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:
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 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 improves SNR above threshold but needs more bandwidth, so more noise enters and the threshold rises; this sets a practical limit on .
- 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 , , plus white Gaussian noise of power spectral density , applied to a linear time-invariant filter with impulse response and transfer function .
x(t) = g(t) + w(t) --> [ h(t) ] --> y(t) = go(t) + n(t)
|
sample at t = T
Signal output at the sampling instant:
Noise output power (white noise through ):
Peak pulse SNR to be maximised:
Schwarz inequality: for any ,
with equality only when . Put and :
Substituting into :
using Rayleigh's energy theorem .
Optimum filter: the maximum is reached when equality holds:
Impulse response: take the inverse Fourier transform:
So the impulse response of the optimum filter is the time-reversed and delayed (by ) version of the input signal, scaled by . 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 depends only on the signal energy, not on its shape.
- : the filter passes strongly where the signal spectrum is strong.
- The output is proportional to the autocorrelation of shifted by , peaking at ; hence a matched filter is equivalent to a correlator (multiply by and integrate over to ).
- The delay makes causal (zero for ).
Example
For a rectangular pulse , , the matched filter is for (an integrate-and-dump filter). The output is a triangle peaking at with value , and .
- 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 , , plus white Gaussian noise of power spectral density , applied to a linear time-invariant filter with impulse response and transfer function .
x(t) = g(t) + w(t) --> [ h(t) ] --> y(t) = go(t) + n(t)
|
sample at t = T
Signal output at the sampling instant:
Noise output power (white noise through ):
Peak pulse SNR to be maximised:
Schwarz inequality: for any ,
with equality only when . Put and :
Substituting into :
using Rayleigh's energy theorem .
Optimum filter: the maximum is reached when equality holds:
Impulse response: take the inverse Fourier transform:
So the impulse response of the optimum filter is the time-reversed and delayed (by ) version of the input signal, scaled by . 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 depends only on the signal energy, not on its shape.
- : the filter passes strongly where the signal spectrum is strong.
- The output is proportional to the autocorrelation of shifted by , peaking at ; hence a matched filter is equivalent to a correlator (multiply by and integrate over to ).
- The delay makes causal (zero for ).
- 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 :
For an energy signal, .
Three (and more) important properties:
- Even symmetry: .
- Value at origin = mean-square value (power): (for an energy signal, ).
- Maximum at origin: for all .
- Fourier pair with PSD (Wiener–Khinchin): and .
- Periodicity: if is periodic with period , is also periodic with .
- DC component: if has mean and no periodic part, as .
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 ( Hz), so the model is accurate for any practical receiver bandwidth. It is usually also assumed Gaussian, zero-mean (AWGN).
PSD (two-sided):
where (W/Hz), = Boltzmann constant and = equivalent noise temperature.
Autocorrelation (inverse Fourier transform of PSD, using ):
S_W(f) R_W(tau)
| ^ (N0/2) delta
N0/2+----------------- |
| |
----+-----------------> f ------+------> tau
0 0
Meaning: for every , so any two samples of white noise, however close, are uncorrelated (and independent if Gaussian). The total power 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 :
For an energy signal, .
It tells how fast a signal changes: a slowly varying signal stays correlated for large , 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:
- Even symmetry: .
- Value at origin = mean-square value (power): (for an energy signal, ).
- Maximum at origin: for all .
- Fourier pair with PSD (Wiener–Khinchin): and .
- Periodicity: if is periodic with period , is also periodic with .
- DC component: if has mean and no periodic part, as .
Example: white noise has ; 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 , (J/Hz). It shows how the energy is spread over frequency:
and is the Fourier transform of the energy autocorrelation .
Power spectral density (PSD): for a power signal, take a truncated version (length ) with transform :
For a WSS random process, . Both ESD and PSD are real, even and non-negative.
Wiener–Khinchin relation (PSD AC): for a power signal, . The time-domain integral is scaled by , and , so
Hence
and at : .
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 ( Hz), so the model is accurate for any practical receiver bandwidth. It is usually also assumed Gaussian, zero-mean (AWGN).
PSD (two-sided):
where (W/Hz), = Boltzmann constant and = equivalent noise temperature.
Autocorrelation (inverse Fourier transform of PSD, using ):
S_W(f) R_W(tau)
| ^ (N0/2) delta
N0/2+----------------- |
| |
----+-----------------> f ------+------> tau
0 0
Meaning: for every , so any two samples of white noise, however close, are uncorrelated (and independent if Gaussian). The total power 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 for and zero elsewhere,
Its power is finite, , and samples spaced 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 and 0 as , each lasting , with equal probability. Channel adds white Gaussian noise of PSD . The receiver filter output is sampled at :
s_i+w +--------+ y(t) sample y +---------+
------>| filter |------>-/ ------->| y > lam?|--> 1/0
+--------+ t=Tb +---------+
The sample is , where are the noise-free outputs () and is Gaussian with zero mean and variance . Conditional pdfs:
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 , else 0. For equal priors the optimum threshold is midway, .
Error when 0 is sent (): with ,
By symmetry (1 sent, ) is the same. The average error probability is
where and .
With a matched filter (matched to ), the maximum of is , where . Therefore
depends only on the energy of the difference signal relative to , not on pulse shape. Making the two signals as different as possible (antipodal) gives the lowest .
| Scheme | , | ||
|---|---|---|---|
| Unipolar / ASK (OOK) | , | ||
| Polar / PSK | |||
| Coherent FSK | , |
( = average energy per bit.)
Threshold effect in FM
Threshold effect in FM. The FM improvement 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 phase jump.
high CNR: noise wobbles low CNR: resultant can swing
the tip slightly around origin -> 2pi jump
.-. .---.
0 -->( * ) 0 --( * )
'-' '---'
Each 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 ) means larger , 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:
- 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.
- 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.
- 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.
- Reducing IF bandwidth / modulation index where possible, or increasing transmitted power (raises CNR above threshold).
