Introduction

Energy spectral density (ESD)

For an energy signal $x(t)$ (finite energy, zero average power) with Fourier transform $X(f)$, the energy spectral density shows how the energy is spread over frequency:

$$\Psi_x(f) = |X(f)|^2 \quad (\text{J/Hz})$$

The total energy is the area under the ESD (Rayleigh's theorem):

$$E = \int_{-\infty}^{\infty} |x(t)|^2\,dt = \int_{-\infty}^{\infty} |X(f)|^2\,df$$

ESD is real, non-negative and even (for real $x(t)$), and it is the Fourier transform of the autocorrelation function $R_x(\tau)$.

Power spectral density (PSD)

For a power signal (finite non-zero average power, infinite energy) the Fourier transform may not exist, so we truncate the signal to $x_T(t)$ over $-T/2 < t < T/2$ and define

$$S_x(f) = \lim_{T\to\infty} \frac{|X_T(f)|^2}{T} \quad (\text{W/Hz})$$

The average power is $P = \int_{-\infty}^{\infty} S_x(f)\,df$. PSD is real, non-negative, even for real signals, and $S_x(f)$ and $R_x(\tau)$ form a Fourier transform pair (Wiener–Khinchine theorem).

Analog (swept-tuned) spectrum analyzer

A spectrum analyzer displays signal amplitude (or power) versus frequency on a screen. The common analog type is the swept-tuned superheterodyne analyzer.

 Input   +-------+   +-----+   +--------+   +-------+
 ------->| Atten |-->| LPF |-->| Mixer  |-->|  IF   |
         +-------+   +-----+   +--------+   | filter|
                                   ^        +-------+
                                   |            |
                              +---------+   +--------+
                              | Swept LO|   |Log amp |
                              |  (VCO)  |   |Env det |
                              +---------+   +--------+
                                   ^            |
                              +---------+   +--------+
                              |  Ramp   |-->| Video  |
                              |generator|   | filter |
                              +---------+   +--------+
                                   |  X         | Y
                                   +-->[ DISPLAY ]<+

Operation:

1. Attenuator and low-pass filter (preselector): set a safe input level and remove image frequencies.
2. Ramp (sweep) generator: produces a sawtooth voltage. It tunes the local oscillator and also moves the display trace horizontally, so the X-axis becomes frequency.
3. Mixer and swept LO: the LO frequency is swept; a component of the input at $f_{in}$ is converted to the fixed IF when $|f_{LO} - f_{in}| = f_{IF}$. So, at each instant, only one input frequency reaches the IF stage.
4. IF filter: a narrow band-pass filter; its bandwidth is the resolution bandwidth (RBW), which decides how closely two spectral lines can be separated.
5. Log amplifier and envelope detector: convert the IF signal amplitude to a DC level, usually on a dB scale.
6. Video filter: smooths the detected output to reduce noise on the display.
7. Display: the vertical (Y) deflection is the detected amplitude and the horizontal (X) deflection is the ramp, so the screen shows amplitude versus frequency.

As the LO sweeps once over its range, each spectral component of the input appears as a peak at its frequency position, giving the amplitude spectrum (or power spectrum) of the signal.

Energy and power signals

• Energy signal: a signal with finite, non-zero total energy, $0 < E < \infty$, where $E = \int_{-\infty}^{\infty} |x(t)|^2\,dt$. Its average power is zero. Example: a single rectangular pulse, $e^{-at}u(t)$.
• Power signal: a signal with finite, non-zero average power, $0 < P < \infty$, where $P = \lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2} |x(t)|^2\,dt$. Its energy is infinite. Example: periodic signals such as sinusoids, the unit step.

A signal cannot be both; some signals (e.g. $e^{t}$, a ramp $t\,u(t)$) are neither.

Is $x(t) = A\cos 2\pi f t$ an energy or power signal?

Let $f = f_0 = 1/T_0$.

Energy:

$$\begin{aligned} E &= \lim_{T\to\infty}\int_{-T/2}^{T/2} A^2\cos^2(2\pi f_0 t)\,dt \\ &= \lim_{T\to\infty}\int_{-T/2}^{T/2} \frac{A^2}{2}\left[1+\cos(4\pi f_0 t)\right]dt \\ &= \lim_{T\to\infty}\left[\frac{A^2 T}{2} + \frac{A^2\sin(2\pi f_0 T)}{4\pi f_0}\right] = \infty \end{aligned}$$

The second term stays bounded while $A^2T/2$ grows without limit, so the energy is infinite. It is not an energy signal.

