2} \\ \frac{\text{SNR}_o}{\text{SNR}_i} &= \frac{3k_f^2P\,B_T}{W^3} \end{aligned}

For a single tone m(t)=Amcos⁡ωmtm(t)=A_m\cos\omega_mt: P=Am2/2P=A_m^2/2, Δf=kfAm\Delta f=k_fA_m, β=Δf/W\beta=\Delta f/W, BT=2(β+1)WB_T=2(\beta+1)W (Carson):

FOM=32β2,SNRoSNRi=3β2(β+1)\text{FOM}=\frac{3}{2}\beta^2,\qquad \frac{\text{SNR}_o}{\text{SNR}_i}=3\beta^2(\beta+1)

So FM noise performance improves with the square of the deviation ratio: FM trades bandwidth for SNR. FM beats AM (FOM=1/3\text{FOM}=1/3) when 32β2>13\frac32\beta^2>\frac13, i.e. β>0.471\beta>0.471 (about 0.5).

Threshold effect

Threshold effect in FM. The FM improvement SNRo∝β2\text{SNR}_o\propto\beta^2 holds only when the carrier is much stronger than the noise at the discriminator input. When the carrier-to-noise ratio (CNR) falls below a threshold, about 10 dB (typically 10–13 dB), the noise phasor sometimes becomes larger than the carrier and the resultant phasor encircles the origin, causing a sudden 2π2\pi phase jump.

  high CNR: noise wobbles     low CNR: resultant can swing
  the tip slightly            around origin -> 2pi jump
        .-.                         .---.
  0 -->( * )                   0 --( *   )
        '-'                         '---'

Each 2π2\pi jump produces an impulse (spike) in the discriminator output, heard as clicks. The click rate rises rapidly as CNR drops, so the output SNR falls much faster than linearly and the signal is mutilated. Above threshold, FM is excellent; below it, FM is worse than AM. Wider deviation (larger β\beta) means larger BTB_T, more noise and a higher threshold.

 SNRo(dB)            . FM above threshold
     |             .   (slope 1, offset 3/2 beta^2)
     |           .
     |         .
     |       . <- threshold knee (CNR ~ 10 dB)
     |     .:
     |   . :     AM / baseband
     |  .  :
     +------------------> SNRi (dB)
  • 2074 Chaitra (CS II) · 2+6+3 marks

What do you mean by optimum detector? Show that the impulse response of the matched filter is reverse delayed version of the input signal. Explain auto correlation function.

Answer

Optimum detector

An optimum detector is a receiver (filter followed by sampler and threshold device) that makes the decision with the minimum probability of error. For a known pulse in additive white Gaussian noise, this is achieved by the filter that maximises the output signal-to-noise ratio at the sampling instant; this optimum filter is the matched filter.

Impulse response of matched filter is reverse delayed input

Set-up. The received signal is x(t)=s(t)+w(t)x(t)=s(t)+w(t), 0≤t≤T0\le t\le T, where s(t)s(t) is a known pulse and w(t)w(t) is white noise of two-sided PSD N0/2N_0/2. It passes through an LTI filter h(t)↔H(f)h(t)\leftrightarrow H(f) and is sampled at t=Tt=T:

 s(t)+w(t)    +--------+   y(t)    sample     y(T)
 ----------->|  h(t)  |---------->--/ ---->  decision
              +--------+           at t=T

Output y(t)=so(t)+no(t)y(t)=s_o(t)+n_o(t) with

so(T)=∫−∞∞H(f)S(f)ej2πfTdf,E[no2]=N02∫−∞∞∣H(f)∣2dfs_o(T)=\int_{-\infty}^{\infty}H(f)S(f)e^{j2\pi fT}df,\qquad E[n_o^2]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df

We want H(f)H(f) that maximises the output peak signal-to-noise ratio

η=∣so(T)∣2E[no2]=∣∫H(f)S(f)ej2πfTdf∣2N02∫∣H(f)∣2df\eta=\frac{\lvert s_o(T)\rvert^2}{E[n_o^2]}=\frac{\left\lvert\int H(f)S(f)e^{j2\pi fT}df\right\rvert^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}

Schwarz inequality: ∣∫ϕ1ϕ2 df∣2≤∫∣ϕ1∣2df∫∣ϕ2∣2df\left\lvert\int\phi_1\phi_2\,df\right\rvert^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df, with equality only when ϕ1(f)=k ϕ2∗(f)\phi_1(f)=k\,\phi_2^*(f). Put ϕ1=H(f)\phi_1=H(f) and ϕ2=S(f)ej2πfT\phi_2=S(f)e^{j2\pi fT}:

η≤∫∣H∣2df∫∣S∣2dfN02∫∣H∣2df=2N0∫−∞∞∣S(f)∣2df=2EN0\eta\le\frac{\int\lvert H\rvert^2df\int\lvert S\rvert^2df}{\frac{N_0}{2}\int\lvert H\rvert^2df}=\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert S(f)\rvert^2df=\frac{2E}{N_0}

(using Rayleigh's theorem, EE = pulse energy). The maximum is reached when

Hopt(f)=k S∗(f) e−j2πfTH_{opt}(f)=k\,S^*(f)\,e^{-j2\pi fT}

Impulse response. Take the inverse transform. For real s(t)s(t), S∗(f)=S(−f)S^*(f)=S(-f), which is the transform of s(−t)s(-t); the factor e−j2πfTe^{-j2\pi fT} is a delay of TT:

hopt(t)=k∫−∞∞S∗(f)e−j2πfTej2πftdf=k∫−∞∞S(−f)e−j2πf(T−t)df=k s(T−t)\begin{aligned} h_{opt}(t) &= k\int_{-\infty}^{\infty}S^*(f)e^{-j2\pi fT}e^{j2\pi ft}df \\ &= k\int_{-\infty}^{\infty}S(-f)e^{-j2\pi f(T-t)}df \\ &= k\,s(T-t) \end{aligned}

So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by TT. Choosing the sampling instant TT equal to the pulse duration makes h(t)h(t) causal.

   s(t)                      h(t) = s(T - t)
    |\                          /|
    | \                        / |
    |  \                      /  |
 ---+---+---> t          ---+---+---> t
    0   T                   0   T

Results: maximum SNR =2E/N0=2E/N_0 depends only on pulse energy, not on pulse shape; the filter's frequency response is the conjugate of the pulse spectrum, so it passes strongly the frequencies where the signal is strong.

Autocorrelation function

The autocorrelation (AC) function measures how similar a signal is to a time-shifted copy of itself. For a power signal or a wide-sense stationary (WSS) random process X(t)X(t):

RX(τ)=lim⁡T→∞1T∫−T/2T/2x(t) x(t+τ) dtorRX(τ)=E[X(t)X(t+τ)]R_X(\tau)=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}x(t)\,x(t+\tau)\,dt \quad\text{or}\quad R_X(\tau)=E[X(t)X(t+\tau)]

For an energy signal, Rg(τ)=∫−∞∞g(t)g(t+τ) dtR_g(\tau)=\int_{-\infty}^{\infty}g(t)g(t+\tau)\,dt.

Main properties: R(−τ)=R(τ)R(-\tau)=R(\tau) (even); R(0)=R(0)= average power (energy); ∣R(τ)∣≤R(0)\lvert R(\tau)\rvert\le R(0); R(τ)↔S(f)R(\tau)\leftrightarrow S(f) (Wiener–Khinchin). Example: white noise has R(τ)=N02δ(τ)R(\tau)=\frac{N_0}{2}\delta(\tau). Note that the matched filter output for the signal is so(t)=kRs(t−T)s_o(t)=kR_s(t-T), the pulse autocorrelation shifted to TT, which peaks at t=Tt=T because RsR_s is maximum at zero shift.

  • 2074 Chaitra (CS II) · 2+10 marks

What do you mean by Ergodic Stochastic Process? Explain with necessary derivation passage of wide-sense random signals through a LTI.

Answer

Ergodic stochastic process

A stationary random process is ergodic if its time averages, taken over one long sample function, equal its ensemble averages, taken across all sample functions at one instant:

lim⁡T→∞1T∫−T/2T/2x(t) dt=E[X(t)]=mX,lim⁡T→∞1T∫−T/2T/2x(t)x(t+τ) dt=RX(τ)\lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}x(t)\,dt=E[X(t)]=m_X,\qquad \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}x(t)x(t+\tau)\,dt=R_X(\tau)

Ergodicity allows the mean (DC value), power and autocorrelation of noise to be measured from a single long record. An ergodic process must be stationary, but a stationary process need not be ergodic.

Passage of WSS random signals through an LTI system

Let a WSS process X(t)X(t) (mean mXm_X, autocorrelation RX(τ)R_X(\tau), PSD SX(f)S_X(f)) be applied to a stable LTI system with impulse response h(t)h(t) and H(f)H(f):

Y(t)=∫−∞∞h(α)X(t−α) dαY(t)=\int_{-\infty}^{\infty}h(\alpha)X(t-\alpha)\,d\alpha
  X(t) (WSS)   +--------------+   Y(t)
 ------------->| h(t) , H(f)  |------------->
  R_X, S_X     +--------------+  R_Y, S_Y

1. Mean of output. Expectation and integration can be interchanged:

E[Y(t)]=∫−∞∞h(α)E[X(t−α)] dα=mX∫−∞∞h(α) dα=mX H(0)(a constant, independent of t)\begin{aligned} E[Y(t)] &= \int_{-\infty}^{\infty}h(\alpha)E[X(t-\alpha)]\,d\alpha = m_X\int_{-\infty}^{\infty}h(\alpha)\,d\alpha \\ &= m_X\,H(0) \quad(\text{a constant, independent of } t) \end{aligned}

2. Autocorrelation of output.

RY(t,t+τ)=E[Y(t)Y(t+τ)]=E[∫h(α)X(t−α)dα∫h(β)X(t+τ−β)dβ]=∫ ⁣ ⁣∫h(α)h(β) E[X(t−α)X(t+τ−β)] dα dβ=∫ ⁣ ⁣∫h(α)h(β) RX(τ+α−β) dα dβ\begin{aligned} R_Y(t,t+\tau) &= E[Y(t)Y(t+\tau)] \\ &= E\left[\int h(\alpha)X(t-\alpha)d\alpha\int h(\beta)X(t+\tau-\beta)d\beta\right] \\ &= \int\!\!\int h(\alpha)h(\beta)\,E[X(t-\alpha)X(t+\tau-\beta)]\,d\alpha\,d\beta \\ &= \int\!\!\int h(\alpha)h(\beta)\,R_X(\tau+\alpha-\beta)\,d\alpha\,d\beta \end{aligned}

because XX is WSS, its autocorrelation depends only on the time difference (t+τ−β)−(t−α)=τ+α−β(t+\tau-\beta)-(t-\alpha)=\tau+\alpha-\beta. The result depends only on τ\tau, not on tt:

RY(τ)=h(τ)∗h(−τ)∗RX(τ)R_Y(\tau)=h(\tau)*h(-\tau)*R_X(\tau)

3. Mean-square value E[Y2]=RY(0)=∫ ⁣ ⁣∫h(α)h(β)RX(α−β)dαdβE[Y^2]=R_Y(0)=\int\!\!\int h(\alpha)h(\beta)R_X(\alpha-\beta)d\alpha d\beta is finite and constant.

Since the mean is constant and the autocorrelation depends only on τ\tau, Y(t)Y(t) is also wide-sense stationary.

4. Output PSD. Taking the Fourier transform of RY(τ)R_Y(\tau) (h(τ)→H(f)h(\tau)\to H(f), h(−τ)→H∗(f)h(-\tau)\to H^*(f)):

SY(f)=H(f)H∗(f)SX(f)=∣H(f)∣2SX(f)S_Y(f)=H(f)H^*(f)S_X(f)=\lvert H(f)\rvert^2S_X(f)

and the output power is PY=RY(0)=∫−∞∞∣H(f)∣2SX(f) dfP_Y=R_Y(0)=\int_{-\infty}^{\infty}\lvert H(f)\rvert^2S_X(f)\,df. The phase of H(f)H(f) has no effect on the output PSD.

Example (white noise through RC low-pass filter). SY(f)=N0/21+(2πfRC)2S_Y(f)=\frac{N_0/2}{1+(2\pi fRC)^2}, RY(τ)=N04RCe−∣τ∣/RCR_Y(\tau)=\frac{N_0}{4RC}e^{-\lvert\tau\rvert/RC}, output power N04RC\frac{N_0}{4RC}.

Cross-correlation: RXY(τ)=h(τ)∗RX(τ)R_{XY}(\tau)=h(\tau)*R_X(\tau); for white-noise input, RXY(τ)=N02h(τ)R_{XY}(\tau)=\frac{N_0}{2}h(\tau), so correlating output with input measures h(t)h(t).

  • 2074 Chaitra (CS II) · 8 marks

What is detecting gain? Prove that for 100% modulation of (DSB-AM), the detection gain is less than 1.

Answer

Detection gain is the ratio of the output signal-to-noise ratio of a demodulator to its input (pre-detection) SNR:

γd=SNRoSNRi\gamma_d=\frac{\text{SNR}_o}{\text{SNR}_i}

It shows whether the detector improves (γd>1\gamma_d>1) or degrades (γd<1\gamma_d<1) the SNR.