- 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 , , where is a known pulse and is white noise of two-sided PSD . It passes through an LTI filter and is sampled at :
s(t)+w(t) +--------+ y(t) sample y(T)
----------->| h(t) |---------->--/ ----> decision
+--------+ at t=T
Output with
We want that maximises the output peak signal-to-noise ratio
Schwarz inequality: , with equality only when . Put and :
(using Rayleigh's theorem, = pulse energy). The maximum is reached when
Impulse response. Take the inverse transform. For real , , which is the transform of ; the factor is a delay of :
So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by . Choosing the sampling instant equal to the pulse duration makes causal.
s(t) h(t) = s(T - t)
|\ /|
| \ / |
| \ / |
---+---+---> t ---+---+---> t
0 T 0 T
Results: maximum SNR 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.
- Asked 2 times
- 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 plus white noise () passes through a band-pass (IF) filter of bandwidth , then a demodulator and a low-pass filter of bandwidth (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
where and are low-pass, each with the same power as : .
Definitions. Message power .
- Input SNR
- Output SNR
- Detection gain ; figure of merit , where is the channel SNR measured in the message band (baseband reference). Both show how much the demodulator improves or worsens SNR.
SSB-SC (coherent detection). , .
- Signal power (since ); noise in , so .
- Multiply by and low-pass filter:
- Signal power , noise power , so
With baseband reference (, same transmitted power), figure of merit .
DSB-SC (coherent detection). , .
- Signal power ; noise in , so .
- Multiply by and low-pass filter:
The quadrature noise is rejected by coherent detection, giving a 3 dB gain. With baseband reference , figure of merit .
Comparison:
| Quantity | DSB-SC | SSB-SC |
|---|---|---|
| Transmission bandwidth | ||
| Noise power at input | ||
| (same ) | ||
| Detection gain | 2 | 1 |
| Figure of merit (baseband reference) | 1 | 1 |
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 :
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 , with :
- Bit 1: (for an integer multiple of ).
- Bit 0: .
- Noise: is Gaussian, mean 0, variance
Threshold midway: . Using the binary result :
Energy of a "1" pulse is ; the average energy per bit (half the bits are 0) is . Hence and
(In terms of peak energy : .)
Remarks: coherent ASK needs 3 dB more average power than BPSK for the same , and the threshold depends on received amplitude, so ASK is sensitive to fading. Example: for (10 dB), .
- 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 , , so that it gives the maximum possible peak signal-to-noise ratio at the sampling instant 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 , , where is a known pulse and is white noise of two-sided PSD . It passes through an LTI filter and is sampled at :
s(t)+w(t) +--------+ y(t) sample y(T)
----------->| h(t) |---------->--/ ----> decision
+--------+ at t=T
Output with
We want that maximises the output peak signal-to-noise ratio
Schwarz inequality: , with equality only when . Put and :
(using Rayleigh's theorem, = pulse energy). The maximum is reached when
Impulse response. Take the inverse transform. For real , , which is the transform of ; the factor is a delay of :
So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by . Choosing the sampling instant equal to the pulse duration makes causal.
s(t) h(t) = s(T - t)
|\ /|
| \ / |
| \ / |
---+---+---> t ---+---+---> t
0 T 0 T
Results: maximum SNR 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 , , for , which is the same rectangle; it behaves as an integrate-and-dump circuit. The output is a triangle with peak at .
- Asked 2 times
- 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 is found assuming high carrier-to-noise ratio (above threshold).
FM signal and receiver. , instantaneous frequency deviation . Receiver: BPF () → limiter → discriminator (output ) → LPF ().
Noise in phasor form. Write the narrowband noise as . At high carrier-to-noise ratio (), the resultant phase is
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 (the phase can be dropped for noise calculation). Discriminator output:
Output signal power: .
Output noise power. has PSD for . Differentiation multiplies the spectrum by , so the noise PSD at discriminator output is
This parabolic noise spectrum is the key property of FM. After the LPF ():
S(f) |\ /|
| \ / | parabolic noise
| \ / | after discriminator
| \. ./ | (only |f|<W kept)
+----+-----+-----+----+--> f
-W 0 W
Output SNR:
Input SNR. Received power (constant envelope); noise in is :
Gain and figure of merit:
For a single tone : , , , (Carson):
So FM noise performance improves with the square of the deviation ratio: FM trades bandwidth for SNR. FM beats AM () when , i.e. (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 is a collection (ensemble) of time functions, one for each outcome of a random experiment. At a fixed time , 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 , autocorrelation 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 ( Hz), so the model is accurate for any practical receiver bandwidth. It is usually also assumed Gaussian, zero-mean (AWGN).
PSD (two-sided):
where (W/Hz), = Boltzmann constant and = equivalent noise temperature.
Autocorrelation (inverse Fourier transform of PSD, using ):
S_W(f) R_W(tau)
| ^ (N0/2) delta
N0/2+----------------- |
| |
----+-----------------> f ------+------> tau
0 0
Meaning: for every , so any two samples of white noise, however close, are uncorrelated (and independent if Gaussian). The total power 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 , where is the channel SNR: average received signal power divided by noise power in the message bandwidth .
Coherent SSB receiver: BPF of width → product modulator with → LPF (). Noise PSD .
Signal: (USB). Since and have equal power and are orthogonal:
Noise after BPF (width ): with .
Detector output (multiply by , LPF):
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 bits. 32 is not a perfect square, so a square grid cannot be used. The efficient arrangement is the cross constellation: a 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 .
Q
|
. * * * * .
* * * * * *
* * * * * *
-----------+------------ I
* * * * * *
* * * * * *
. * * * * .
|
* = point (32), . = removed corner
Points lie at except the four corners .
- 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 , , where is a known pulse and is white noise of two-sided PSD . It passes through an LTI filter and is sampled at :
s(t)+w(t) +--------+ y(t) sample y(T)
----------->| h(t) |---------->--/ ----> decision
+--------+ at t=T
Output with
We want that maximises the output peak signal-to-noise ratio
Schwarz inequality: , with equality only when . Put and :
(using Rayleigh's theorem, = pulse energy). The maximum is reached when
Impulse response. Take the inverse transform. For real , , which is the transform of ; the factor is a delay of :
So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by . Choosing the sampling instant equal to the pulse duration makes causal.
s(t) h(t) = s(T - t)
|\ /|
| \ / |
| \ / |
---+---+---> t ---+---+---> t
0 T 0 T
Results: maximum SNR 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 of is obtained by shifting the phase of every frequency component by (positive frequencies) and (negative frequencies), without changing amplitudes:
Example: the Hilbert transform of is . Properties: and have the same amplitude spectrum and energy, and are orthogonal. Uses: SSB generation by the phase-shift method , analytic signals , 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 :
- Energy signal (Rayleigh form):
Relation between PSD and autocorrelation
Wiener–Khinchin relation (PSD AC): for a power signal, . The time-domain integral is scaled by , and , so
Hence
and at : .