Power:

$$\begin{aligned} P &= \lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2} A^2\cos^2(2\pi f_0 t)\,dt \\ &= \lim_{T\to\infty}\frac{1}{T}\left[\frac{A^2 T}{2} + \frac{A^2\sin(2\pi f_0 T)}{4\pi f_0}\right] \\ &= \frac{A^2}{2} + \lim_{T\to\infty}\frac{A^2\sin(2\pi f_0 T)}{4\pi f_0 T} = \frac{A^2}{2} \end{aligned}$$

(Same result by averaging over one period $T_0$, since the signal is periodic.)

Since $0 < P = A^2/2 < \infty$ and $E = \infty$:

Answer: $x(t) = A\cos 2\pi f t$ is a power signal with average power $P = \dfrac{A^2}{2}$ (normalised to a 1 $\Omega$ load); its energy is infinite.

An analog communication system sends a continuously varying message (voice, music, video) from a source to a destination through a channel, usually using analog modulation such as AM or FM.

+--------+  +----------+  +-----------+
| Info   |->|  Input   |->|Transmitter|
| source |  |transducer|  |(modulator)|
+--------+  +----------+  +-----------+
                                |
                                v
                         +-----------+
              Noise ---->|  Channel  |
                         +-----------+
                                |
                                v
+--------+  +----------+  +-----------+
| Desti- |<-|  Output  |<-| Receiver  |
| nation |  |transducer|  |(demod.)   |
+--------+  +----------+  +-----------+

1. Information source

Produces the message to be sent, e.g. human speech, music, a picture or a sensor reading. In analog systems the message takes continuous values.

2. Input transducer

Converts the non-electrical message into an electrical signal called the baseband or message signal $m(t)$. Examples: microphone (sound to voltage), camera (light to voltage).

3. Transmitter

Makes the message suitable for the channel. Main parts:

• Modulator: varies a parameter (amplitude, frequency or phase) of a high-frequency carrier $A_c\cos 2\pi f_c t$ according to $m(t)$, giving AM, FM or PM.
• Filters and amplifiers: limit the bandwidth and raise the power to the required level.
• Antenna (in radio systems): radiates the signal.

4. Channel

The physical medium between transmitter and receiver: free space, coaxial cable, twisted pair, optical fibre, waveguide. While passing through the channel the signal is:

• attenuated (loses power),
• distorted (by limited bandwidth, non-linearity, multipath),
• corrupted by noise and interference, added mostly here and at the receiver front end.

5. Receiver

Recovers the message from the weak, noisy received signal. It includes an RF amplifier, mixer and IF amplifier (in superheterodyne receivers), filters that reject noise and adjacent channels, and the demodulator/detector, which extracts $m(t)$ from the carrier.

6. Output transducer

Converts the recovered electrical signal back into its original form, e.g. loudspeaker (voltage to sound), picture tube/LCD (voltage to image).

7. Destination

The person or device that uses the message.

Example: in AM radio broadcasting, the studio microphone is the input transducer, the AM transmitter and antenna form the transmitter, free space is the channel, and the home radio (superheterodyne receiver with envelope detector and loudspeaker) is the receiver and output transducer.

The quality of an analog system is measured mainly by the output signal-to-noise ratio and the fidelity (how closely the output follows the original message).

Distortionless transmission means the output of a system (channel, filter or amplifier) has exactly the same shape as the input; only its amplitude may be scaled by a constant and it may be delayed by a fixed time:

$$y(t) = K\,x(t - t_d)$$

where $K$ is a constant gain and $t_d$ a constant delay.

Transfer function: taking the Fourier transform,

$$Y(f) = K X(f)\,e^{-j2\pi f t_d} \;\Rightarrow\; H(f) = K\,e^{-j2\pi f t_d}$$

and the impulse response is $h(t) = K\,\delta(t - t_d)$.