Receiver model. The received signal s(t)s(t) plus white noise (N0/2N_0/2) passes through a band-pass (IF) filter of bandwidth BTB_T, then a demodulator and a low-pass filter of bandwidth WW (message bandwidth).

 s(t)  +   +-------+    +-------+   +-----+
 ---->(+)-->|  BPF  |--->| demod |-->| LPF |--> y(t)
       ^    |  B_T  |    |       |   |  W  |
      w(t)  +-------+    +-------+   +-----+

The filtered noise is narrowband and is written in in-phase/quadrature form

n(t)=nc(t)cos⁡ωct−ns(t)sin⁡ωctn(t)=n_c(t)\cos\omega_ct-n_s(t)\sin\omega_ct

where ncn_c and nsn_s are low-pass, each with the same power as n(t)n(t): nc2‾=ns2‾=n2‾=N0BT\overline{n_c^2}=\overline{n_s^2}=\overline{n^2}=N_0B_T.

DSB-FC (conventional AM). s(t)=Ac[1+kam(t)]cos⁡ωcts(t)=A_c[1+k_am(t)]\cos\omega_ct, BT=2WB_T=2W. (Coherent and envelope detection give the same result at high SNR.)

  • Signal power =Ac22(1+ka2P)=\frac{A_c^2}{2}(1+k_a^2P); noise in 2W2W =2N0W=2N_0W:
SNRi=Ac2(1+ka2P)4N0W\text{SNR}_i=\frac{A_c^2(1+k_a^2P)}{4N_0W}
  • Coherent detection (or envelope detection at high SNR) and DC blocking:
y(t)=Ac2kam(t)+12nc(t) ⇒ SNRo=Ac2ka2P/42N0W/4=Ac2ka2P2N0Wy(t)=\frac{A_c}{2}k_am(t)+\frac12n_c(t)\ \Rightarrow\ \text{SNR}_o=\frac{A_c^2k_a^2P/4}{2N_0W/4}=\frac{A_c^2k_a^2P}{2N_0W} γDSB-FC=SNRoSNRi=2ka2P1+ka2P,FOM=ka2P1+ka2P\gamma_{DSB\text{-}FC}=\frac{\text{SNR}_o}{\text{SNR}_i}=\frac{2k_a^2P}{1+k_a^2P},\qquad \text{FOM}=\frac{k_a^2P}{1+k_a^2P}

The carrier uses power but carries no information, so the gain is reduced.

Proof that γd<1\gamma_d<1. To avoid over-modulation, ∣kam(t)∣≤1\lvert k_am(t)\rvert\le1, so ka2P=ka2m2(t)‾≤1k_a^2P=\overline{k_a^2m^2(t)}\le1, with equality only for a square-wave message at 100% modulation. Hence

γd=2ka2P1+ka2P≤2(1)1+1=1\gamma_d=\frac{2k_a^2P}{1+k_a^2P}\le\frac{2(1)}{1+1}=1

For any practical message (sinusoid, speech) ka2P<1k_a^2P<1 and γd<1\gamma_d<1.

Single tone, 100% modulation. m(t)=Amcos⁡ωmtm(t)=A_m\cos\omega_mt, μ=kaAm=1\mu=k_aA_m=1, so ka2P=μ22=12k_a^2P=\frac{\mu^2}{2}=\frac12:

γDSB-FC=2(μ2/2)1+μ2/2=2μ22+μ2=23≈0.667FOM=μ22+μ2=13\begin{aligned} \gamma_{DSB\text{-}FC} &= \frac{2(\mu^2/2)}{1+\mu^2/2}=\frac{2\mu^2}{2+\mu^2}=\frac{2}{3}\approx0.667 \\ \text{FOM} &= \frac{\mu^2}{2+\mu^2}=\frac13 \end{aligned}

So for 100% single-tone DSB-AM the detection gain is 2/3<12/3<1: the detector output SNR is about 1.761.76 dB lower than the input SNR. The reason is that two-thirds of the transmitted power is in the carrier, which is counted in the input signal power but contributes nothing to the output message. (Compare DSB-SC, γd=2\gamma_d=2.)

  • 2073 Shrawan (CS II) · 8 marks

Derive the general expression for evaluating error probability for binary ASK system and extend it to M-ary.

Answer

Binary ASK

ASK (on–off keying) signals over 0≤t≤Tb0\le t\le T_b:

s1(t)=Acos⁡ωct  (bit 1),s2(t)=0  (bit 0)s_1(t)=A\cos\omega_ct\ \ (\text{bit }1),\qquad s_2(t)=0\ \ (\text{bit }0)

Coherent receiver (correlator = matched filter):

 r(t) -->(x)--> integrate --> sample --> y > lam ? 1 : 0
          ^       0..Tb       t=Tb
          |
       cos(wc t)  (local carrier, in phase)

Correlator output at t=Tbt=T_b, with r(t)=si(t)+w(t)r(t)=s_i(t)+w(t):

y=∫0Tbr(t)cos⁡ωct dt=ai+ny=\int_0^{T_b}r(t)\cos\omega_ct\,dt=a_i+n
  • Bit 1: a1=∫0TbAcos⁡2ωct dt=ATb2a_1=\int_0^{T_b}A\cos^2\omega_ct\,dt=\frac{AT_b}{2} (for fcf_c an integer multiple of 1/Tb1/T_b).
  • Bit 0: a2=0a_2=0.
  • Noise: n=∫0Tbw(t)cos⁡ωct dtn=\int_0^{T_b}w(t)\cos\omega_ct\,dt is Gaussian, mean 0, variance
σ2=∫0Tb ⁣ ⁣∫0TbN02δ(t−u)cos⁡ωctcos⁡ωcu dt du=N02⋅Tb2=N0Tb4\sigma^2=\int_0^{T_b}\!\!\int_0^{T_b}\frac{N_0}{2}\delta(t-u)\cos\omega_ct\cos\omega_cu\,dt\,du=\frac{N_0}{2}\cdot\frac{T_b}{2}=\frac{N_0T_b}{4}

Threshold midway: λ=ATb4\lambda=\frac{AT_b}{4}. Using the binary result Pe=12erfc⁡(a1−a222σ)P_e=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\sigma}\right):

a1−a222 σ=ATb/222N0Tb/4=ATb22N0=A2Tb8N0\begin{aligned} \frac{a_1-a_2}{2\sqrt2\,\sigma} &= \frac{AT_b/2}{2\sqrt2\sqrt{N_0T_b/4}}=\frac{A\sqrt{T_b}}{2\sqrt{2N_0}}=\sqrt{\frac{A^2T_b}{8N_0}} \end{aligned}

Energy of a "1" pulse is E1=A2Tb2E_1=\frac{A^2T_b}{2}; the average energy per bit (half the bits are 0) is Eb=E12=A2Tb4E_b=\frac{E_1}{2}=\frac{A^2T_b}{4}. Hence A2Tb8N0=Eb2N0\frac{A^2T_b}{8N_0}=\frac{E_b}{2N_0} and

Pe(ASK)=12erfc⁡(Eb2N0)=Q(EbN0)P_e(\text{ASK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)

(In terms of peak energy E1E_1: Pe=12erfc⁡E1/4N0P_e=\frac12\operatorname{erfc}\sqrt{E_1/4N_0}.)

Extension to M-ary ASK

In M-ary ASK the carrier takes MM amplitudes Ai=iAA_i=iA, i=0,1,…,M−1i=0,1,\dots,M-1, each symbol of duration TsT_s carrying log⁡2M\log_2M bits. The same coherent correlator gives

y=ai+n,ai=iATs2,σ2=N0Ts4y=a_i+n,\qquad a_i=\frac{iAT_s}{2},\qquad \sigma^2=\frac{N_0T_s}{4}

Levels are spaced Δ=ATs2\Delta=\frac{AT_s}{2} apart, and thresholds are placed midway. The two end levels can be mistaken on one side only; the M−2M-2 inner levels on both sides:

Pe=1M[(M−2)⋅2Q(Δ2σ)+2⋅Q(Δ2σ)]=2(1−1M)Q(Δ2σ)Δ2σ=ATs/22N0Ts/4=A2Ts4N0\begin{aligned} P_e &= \frac1M\left[(M-2)\cdot2Q\left(\frac{\Delta}{2\sigma}\right)+2\cdot Q\left(\frac{\Delta}{2\sigma}\right)\right]=2\left(1-\frac1M\right)Q\left(\frac{\Delta}{2\sigma}\right) \\ \frac{\Delta}{2\sigma} &= \frac{AT_s/2}{2\sqrt{N_0T_s/4}}=\sqrt{\frac{A^2T_s}{4N_0}} \end{aligned}

Average symbol energy: Ei=i2A2Ts2E_i=\frac{i^2A^2T_s}{2}, so

Es=A2Ts2M∑i=0M−1i2=A2Ts(M−1)(2M−1)12E_s=\frac{A^2T_s}{2M}\sum_{i=0}^{M-1}i^2=\frac{A^2T_s(M-1)(2M-1)}{12}

Substituting A2Ts=12Es(M−1)(2M−1)A^2T_s=\frac{12E_s}{(M-1)(2M-1)}:

Pe(M-ASK)=2(1−1M)Q(3Es(M−1)(2M−1)N0)P_e(\text{M-ASK})=2\left(1-\frac1M\right)Q\left(\sqrt{\frac{3E_s}{(M-1)(2M-1)N_0}}\right)

Check M=2M=2: Pe=Q(Es/N0)P_e=Q\left(\sqrt{E_s/N_0}\right) with Es=EbE_s=E_b, the binary ASK result. For larger MM the levels crowd together for the same average energy, so PeP_e grows, while the bandwidth per bit falls by log⁡2M\log_2M.

  • 2073 Chaitra (CS II) · 10+2 marks

Explain noise equivalent bandwidth. Prove that the impulse response of matched filter is a time reversed delayed version of input signal Si(t).

Answer

Noise equivalent bandwidth

The noise equivalent bandwidth (NEB) BNB_N of a filter is the bandwidth of an ideal rectangular filter, with the same peak (usually zero-frequency or centre) gain ∣H(0)∣\lvert H(0)\rvert, that passes the same total noise power as the actual filter when both are fed with white noise.

White noise N0/2N_0/2 into H(f)H(f) gives output power

Pactual=N02∫−∞∞∣H(f)∣2df=N0∫0∞∣H(f)∣2dfP_{actual}=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df=N_0\int_0^{\infty}\lvert H(f)\rvert^2df

The ideal filter of gain ∣H(0)∣\lvert H(0)\rvert and bandwidth BNB_N gives Pideal=N0BN∣H(0)∣2P_{ideal}=N_0B_N\lvert H(0)\rvert^2. Equating:

BN=∫0∞∣H(f)∣2df∣H(0)∣2B_N=\frac{\int_0^{\infty}\lvert H(f)\rvert^2df}{\lvert H(0)\rvert^2}
 |H(f)|^2
 |H(0)|^2 +---------+     equal areas
          |  ...    |
          |      .. |
          |        .|..
          |         |  ......
        --+---------+----------> f
          0        B_N

NEB lets us write the output noise power simply as N=N0BN∣H(0)∣2N=N_0B_N\lvert H(0)\rvert^2 (or kTBNkTB_N per unit gain), which is used in noise figure and SNR calculations. For a band-pass filter ∣H(0)∣\lvert H(0)\rvert is replaced by the centre-frequency gain ∣H(f0)∣\lvert H(f_0)\rvert.

Example: RC low-pass filter.

For the RC low-pass filter (series RR, shunt CC):

H(f)=11+j2πfRC,∣H(f)∣2=11+(2πfRC)2,∣H(0)∣=1H(f)=\frac{1}{1+j2\pi fRC},\qquad \lvert H(f)\rvert^2=\frac{1}{1+(2\pi fRC)^2},\qquad \lvert H(0)\rvert=1 BN=∫0∞df1+(2πfRC)2(let u=2πfRC, df=du2πRC)=12πRC∫0∞du1+u2=12πRC⋅π2=14RC\begin{aligned} B_N &= \int_0^{\infty}\frac{df}{1+(2\pi fRC)^2}\qquad (\text{let } u=2\pi fRC,\ df=\tfrac{du}{2\pi RC}) \\ &= \frac{1}{2\pi RC}\int_0^{\infty}\frac{du}{1+u^2}=\frac{1}{2\pi RC}\cdot\frac{\pi}{2} \\ &= \frac{1}{4RC} \end{aligned}

Since the 3 dB bandwidth is f3dB=12πRCf_{3dB}=\frac{1}{2\pi RC},

BN=π2f3dB≈1.57 f3dBB_N=\frac{\pi}{2}f_{3dB}\approx1.57\,f_{3dB}

The NEB is larger than the 3 dB bandwidth because the RC response rolls off slowly (only −20-20 dB/decade) and passes noise well beyond f3dBf_{3dB}. Output noise power =N0BN=N04RC=N_0B_N=\frac{N_0}{4RC}.

Example: ideal filter. For an ideal LPF of bandwidth BB, BN=BB_N=B exactly; for practical filters with steep skirts (e.g. high-order Butterworth), BNB_N approaches the 3 dB bandwidth.