Example: white noise. The PSD is flat, . Then by the relation above:
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 , spreads out (with ); the narrower the spectrum, the wider the autocorrelation.
Link with Parseval: putting in gives , 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 ( 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:
PSD (two-sided):
where (W/Hz), = Boltzmann constant and = equivalent noise temperature.
Autocorrelation (inverse Fourier transform of PSD, using ):
S_W(f) R_W(tau)
| ^ (N0/2) delta
N0/2+----------------- |
| |
----+-----------------> f ------+------> tau
0 0
Meaning: for every , so any two samples of white noise, however close, are uncorrelated (and independent if Gaussian). The total power 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 ( Hz), so the model is accurate for any practical receiver bandwidth. It is usually also assumed Gaussian, zero-mean (AWGN).
PSD (two-sided):
where (W/Hz), = Boltzmann constant and = equivalent noise temperature.
Autocorrelation (inverse Fourier transform of PSD, using ):
S_W(f) R_W(tau)
| ^ (N0/2) delta
N0/2+----------------- |
| |
----+-----------------> f ------+------> tau
0 0
Meaning: for every , so any two samples of white noise, however close, are uncorrelated (and independent if Gaussian). The total power 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
- Even symmetry: .
- Value at origin = mean-square value (power): (for an energy signal, ).
- Maximum at origin: for all .
- Fourier pair with PSD (Wiener–Khinchin): and .
- Periodicity: if is periodic with period , is also periodic with .
- DC component: if has mean and no periodic part, as .
White noise satisfies these: is even, maximum at , 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) of a filter is the bandwidth of an ideal rectangular filter, with the same peak (usually zero-frequency or centre) gain , that passes the same total noise power as the actual filter when both are fed with white noise.
White noise into gives output power
The ideal filter of gain and bandwidth gives . Equating:
|H(f)|^2
|H(0)|^2 +---------+ equal areas
| ... |
| .. |
| .|..
| | ......
--+---------+----------> f
0 B_N
NEB lets us write the output noise power simply as (or per unit gain), which is used in noise figure and SNR calculations. For a band-pass filter is replaced by the centre-frequency gain .
Example: an RC low-pass filter has .
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 , , where is a known pulse and is white noise of two-sided PSD . It passes through an LTI filter and is sampled at :
s(t)+w(t) +--------+ y(t) sample y(T)
----------->| h(t) |---------->--/ ----> decision
+--------+ at t=T
Output with
We want that maximises the output peak signal-to-noise ratio
Schwarz inequality: , with equality only when . Put and :
(using Rayleigh's theorem, = pulse energy). The maximum is reached when
Impulse response. Take the inverse transform. For real , , which is the transform of ; the factor is a delay of :
So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by . Choosing the sampling instant equal to the pulse duration makes causal.
s(t) h(t) = s(T - t)
|\ /|
| \ / |
| \ / |
---+---+---> t ---+---+---> t
0 T 0 T
Results: maximum SNR 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 and 0 as , each lasting , with equal probability. Channel adds white Gaussian noise of PSD . The receiver filter output is sampled at :
s_i+w +--------+ y(t) sample y +---------+
------>| filter |------>-/ ------->| y > lam?|--> 1/0
+--------+ t=Tb +---------+
The sample is , where are the noise-free outputs () and is Gaussian with zero mean and variance . Conditional pdfs:
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 , else 0. For equal priors the optimum threshold is midway, .
Error when 0 is sent (): with ,
By symmetry (1 sent, ) is the same. The average error probability is
where and .
With a matched filter (matched to ), the maximum of is , where . Therefore
depends only on the energy of the difference signal relative to , not on pulse shape. Making the two signals as different as possible (antipodal) gives the lowest .
Special cases:
| Scheme | , | ||
|---|---|---|---|
| Unipolar / ASK (OOK) | , | ||
| Polar / PSK | |||
| Coherent FSK | , |
( = 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):
and similarly . 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 , , where is a known pulse and is white noise of two-sided PSD . It passes through an LTI filter and is sampled at :
s(t)+w(t) +--------+ y(t) sample y(T)
----------->| h(t) |---------->--/ ----> decision
+--------+ at t=T
Output with
We want that maximises the output peak signal-to-noise ratio
Schwarz inequality: , with equality only when . Put and :
(using Rayleigh's theorem, = pulse energy). The maximum is reached when
Impulse response. Take the inverse transform. For real , , which is the transform of ; the factor is a delay of :
So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by . Choosing the sampling instant equal to the pulse duration makes causal.
s(t) h(t) = s(T - t)
|\ /|
| \ / |
| \ / |
---+---+---> t ---+---+---> t
0 T 0 T
Results: maximum SNR 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 (mean , autocorrelation , PSD ) be applied to a stable LTI system with impulse response and :
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:
2. Autocorrelation of output.
because is WSS, its autocorrelation depends only on the time difference . The result depends only on , not on :
3. Mean-square value is finite and constant.
Since the mean is constant and the autocorrelation depends only on , is also wide-sense stationary.
4. Output PSD. Taking the Fourier transform of (, ):
and the output power is . The phase of has no effect on the output PSD.
Example. White noise (, WSS) through an RC low-pass filter, :
depends only on , confirming that the output is WSS; its power 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 (or figure of merit, with SNR referred to the message band ).
s(t)+w(t) +-----+ +-----+ +-------+
--------->| BPF |-->(x)-->| LPF |-->| DC |--> y(t)
| 2W | ^ | W | | block |
+-----+ | +-----+ +-------+
cos(wc t)
Receiver: BPF () → product detector with → LPF () → DC block. Noise PSD , narrowband noise with .
- Signal power ; noise in :
- Coherent detection (or envelope detection at high SNR) and DC blocking:
The carrier uses power but carries no information, so the gain is reduced.
Single-tone DSB-FC with 100% modulation
Single tone, 100% modulation. , , so :
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 ( 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 and with the same amplitude.
Coherent BFSK signals over :
The tones are chosen orthogonal over : (e.g. ).
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 .
- 1 sent: , (orthogonality) so .
- 0 sent: , so .