Conditions (over the band occupied by the signal):

1. Constant amplitude response: $|H(f)| = K$. All frequency components are amplified or attenuated equally.
2. Linear phase response: $\theta(f) = -2\pi f t_d$ (a straight line through the origin, or through $\pm n\pi$). All components are delayed by the same time $t_d$; the group delay $-\frac{1}{2\pi}\frac{d\theta}{df} = t_d$ is constant.

 |H(f)|              theta(f)
   |                   |\
 K +-----------        | \  slope = -2*pi*td
   |                   |  \
   +-----------> f   --+---\------> f
                       |    \

Types of distortion when conditions fail:

• Amplitude distortion: $|H(f)|$ not constant, so some frequencies are boosted or cut more than others.
• Phase (delay) distortion: phase not linear, so components arrive at different times; serious for video and data pulses (causes ISI).
• Non-linear distortion: the system is non-linear, creating harmonics and intermodulation products.

In practice exact distortionless transmission is impossible over all frequencies, but it is enough for $|H(f)|$ to be flat and the phase linear over the signal bandwidth. Equalizers are used to correct channel amplitude and phase response.

Modulation

Modulation is the process of varying some characteristic (amplitude, frequency or phase) of a high-frequency carrier wave in accordance with the instantaneous value of the message (modulating) signal. With carrier $c(t) = A_c\cos(2\pi f_c t + \phi)$, varying $A_c$ gives AM, varying $f_c$ gives FM and varying $\phi$ gives PM. The reverse process at the receiver is demodulation.

Reasons (needs) for modulation

1. Practical antenna height: an antenna must be about $\lambda/4$ long to radiate efficiently. For a 1 kHz audio signal $\lambda = c/f = 3\times10^8/10^3 = 300$ km, so $\lambda/4 = 75$ km, which is impossible. With a 1 MHz carrier, $\lambda/4 = 75$ m, which is practical.
2. Multiplexing: many messages occupy the same baseband (e.g. 0–4 kHz for speech). Modulating them onto different carriers places them in different frequency slots, so many signals share one channel (FDM), e.g. many radio stations.
3. Reduce noise and interference: wideband modulation such as FM trades bandwidth for better signal-to-noise ratio.
4. Overcome equipment limitations / narrowbanding: the ratio of highest to lowest frequency of audio (20 Hz–20 kHz) is 1000; after shifting to a 1 MHz carrier the ratio is about 1.04, so one antenna and amplifier design works well for the whole band.
5. Match the channel: shift the signal to a frequency band where the channel passes signals well (e.g. optical fibre, satellite bands) and assign frequencies by regulation.
6. Long-distance propagation: low-frequency baseband signals are absorbed quickly; high-frequency carriers travel far by ground, sky or line-of-sight waves.

Functional block diagram of analog communication system

+------+  +-------+  +---------+  +-------+
|Source|->| Input |->|  Trans- |->|Channel|<-- Noise
|      |  |transd.|  |  mitter |  |       |
+------+  +-------+  +---------+  +-------+
                                      |
+------+  +-------+  +---------+      |
| Dest.|<-|Output |<-| Receiver|<-----+
|      |  |transd.|  |         |
+------+  +-------+  +---------+

• Information source: produces the message (speech, music, picture).
• Input transducer: converts it into an electrical baseband signal $m(t)$, e.g. microphone.
• Transmitter: modulates $m(t)$ onto a carrier, filters and amplifies it, and feeds it to the antenna or line.
• Channel: medium (free space, cable, fibre) that carries the signal; it attenuates and distorts the signal, and noise is added here.
• Receiver: amplifies and filters the received signal and demodulates it to recover $m(t)$.
• Output transducer: converts the electrical signal back to the original form, e.g. loudspeaker.
• Destination: the user of the message.

In duplex (full-duplex) mode both terminals send and receive at the same time, e.g. a mobile phone call. Each terminal therefore has a complete digital transmitter and receiver, and the two directions share the channel by separate frequencies (FDD) or time slots (TDD) through a duplexer.