Impulse response of matched filter is time-reversed, delayed input

Set-up. The received signal is x(t)=s(t)+w(t)x(t)=s(t)+w(t), 0≤t≤T0\le t\le T, where s(t)s(t) is a known pulse and w(t)w(t) is white noise of two-sided PSD N0/2N_0/2. It passes through an LTI filter h(t)↔H(f)h(t)\leftrightarrow H(f) and is sampled at t=Tt=T:

 s(t)+w(t)    +--------+   y(t)    sample     y(T)
 ----------->|  h(t)  |---------->--/ ---->  decision
              +--------+           at t=T

Output y(t)=so(t)+no(t)y(t)=s_o(t)+n_o(t) with

so(T)=∫−∞∞H(f)S(f)ej2πfTdf,E[no2]=N02∫−∞∞∣H(f)∣2dfs_o(T)=\int_{-\infty}^{\infty}H(f)S(f)e^{j2\pi fT}df,\qquad E[n_o^2]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df

We want H(f)H(f) that maximises the output peak signal-to-noise ratio

η=∣so(T)∣2E[no2]=∣∫H(f)S(f)ej2πfTdf∣2N02∫∣H(f)∣2df\eta=\frac{\lvert s_o(T)\rvert^2}{E[n_o^2]}=\frac{\left\lvert\int H(f)S(f)e^{j2\pi fT}df\right\rvert^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}

Schwarz inequality: ∣∫ϕ1ϕ2 df∣2≤∫∣ϕ1∣2df∫∣ϕ2∣2df\left\lvert\int\phi_1\phi_2\,df\right\rvert^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df, with equality only when ϕ1(f)=k ϕ2∗(f)\phi_1(f)=k\,\phi_2^*(f). Put ϕ1=H(f)\phi_1=H(f) and ϕ2=S(f)ej2πfT\phi_2=S(f)e^{j2\pi fT}:

η≤∫∣H∣2df∫∣S∣2dfN02∫∣H∣2df=2N0∫−∞∞∣S(f)∣2df=2EN0\eta\le\frac{\int\lvert H\rvert^2df\int\lvert S\rvert^2df}{\frac{N_0}{2}\int\lvert H\rvert^2df}=\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert S(f)\rvert^2df=\frac{2E}{N_0}

(using Rayleigh's theorem, EE = pulse energy). The maximum is reached when

Hopt(f)=k S∗(f) e−j2πfTH_{opt}(f)=k\,S^*(f)\,e^{-j2\pi fT}

Impulse response. Take the inverse transform. For real s(t)s(t), S∗(f)=S(−f)S^*(f)=S(-f), which is the transform of s(−t)s(-t); the factor e−j2πfTe^{-j2\pi fT} is a delay of TT:

hopt(t)=k∫−∞∞S∗(f)e−j2πfTej2πftdf=k∫−∞∞S(−f)e−j2πf(T−t)df=k s(T−t)\begin{aligned} h_{opt}(t) &= k\int_{-\infty}^{\infty}S^*(f)e^{-j2\pi fT}e^{j2\pi ft}df \\ &= k\int_{-\infty}^{\infty}S(-f)e^{-j2\pi f(T-t)}df \\ &= k\,s(T-t) \end{aligned}

So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by TT. Choosing the sampling instant TT equal to the pulse duration makes h(t)h(t) causal.

   s(t)                      h(t) = s(T - t)
    |\                          /|
    | \                        / |
    |  \                      /  |
 ---+---+---> t          ---+---+---> t
    0   T                   0   T

Results: maximum SNR =2E/N0=2E/N_0 depends only on pulse energy, not on pulse shape; the filter's frequency response is the conjugate of the pulse spectrum, so it passes strongly the frequencies where the signal is strong.

  • 2073 Chaitra (CS II) · 8 marks

Calculate detection gains for DSB-FC, DSB-SC and SSB and also compare them.

Answer

Receiver model. The received signal s(t)s(t) plus white noise (N0/2N_0/2) passes through a band-pass (IF) filter of bandwidth BTB_T, then a demodulator and a low-pass filter of bandwidth WW (message bandwidth).

 s(t)  +   +-------+    +-------+   +-----+
 ---->(+)-->|  BPF  |--->| demod |-->| LPF |--> y(t)
       ^    |  B_T  |    |       |   |  W  |
      w(t)  +-------+    +-------+   +-----+

The filtered noise is narrowband and is written in in-phase/quadrature form

n(t)=nc(t)cos⁡ωct−ns(t)sin⁡ωctn(t)=n_c(t)\cos\omega_ct-n_s(t)\sin\omega_ct

where ncn_c and nsn_s are low-pass, each with the same power as n(t)n(t): nc2‾=ns2‾=n2‾=N0BT\overline{n_c^2}=\overline{n_s^2}=\overline{n^2}=N_0B_T.

Definitions. Message power P=m2(t)‾P=\overline{m^2(t)}.

  • Input SNR SNRi=received signal powernoise power in BT\text{SNR}_i=\dfrac{\text{received signal power}}{\text{noise power in } B_T}
  • Output SNR SNRo=message power at outputnoise power at output\text{SNR}_o=\dfrac{\text{message power at output}}{\text{noise power at output}}
  • Detection gain γd=SNRoSNRi\gamma_d=\dfrac{\text{SNR}_o}{\text{SNR}_i}; figure of merit =SNRoSNRref=\dfrac{\text{SNR}_o}{\text{SNR}_{ref}}, where SNRref\text{SNR}_{ref} is the channel SNR measured in the message band WW (baseband reference). Both show how much the demodulator improves or worsens SNR.

DSB-FC (conventional AM). s(t)=Ac[1+kam(t)]cos⁡ωcts(t)=A_c[1+k_am(t)]\cos\omega_ct, BT=2WB_T=2W. (Coherent and envelope detection give the same result at high SNR.)

  • Signal power =Ac22(1+ka2P)=\frac{A_c^2}{2}(1+k_a^2P); noise in 2W2W =2N0W=2N_0W:
SNRi=Ac2(1+ka2P)4N0W\text{SNR}_i=\frac{A_c^2(1+k_a^2P)}{4N_0W}
  • Coherent detection (or envelope detection at high SNR) and DC blocking:
y(t)=Ac2kam(t)+12nc(t) ⇒ SNRo=Ac2ka2P/42N0W/4=Ac2ka2P2N0Wy(t)=\frac{A_c}{2}k_am(t)+\frac12n_c(t)\ \Rightarrow\ \text{SNR}_o=\frac{A_c^2k_a^2P/4}{2N_0W/4}=\frac{A_c^2k_a^2P}{2N_0W} γDSB-FC=SNRoSNRi=2ka2P1+ka2P,FOM=ka2P1+ka2P\gamma_{DSB\text{-}FC}=\frac{\text{SNR}_o}{\text{SNR}_i}=\frac{2k_a^2P}{1+k_a^2P},\qquad \text{FOM}=\frac{k_a^2P}{1+k_a^2P}

The carrier uses power but carries no information, so the gain is reduced.

Single tone, 100% modulation. m(t)=Amcos⁡ωmtm(t)=A_m\cos\omega_mt, μ=kaAm=1\mu=k_aA_m=1, so ka2P=μ22=12k_a^2P=\frac{\mu^2}{2}=\frac12:

γDSB-FC=2(μ2/2)1+μ2/2=2μ22+μ2=23≈0.667FOM=μ22+μ2=13\begin{aligned} \gamma_{DSB\text{-}FC} &= \frac{2(\mu^2/2)}{1+\mu^2/2}=\frac{2\mu^2}{2+\mu^2}=\frac{2}{3}\approx0.667 \\ \text{FOM} &= \frac{\mu^2}{2+\mu^2}=\frac13 \end{aligned}

DSB-SC (coherent detection). s(t)=Acm(t)cos⁡ωcts(t)=A_cm(t)\cos\omega_ct, BT=2WB_T=2W.

  • Signal power =Ac2P2=\frac{A_c^2P}{2}; noise in 2W2W =2N0W=2N_0W, so SNRi=Ac2P4N0W\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}.
  • Multiply x(t)=s(t)+n(t)x(t)=s(t)+n(t) by cos⁡ωct\cos\omega_ct and low-pass filter:
y(t)=Ac2m(t)+12nc(t)y(t)=\frac{A_c}{2}m(t)+\frac12n_c(t)
  • SNRo=Ac2P/42N0W/4=Ac2P2N0W\text{SNR}_o=\dfrac{A_c^2P/4}{2N_0W/4}=\dfrac{A_c^2P}{2N_0W}
γDSB-SC=SNRoSNRi=2\gamma_{DSB\text{-}SC}=\frac{\text{SNR}_o}{\text{SNR}_i}=2

The quadrature noise nsn_s is rejected by coherent detection, giving a 3 dB gain. With baseband reference SNRref=Ac2P2N0W\text{SNR}_{ref}=\frac{A_c^2P}{2N_0W}, figure of merit =1=1.

SSB-SC (coherent detection). s(t)=Ac2[m(t)cos⁡ωct∓m^(t)sin⁡ωct]s(t)=\frac{A_c}{2}[m(t)\cos\omega_ct\mp\hat m(t)\sin\omega_ct], BT=WB_T=W.

  • Signal power =Ac24⋅P+P2=Ac2P4=\frac{A_c^2}{4}\cdot\frac{P+P}{2}=\frac{A_c^2P}{4} (since m^2‾=P\overline{\hat m^2}=P); noise in WW =N0W=N_0W, so SNRi=Ac2P4N0W\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}.
  • Multiply by cos⁡ωct\cos\omega_ct and low-pass filter:
y(t)=Ac4m(t)+12nc(t)y(t)=\frac{A_c}{4}m(t)+\frac12n_c(t)
  • Signal power Ac2P16\frac{A_c^2P}{16}, noise power N0W4\frac{N_0W}{4}, so SNRo=Ac2P4N0W\text{SNR}_o=\dfrac{A_c^2P}{4N_0W}
γSSB=SNRoSNRi=1\gamma_{SSB}=\frac{\text{SNR}_o}{\text{SNR}_i}=1

With baseband reference (SNRref=Ac2P4N0W\text{SNR}_{ref}=\frac{A_c^2P}{4N_0W}, same transmitted power), figure of merit =1=1.

Comparison:

ParameterDSB-FC (AM)DSB-SCSSB-SC
Transmission bandwidth BTB_T2W2W2W2WWW
DetectorEnvelope or coherentCoherentCoherent
Detection gain SNRo/SNRi\text{SNR}_o/\text{SNR}_i2ka2P1+ka2P\frac{2k_a^2P}{1+k_a^2P} (=23=\frac23 at μ=1\mu=1)21
Figure of merit (baseband ref.)ka2P1+ka2P\frac{k_a^2P}{1+k_a^2P} (=13=\frac13 at μ=1\mu=1)11
Threshold effectYes (envelope detector)NoNo
Power wasted in carrierYes (at least 2/3)NoNo

Conclusions: for the same transmitted power and noise PSD, DSB-SC and SSB give the same output SNR (equal to baseband); DSB-SC's detection gain of 2 is offset by its double noise bandwidth. SSB achieves this in half the bandwidth. DSB-FC is at least 4.77 dB worse (factor 3 at 100% tone modulation) because the carrier carries no information, but it allows a cheap envelope detector.

  • 2072 Kartik (CS II) · 2+6 marks

Define Matched filter. Explain the approximation of the matched filter for a rectangular pulse using a single pole RC low pass filter with variable bandwidth.

Answer

Matched filter

A matched filter is a linear filter whose impulse response is matched to (a time-reversed, delayed copy of) a known input pulse s(t)s(t), h(t)=k s(T−t)h(t)=k\,s(T-t), so that it gives the maximum possible peak signal-to-noise ratio at the sampling instant t=Tt=T when the pulse is received in additive white noise. It is the optimum detector for digital signals in AWGN; in practice it is built as a correlator.

RC low-pass filter as approximate matched filter

The exact matched filter for a rectangular pulse is an integrate-and-dump circuit, which is not always convenient. A simple single-pole RC low-pass filter can approximate it; we find the best bandwidth and the loss.

Input pulse: s(t)=As(t)=A for 0≤t≤T0\le t\le T (energy E=A2TE=A^2T), white noise PSD N0/2N_0/2.

Filter: H(f)=11+jf/fcH(f)=\dfrac{1}{1+jf/f_c}, fc=12πRCf_c=\dfrac{1}{2\pi RC} (3 dB bandwidth, variable).