Noise part. and each have variance and are uncorrelated (orthogonal tones), so has
Error probability. is Gaussian with mean and variance ; threshold 0. For 0 sent, error if :
With :
Check with the general formula: , so .
Comparison: BFSK has the same as ASK (for equal average energy) and needs 3 dB more than BPSK, because orthogonal signals are apart while antipodal ones are apart. Non-coherent FSK gives , 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 ) 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 and 0 as , each lasting , with equal probability. Channel adds white Gaussian noise of PSD . The receiver filter output is sampled at :
s_i+w +--------+ y(t) sample y +---------+
------>| filter |------>-/ ------->| y > lam?|--> 1/0
+--------+ t=Tb +---------+
The sample is , where are the noise-free outputs () and is Gaussian with zero mean and variance . Conditional pdfs:
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 , else 0. For equal priors the optimum threshold is midway, .
Error when 0 is sent (): with ,
By symmetry (1 sent, ) is the same. The average error probability is
where and .
With a matched filter (matched to ), the maximum of is , where . Therefore
depends only on the energy of the difference signal relative to , not on pulse shape. Making the two signals as different as possible (antipodal) gives the lowest .
Polar NRZ example. Levels , , threshold 0: ; with matched filter, . Unipolar (, 0) needs twice the average power for the same .
- 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) of a filter is the bandwidth of an ideal rectangular filter, with the same peak (usually zero-frequency or centre) gain , that passes the same total noise power as the actual filter when both are fed with white noise.
White noise into gives output power
The ideal filter of gain and bandwidth gives . Equating:
|H(f)|^2
|H(0)|^2 +---------+ equal areas
| ... |
| .. |
| .|..
| | ......
--+---------+----------> f
0 B_N
NEB lets us write the output noise power simply as (or per unit gain), which is used in noise figure and SNR calculations. For a band-pass filter is replaced by the centre-frequency gain .
NEB of first-order RC low-pass filter
vin o---/\/\/\---+---o vout
R |
=== C
|
gnd o------------+---o
For the RC low-pass filter (series , shunt ):
Since the 3 dB bandwidth is ,
The NEB is larger than the 3 dB bandwidth because the RC response rolls off slowly (only dB/decade) and passes noise well beyond . Output noise power .
Example: , : Hz, 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
| Point | AM (DSB-FC) | FM |
|---|---|---|
| Transmitted power | Varies with modulation, | Constant, , independent of modulation |
| Useful (sideband) power | At most 1/3 of total () | All power is useful; carrier power is redistributed to sidebands |
| Power amplifier | Must be linear (class A/B), lower efficiency | Class C (non-linear), high efficiency, since envelope is constant |
| Noise performance | FOM | FOM ; much better for |
| Bandwidth | (Carson), much wider | |
| Transmitter complexity | Simple (high-level collector/plate modulation) | More complex (VCO, AFC/frequency stabilisation, multipliers) |
| Receiver complexity | Very simple (diode envelope detector) | Needs limiter, discriminator/PLL, de-emphasis |
| Noise immunity | Amplitude noise directly affects output | Limiter 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 :
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 , with :
- Bit 1: (for an integer multiple of ).
- Bit 0: .
- Noise: is Gaussian, mean 0, variance
Threshold midway: . Using the binary result :
Energy of a "1" pulse is ; the average energy per bit (half the bits are 0) is . Hence and
(In terms of peak energy : .)
- 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 , , so that it gives the maximum possible peak signal-to-noise ratio at the sampling instant 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 ; ; 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 , , where is a known pulse and is white noise of two-sided PSD . It passes through an LTI filter and is sampled at :
s(t)+w(t) +--------+ y(t) sample y(T)
----------->| h(t) |---------->--/ ----> decision
+--------+ at t=T
Output with
We want that maximises the output peak signal-to-noise ratio
Schwarz inequality: , with equality only when . Put and :
(using Rayleigh's theorem, = pulse energy). The maximum is reached when
Impulse response. Take the inverse transform. For real , , which is the transform of ; the factor is a delay of :
So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by . Choosing the sampling instant equal to the pulse duration makes causal.
s(t) h(t) = s(T - t)
|\ /|
| \ / |
| \ / |
---+---+---> t ---+---+---> t
0 T 0 T
Results: maximum SNR 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 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 phase jump.
high CNR: noise wobbles low CNR: resultant can swing
the tip slightly around origin -> 2pi jump
.-. .---.
0 -->( * ) 0 --( * )
'-' '---'
Each 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 ) means larger , 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 (antipodal):
Coherent receiver: multiply by the locally generated , integrate over , sample, compare with threshold .
r(t) -->(x)--> integrate --> sample --> y > 0 ? 1 : 0
^ 0..Tb t=Tb
|
cos(wc t) (from carrier recovery)
Correlator output with
The pdfs of are Gaussian centred at . Error when 0 is sent ():
By symmetry the error for a sent 1 is the same, so
BPSK is the best binary scheme: it needs 3 dB less than coherent ASK or FSK for the same . Example: dB gives .
- 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 is an ensemble (collection) of time functions, one sample function for each outcome of a random experiment. At any fixed time , is a random variable. Example: the noise voltage at the output of many identical receivers. It is described by its mean and autocorrelation .
A process is wide-sense stationary (WSS) if (i) its mean is constant and (ii) its autocorrelation depends only on the time difference : .
Passage of WSS random signals through an LTI system
Let a WSS process (mean , autocorrelation , PSD ) be applied to a stable LTI system with impulse response and :
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:
2. Autocorrelation of output.
because is WSS, its autocorrelation depends only on the time difference . The result depends only on , not on :
3. Mean-square value is finite and constant.
Since the mean is constant and the autocorrelation depends only on , is also wide-sense stationary.
4. Output PSD. Taking the Fourier transform of (, ):
and the output power is . The phase of has no effect on the output PSD.
Example (white noise through RC low-pass filter). Input , :
The filter "colours" the noise: the output is correlated over about one time constant , and its power becomes finite.
Cross-correlation between input and output (useful for system identification): ; with white-noise input , 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 and 0 as , each lasting , with equal probability. Channel adds white Gaussian noise of PSD . The receiver filter output is sampled at :
s_i+w +--------+ y(t) sample y +---------+
------>| filter |------>-/ ------->| y > lam?|--> 1/0
+--------+ t=Tb +---------+
The sample is , where are the noise-free outputs () and is Gaussian with zero mean and variance . Conditional pdfs:
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 , else 0. For equal priors the optimum threshold is midway, .