Functional digital communication system (duplex)

  TERMINAL A                         TERMINAL B
 +-----------+                      +-----------+
 | Source    |                      | Sink      |
 | Formatting|                      | Formatting|
 | Src encode|                      | Src decode|
 | Encrypt   |                      | Decrypt   |
 | Chan code |                      | Chan dec. |
 | Modulator |                      | Demod.    |
 +-----+-----+                      +-----^-----+
       |   +-------+  CHANNEL  +-------+  |
       +-->|       |==========>|       |--+
           |Duplex-|  (noise,  |Duplex-|
       +<--|  er   |<==========|  er   |<-+
       |   +-------+  fading)  +-------+  |
 +-----v-----+                      +-----+-----+
 | Demod.    |                      | Modulator |
 | Chan dec. |                      | Chan code |
 | Decrypt   |                      | Encrypt   |
 | Src decode|                      | Src encode|
 | Sink      |                      | Source    |
 +-----------+                      +-----------+

Main blocks of each direction:

1. Formatting: converts the source output to digital form (sampling, quantization, PCM).
2. Source encoding: removes redundancy to reduce bit rate (e.g. Huffman, speech codecs).
3. Encryption: secures the data.
4. Channel encoding: adds controlled redundancy for error detection/correction (parity, Hamming, CRC, convolutional codes).
5. Multiplexing / multiple access: combines several streams or users (TDM, FDMA, CDMA).
6. Modulation: maps bits to waveforms (ASK, FSK, PSK, QAM); pulse shaping limits bandwidth.
7. Duplexer and channel: separates the transmit and receive paths over the shared medium.
8. At the receiver, the blocks work in reverse: demodulation and detection, channel decoding, decryption, source decoding, and formatting back to the user's form.

Contrast with analog communication

Impairments

• Additive noise (thermal, shot, atmospheric) causes bit errors.
• Attenuation reduces signal strength with distance.
• Bandwidth limitation and dispersion cause inter-symbol interference (ISI).
• Interference: co-channel, adjacent channel, crosstalk, echo in duplex links.
• Fading and multipath in wireless channels.
• Timing jitter and carrier/clock synchronization errors.

Performance parameters

• Probability of bit error (BER) at a given $E_b/N_0$.
• Data rate (bit/s) and bandwidth efficiency (bit/s/Hz).
• Power efficiency: $E_b/N_0$ needed for a target BER.
• Channel capacity limit $C = B\log_2(1 + S/N)$.
• Latency/delay and system complexity and cost.

A system is linear if it obeys superposition: for inputs $x_1(t)\to y_1(t)$ and $x_2(t)\to y_2(t)$, the input $a x_1(t) + b x_2(t)$ must give $a y_1(t) + b y_2(t)$ for all constants $a$, $b$.

Given: $y(t) = 2t^2 x(t) + 3t\,x(t-4)$

Step 1: outputs for the individual inputs

$$\begin{aligned} y_1(t) &= 2t^2 x_1(t) + 3t\,x_1(t-4) \\ y_2(t) &= 2t^2 x_2(t) + 3t\,x_2(t-4) \end{aligned}$$

Step 2: output for the combined input $x_3(t) = a x_1(t) + b x_2(t)$

$$\begin{aligned} y_3(t) &= 2t^2\left[a x_1(t) + b x_2(t)\right] + 3t\left[a x_1(t-4) + b x_2(t-4)\right] \\ &= a\left[2t^2 x_1(t) + 3t\,x_1(t-4)\right] + b\left[2t^2 x_2(t) + 3t\,x_2(t-4)\right] \\ &= a\,y_1(t) + b\,y_2(t) \end{aligned}$$

Step 3: compare. The response to the weighted sum equals the weighted sum of responses, so superposition (additivity and homogeneity) holds. Also, zero input gives zero output.

The multipliers $2t^2$ and $3t$ depend on time, which makes the system time-varying, but they do not affect linearity, because the input appears only to the first power and there is no constant term.

Answer: $y(t) = 2t^2x(t) + 3t\,x(t-4)$ is a linear system (it is, however, time-variant).

A system is causal if the output at any time $t$ depends only on present and past values of the input, never on future values, i.e. $y(t_0)$ depends only on $x(t)$ for $t \le t_0$.

Given: $y(t) = x^2(t) + x(t+2)$

Check the output at a few time instants:

• The term $x^2(t)$ uses the present input, which is allowed.
• The term $x(t+2)$ uses the input 2 seconds ahead, i.e. a future value, at every instant $t$.

Since the output at time $t$ needs $x(t+2)$, which has not yet arrived, the system cannot be built to work in real time.

Answer: $y(t) = x^2(t) + x(t+2)$ is non-causal (anticipative). (Note also that, because of the $x^2(t)$ term, it is a non-linear system.)