Signal output. The step response of the RC filter rises as A(1−e−t/RC)A(1-e^{-t/RC}); it is maximum at the end of the pulse, t=Tt=T:

so(T)=A(1−e−T/RC)=A(1−e−2πfcT)s_o(T)=A\left(1-e^{-T/RC}\right)=A\left(1-e^{-2\pi f_cT}\right)
 s(t)                 s_o(t)
 A +-----+          A |      ..-.  peak at t=T
   |     |            |    .'    '.
   |     |            |  .'        '..
 --+-----+--> t     --+-'-----+-------'--> t
   0     T            0       T

Noise output power:

No=N02∫−∞∞df1+(f/fc)2=N02⋅πfcN_o=\frac{N_0}{2}\int_{-\infty}^{\infty}\frac{df}{1+(f/f_c)^2}=\frac{N_0}{2}\cdot\pi f_c

Output SNR:

SNRo=A2(1−e−2πfcT)2πN0fc/2=2A2TN0⋅2(1−e−x)2x,x=2πfcT\text{SNR}_o=\frac{A^2\left(1-e^{-2\pi f_cT}\right)^2}{\pi N_0f_c/2}=\frac{2A^2T}{N_0}\cdot\frac{2\left(1-e^{-x}\right)^2}{x},\qquad x=2\pi f_cT

Compared with the matched filter (SNRmax=2E/N0=2A2T/N0\text{SNR}_{max}=2E/N_0=2A^2T/N_0):

ρ(x)=SNRoSNRmax=2(1−e−x)2x\rho(x)=\frac{\text{SNR}_o}{\text{SNR}_{max}}=\frac{2(1-e^{-x})^2}{x}

Effect of variable bandwidth:

  • fcf_c too small: the filter cannot follow the pulse, so(T)s_o(T) is small, SNR low.
  • fcf_c too large: signal fully passes but much extra noise enters, SNR low.
  • An optimum lies between. Setting dρ/dx=0d\rho/dx=0 gives (1+2x)e−x=1(1+2x)e^{-x}=1, solved numerically: x≈1.256x\approx1.256.
fcT=1.2562π≈0.2,ρmax=2(1−e−1.256)21.256≈0.815f_cT=\frac{1.256}{2\pi}\approx0.2,\qquad \rho_{max}=\frac{2(1-e^{-1.256})^2}{1.256}\approx0.815

Result: the best RC filter has fc≈0.2Tf_c\approx\dfrac{0.2}{T} (i.e. RC≈0.8TRC\approx0.8T) and gives an SNR only 10log⁡100.815≈−0.8910\log_{10}0.815\approx-0.89 dB, i.e. less than 1 dB below the ideal matched filter. Hence a simple RC filter is a good practical substitute for the matched filter of a rectangular pulse.

  • 2072 Kartik (CS II) · 7 marks

With necessary derivations, explain the threshold effect in envelope detector for DSB-FC modulation in analog communication system.

Answer

An envelope detector is a non-linear device whose output follows the magnitude of the resultant of signal plus noise. Its noise performance changes sharply when the carrier-to-noise ratio drops below a certain level; this is the threshold effect.

Envelope detector input. DSB-FC signal plus narrowband noise:

x(t)=Ac[1+kam(t)]cos⁡ωct+nc(t)cos⁡ωct−ns(t)sin⁡ωctx(t)=A_c[1+k_am(t)]\cos\omega_ct+n_c(t)\cos\omega_ct-n_s(t)\sin\omega_ct

The envelope detector output is the magnitude of the resultant phasor:

y(t)={Ac[1+kam(t)]+nc(t)}2+ns2(t)y(t)=\sqrt{\{A_c[1+k_am(t)]+n_c(t)\}^2+n_s^2(t)}
              n_s
          .---->           resultant length = y(t)
         /    ^
        /     | n_c
  ----------->+
  Ac[1+ka m(t)]

Case 1: high carrier-to-noise ratio (Ac[1+kam]≫A_c[1+k_am]\gg noise). The quadrature term is negligible:

y(t)≈Ac+Ackam(t)+nc(t)y(t)\approx A_c+A_ck_am(t)+n_c(t)

The message appears additively with the noise. After removing DC, SNRo=Ac2ka2P2N0W\text{SNR}_o=\dfrac{A_c^2k_a^2P}{2N_0W}, the same as coherent detection. Figure of merit ka2P1+ka2P\frac{k_a^2P}{1+k_a^2P} (=1/3=1/3 for a single tone with μ=1\mu=1).

Case 2: low carrier-to-noise ratio (noise ≫\gg carrier). Write the noise as n(t)=r(t)cos⁡[ωct+ψ(t)]n(t)=r(t)\cos[\omega_ct+\psi(t)], with Rayleigh envelope rr and uniform phase ψ\psi. Then

y(t)=r2+2rAc[1+kam]cos⁡ψ+Ac2[1+kam]2≈r(t)1+2Acr(t)[1+kam(t)]cos⁡ψ(t)≈r(t)+Accos⁡ψ(t)+Ackam(t)cos⁡ψ(t)\begin{aligned} y(t) &= \sqrt{r^2+2rA_c[1+k_am]\cos\psi+A_c^2[1+k_am]^2} \\ &\approx r(t)\sqrt{1+\frac{2A_c}{r(t)}[1+k_am(t)]\cos\psi(t)} \\ &\approx r(t)+A_c\cos\psi(t)+A_ck_am(t)\cos\psi(t) \end{aligned}

using 1+x≈1+x/2\sqrt{1+x}\approx1+x/2. Now the message m(t)m(t) is multiplied by the random noise cos⁡ψ(t)\cos\psi(t); there is no term proportional to m(t)m(t) alone. The message is mutilated and cannot be recovered; output SNR collapses.

Threshold effect. The change from Case 1 to Case 2 is not gradual: below a certain carrier-to-noise ratio, the threshold (roughly 10 dB carrier-to-noise ratio), the output SNR falls much faster than the input SNR. The threshold is a property of non-linear (envelope) detection; coherent detection of DSB-FC, DSB-SC or SSB has no threshold, because the product detector is linear and the message always stays additive with noise.

 SNRo (dB)                 coherent / envelope
     |                   . (above threshold)
     |                 .
     |               .
     |             .  <- threshold
     |          ..'
     |      ..''    envelope detector
     |  ..''        (below threshold)
     +-------------------------> CNR (dB)
  • 2072 Kartik (CS II) · 5 marks

Derive the expression of error probability for binary PAM signal.

Answer

In binary (polar) PAM bit 1 is sent as a pulse of amplitude +A+A and bit 0 as −A-A. White Gaussian noise is added; the receiver samples once per bit and decides with threshold λ=0\lambda=0.

The sample is y=±A+ny=\pm A+n, n∼N(0,σ2)n\sim\mathcal{N}(0,\sigma^2):

f(y∣1)=12πσe−(y−A)2/2σ2,f(y∣0)=12πσe−(y+A)2/2σ2f(y\mid1)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-A)^2/2\sigma^2},\qquad f(y\mid0)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y+A)^2/2\sigma^2}
    f(y|0)       f(y|1)
     .--.    |    .--.
    /    \   |   /    \
 --/------\..|../------\---> y
  -A         0        +A

Error when 0 is sent (y>0y>0): let u=y+A2σu=\frac{y+A}{\sqrt2\sigma}

Pe0=∫0∞12πσe−(y+A)2/2σ2dy=1π∫A/2σ∞e−u2du=12erfc⁡(A2 σ)\begin{aligned} P_{e0} &= \int_0^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-(y+A)^2/2\sigma^2}dy=\frac{1}{\sqrt\pi}\int_{A/\sqrt2\sigma}^{\infty}e^{-u^2}du \\ &= \frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\,\sigma}\right) \end{aligned}

By symmetry Pe1=Pe0P_{e1}=P_{e0}. With equal priors:

Pe=12erfc⁡(A2 σ)=Q(Aσ)P_e=\frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\,\sigma}\right)=Q\left(\frac{A}{\sigma}\right)

With a matched filter, A2σ2=2EbN0\frac{A^2}{\sigma^2}=\frac{2E_b}{N_0}, so Pe=12erfc⁡Eb/N0P_e=\frac12\operatorname{erfc}\sqrt{E_b/N_0}. For unipolar PAM (AA, 0, threshold A/2A/2): Pe=12erfc⁡(A22σ)P_e=\frac12\operatorname{erfc}\left(\frac{A}{2\sqrt2\sigma}\right), which needs twice the average power for the same PeP_e.

  • 2072 Chaitra (CS II) · 4+3 marks

Find the error probability in coherent ASK and PSK detections and show that ASK requires double the average signal power than PSK for same error probability.

Answer

Coherent ASK

ASK (on–off keying) signals over 0≤t≤Tb0\le t\le T_b:

s1(t)=Acos⁡ωct  (bit 1),s2(t)=0  (bit 0)s_1(t)=A\cos\omega_ct\ \ (\text{bit }1),\qquad s_2(t)=0\ \ (\text{bit }0)

Coherent receiver (correlator = matched filter):

 r(t) -->(x)--> integrate --> sample --> y > lam ? 1 : 0
          ^       0..Tb       t=Tb
          |
       cos(wc t)  (local carrier, in phase)

Correlator output at t=Tbt=T_b, with r(t)=si(t)+w(t)r(t)=s_i(t)+w(t):

y=∫0Tbr(t)cos⁡ωct dt=ai+ny=\int_0^{T_b}r(t)\cos\omega_ct\,dt=a_i+n
  • Bit 1: a1=∫0TbAcos⁡2ωct dt=ATb2a_1=\int_0^{T_b}A\cos^2\omega_ct\,dt=\frac{AT_b}{2} (for fcf_c an integer multiple of 1/Tb1/T_b).
  • Bit 0: a2=0a_2=0.
  • Noise: n=∫0Tbw(t)cos⁡ωct dtn=\int_0^{T_b}w(t)\cos\omega_ct\,dt is Gaussian, mean 0, variance
σ2=∫0Tb ⁣ ⁣∫0TbN02δ(t−u)cos⁡ωctcos⁡ωcu dt du=N02⋅Tb2=N0Tb4\sigma^2=\int_0^{T_b}\!\!\int_0^{T_b}\frac{N_0}{2}\delta(t-u)\cos\omega_ct\cos\omega_cu\,dt\,du=\frac{N_0}{2}\cdot\frac{T_b}{2}=\frac{N_0T_b}{4}

Threshold midway: λ=ATb4\lambda=\frac{AT_b}{4}. Using the binary result Pe=12erfc⁡(a1−a222σ)P_e=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\sigma}\right):

a1−a222 σ=ATb/222N0Tb/4=ATb22N0=A2Tb8N0\begin{aligned} \frac{a_1-a_2}{2\sqrt2\,\sigma} &= \frac{AT_b/2}{2\sqrt2\sqrt{N_0T_b/4}}=\frac{A\sqrt{T_b}}{2\sqrt{2N_0}}=\sqrt{\frac{A^2T_b}{8N_0}} \end{aligned}

Energy of a "1" pulse is E1=A2Tb2E_1=\frac{A^2T_b}{2}; the average energy per bit (half the bits are 0) is Eb=E12=A2Tb4E_b=\frac{E_1}{2}=\frac{A^2T_b}{4}. Hence A2Tb8N0=Eb2N0\frac{A^2T_b}{8N_0}=\frac{E_b}{2N_0} and

Pe(ASK)=12erfc⁡(Eb2N0)=Q(EbN0)P_e(\text{ASK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)

(In terms of peak energy E1E_1: Pe=12erfc⁡E1/4N0P_e=\frac12\operatorname{erfc}\sqrt{E_1/4N_0}.)

Coherent PSK

BPSK signals over 0≤t≤Tb0\le t\le T_b (antipodal):

s1(t)=Acos⁡ωct  (1),s2(t)=−Acos⁡ωct  (0),Eb=A2Tb2s_1(t)=A\cos\omega_ct\ \ (1),\qquad s_2(t)=-A\cos\omega_ct\ \ (0),\qquad E_b=\frac{A^2T_b}{2}

Coherent receiver: multiply by the locally generated cos⁡ωct\cos\omega_ct, integrate over TbT_b, sample, compare with threshold λ=0\lambda=0.

 r(t) -->(x)--> integrate --> sample --> y > 0 ? 1 : 0
          ^       0..Tb       t=Tb
          |
       cos(wc t)  (from carrier recovery)

Correlator output y=ai+ny=a_i+n with

a1=+ATb2,a2=−ATb2,σ2=N0Tb4a_1=+\frac{AT_b}{2},\qquad a_2=-\frac{AT_b}{2},\qquad \sigma^2=\frac{N_0T_b}{4}

The pdfs of yy are Gaussian centred at ±ATb/2\pm AT_b/2. Error when 0 is sent (y>0y>0):

Pe=∫0∞12πσe−(y+ATb/2)2/2σ2dy=12erfc⁡(ATb/22 σ)=12erfc⁡(ATb/2N0Tb/2)=12erfc⁡(A2Tb2N0)\begin{aligned} P_e &= \int_0^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-(y+AT_b/2)^2/2\sigma^2}dy=\frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt2\,\sigma}\right) \\ &= \frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt{N_0T_b/2}}\right)=\frac12\operatorname{erfc}\left(\sqrt{\frac{A^2T_b}{2N_0}}\right) \end{aligned}

By symmetry the error for a sent 1 is the same, so

Pe(BPSK)=12erfc⁡(EbN0)=Q(2EbN0)P_e(\text{BPSK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{N_0}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)

ASK needs double the power of PSK

Comparison (same PeP_e). ASK: Pe=12erfc⁡Eb/2N0P_e=\frac12\operatorname{erfc}\sqrt{E_b/2N_0}; PSK: Pe=12erfc⁡Eb/N0P_e=\frac12\operatorname{erfc}\sqrt{E_b/N_0}. For equal PeP_e the arguments must be equal:

Eb,ASK2N0=Eb,PSKN0 ⇒ Eb,ASK=2Eb,PSK\frac{E_{b,ASK}}{2N_0}=\frac{E_{b,PSK}}{N_0}\ \Rightarrow\ E_{b,ASK}=2E_{b,PSK}

Since average power =Eb/Tb=E_b/T_b, ASK needs twice (3 dB more) average signal power than PSK for the same error probability. Physically, the PSK signal points are 2Eb2\sqrt{E_b} apart, while ASK points are only 2Eb\sqrt{2E_b} apart for the same average energy.