Error when 0 is sent (): with ,
By symmetry (1 sent, ) is the same. The average error probability is
where and .
With a matched filter (matched to ), the maximum of is , where . Therefore
depends only on the energy of the difference signal relative to , not on pulse shape. Making the two signals as different as possible (antipodal) gives the lowest .
For polar NRZ ( at the sampler): .
Comparison with M-ary system
For M-ary PAM with levels (spacing ), inner levels can err on both sides and outer levels on one side, giving
which reduces to the binary result for .
| Point | Binary () | M-ary () |
|---|---|---|
| Bits per symbol | 1 | |
| Bandwidth for same bit rate | (Nyquist) | , i.e. times less |
| Level spacing for same peak power | Large | Smaller, by factor |
| for same average power | Lowest | Higher; needs about times more power |
| Thresholds | One | |
| Receiver complexity | Simple | More complex |
| Typical use | Noisy, power-limited links | Band-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 (, ) has infinite power and zero correlation time. Passing it through a filter gives and : the noise becomes band-limited (coloured) and its samples become correlated.
Ideal low-pass filtering
Ideal LPF, gain 1 for , zero elsewhere:
- Power .
- is a sinc with zeros at : samples taken at the Nyquist rate are uncorrelated (independent if Gaussian).
- It oscillates (takes negative values) and decays slowly as .
Practical RC low-pass filtering
, so
Using the pair with :
- Power , i.e. NEB .
- is a decaying exponential: always positive, never exactly zero, falls to at . Correlation time .
R(tau) ideal LPF: sinc RC LPF: exponential
N0B . N0/4RC .
/ \ /|\
. / \ . / | \
-.-/-----\-.-- tau ----'--+--'---- tau
-1/2B 1/2B -RC 0 RC
Comparison
| Point | Ideal LPF (bandwidth ) | RC LPF () |
|---|---|---|
| Output PSD | Flat up to , sharp cut-off | Lorentzian, gradual roll-off |
| Autocorrelation | ||
| Output noise power | ||
| Zero crossings of | At | None |
| Sign of | Positive and negative | Always positive |
| Decay of | Slow, | Fast, exponential |
| Realisable | No (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 bits. For baseband M-ary PAM the receiver samples the filtered signal and decides which of the levels was sent using thresholds.
M-ary baseband PAM. Each symbol carries bits and takes one of equally likely levels spaced apart, symmetric about zero:
The received sample is , . Thresholds are placed midway between adjacent levels (at ).
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 ): error if :
- Outer level (there are 2): error only if noise pushes it inward by more than :
Average symbol error probability:
In terms of signal power. Average symbol power
so
For this reduces to the binary polar result .
Observations:
- For fixed , rises slowly with (factor ), but for fixed average power the spacing shrinks as , so rises sharply.
- With Gray coding, bit error probability .
- M-ary signalling reduces bandwidth by 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. , instantaneous frequency deviation . Receiver: BPF () → limiter → discriminator (output ) → LPF ().
Noise in phasor form. Write the narrowband noise as . At high carrier-to-noise ratio (), the resultant phase is
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 (the phase can be dropped for noise calculation). Discriminator output:
Output signal power: .
Output noise power. has PSD for . Differentiation multiplies the spectrum by , so the noise PSD at discriminator output is
This parabolic noise spectrum is the key property of FM. After the LPF ():
S(f) |\ /|
| \ / | parabolic noise
| \ / | after discriminator
| \. ./ | (only |f|<W kept)
+----+-----+-----+----+--> f
-W 0 W
Output SNR:
Input SNR. Received power (constant envelope); noise in is :
Gain and figure of merit:
For a single tone : , , , (Carson):
So FM noise performance improves with the square of the deviation ratio: FM trades bandwidth for SNR. FM beats AM () when , i.e. (about 0.5).
Threshold effect
Threshold effect in FM. The FM improvement 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 phase jump.
high CNR: noise wobbles low CNR: resultant can swing
the tip slightly around origin -> 2pi jump
.-. .---.
0 -->( * ) 0 --( * )
'-' '---'
Each 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 ) means larger , 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 , , where is a known pulse and is white noise of two-sided PSD . It passes through an LTI filter and is sampled at :
s(t)+w(t) +--------+ y(t) sample y(T)
----------->| h(t) |---------->--/ ----> decision
+--------+ at t=T
Output with
We want that maximises the output peak signal-to-noise ratio
Schwarz inequality: , with equality only when . Put and :
(using Rayleigh's theorem, = pulse energy). The maximum is reached when
Impulse response. Take the inverse transform. For real , , which is the transform of ; the factor is a delay of :
So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by . Choosing the sampling instant equal to the pulse duration makes causal.
s(t) h(t) = s(T - t)
|\ /|
| \ / |
| \ / |
---+---+---> t ---+---+---> t
0 T 0 T
Results: maximum SNR 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 :
For an energy signal, .
Main properties: (even); average power (energy); ; (Wiener–Khinchin). Example: white noise has . Note that the matched filter output for the signal is , the pulse autocorrelation shifted to , which peaks at because 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:
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 (mean , autocorrelation , PSD ) be applied to a stable LTI system with impulse response and :
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:
2. Autocorrelation of output.
because is WSS, its autocorrelation depends only on the time difference . The result depends only on , not on :
3. Mean-square value is finite and constant.
Since the mean is constant and the autocorrelation depends only on , is also wide-sense stationary.
4. Output PSD. Taking the Fourier transform of (, ):
and the output power is . The phase of has no effect on the output PSD.
Example (white noise through RC low-pass filter). , , output power .
Cross-correlation: ; for white-noise input, , so correlating output with input measures .
- 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:
It shows whether the detector improves () or degrades () the SNR.
Receiver model. The received signal plus white noise () passes through a band-pass (IF) filter of bandwidth , then a demodulator and a low-pass filter of bandwidth (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
where and are low-pass, each with the same power as : .
DSB-FC (conventional AM). , . (Coherent and envelope detection give the same result at high SNR.)
- Signal power ; noise in :
- Coherent detection (or envelope detection at high SNR) and DC blocking:
The carrier uses power but carries no information, so the gain is reduced.