Signals (from the figure):

• $x_1(t) = 2$ for $1 \le t \le 2$, zero elsewhere (rectangular pulse).
• $x_2(t) = 2(2 - t)$ for $1 \le t \le 2$, zero elsewhere (jumps to 2 at $t=1$, falls linearly to 0 at $t=2$).

Definitions used (real signals): $R_{11}(\tau) = \int x_1(t)x_1(t+\tau)\,dt$, $R_{12}(\tau) = \int x_1(t)x_2(t+\tau)\,dt$, $y(t) = \int x_1(\lambda)x_2(t-\lambda)\,d\lambda$.

Auto-correlation

Graphically, keep $x(t)$ fixed, slide a copy by $\tau$ (left for $\tau > 0$), multiply and find the area of the overlap. Overlap exists only for $|\tau| < 1$.

Rectangle $x_1$: for $0 \le \tau \le 1$ the overlap is $1 \le t \le 2-\tau$, of width $1-\tau$, height $2\times2 = 4$:

$$R_{11}(\tau) = \int_{1}^{2-\tau} 4\,dt = 4(1-|\tau|), \quad |\tau| \le 1$$

It is a triangle of peak $R_{11}(0) = 4$ = energy of $x_1$, base $-1$ to $1$.

Triangle $x_2$: for $0 \le \tau \le 1$,

$$\begin{aligned} R_{22}(\tau) &= \int_{1}^{2-\tau} 2(2-t)\cdot 2(2-t-\tau)\,dt \\ &= \frac{4}{3} - 2|\tau| + \frac{2}{3}|\tau|^3, \quad |\tau| \le 1 \end{aligned}$$

Peak $R_{22}(0) = 4/3$ = energy of $x_2$; it is even, smooth and reaches 0 at $\tau = \pm1$ (e.g. $R_{22}(0.5) = 5/12$).

Cross-correlation $R_{12}(\tau)$

Slide $x_2$ relative to $x_1$; overlap again only for $|\tau| < 1$.

• $0 \le \tau \le 1$ ($x_2$ moved left, its low tail overlaps): $R_{12}(\tau) = \int_{1}^{2-\tau} 2\cdot 2(2-t-\tau)\,dt = 2(1-\tau)^2$
• $-1 \le \tau \le 0$ (its tall front edge overlaps): $R_{12}(\tau) = \int_{1-\tau}^{2} 2\cdot 2(2-t-\tau)\,dt = 2(1-\tau^2)$

Values: $R_{12}(-1) = 0$, $R_{12}(-0.5) = 1.5$, $R_{12}(0) = 2$, $R_{12}(0.5) = 0.5$, $R_{12}(1) = 0$. The curve is not symmetric (cross-correlation is not even); its peak is at $\tau = 0$ with value 2.

Convolution $y(t) = x_1(t) * x_2(t)$

Fold $x_2(\lambda)$ to $x_2(t-\lambda)$ (now non-zero for $t-2 \le \lambda \le t-1$, rising from 0 to 2) and slide it over $x_1(\lambda)$ on $[1,2]$. Output exists from $1+1 = 2$ to $2+2 = 4$.

• $t < 2$: no overlap, $y = 0$.
• $2 \le t \le 3$ (partial overlap $1 \le \lambda \le t-1$):

$$y(t) = \int_{1}^{t-1} 2\cdot 2(2-t+\lambda)\,d\lambda = -2t^2 + 12t - 16 = 2(t-2)(4-t)$$

• $3 \le t \le 4$ (overlap $t-2 \le \lambda \le 2$):

$$y(t) = \int_{t-2}^{2} 2\cdot 2(2-t+\lambda)\,d\lambda = 2(t-4)^2$$

• $t > 4$: $y = 0$.

Check: $y(2) = 0$, $y(3) = 2$ from both pieces (continuous), $y(4) = 0$; area $= 2 \times 1 = 2$ = (area of $x_1$) $\times$ (area of $x_2$) = $2 \times 1$.

 y(t)
  2 |          *
    |       *     *
  1 |     *          *
    |   *               *
  0 +--*------------------*----> t
       2         3        4

Answer: $R_{11} = 4(1-|\tau|)$; $R_{22} = \frac43 - 2|\tau| + \frac23|\tau|^3$; $R_{12} = 2(1-\tau^2)$ for $-1\le\tau\le0$ and $2(1-\tau)^2$ for $0\le\tau\le1$; $y(t) = 2(t-2)(4-t)$ for $2\le t\le3$ and $2(t-4)^2$ for $3\le t\le4$ (all zero elsewhere).