 ASK:  *0 ------------ *sqrt(2Eb)   d = sqrt(2Eb)
 PSK:  *-sqrt(Eb) -- 0 -- *+sqrt(Eb) d = 2 sqrt(Eb)
  • 2072 Chaitra (CS II) · 8 marks

With necessary derivation, compare noise performance of DSB-AM, DSB-SC, SSB-SC.

Answer

Receiver model. The received signal s(t)s(t) plus white noise (N0/2N_0/2) passes through a band-pass (IF) filter of bandwidth BTB_T, then a demodulator and a low-pass filter of bandwidth WW (message bandwidth).

 s(t)  +   +-------+    +-------+   +-----+
 ---->(+)-->|  BPF  |--->| demod |-->| LPF |--> y(t)
       ^    |  B_T  |    |       |   |  W  |
      w(t)  +-------+    +-------+   +-----+

The filtered noise is narrowband and is written in in-phase/quadrature form

n(t)=nc(t)cos⁡ωct−ns(t)sin⁡ωctn(t)=n_c(t)\cos\omega_ct-n_s(t)\sin\omega_ct

where ncn_c and nsn_s are low-pass, each with the same power as n(t)n(t): nc2‾=ns2‾=n2‾=N0BT\overline{n_c^2}=\overline{n_s^2}=\overline{n^2}=N_0B_T.

Definitions. Message power P=m2(t)‾P=\overline{m^2(t)}.

  • Input SNR SNRi=received signal powernoise power in BT\text{SNR}_i=\dfrac{\text{received signal power}}{\text{noise power in } B_T}
  • Output SNR SNRo=message power at outputnoise power at output\text{SNR}_o=\dfrac{\text{message power at output}}{\text{noise power at output}}
  • Detection gain γd=SNRoSNRi\gamma_d=\dfrac{\text{SNR}_o}{\text{SNR}_i}; figure of merit =SNRoSNRref=\dfrac{\text{SNR}_o}{\text{SNR}_{ref}}, where SNRref\text{SNR}_{ref} is the channel SNR measured in the message band WW (baseband reference). Both show how much the demodulator improves or worsens SNR.

DSB-FC (conventional AM). s(t)=Ac[1+kam(t)]cos⁡ωcts(t)=A_c[1+k_am(t)]\cos\omega_ct, BT=2WB_T=2W. (Coherent and envelope detection give the same result at high SNR.)

  • Signal power =Ac22(1+ka2P)=\frac{A_c^2}{2}(1+k_a^2P); noise in 2W2W =2N0W=2N_0W:
SNRi=Ac2(1+ka2P)4N0W\text{SNR}_i=\frac{A_c^2(1+k_a^2P)}{4N_0W}
  • Coherent detection (or envelope detection at high SNR) and DC blocking:
y(t)=Ac2kam(t)+12nc(t) ⇒ SNRo=Ac2ka2P/42N0W/4=Ac2ka2P2N0Wy(t)=\frac{A_c}{2}k_am(t)+\frac12n_c(t)\ \Rightarrow\ \text{SNR}_o=\frac{A_c^2k_a^2P/4}{2N_0W/4}=\frac{A_c^2k_a^2P}{2N_0W} γDSB-FC=SNRoSNRi=2ka2P1+ka2P,FOM=ka2P1+ka2P\gamma_{DSB\text{-}FC}=\frac{\text{SNR}_o}{\text{SNR}_i}=\frac{2k_a^2P}{1+k_a^2P},\qquad \text{FOM}=\frac{k_a^2P}{1+k_a^2P}

The carrier uses power but carries no information, so the gain is reduced.

Single tone, 100% modulation. m(t)=Amcos⁡ωmtm(t)=A_m\cos\omega_mt, μ=kaAm=1\mu=k_aA_m=1, so ka2P=μ22=12k_a^2P=\frac{\mu^2}{2}=\frac12:

γDSB-FC=2(μ2/2)1+μ2/2=2μ22+μ2=23≈0.667FOM=μ22+μ2=13\begin{aligned} \gamma_{DSB\text{-}FC} &= \frac{2(\mu^2/2)}{1+\mu^2/2}=\frac{2\mu^2}{2+\mu^2}=\frac{2}{3}\approx0.667 \\ \text{FOM} &= \frac{\mu^2}{2+\mu^2}=\frac13 \end{aligned}

DSB-SC (coherent detection). s(t)=Acm(t)cos⁡ωcts(t)=A_cm(t)\cos\omega_ct, BT=2WB_T=2W.

  • Signal power =Ac2P2=\frac{A_c^2P}{2}; noise in 2W2W =2N0W=2N_0W, so SNRi=Ac2P4N0W\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}.
  • Multiply x(t)=s(t)+n(t)x(t)=s(t)+n(t) by cos⁡ωct\cos\omega_ct and low-pass filter:
y(t)=Ac2m(t)+12nc(t)y(t)=\frac{A_c}{2}m(t)+\frac12n_c(t)
  • SNRo=Ac2P/42N0W/4=Ac2P2N0W\text{SNR}_o=\dfrac{A_c^2P/4}{2N_0W/4}=\dfrac{A_c^2P}{2N_0W}
γDSB-SC=SNRoSNRi=2\gamma_{DSB\text{-}SC}=\frac{\text{SNR}_o}{\text{SNR}_i}=2

The quadrature noise nsn_s is rejected by coherent detection, giving a 3 dB gain. With baseband reference SNRref=Ac2P2N0W\text{SNR}_{ref}=\frac{A_c^2P}{2N_0W}, figure of merit =1=1.

SSB-SC (coherent detection). s(t)=Ac2[m(t)cos⁡ωct∓m^(t)sin⁡ωct]s(t)=\frac{A_c}{2}[m(t)\cos\omega_ct\mp\hat m(t)\sin\omega_ct], BT=WB_T=W.

  • Signal power =Ac24⋅P+P2=Ac2P4=\frac{A_c^2}{4}\cdot\frac{P+P}{2}=\frac{A_c^2P}{4} (since m^2‾=P\overline{\hat m^2}=P); noise in WW =N0W=N_0W, so SNRi=Ac2P4N0W\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}.
  • Multiply by cos⁡ωct\cos\omega_ct and low-pass filter:
y(t)=Ac4m(t)+12nc(t)y(t)=\frac{A_c}{4}m(t)+\frac12n_c(t)
  • Signal power Ac2P16\frac{A_c^2P}{16}, noise power N0W4\frac{N_0W}{4}, so SNRo=Ac2P4N0W\text{SNR}_o=\dfrac{A_c^2P}{4N_0W}
γSSB=SNRoSNRi=1\gamma_{SSB}=\frac{\text{SNR}_o}{\text{SNR}_i}=1

With baseband reference (SNRref=Ac2P4N0W\text{SNR}_{ref}=\frac{A_c^2P}{4N_0W}, same transmitted power), figure of merit =1=1.

Comparison:

ParameterDSB-FC (AM)DSB-SCSSB-SC
Transmission bandwidth BTB_T2W2W2W2WWW
DetectorEnvelope or coherentCoherentCoherent
Detection gain SNRo/SNRi\text{SNR}_o/\text{SNR}_i2ka2P1+ka2P\frac{2k_a^2P}{1+k_a^2P} (=23=\frac23 at μ=1\mu=1)21
Figure of merit (baseband ref.)ka2P1+ka2P\frac{k_a^2P}{1+k_a^2P} (=13=\frac13 at μ=1\mu=1)11
Threshold effectYes (envelope detector)NoNo
Power wasted in carrierYes (at least 2/3)NoNo

Conclusions: for the same transmitted power and noise PSD, DSB-SC and SSB give the same output SNR (equal to baseband); DSB-SC's detection gain of 2 is offset by its double noise bandwidth. SSB achieves this in half the bandwidth. DSB-FC is at least 4.77 dB worse (factor 3 at 100% tone modulation) because the carrier carries no information, but it allows a cheap envelope detector.

  • 2072 Chaitra (CS II) · 2.5 marks

Write a short note on noise equivalent bandwidth.

Answer

The noise equivalent bandwidth (NEB) BNB_N of a filter is the bandwidth of an ideal rectangular filter, with the same maximum gain ∣H(0)∣\lvert H(0)\rvert, that passes the same noise power as the real filter when both are driven by white noise:

BN=1∣H(0)∣2∫0∞∣H(f)∣2dfB_N=\frac{1}{\lvert H(0)\rvert^2}\int_0^{\infty}\lvert H(f)\rvert^2df
  • Output noise power can then be written simply as N=N0BN∣H(0)∣2N=N_0B_N\lvert H(0)\rvert^2 (or kTBNkTB_N).
  • It is used in noise figure, noise temperature and receiver SNR calculations.
  • Example: RC low-pass filter, BN=14RC=π2f3dB≈1.57f3dBB_N=\frac{1}{4RC}=\frac\pi2f_{3dB}\approx1.57f_{3dB}.
  • For sharp-cut-off filters BN≈B_N\approx 3 dB bandwidth; for gentle roll-off filters BNB_N is larger.
  • 2071 Shrawan (CS II) · 6 marks

With necessary derivations, compare the noise performance of DSB-SC and SSB-SC modulations in analog communication system.

Answer

Receiver model. The received signal s(t)s(t) plus white noise (N0/2N_0/2) passes through a band-pass (IF) filter of bandwidth BTB_T, then a demodulator and a low-pass filter of bandwidth WW (message bandwidth).

 s(t)  +   +-------+    +-------+   +-----+
 ---->(+)-->|  BPF  |--->| demod |-->| LPF |--> y(t)
       ^    |  B_T  |    |       |   |  W  |
      w(t)  +-------+    +-------+   +-----+

The filtered noise is narrowband and is written in in-phase/quadrature form

n(t)=nc(t)cos⁡ωct−ns(t)sin⁡ωctn(t)=n_c(t)\cos\omega_ct-n_s(t)\sin\omega_ct

where ncn_c and nsn_s are low-pass, each with the same power as n(t)n(t): nc2‾=ns2‾=n2‾=N0BT\overline{n_c^2}=\overline{n_s^2}=\overline{n^2}=N_0B_T.

Definitions. Message power P=m2(t)‾P=\overline{m^2(t)}.

  • Input SNR SNRi=received signal powernoise power in BT\text{SNR}_i=\dfrac{\text{received signal power}}{\text{noise power in } B_T}
  • Output SNR SNRo=message power at outputnoise power at output\text{SNR}_o=\dfrac{\text{message power at output}}{\text{noise power at output}}
  • Detection gain γd=SNRoSNRi\gamma_d=\dfrac{\text{SNR}_o}{\text{SNR}_i}; figure of merit =SNRoSNRref=\dfrac{\text{SNR}_o}{\text{SNR}_{ref}}, where SNRref\text{SNR}_{ref} is the channel SNR measured in the message band WW (baseband reference). Both show how much the demodulator improves or worsens SNR.

DSB-SC (coherent detection). s(t)=Acm(t)cos⁡ωcts(t)=A_cm(t)\cos\omega_ct, BT=2WB_T=2W.

  • Signal power =Ac2P2=\frac{A_c^2P}{2}; noise in 2W2W =2N0W=2N_0W, so SNRi=Ac2P4N0W\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}.
  • Multiply x(t)=s(t)+n(t)x(t)=s(t)+n(t) by cos⁡ωct\cos\omega_ct and low-pass filter:
y(t)=Ac2m(t)+12nc(t)y(t)=\frac{A_c}{2}m(t)+\frac12n_c(t)
  • SNRo=Ac2P/42N0W/4=Ac2P2N0W\text{SNR}_o=\dfrac{A_c^2P/4}{2N_0W/4}=\dfrac{A_c^2P}{2N_0W}
γDSB-SC=SNRoSNRi=2\gamma_{DSB\text{-}SC}=\frac{\text{SNR}_o}{\text{SNR}_i}=2

The quadrature noise nsn_s is rejected by coherent detection, giving a 3 dB gain. With baseband reference SNRref=Ac2P2N0W\text{SNR}_{ref}=\frac{A_c^2P}{2N_0W}, figure of merit =1=1.

SSB-SC (coherent detection). s(t)=Ac2[m(t)cos⁡ωct∓m^(t)sin⁡ωct]s(t)=\frac{A_c}{2}[m(t)\cos\omega_ct\mp\hat m(t)\sin\omega_ct], BT=WB_T=W.

  • Signal power =Ac24⋅P+P2=Ac2P4=\frac{A_c^2}{4}\cdot\frac{P+P}{2}=\frac{A_c^2P}{4} (since m^2‾=P\overline{\hat m^2}=P); noise in WW =N0W=N_0W, so SNRi=Ac2P4N0W\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}.
  • Multiply by cos⁡ωct\cos\omega_ct and low-pass filter:
y(t)=Ac4m(t)+12nc(t)y(t)=\frac{A_c}{4}m(t)+\frac12n_c(t)
  • Signal power Ac2P16\frac{A_c^2P}{16}, noise power N0W4\frac{N_0W}{4}, so SNRo=Ac2P4N0W\text{SNR}_o=\dfrac{A_c^2P}{4N_0W}
γSSB=SNRoSNRi=1\gamma_{SSB}=\frac{\text{SNR}_o}{\text{SNR}_i}=1

With baseband reference (SNRref=Ac2P4N0W\text{SNR}_{ref}=\frac{A_c^2P}{4N_0W}, same transmitted power), figure of merit =1=1.