Proof that . To avoid over-modulation, , so , with equality only for a square-wave message at 100% modulation. Hence
For any practical message (sinusoid, speech) and .
Single tone, 100% modulation. , , so :
So for 100% single-tone DSB-AM the detection gain is : the detector output SNR is about 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, .)
- 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 :
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 , with :
- Bit 1: (for an integer multiple of ).
- Bit 0: .
- Noise: is Gaussian, mean 0, variance
Threshold midway: . Using the binary result :
Energy of a "1" pulse is ; the average energy per bit (half the bits are 0) is . Hence and
(In terms of peak energy : .)
Extension to M-ary ASK
In M-ary ASK the carrier takes amplitudes , , each symbol of duration carrying bits. The same coherent correlator gives
Levels are spaced apart, and thresholds are placed midway. The two end levels can be mistaken on one side only; the inner levels on both sides:
Average symbol energy: , so
Substituting :
Check : with , the binary ASK result. For larger the levels crowd together for the same average energy, so grows, while the bandwidth per bit falls by .
- 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) of a filter is the bandwidth of an ideal rectangular filter, with the same peak (usually zero-frequency or centre) gain , that passes the same total noise power as the actual filter when both are fed with white noise.
White noise into gives output power
The ideal filter of gain and bandwidth gives . Equating:
|H(f)|^2
|H(0)|^2 +---------+ equal areas
| ... |
| .. |
| .|..
| | ......
--+---------+----------> f
0 B_N
NEB lets us write the output noise power simply as (or per unit gain), which is used in noise figure and SNR calculations. For a band-pass filter is replaced by the centre-frequency gain .
Example: RC low-pass filter.
For the RC low-pass filter (series , shunt ):
Since the 3 dB bandwidth is ,
The NEB is larger than the 3 dB bandwidth because the RC response rolls off slowly (only dB/decade) and passes noise well beyond . Output noise power .
Example: ideal filter. For an ideal LPF of bandwidth , exactly; for practical filters with steep skirts (e.g. high-order Butterworth), approaches the 3 dB bandwidth.
Impulse response of matched filter is time-reversed, delayed input
Set-up. The received signal is , , where is a known pulse and is white noise of two-sided PSD . It passes through an LTI filter and is sampled at :
s(t)+w(t) +--------+ y(t) sample y(T)
----------->| h(t) |---------->--/ ----> decision
+--------+ at t=T
Output with
We want that maximises the output peak signal-to-noise ratio
Schwarz inequality: , with equality only when . Put and :
(using Rayleigh's theorem, = pulse energy). The maximum is reached when
Impulse response. Take the inverse transform. For real , , which is the transform of ; the factor is a delay of :
So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by . Choosing the sampling instant equal to the pulse duration makes causal.
s(t) h(t) = s(T - t)
|\ /|
| \ / |
| \ / |
---+---+---> t ---+---+---> t
0 T 0 T
Results: maximum SNR 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 plus white noise () passes through a band-pass (IF) filter of bandwidth , then a demodulator and a low-pass filter of bandwidth (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
where and are low-pass, each with the same power as : .
Definitions. Message power .
- Input SNR
- Output SNR
- Detection gain ; figure of merit , where is the channel SNR measured in the message band (baseband reference). Both show how much the demodulator improves or worsens SNR.
DSB-FC (conventional AM). , . (Coherent and envelope detection give the same result at high SNR.)
- Signal power ; noise in :
- Coherent detection (or envelope detection at high SNR) and DC blocking:
The carrier uses power but carries no information, so the gain is reduced.
Single tone, 100% modulation. , , so :
DSB-SC (coherent detection). , .
- Signal power ; noise in , so .
- Multiply by and low-pass filter:
The quadrature noise is rejected by coherent detection, giving a 3 dB gain. With baseband reference , figure of merit .
SSB-SC (coherent detection). , .
- Signal power (since ); noise in , so .
- Multiply by and low-pass filter:
- Signal power , noise power , so
With baseband reference (, same transmitted power), figure of merit .
Comparison:
| Parameter | DSB-FC (AM) | DSB-SC | SSB-SC |
|---|---|---|---|
| Transmission bandwidth | |||
| Detector | Envelope or coherent | Coherent | Coherent |
| Detection gain | ( at ) | 2 | 1 |
| Figure of merit (baseband ref.) | ( at ) | 1 | 1 |
| Threshold effect | Yes (envelope detector) | No | No |
| Power wasted in carrier | Yes (at least 2/3) | No | No |
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 , , so that it gives the maximum possible peak signal-to-noise ratio at the sampling instant 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: for (energy ), white noise PSD .
Filter: , (3 dB bandwidth, variable).
Signal output. The step response of the RC filter rises as ; it is maximum at the end of the pulse, :
s(t) s_o(t)
A +-----+ A | ..-. peak at t=T
| | | .' '.
| | | .' '..
--+-----+--> t --+-'-----+-------'--> t
0 T 0 T
Noise output power:
Output SNR:
Compared with the matched filter ():
Effect of variable bandwidth:
- too small: the filter cannot follow the pulse, is small, SNR low.
- too large: signal fully passes but much extra noise enters, SNR low.
- An optimum lies between. Setting gives , solved numerically: .
Result: the best RC filter has (i.e. ) and gives an SNR only 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:
The envelope detector output is the magnitude of the resultant phasor:
n_s
.----> resultant length = y(t)
/ ^
/ | n_c
----------->+
Ac[1+ka m(t)]
Case 1: high carrier-to-noise ratio ( noise). The quadrature term is negligible:
The message appears additively with the noise. After removing DC, , the same as coherent detection. Figure of merit ( for a single tone with ).
Case 2: low carrier-to-noise ratio (noise carrier). Write the noise as , with Rayleigh envelope and uniform phase . Then
using . Now the message is multiplied by the random noise ; there is no term proportional to 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 and bit 0 as . White Gaussian noise is added; the receiver samples once per bit and decides with threshold .
The sample is , :
f(y|0) f(y|1)
.--. | .--.
/ \ | / \
--/------\..|../------\---> y
-A 0 +A
Error when 0 is sent (): let
By symmetry . With equal priors:
With a matched filter, , so . For unipolar PAM (, 0, threshold ): , which needs twice the average power for the same .
- 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 :
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 , with :
- Bit 1: (for an integer multiple of ).