Shannon–Hartley channel capacity theorem

The theorem gives the maximum error-free data rate (channel capacity) of a band-limited channel disturbed by additive white Gaussian noise (AWGN):

$$C = B\log_2\left(1 + \frac{S}{N}\right) \ \text{bit/s}$$

where $B$ = channel bandwidth (Hz), $S$ = average signal power, $N = N_0B$ = noise power in the band.

Meaning:

• If the information rate $R \le C$, there exists a coding scheme that makes the error probability as small as desired.
• If $R > C$, error-free transmission is impossible whatever coding is used.
• Bandwidth and SNR can be traded: the same capacity can be obtained with less bandwidth and more power, or vice versa.
• As $B \to \infty$, capacity does not become infinite; it approaches $C_\infty = 1.44\,S/N_0$.

Example: a telephone channel with $B = 3.1$ kHz and SNR $= 30$ dB ($S/N = 1000$):

$$C = 3100\log_2(1 + 1000) = 3100 \times 9.967 \approx 30.9\ \text{kbit/s}$$

So no modem can reliably send more than about 30.9 kbit/s over such a line.

Is $x(t) = e^{-2t}u(t)$ an energy or power signal?

Energy:

$$\begin{aligned} E &= \int_{-\infty}^{\infty} |x(t)|^2\,dt = \int_{0}^{\infty} e^{-4t}\,dt \\ &= \left[\frac{e^{-4t}}{-4}\right]_0^{\infty} = 0 - \left(-\frac14\right) = \frac14\ \text{J} \end{aligned}$$

Power:

$$P = \lim_{T\to\infty}\frac{1}{T}\int_{0}^{T/2} e^{-4t}\,dt = \lim_{T\to\infty}\frac{1 - e^{-2T}}{4T} = 0$$

Answer: Energy $E = 0.25$ J (finite, non-zero) and power $P = 0$, so $x(t) = e^{-2t}u(t)$ is an energy signal.

A system is linear if it satisfies superposition: $a x_1(t) + b x_2(t) \to a y_1(t) + b y_2(t)$.

Given: $y(t) = x^2(t)$

For individual inputs: $y_1(t) = x_1^2(t)$, $y_2(t) = x_2^2(t)$.

For the combined input $x_3(t) = a x_1(t) + b x_2(t)$:

$$\begin{aligned} y_3(t) &= \left[a x_1(t) + b x_2(t)\right]^2 \\ &= a^2x_1^2(t) + b^2x_2^2(t) + 2ab\,x_1(t)x_2(t) \end{aligned}$$

Required for linearity: $a y_1(t) + b y_2(t) = a x_1^2(t) + b x_2^2(t)$.

Clearly $y_3(t) \ne a y_1(t) + b y_2(t)$ (the coefficients become $a^2$, $b^2$ and a cross term $2ab\,x_1x_2$ appears).

Simple check with homogeneity: doubling the input, $x \to 2x$, gives $y = 4x^2$, i.e. the output becomes four times, not twice.

Answer: $y(t) = x^2(t)$ is a non-linear system (a square-law device, such as that used in square-law modulators and detectors).

A system is time-invariant if a time shift of the input causes the same time shift of the output: if $x(t) \to y(t)$, then $x(t-t_0) \to y(t-t_0)$.

Given: $y(t) = y(t-4) + x(t-4)$ (a recursive system; assume it starts at rest).

Step 1: apply a delayed input $x_1(t) = x(t-t_0)$. Its output $y_1(t)$ satisfies

$$y_1(t) = y_1(t-4) + x_1(t-4) = y_1(t-4) + x(t-t_0-4)$$

Step 2: shift the original equation by $t_0$ (replace $t$ by $t-t_0$):

$$y(t-t_0) = y(t-t_0-4) + x(t-t_0-4)$$

Step 3: compare. The delayed output $y(t-t_0)$ satisfies exactly the same equation, driven by the same input $x(t-t_0-4)$, as $y_1(t)$. With the same (rest) initial condition, the solutions are equal:

$$y_1(t) = y(t-t_0)$$

The reason is that the coefficients (both equal to 1) and the delay (4 s) are constants; time $t$ does not appear explicitly in the equation.