Comparison:

QuantityDSB-SCSSB-SC
Bandwidth BTB_T2W2WWW
Input noise power2N0W2N_0WN0WN_0W
Detection gain21
Figure of merit11
Output SNR (same transmitted power)EqualEqual

DSB-SC has twice the detection gain because its two sidebands add coherently, but it collects twice the noise. For the same transmitted power both give identical output SNR; SSB is preferred because it needs half the bandwidth.

  • 2071 Shrawan (CS II) · 2+6 marks

Define random process? Show that the output of LTI is WSSP if the input is also a WSSP.

Answer

Random process

A random process X(t,s)X(t, s) is a collection (ensemble) of time functions, one for each outcome ss of a random experiment. At a fixed time tkt_k, X(tk)X(t_k) is a random variable; for a fixed outcome sjs_j, xj(t)x_j(t) is one sample function. Example: the thermal-noise voltage across a resistor, observed on many identical resistors.

 x1(t) ~~~/\~~~/\/~~~~   <- sample function 1
 x2(t) ~/\~~~~\/~~~/\~   <- sample function 2
 x3(t) ~~~\/~/\~~~~~/\   <- sample function 3
          |
          t = tk  ->  X(tk) is a random variable

WSS input gives WSS output

A process is wide-sense stationary (WSS) if (i) its mean is constant and (ii) its autocorrelation depends only on the time difference τ=t1−t2\tau = t_1 - t_2:

E[X(t)]=μX,E[X(t1)X(t2)]=RX(t1−t2)E[X(t)] = \mu_X, \qquad E[X(t_1)X(t_2)] = R_X(t_1 - t_2)

Let a WSS process X(t)X(t) be the input of a stable LTI system with impulse response h(t)h(t). The output is

Y(t)=∫−∞∞h(α) X(t−α) dαY(t) = \int_{-\infty}^{\infty} h(\alpha)\,X(t-\alpha)\,d\alpha

1. Mean of the output. Expectation and integration are linear, so they can be interchanged:

E[Y(t)]=∫−∞∞h(α) E[X(t−α)] dα=μX∫−∞∞h(α) dα=μX H(0)\begin{aligned} E[Y(t)] &= \int_{-\infty}^{\infty} h(\alpha)\,E[X(t-\alpha)]\,d\alpha \\ &= \mu_X\int_{-\infty}^{\infty} h(\alpha)\,d\alpha = \mu_X\,H(0) \end{aligned}

H(0)H(0) is the DC gain of the system, a constant. So the output mean is constant, independent of tt.

2. Autocorrelation of the output.

RY(t1,t2)=E[Y(t1)Y(t2)]=E[∫h(α)X(t1−α)dα∫h(β)X(t2−β)dβ]=∫ ⁣ ⁣∫h(α)h(β) E[X(t1−α)X(t2−β)] dα dβ=∫ ⁣ ⁣∫h(α)h(β) RX(t1−t2−α+β) dα dβ\begin{aligned} R_Y(t_1,t_2) &= E[Y(t_1)Y(t_2)] \\ &= E\left[\int h(\alpha)X(t_1-\alpha)d\alpha \int h(\beta)X(t_2-\beta)d\beta\right] \\ &= \int\!\!\int h(\alpha)h(\beta)\,E[X(t_1-\alpha)X(t_2-\beta)]\,d\alpha\,d\beta \\ &= \int\!\!\int h(\alpha)h(\beta)\,R_X(t_1-t_2-\alpha+\beta)\,d\alpha\,d\beta \end{aligned}

The last step uses the WSS property of X(t)X(t): the input autocorrelation depends only on the difference of its two time arguments, (t1−α)−(t2−β)(t_1-\alpha)-(t_2-\beta). Putting τ=t1−t2\tau = t_1 - t_2:

RY(τ)=∫−∞∞∫−∞∞h(α)h(β) RX(τ−α+β) dα dβR_Y(\tau) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} h(\alpha)h(\beta)\,R_X(\tau-\alpha+\beta)\,d\alpha\,d\beta

The right side depends only on τ\tau, not on t1t_1 or t2t_2 separately.

3. Mean square value E[Y2(t)]=RY(0)E[Y^2(t)] = R_Y(0) is a constant, finite for a stable system with finite input power.

Since the output mean is constant and the output autocorrelation is a function of τ\tau only, Y(t)Y(t) is also WSS.

Taking the Fourier transform of RY(τ)R_Y(\tau) gives the useful result

SY(f)=∣H(f)∣2 SX(f)S_Y(f) = |H(f)|^2\,S_X(f)

i.e. the output PSD equals the input PSD times the squared magnitude response.

  • 2071 Shrawan (CS II) · 5+3 marks

Derive the expression of error probability in case of M-ary system. Compare binary and M-ary scheme in terms of bandwidth efficiency and system complexity.

Answer

Error probability of an M-ary (PAM) system

In an M-ary system each symbol carries n=log⁡2Mn = \log_2 M bits. Take M-ary baseband PAM: the receiver output sample (after the matched filter) takes one of MM equally spaced levels

ak∈{±A, ±3A, …, ±(M−1)A}a_k \in \{\pm A,\ \pm 3A,\ \dots,\ \pm(M-1)A\}

so adjacent levels are 2A2A apart. Assumptions: symbols equally likely, additive white Gaussian noise of PSD N0/2N_0/2, so the sampled noise nn is Gaussian with zero mean and variance σ2=N0/2\sigma^2 = N_0/2 (unit-energy pulse, matched filter). Decision thresholds are placed midway between levels: 0,±2A,±4A,…0, \pm 2A, \pm 4A, \dots

  -3A     -A      +A     +3A      (M = 4)
   x  |    x   |   x   |   x
     -2A       0      +2A        <- thresholds
 outer         inner        outer

Inner level (two neighbours): an error occurs if ∣n∣>A|n| > A:

Pe,inner=P(n>A)+P(n<−A)=2Q ⁣(Aσ)P_{e,\text{inner}} = P(n>A) + P(n<-A) = 2Q\!\left(\frac{A}{\sigma}\right)

Outer level (one neighbour, e.g. +(M−1)A+(M-1)A): an error occurs only if n<−An < -A:

Pe,outer=Q ⁣(Aσ)P_{e,\text{outer}} = Q\!\left(\frac{A}{\sigma}\right)

There are M−2M-2 inner and 22 outer levels, so the average symbol error probability is

Pe=1M[(M−2) 2Q ⁣(Aσ)+2 Q ⁣(Aσ)]=2(1−1M)Q ⁣(Aσ)\begin{aligned} P_e &= \frac{1}{M}\left[(M-2)\,2Q\!\left(\frac{A}{\sigma}\right) + 2\,Q\!\left(\frac{A}{\sigma}\right)\right] \\ &= 2\left(1-\frac{1}{M}\right)Q\!\left(\frac{A}{\sigma}\right) \end{aligned}

In terms of energy. The average symbol energy is

Es=2M∑i=1M/2(2i−1)2A2=(M2−1)A23  ⇒  A2=3EsM2−1E_s = \frac{2}{M}\sum_{i=1}^{M/2}(2i-1)^2A^2 = \frac{(M^2-1)A^2}{3} \;\Rightarrow\; A^2 = \frac{3E_s}{M^2-1}

With σ2=N0/2\sigma^2 = N_0/2:

Pe=2(1−1M)Q ⁣(6Es(M2−1)N0),Es=Eblog⁡2MP_e = 2\left(1-\frac{1}{M}\right)Q\!\left(\sqrt{\frac{6E_s}{(M^2-1)N_0}}\right), \qquad E_s = E_b\log_2 M

Check, M = 2: Es=EbE_s = E_b and Pe=Q(2Eb/N0)P_e = Q\left(\sqrt{2E_b/N_0}\right), the polar binary result.

With Gray coding, bit error rate ≈Pe/log⁡2M\approx P_e/\log_2 M.

Binary vs M-ary

PointBinaryM-ary
Bits per symbol1log⁡2M\log_2 M
Symbol rate for bit rate RbR_bRbR_bRb/log⁡2MR_b/\log_2 M
Bandwidth needed≈Rb/2\approx R_b/2 (Nyquist)≈Rb/(2log⁡2M)\approx R_b/(2\log_2 M)
Bandwidth efficiency2 bit/s/Hz (ideal)2log⁡2M2\log_2 M bit/s/Hz
Power for same PeP_elowerhigher (levels closer)
Noise immunitybetterpoorer
Transmitter/receiversimple (one threshold)complex (M−1M-1 thresholds, AGC, precise level control)

So M-ary signalling saves bandwidth by a factor log⁡2M\log_2 M but needs more signal power and a more complex system for the same error rate. It is used where bandwidth is scarce (telephone-line modems, digital radio).

  • 2071 Chaitra (CS II) · 5+1 marks

Define moment and central moment of continuous random variable. Show that first central moment is always zero. Determine the noise equivalent bandwidth of RC-LPF and that of ideal LPF of zero frequency response one. Also, find output noise power of this RC-LPF when input is white noise.

Answer

Moments and central moments

For a continuous random variable XX with PDF fX(x)f_X(x):

  • nnth moment (about the origin):
E[Xn]=∫−∞∞xnfX(x) dxE[X^n] = \int_{-\infty}^{\infty} x^n f_X(x)\,dx

The first moment is the mean μ=E[X]\mu = E[X]; the second is the mean-square value E[X2]E[X^2].

  • nnth central moment (about the mean):
E[(X−μ)n]=∫−∞∞(x−μ)nfX(x) dxE[(X-\mu)^n] = \int_{-\infty}^{\infty} (x-\mu)^n f_X(x)\,dx

The second central moment is the variance σ2=E[X2]−μ2\sigma^2 = E[X^2]-\mu^2.

First central moment is always zero:

E[X−μ]=∫−∞∞(x−μ)fX(x) dx=∫−∞∞xfX(x) dx−μ∫−∞∞fX(x) dx=μ−μ(1)=0\begin{aligned} E[X-\mu] &= \int_{-\infty}^{\infty}(x-\mu)f_X(x)\,dx \\ &= \int_{-\infty}^{\infty}x f_X(x)\,dx - \mu\int_{-\infty}^{\infty}f_X(x)\,dx \\ &= \mu - \mu(1) = 0 \end{aligned}

Noise equivalent bandwidth of RC low-pass filter

The NEB BNB_N is the bandwidth of an ideal LPF, with the same zero-frequency gain H(0)H(0), that passes the same white-noise power:

BN=1∣H(0)∣2∫0∞∣H(f)∣2 dfB_N = \frac{1}{|H(0)|^2}\int_0^{\infty}|H(f)|^2\,df

For the RC-LPF, H(f)=11+j2πfRCH(f) = \dfrac{1}{1+j2\pi fRC}, so H(0)=1H(0) = 1 and

BN=∫0∞df1+(2πfRC)2=12πRC[tan⁡−1(2πfRC)]0∞=12πRC⋅π2=14RC\begin{aligned} B_N &= \int_0^{\infty}\frac{df}{1+(2\pi fRC)^2} \\ &= \frac{1}{2\pi RC}\Big[\tan^{-1}(2\pi fRC)\Big]_0^{\infty} \\ &= \frac{1}{2\pi RC}\cdot\frac{\pi}{2} = \frac{1}{4RC} \end{aligned}

Since the 3 dB bandwidth is f3dB=1/(2πRC)f_{3dB} = 1/(2\pi RC), BN=π2f3dB≈1.57f3dBB_N = \frac{\pi}{2}f_{3dB} \approx 1.57 f_{3dB}.

Ideal LPF with ∣H(f)∣=1|H(f)| = 1 for ∣f∣≤B|f| \le B and 0 elsewhere:

BN=11∫0B1 df=BB_N = \frac{1}{1}\int_0^{B}1\,df = B

So the NEB of an ideal LPF equals its actual bandwidth.

Output noise power of the RC-LPF

Input white noise has two-sided PSD SX(f)=N0/2S_X(f) = N_0/2. Output PSD is SY(f)=∣H(f)∣2N0/2S_Y(f) = |H(f)|^2 N_0/2, so

Pout=∫−∞∞N02⋅df1+(2πfRC)2=N02⋅2BN=N02⋅12RC=N04RC\begin{aligned} P_{out} &= \int_{-\infty}^{\infty}\frac{N_0}{2}\cdot\frac{df}{1+(2\pi fRC)^2} \\ &= \frac{N_0}{2}\cdot 2B_N = \frac{N_0}{2}\cdot\frac{1}{2RC} = \frac{N_0}{4RC} \end{aligned}

Answer: BN,RC=14RCB_{N,RC} = \dfrac{1}{4RC}, BN,ideal=BB_{N,ideal} = B, output noise power =N04RC= \dfrac{N_0}{4RC} (watts, on a 1 Ω\Omega basis).

  • 2071 Chaitra (CS II) · 6 marks

With necessary assumption, derive the expression for bit error probability for binary ASK system.