- Bit 0: .
- Noise: is Gaussian, mean 0, variance
Threshold midway: . Using the binary result :
Energy of a "1" pulse is ; the average energy per bit (half the bits are 0) is . Hence and
(In terms of peak energy : .)
Coherent PSK
BPSK signals over (antipodal):
Coherent receiver: multiply by the locally generated , integrate over , sample, compare with threshold .
r(t) -->(x)--> integrate --> sample --> y > 0 ? 1 : 0
^ 0..Tb t=Tb
|
cos(wc t) (from carrier recovery)
Correlator output with
The pdfs of are Gaussian centred at . Error when 0 is sent ():
By symmetry the error for a sent 1 is the same, so
ASK needs double the power of PSK
Comparison (same ). ASK: ; PSK: . For equal the arguments must be equal:
Since average power , ASK needs twice (3 dB more) average signal power than PSK for the same error probability. Physically, the PSK signal points are apart, while ASK points are only 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 plus white noise () passes through a band-pass (IF) filter of bandwidth , then a demodulator and a low-pass filter of bandwidth (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
where and are low-pass, each with the same power as : .
Definitions. Message power .
- Input SNR
- Output SNR
- Detection gain ; figure of merit , where is the channel SNR measured in the message band (baseband reference). Both show how much the demodulator improves or worsens SNR.
DSB-FC (conventional AM). , . (Coherent and envelope detection give the same result at high SNR.)
- Signal power ; noise in :
- Coherent detection (or envelope detection at high SNR) and DC blocking:
The carrier uses power but carries no information, so the gain is reduced.
Single tone, 100% modulation. , , so :
DSB-SC (coherent detection). , .
- Signal power ; noise in , so .
- Multiply by and low-pass filter:
The quadrature noise is rejected by coherent detection, giving a 3 dB gain. With baseband reference , figure of merit .
SSB-SC (coherent detection). , .
- Signal power (since ); noise in , so .
- Multiply by and low-pass filter:
- Signal power , noise power , so
With baseband reference (, same transmitted power), figure of merit .
Comparison:
| Parameter | DSB-FC (AM) | DSB-SC | SSB-SC |
|---|---|---|---|
| Transmission bandwidth | |||
| Detector | Envelope or coherent | Coherent | Coherent |
| Detection gain | ( at ) | 2 | 1 |
| Figure of merit (baseband ref.) | ( at ) | 1 | 1 |
| Threshold effect | Yes (envelope detector) | No | No |
| Power wasted in carrier | Yes (at least 2/3) | No | No |
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) of a filter is the bandwidth of an ideal rectangular filter, with the same maximum gain , that passes the same noise power as the real filter when both are driven by white noise:
- Output noise power can then be written simply as (or ).
- It is used in noise figure, noise temperature and receiver SNR calculations.
- Example: RC low-pass filter, .
- For sharp-cut-off filters 3 dB bandwidth; for gentle roll-off filters 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 plus white noise () passes through a band-pass (IF) filter of bandwidth , then a demodulator and a low-pass filter of bandwidth (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
where and are low-pass, each with the same power as : .
Definitions. Message power .
- Input SNR
- Output SNR
- Detection gain ; figure of merit , where is the channel SNR measured in the message band (baseband reference). Both show how much the demodulator improves or worsens SNR.
DSB-SC (coherent detection). , .
- Signal power ; noise in , so .
- Multiply by and low-pass filter:
The quadrature noise is rejected by coherent detection, giving a 3 dB gain. With baseband reference , figure of merit .
SSB-SC (coherent detection). , .
- Signal power (since ); noise in , so .
- Multiply by and low-pass filter:
- Signal power , noise power , so
With baseband reference (, same transmitted power), figure of merit .
Comparison:
| Quantity | DSB-SC | SSB-SC |
|---|---|---|
| Bandwidth | ||
| Input noise power | ||
| Detection gain | 2 | 1 |
| Figure of merit | 1 | 1 |
| Output SNR (same transmitted power) | Equal | Equal |
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 is a collection (ensemble) of time functions, one for each outcome of a random experiment. At a fixed time , is a random variable; for a fixed outcome , 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 :
Let a WSS process be the input of a stable LTI system with impulse response . The output is
1. Mean of the output. Expectation and integration are linear, so they can be interchanged:
is the DC gain of the system, a constant. So the output mean is constant, independent of .
2. Autocorrelation of the output.
The last step uses the WSS property of : the input autocorrelation depends only on the difference of its two time arguments, . Putting :
The right side depends only on , not on or separately.
3. Mean square value 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 only, is also WSS.
Taking the Fourier transform of gives the useful result
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 bits. Take M-ary baseband PAM: the receiver output sample (after the matched filter) takes one of equally spaced levels
so adjacent levels are apart. Assumptions: symbols equally likely, additive white Gaussian noise of PSD , so the sampled noise is Gaussian with zero mean and variance (unit-energy pulse, matched filter). Decision thresholds are placed midway between levels:
-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 :
Outer level (one neighbour, e.g. ): an error occurs only if :
There are inner and outer levels, so the average symbol error probability is
In terms of energy. The average symbol energy is
With :
Check, M = 2: and , the polar binary result.
With Gray coding, bit error rate .
Binary vs M-ary
| Point | Binary | M-ary |
|---|---|---|
| Bits per symbol | 1 | |
| Symbol rate for bit rate | ||
| Bandwidth needed | (Nyquist) | |
| Bandwidth efficiency | 2 bit/s/Hz (ideal) | bit/s/Hz |
| Power for same | lower | higher (levels closer) |
| Noise immunity | better | poorer |
| Transmitter/receiver | simple (one threshold) | complex ( thresholds, AGC, precise level control) |
So M-ary signalling saves bandwidth by a factor 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 with PDF :
- th moment (about the origin):
The first moment is the mean ; the second is the mean-square value .
- th central moment (about the mean):
The second central moment is the variance .
First central moment is always zero:
Noise equivalent bandwidth of RC low-pass filter
The NEB is the bandwidth of an ideal LPF, with the same zero-frequency gain , that passes the same white-noise power:
For the RC-LPF, , so and
Since the 3 dB bandwidth is , .
Ideal LPF with for and 0 elsewhere:
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 . Output PSD is , so
Answer: , , output noise power (watts, on a 1 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 , where is the average energy per bit.
Assumptions
- Signals over :
with an integer multiple of . 2. Channel adds white Gaussian noise of two-sided PSD . 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
- For "1": signal part
- For "0": signal part
Noise at the output: is Gaussian, zero mean, with variance
Threshold. For equal priors and equal variances, the optimum threshold is midway: .
Error probabilities
In terms of energy. Energy of the "1" pulse ; the "0" has zero energy, so average energy per bit . Hence
Remarks
- Coherent BPSK gives , so ASK needs 3 dB more average energy for the same .
- Non-coherent (envelope) ASK gives approximately , 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) 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 is applied, actual output power and ideal output power . Equating:
|H(f)|^2
|H(0)|^2 +-------+ area of rectangle
|\ | = area under |H(f)|^2
| \ |
| \__ |
| ~~|~~~___
+-------+------------> f
0 BN
Example: RC-LPF has . Then output noise power is simply .
Mean and autocorrelation at the output
Let a WSS process (mean , autocorrelation ) drive a stable LTI system with impulse response :
Mean
The output mean is the input mean times the DC gain, a constant.
Autocorrelation
This depends only on , so is also WSS. In compact form:
Power spectral density (Fourier transform, using ):
Output power
For white input, , this gives , 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 for a known pulse in white noise. Its impulse response is .
Derivation (brief). Input , noise PSD . At :
by Schwarz's inequality. Equality holds when , i.e. , giving .
Realization 1: Integrate-and-dump (rectangular pulse). For , : for , a rectangular impulse response, which is an integrator over the last seconds:
R S1 (dump)
r(t)--/\/\--+--(-A+)--+-------+---> y(T)
| op-amp| | |
+---||----+ +-/-+ short C
C at end of T
An op-amp integrator (or RC with ) integrates over one bit; the output is sampled at , then the capacitor is shorted (dumped) so the next bit starts from zero.
Realization 2: Correlator. Since
the matched filter output at equals the correlation of with a stored replica of :
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 , and a summer, giving a digital (FIR) matched filter.
Note: correlator and matched filter give the same value only at ; 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 of bandwidth and power ; white noise of PSD ; IF filter bandwidth ; high carrier-to-noise ratio.
Transmitted signal:
Channel (input) SNR. Signal power ; noise power in bandwidth (reference baseband) :
Received signal with narrowband noise
where each have power . The envelope is
For high SNR, , so
The DC term is removed by a blocking capacitor.
Output SNR
Figure of merit (gain parameter)
which is always less than 1.
Single tone , : , so
For 100 % modulation (): . So DSB-FC with envelope detection needs three times (4.77 dB) more transmitted power than DSB-SC/SSB with coherent detection () 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
| Point | AM (DSB-FC) | FM |
|---|---|---|
| Power efficiency | Poor: carrier carries of power; | Good: constant envelope; (tone), can be |
| Noise performance | Noise adds directly to amplitude | Limiter removes amplitude noise; SNR rises with |
| Bandwidth | (narrow) | (Carson), much wider |
| Bandwidth efficiency | Better | Poorer; trades bandwidth for SNR |
| Threshold | Envelope detector threshold at low SNR | Sharper threshold (~10 dB CNR) |
| Transmitter | Needs linear power amplifier at high level | Class C amplifiers usable (constant envelope) |
| Receiver complexity | Simple (diode envelope detector) | More complex (limiter, discriminator/PLL, de-emphasis) |
Error probability of coherent ASK
Signals over a bit: for "1", for "0"; AWGN of PSD ; equal priors.
Correlator output has mean for "1" and 0 for "0", with Gaussian noise variance
Optimum threshold is midway, , so
With average bit energy :
This is 3 dB worse than coherent BPSK, .
- 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 . A simpler practical receiver uses an ideal LPF whose bandwidth can be chosen; the question is how close it can come.
Set-up. Pulse for (energy ), white noise PSD , ideal LPF of unit gain and bandwidth . Pulse spectrum:
|H(f)|
1 +-----------+
| |
----+-----+-----+-----> f
-B 0 B
Peak output signal (at the pulse centre ):
Output noise power:
Output SNR
Compare with the matched filter, :
Effect of bandwidth
- too small: pulse energy is cut off, peak signal is small.
- too large: little extra signal but noise grows in proportion to .
- An optimum exists in between.
Evaluating numerically:
| 0.5 | 0.685 | 1.0 | 1.5 | |
|---|---|---|---|---|
| 0.762 | 0.825 | 0.695 | 0.349 |
Result: the best choice is (about ), giving , i.e. only about 0.84 dB worse than the true matched filter. Hence an ideal LPF with bandwidth about 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
Its autocorrelation is the inverse Fourier transform, an impulse:
so any two different samples are uncorrelated (and independent if Gaussian).
Sn(f) Rn(tau)
| N0/2 ^ (N0/2) delta
--+---------------- f ------+------ tau
- Total power is infinite, so it is only a model; real noise (thermal, shot) is flat up to very high frequencies ( Hz), so it acts white in any practical bandwidth.
- For thermal noise . Noise through a filter of bandwidth has power .
- 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 for "1" and for "0"; additive Gaussian noise of zero mean and variance ; equal priors, so the threshold is .
f(y|0) f(y|1)
/\ /\
/ \ | / \
__/ \____|_______/ \__ y
-A 0 threshold +A
shaded tails = errors
Error when "1" is sent: :
Error when "0" is sent: , by symmetry .
With a matched filter, , so
(For unipolar on-off PAM with levels and , threshold : .)
Extension to M-ary PAM
Levels (spacing ), thresholds midway.
- Each of the inner levels errs if : probability .
- Each of the 2 outer levels errs only on one side: .
Average symbol energy , so with a matched filter ():
For this reduces to the binary result. For fixed , larger gives larger , but bandwidth falls by .
- 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
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 and noise :
noise r(t)
,--->
/
0 ------+------> Ac (carrier)
resultant phase = phi(t) + noise angle
- High CNR (): noise only slightly perturbs the phase; output noise is small and smooth (parabolic PSD).
- Near/below threshold ( occasionally): the resultant phasor can swing around the origin, so its phase changes suddenly by . The discriminator output (derivative of phase) then contains sharp impulses ("clicks") of area .
- 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 ) 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:
- 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.
- 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.
- 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.
- 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.
Chapter titles and hours from the IOE syllabus ↗