Answer

Binary ASK (on-off keying) sends a carrier burst for "1" and nothing for "0". With coherent (matched-filter/correlator) detection in AWGN, the bit error probability is Pe=Q(Eb/N0)P_e = Q\left(\sqrt{E_b/N_0}\right), where EbE_b is the average energy per bit.

Assumptions

  1. Signals over 0≤t≤Tb0 \le t \le T_b:
s1(t)=Acos⁡ωct  ("1"),s0(t)=0  ("0")s_1(t) = A\cos\omega_c t \ \ (\text{"1"}), \qquad s_0(t) = 0 \ \ (\text{"0"})

with fcf_c an integer multiple of 1/Tb1/T_b. 2. Channel adds white Gaussian noise n(t)n(t) of two-sided PSD N0/2N_0/2. 3. Symbols are equally likely; carrier phase and bit timing are known (coherent detection).

Receiver

 r(t) --->(X)--->[ integrate 0..Tb ]--> y --> [ y > Vth ? ] --> 1/0
           ^                                   sample at Tb
     cos(wc t)  (local coherent carrier)

Correlator output

y=∫0Tbr(t)cos⁡ωct dty = \int_0^{T_b} r(t)\cos\omega_c t\,dt
  • For "1": signal part =∫0TbAcos⁡2ωct dt=ATb2= \int_0^{T_b}A\cos^2\omega_c t\,dt = \dfrac{AT_b}{2}
  • For "0": signal part =0= 0

Noise at the output: no=∫0Tbn(t)cos⁡ωct dtn_o = \int_0^{T_b} n(t)\cos\omega_c t\,dt is Gaussian, zero mean, with variance

σ2=N02∫0Tbcos⁡2ωct dt=N0Tb4\sigma^2 = \frac{N_0}{2}\int_0^{T_b}\cos^2\omega_c t\,dt = \frac{N_0T_b}{4}

Threshold. For equal priors and equal variances, the optimum threshold is midway: Vth=ATb/4V_{th} = AT_b/4.

Error probabilities

P(e∣0)=P(no>ATb4)=Q ⁣(ATb/4σ)P(e∣1)=P(no<−ATb4)=Q ⁣(ATb/4σ)\begin{aligned} P(e|0) &= P\left(n_o > \tfrac{AT_b}{4}\right) = Q\!\left(\frac{AT_b/4}{\sigma}\right) \\ P(e|1) &= P\left(n_o < -\tfrac{AT_b}{4}\right) = Q\!\left(\frac{AT_b/4}{\sigma}\right) \end{aligned} Pe=12P(e∣0)+12P(e∣1)=Q ⁣(ATb/4N0Tb/4)=Q ⁣(A2Tb4N0)P_e = \tfrac12 P(e|0) + \tfrac12 P(e|1) = Q\!\left(\frac{AT_b/4}{\sqrt{N_0T_b/4}}\right) = Q\!\left(\sqrt{\frac{A^2T_b}{4N_0}}\right)

In terms of energy. Energy of the "1" pulse E1=A2Tb/2E_1 = A^2T_b/2; the "0" has zero energy, so average energy per bit Eb=A2Tb/4E_b = A^2T_b/4. Hence

Pe=Q ⁣(EbN0)=Q ⁣(E12N0)=12 erfc ⁣(Eb2N0)P_e = Q\!\left(\sqrt{\frac{E_b}{N_0}}\right) = Q\!\left(\sqrt{\frac{E_1}{2N_0}}\right) = \frac12\,\text{erfc}\!\left(\sqrt{\frac{E_b}{2N_0}}\right)

Remarks

  • Coherent BPSK gives Q(2Eb/N0)Q(\sqrt{2E_b/N_0}), so ASK needs 3 dB more average energy for the same PeP_e.
  • Non-coherent (envelope) ASK gives approximately Pe≈12e−Eb/2N0P_e \approx \tfrac12 e^{-E_b/2N_0}, slightly worse but simpler.
  • 2070 Asar (CS II) · 3+5 marks

Define noise equivalent bandwidth. Find mean and AC function at the output when a WSSP signal is passed through the LTI system.

Answer

Noise equivalent bandwidth

The noise equivalent bandwidth (NEB) BNB_N of a filter is the bandwidth of an ideal rectangular filter, with the same peak (or zero-frequency) gain, that passes the same noise power as the actual filter when both are driven by the same white noise.

If white noise of PSD N0/2N_0/2 is applied, actual output power =N02∫−∞∞∣H(f)∣2df= \frac{N_0}{2}\int_{-\infty}^{\infty}|H(f)|^2df and ideal output power =N02∣H(0)∣2⋅2BN= \frac{N_0}{2}|H(0)|^2 \cdot 2B_N. Equating:

BN=∫0∞∣H(f)∣2 df∣H(0)∣2B_N = \frac{\int_0^{\infty}|H(f)|^2\,df}{|H(0)|^2}
 |H(f)|^2
  |H(0)|^2 +-------+         area of rectangle
           |\      |         = area under |H(f)|^2
           | \     |
           |  \__  |
           |     ~~|~~~___
           +-------+------------> f
           0       BN

Example: RC-LPF has BN=1/(4RC)=1.57f3dBB_N = 1/(4RC) = 1.57 f_{3dB}. Then output noise power is simply N0BN∣H(0)∣2N_0 B_N |H(0)|^2.

Mean and autocorrelation at the output

Let a WSS process X(t)X(t) (mean μX\mu_X, autocorrelation RX(τ)R_X(\tau)) drive a stable LTI system with impulse response h(t)h(t):

Y(t)=∫−∞∞h(α)X(t−α) dαY(t) = \int_{-\infty}^{\infty} h(\alpha)X(t-\alpha)\,d\alpha

Mean

μY=E[Y(t)]=∫−∞∞h(α)E[X(t−α)] dα=μX∫−∞∞h(α) dα=μXH(0)\mu_Y = E[Y(t)] = \int_{-\infty}^{\infty} h(\alpha)E[X(t-\alpha)]\,d\alpha = \mu_X\int_{-\infty}^{\infty}h(\alpha)\,d\alpha = \mu_X H(0)

The output mean is the input mean times the DC gain, a constant.

Autocorrelation

RY(t,t+τ)=E[Y(t)Y(t+τ)]=∫ ⁣ ⁣∫h(α)h(β) E[X(t−α)X(t+τ−β)] dα dβ=∫−∞∞∫−∞∞h(α)h(β) RX(τ+α−β) dα dβ\begin{aligned} R_Y(t, t+\tau) &= E[Y(t)Y(t+\tau)] \\ &= \int\!\!\int h(\alpha)h(\beta)\,E[X(t-\alpha)X(t+\tau-\beta)]\,d\alpha\,d\beta \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} h(\alpha)h(\beta)\,R_X(\tau+\alpha-\beta)\,d\alpha\,d\beta \end{aligned}

This depends only on τ\tau, so Y(t)Y(t) is also WSS. In compact form:

RY(τ)=h(τ)∗h(−τ)∗RX(τ)R_Y(\tau) = h(\tau) * h(-\tau) * R_X(\tau)

Power spectral density (Fourier transform, using h(−τ)↔H∗(f)h(-\tau) \leftrightarrow H^*(f)):

SY(f)=H(f)H∗(f)SX(f)=∣H(f)∣2SX(f)S_Y(f) = H(f)H^*(f)S_X(f) = |H(f)|^2 S_X(f)

Output power

E[Y2(t)]=RY(0)=∫−∞∞∣H(f)∣2SX(f) dfE[Y^2(t)] = R_Y(0) = \int_{-\infty}^{\infty}|H(f)|^2S_X(f)\,df

For white input, SX=N0/2S_X = N_0/2, this gives N0∣H(0)∣2BNN_0|H(0)|^2B_N, which links back to the NEB.

  • 2070 Asar (CS II) · 4 marks

Realize the matched filter with relevant mathematical support.

Answer

A matched filter is the linear filter that maximises the output peak-signal-to-noise ratio at the sampling instant TT for a known pulse s(t)s(t) in white noise. Its impulse response is h(t)=k s(T−t)h(t) = k\,s(T-t).

Derivation (brief). Input s(t)+n(t)s(t) + n(t), noise PSD N0/2N_0/2. At t=Tt = T:

(SN)o=∣∫H(f)S(f)ej2πfTdf∣2N02∫∣H(f)∣2df≤∫∣H∣2df∫∣S∣2dfN02∫∣H∣2df=2EN0\left(\frac{S}{N}\right)_o = \frac{\left|\int H(f)S(f)e^{j2\pi fT}df\right|^2}{\frac{N_0}{2}\int|H(f)|^2df} \le \frac{\int|H|^2df\int|S|^2df}{\frac{N_0}{2}\int|H|^2df} = \frac{2E}{N_0}

by Schwarz's inequality. Equality holds when H(f)=kS∗(f)e−j2πfTH(f) = kS^*(f)e^{-j2\pi fT}, i.e. h(t)=k s(T−t)h(t) = k\,s(T-t), giving (S/N)max=2E/N0(S/N)_{max} = 2E/N_0.

Realization 1: Integrate-and-dump (rectangular pulse). For s(t)=As(t) = A, 0≤t≤T0 \le t \le T: h(t)=kAh(t) = kA for 0≤t≤T0 \le t \le T, a rectangular impulse response, which is an integrator over the last TT seconds:

y(T)=∫0Tr(t) dty(T) = \int_0^{T} r(t)\,dt
        R                   S1 (dump)
 r(t)--/\/\--+--(-A+)--+-------+---> y(T)
             |   op-amp|   |   |
             +---||----+   +-/-+  short C
                 C              at end of T

An op-amp integrator (or RC with RC≫TRC \gg T) integrates over one bit; the output is sampled at TT, then the capacitor is shorted (dumped) so the next bit starts from zero.

Realization 2: Correlator. Since

y(T)=∫0Tr(τ) h(T−τ) dτ=k∫0Tr(τ) s(τ) dτy(T) = \int_0^T r(\tau)\,h(T-\tau)\,d\tau = k\int_0^T r(\tau)\,s(\tau)\,d\tau

the matched filter output at t=Tt=T equals the correlation of r(t)r(t) with a stored replica of s(t)s(t):

 r(t) --->(X)--->[ integrate 0..T ]--> sample at T --> y(T)
           ^
          s(t)  (local replica)

Realization 3: Tapped delay line (for general pulses): delays, tap weights equal to samples of s(T−t)s(T-t), and a summer, giving a digital (FIR) matched filter.

Note: correlator and matched filter give the same value only at t=Tt = T; their outputs differ at other times.

  • 2070 Asar (CS II) · 2+5 marks

What is capture effect? Calculate the gain parameter in DSB-FC with envelope detection.

Answer

Capture effect

Capture effect is the property of an FM receiver whereby, when two FM signals on (nearly) the same frequency are received, the stronger one is reproduced and the weaker one is almost completely suppressed. The limiter and discriminator respond to the dominant signal's phase, so the weaker signal appears only as small noise. If the two strengths are nearly equal (within about 1 dB), the receiver may switch between them. AM receivers do not show this; both signals are heard together.

Gain parameter (figure of merit) of DSB-FC with envelope detection

Assumptions: message m(t)m(t) of bandwidth WW and power P=m2(t)‾P = \overline{m^2(t)}; white noise of PSD N0/2N_0/2; IF filter bandwidth 2W2W; high carrier-to-noise ratio.

Transmitted signal: s(t)=Ac[1+kam(t)]cos⁡ωcts(t) = A_c[1 + k_a m(t)]\cos\omega_c t

Channel (input) SNR. Signal power =Ac2(1+ka2P)2= \dfrac{A_c^2(1+k_a^2P)}{2}; noise power in bandwidth WW (reference baseband) =WN0= WN_0:

(SNR)C=Ac2(1+ka2P)2WN0(SNR)_C = \frac{A_c^2(1+k_a^2P)}{2WN_0}

Received signal with narrowband noise

x(t)=[Ac+Ackam(t)+nI(t)]cos⁡ωct−nQ(t)sin⁡ωct\begin{aligned} x(t) &= [A_c + A_ck_am(t) + n_I(t)]\cos\omega_c t - n_Q(t)\sin\omega_c t \end{aligned}

where nI,nQn_I, n_Q each have power 2WN02WN_0. The envelope is

y(t)=[Ac+Ackam(t)+nI(t)]2+nQ2(t)y(t) = \sqrt{[A_c + A_ck_am(t) + n_I(t)]^2 + n_Q^2(t)}

For high SNR, Ac[1+kam(t)]≫∣nQ(t)∣A_c[1+k_am(t)] \gg |n_Q(t)|, so

y(t)≈Ac+Ackam(t)+nI(t)y(t) \approx A_c + A_ck_am(t) + n_I(t)

The DC term AcA_c is removed by a blocking capacitor.

Output SNR

(SNR)O=Ac2ka2P2WN0(SNR)_O = \frac{A_c^2k_a^2P}{2WN_0}

Figure of merit (gain parameter)

γ=(SNR)O(SNR)C=ka2P1+ka2P\gamma = \frac{(SNR)_O}{(SNR)_C} = \frac{k_a^2P}{1+k_a^2P}

which is always less than 1.

Single tone m(t)=Amcos⁡ωmtm(t) = A_m\cos\omega_m t, μ=kaAm\mu = k_aA_m: ka2P=μ2/2k_a^2P = \mu^2/2, so

γ=μ22+μ2\gamma = \frac{\mu^2}{2+\mu^2}

For 100 % modulation (μ=1\mu = 1): γ=1/3\gamma = 1/3. So DSB-FC with envelope detection needs three times (4.77 dB) more transmitted power than DSB-SC/SSB with coherent detection (γ=1\gamma = 1) for the same output SNR, because much power goes into the carrier.

Low SNR: when noise dominates the carrier, the message is no longer separable in the envelope and output SNR falls sharply: this is the threshold effect of the envelope detector.

  • 2070 Asar (CS II) · 3+4 marks

Compare AM and FM in terms of power efficiency, band width efficiency and system complexity. Calculate the error probability of coherent ASK.

Answer

AM vs FM

PointAM (DSB-FC)FM
Power efficiencyPoor: carrier carries ≥2/3\ge 2/3 of power; γ=μ2/(2+μ2)≤1/3\gamma = \mu^2/(2+\mu^2) \le 1/3Good: constant envelope; γ=32β2\gamma = \tfrac32\beta^2 (tone), can be ≫1\gg 1
Noise performanceNoise adds directly to amplitudeLimiter removes amplitude noise; SNR rises with β2\beta^2
Bandwidth2W2W (narrow)2(β+1)W2(\beta+1)W (Carson), much wider
Bandwidth efficiencyBetterPoorer; trades bandwidth for SNR
ThresholdEnvelope detector threshold at low SNRSharper threshold (~10 dB CNR)
TransmitterNeeds linear power amplifier at high levelClass C amplifiers usable (constant envelope)
Receiver complexitySimple (diode envelope detector)More complex (limiter, discriminator/PLL, de-emphasis)

Error probability of coherent ASK

Signals over a bit: s1(t)=Acos⁡ωcts_1(t) = A\cos\omega_c t for "1", s0(t)=0s_0(t) = 0 for "0"; AWGN of PSD N0/2N_0/2; equal priors.

Correlator output y=∫0Tbr(t)cos⁡ωct dty = \int_0^{T_b} r(t)\cos\omega_c t\,dt has mean ATb/2AT_b/2 for "1" and 0 for "0", with Gaussian noise variance

σ2=N02⋅Tb2=N0Tb4\sigma^2 = \frac{N_0}{2}\cdot\frac{T_b}{2} = \frac{N_0T_b}{4}

Optimum threshold is midway, ATb/4AT_b/4, so

Pe=Q ⁣(ATb/4N0Tb/4)=Q ⁣(A2Tb4N0)\begin{aligned} P_e &= Q\!\left(\frac{AT_b/4}{\sqrt{N_0T_b/4}}\right) = Q\!\left(\sqrt{\frac{A^2T_b}{4N_0}}\right) \end{aligned}

With average bit energy Eb=12⋅A2Tb2=A2Tb4E_b = \tfrac12\cdot\tfrac{A^2T_b}{2} = \tfrac{A^2T_b}{4}:

Pe=Q ⁣(EbN0)P_e = Q\!\left(\sqrt{\frac{E_b}{N_0}}\right)

This is 3 dB worse than coherent BPSK, Q(2Eb/N0)Q(\sqrt{2E_b/N_0}).

  • 2070 Chaitra (CS II) · 6 marks

Explain the approximation of the matched filter for a rectangular pulse using an Ideal low pass filter with variable bandwidth.

Answer

The true matched filter for a rectangular pulse is an integrate-and-dump filter, giving (S/N)max=2E/N0(S/N)_{max} = 2E/N_0. A simpler practical receiver uses an ideal LPF whose bandwidth BB can be chosen; the question is how close it can come.

Set-up. Pulse p(t)=Ap(t) = A for ∣t∣≤T/2|t| \le T/2 (energy E=A2TE = A^2T), white noise PSD N0/2N_0/2, ideal LPF of unit gain and bandwidth BB. Pulse spectrum:

P(f)=AT sin⁡πfTπfTP(f) = AT\,\frac{\sin \pi fT}{\pi fT}
 |H(f)|
   1 +-----------+
     |           |
 ----+-----+-----+-----> f
    -B     0     B

Peak output signal (at the pulse centre t=0t=0):

so(0)=∫−BBATsin⁡πfTπfT df=2Aπ∫0πBTsin⁡xx dx=2Aπ Si(πBT)\begin{aligned} s_o(0) &= \int_{-B}^{B} AT\frac{\sin\pi fT}{\pi fT}\,df = \frac{2A}{\pi}\int_0^{\pi BT}\frac{\sin x}{x}\,dx \\ &= \frac{2A}{\pi}\,\text{Si}(\pi BT) \end{aligned}

Output noise power: No=N02⋅2B=N0BN_o = \dfrac{N_0}{2}\cdot 2B = N_0B

Output SNR

(SN)LPF=4A2 Si2(πBT)π2N0B\left(\frac{S}{N}\right)_{LPF} = \frac{4A^2\,\text{Si}^2(\pi BT)}{\pi^2N_0B}

Compare with the matched filter, (S/N)MF=2A2T/N0(S/N)_{MF} = 2A^2T/N_0:

ρ=(S/N)LPF(S/N)MF=2 Si2(πBT)π2 BT\rho = \frac{(S/N)_{LPF}}{(S/N)_{MF}} = \frac{2\,\text{Si}^2(\pi BT)}{\pi^2\,BT}

Effect of bandwidth

  • BB too small: pulse energy is cut off, peak signal is small.
  • BB too large: little extra signal but noise grows in proportion to BB.
  • An optimum exists in between.

Evaluating ρ\rho numerically:

BTBT0.50.6851.01.5
ρ\rho0.7620.8250.6950.349

Result: the best choice is B≈0.685/TB \approx 0.685/T (about 0.7/T0.7/T), giving ρ≈0.825\rho \approx 0.825, i.e. only about 0.84 dB worse than the true matched filter. Hence an ideal LPF with bandwidth about 0.7×0.7 \times the bit rate is a good, simple approximation to the matched filter for rectangular pulses.

  • 2070 Chaitra (CS II) · 2 marks

Write a short note on white noise and its psdf.

Answer

White noise is an idealised random noise whose power spectral density is constant at all frequencies, just as white light contains all colours equally. Its two-sided PSD is

Sn(f)=N02for all fS_n(f) = \frac{N_0}{2} \quad \text{for all } f

Its autocorrelation is the inverse Fourier transform, an impulse:

Rn(τ)=N02 δ(τ)R_n(\tau) = \frac{N_0}{2}\,\delta(\tau)

so any two different samples are uncorrelated (and independent if Gaussian).

 Sn(f)                      Rn(tau)
   | N0/2                      ^ (N0/2) delta
 --+---------------- f   ------+------ tau
  • Total power ∫Sndf\int S_n df is infinite, so it is only a model; real noise (thermal, shot) is flat up to very high frequencies (∼1012\sim 10^{12} Hz), so it acts white in any practical bandwidth.
  • For thermal noise N0=kTN_0 = kT. Noise through a filter of bandwidth BB has power N0BN_0B.
  • Usually also assumed Gaussian: AWGN.
  • 2069 Chaitra (CS II) · 6+2 marks

Derive the expression for error probability for binary PAM system and extend it to M-ary system.

Answer

Binary PAM error probability

Assumptions: polar binary PAM; the sampled receiver output is +A+A for "1" and −A-A for "0"; additive Gaussian noise nn of zero mean and variance σ2\sigma^2; equal priors, so the threshold is 00.

y=±A+n,fN(n)=12πσe−n2/2σ2y = \pm A + n, \qquad f_N(n) = \frac{1}{\sqrt{2\pi}\sigma}e^{-n^2/2\sigma^2}
   f(y|0)            f(y|1)
     /\                /\
    /  \     |        /  \
 __/    \____|_______/    \__ y
    -A       0  threshold +A
          shaded tails = errors

Error when "1" is sent: y<0⇒n<−Ay < 0 \Rightarrow n < -A:

P(e∣1)=∫−∞−A12πσe−n2/2σ2dn=Q ⁣(Aσ)P(e|1) = \int_{-\infty}^{-A}\frac{1}{\sqrt{2\pi}\sigma}e^{-n^2/2\sigma^2}dn = Q\!\left(\frac{A}{\sigma}\right)

Error when "0" is sent: n>An > A, by symmetry P(e∣0)=Q(A/σ)P(e|0) = Q(A/\sigma).

Pe=12P(e∣1)+12P(e∣0)=Q ⁣(Aσ)=12erfc ⁣(A2σ)P_e = \tfrac12P(e|1) + \tfrac12P(e|0) = Q\!\left(\frac{A}{\sigma}\right) = \frac12\text{erfc}\!\left(\frac{A}{\sqrt2\sigma}\right)

With a matched filter, A2/σ2=2Eb/N0A^2/\sigma^2 = 2E_b/N_0, so

Pe=Q ⁣(2EbN0)P_e = Q\!\left(\sqrt{\frac{2E_b}{N_0}}\right)

(For unipolar on-off PAM with levels 00 and AA, threshold A/2A/2: Pe=Q(A/2σ)P_e = Q(A/2\sigma).)

Extension to M-ary PAM

Levels ±A,±3A,…,±(M−1)A\pm A, \pm3A, \dots, \pm(M-1)A (spacing 2A2A), thresholds midway.

  • Each of the M−2M-2 inner levels errs if ∣n∣>A|n| > A: probability 2Q(A/σ)2Q(A/\sigma).
  • Each of the 2 outer levels errs only on one side: Q(A/σ)Q(A/\sigma).
Pe=(M−2) 2Q(A/σ)+2Q(A/σ)M=2(1−1M)Q ⁣(Aσ)P_e = \frac{(M-2)\,2Q(A/\sigma) + 2Q(A/\sigma)}{M} = 2\left(1-\frac1M\right)Q\!\left(\frac{A}{\sigma}\right)

Average symbol energy Es=(M2−1)A2/3E_s = (M^2-1)A^2/3, so with a matched filter (σ2=N0/2\sigma^2 = N_0/2):

Pe=2(1−1M)Q ⁣(6Es(M2−1)N0)P_e = 2\left(1-\frac1M\right)Q\!\left(\sqrt{\frac{6E_s}{(M^2-1)N_0}}\right)

For M=2M=2 this reduces to the binary result. For fixed EsE_s, larger MM gives larger PeP_e, but bandwidth falls by log⁡2M\log_2 M.

  • 2069 Chaitra (CS II) · 4+3 marks

Explain the threshold effect in non coherent detection of FM signal. How can it be corrected?

Answer

Threshold effect in FM

In FM with non-coherent detection (limiter + discriminator), the output SNR follows

(SNR)O=32β2(SNR)C(tone modulation)(SNR)_O = \tfrac32\beta^2 (SNR)_C \quad (\text{tone modulation})

only when the carrier-to-noise ratio (CNR) at the detector is high. When the CNR falls below a certain value, called the threshold (about 10 dB, often quoted 10 to 13 dB), the output SNR drops much faster than the input SNR. This sharp breakdown is the threshold effect.

Cause. The received signal is the resultant phasor of carrier AcA_c and noise r(t)r(t):

            noise r(t)
           ,--->
          /
 0 ------+------> Ac (carrier)
     resultant phase = phi(t) + noise angle
  • High CNR (Ac≫rA_c \gg r): noise only slightly perturbs the phase; output noise is small and smooth (parabolic PSD).
  • Near/below threshold (r≈Acr \approx A_c occasionally): the resultant phasor can swing around the origin, so its phase changes suddenly by ±2π\pm 2\pi. The discriminator output (derivative of phase) then contains sharp impulses ("clicks") of area 2π2\pi.
  • Click rate rises rapidly as CNR falls; clicks have large power and cause the output SNR to collapse. At very low CNR the signal is "captured" by noise (the message is mutilated).
 (SNR)o dB
     |            /  slope = FM gain
     |           /
     |          /
     |      ___/
     |   __/ <- threshold (~10 dB CNR)
     +--------------------- (SNR)c dB

A wider deviation (larger β\beta) increases noise bandwidth and so raises the threshold point.

Correction (threshold extension)

The aim is to lower the CNR at which threshold occurs, i.e. reduce the effective noise bandwidth seen by the detector:

  1. FM demodulator with feedback (FMFB): the VCO output, driven by the demodulated signal, is mixed with the input, compressing the deviation; the IF filter can then be narrow, so noise is reduced and threshold is lowered by about 5 to 7 dB.
  2. Phase-locked loop (PLL) demodulator: the loop bandwidth only follows the message, so it rejects much of the noise; threshold extension of a few dB over a discriminator.
  3. Pre-emphasis and de-emphasis: boosts high message frequencies before modulation and cuts them after detection, reducing high-frequency output noise and improving SNR (about 13 dB for broadcast FM); it improves performance above threshold.
  4. Practical measures: increase transmitter power or antenna gain, or use smaller deviation where bandwidth allows, to keep CNR above threshold.

Questions from Old Question Collection (BEI EX 656) (BEI Communication Systems (EX 656) exam papers, 2078 to 2081 Chaitra), Communication System I (EX 652) (BEX Communication System I (EX 652) papers 2064 to 2080, plus two old BCT Communication Systems papers (2068, 2071)) and Communication System II (EX 702) (BEX Communication System II (EX 702) exam papers, 2069 to 2081). Answers are written for this site; check them against your class notes.

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