Answer: the system $y(t) = y(t-4) + x(t-4)$ is time-invariant.

Reasons for modulation

Modulation shifts the message onto a high-frequency carrier. It is needed because:

1. Antenna size: efficient radiation needs an antenna of about $\lambda/4$. A 3 kHz voice signal has $\lambda = 100$ km, i.e. a 25 km antenna; on a 100 MHz carrier the antenna is only 0.75 m.
2. Multiplexing: different messages are put on different carriers so many users share one medium (FDM in radio, TV, telephone trunks).
3. Noise reduction: wideband schemes (FM, spread spectrum) improve SNR at the cost of bandwidth.
4. Narrowbanding: reduces the ratio of highest to lowest frequency, making antenna and amplifier design easier.
5. Channel matching and long-range propagation: places the signal in a band the channel passes well and in the band allotted by regulation.

Advantages of digital over analog communication

1. Noise immunity: only a decision between a few levels is needed; regenerative repeaters remove accumulated noise, so quality does not fall with distance.
2. Error detection and correction coding is possible.
3. Security: encryption is easy.
4. Integration: voice, data, image and video all become bits and share the same network (TDM, packet switching).
5. Storage and processing: easy to store, compress and process with DSP and computers.
6. Hardware: cheap, reliable, flexible VLSI circuits; easy to reconfigure (software radio).
7. Bandwidth/power trade-off and better performance in low SNR via coding.

Generic digital communication system (full-duplex)

In full-duplex both ends transmit and receive simultaneously. Each end has a transmitter chain and a receiver chain, joined to the channel by a duplexer (FDD or TDD).

   END A (transmit path)          END B (receive path)
 Source                              Sink
   |                                   ^
 Formatting                        Formatting
   |                                   ^
 Source enc.                       Source dec.
   |                                   ^
 Channel enc.                      Channel dec.
   |                                   ^
 Modulator                         Demodulator
   |                                   ^
 [Duplexer]====== CHANNEL + noise ==[Duplexer]
   |                                   ^
 Demodulator ... (same chain in reverse) ...
   v                                   |
 Sink   <---- B transmits to A ----  Source

• Formatting: sampling, quantization, PCM to obtain bits.
• Source encoder/decoder: compression.
• Channel encoder/decoder: adds/uses redundancy to correct errors.
• Modulator/demodulator: maps bits to waveforms (ASK, FSK, PSK, QAM) and back; the receiver also performs synchronization and detection.
• Duplexer: lets one antenna or line carry both directions (different frequency bands in FDD, alternate time slots in TDD).

Unit step in terms of signum

The signum function is

$$\text{sgn}(t) = \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}$$

and the unit step is $u(t) = 1$ for $t > 0$, $0$ for $t < 0$.

Adding 1 to sgn$(t)$ gives $2$ for $t>0$ and $0$ for $t<0$; halving it gives the unit step:

$$u(t) = \frac{1}{2}\left[1 + \text{sgn}(t)\right]$$

Check: $t>0$: $\frac12(1+1) = 1$; $t<0$: $\frac12(1-1) = 0$; $t=0$: $\frac12$ (the mid-value used at the jump).

 sgn(t)              u(t)
  1 |------          1 |------
    |                  |
 ---+------> t      ---+------> t
-1--|                  |

This relation is used to find the Fourier transform of $u(t)$: since $\mathcal{F}\{\text{sgn}(t)\} = \frac{1}{j\pi f}$ and $\mathcal{F}\{1\} = \delta(f)$,

$$U(f) = \frac12\delta(f) + \frac{1}{j2\pi f}$$

Energy or power?

Energy:

$$E = \int_{-\infty}^{\infty} u^2(t)\,dt = \int_{0}^{\infty} 1\,dt = \infty$$

So it is not an energy signal.

Power:

$$\begin{aligned} P &= \lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2} u^2(t)\,dt \\ &= \lim_{T\to\infty}\frac{1}{T}\int_{0}^{T/2} 1\,dt = \lim_{T\to\infty}\frac{1}{T}\cdot\frac{T}{2} = \frac12 \end{aligned}$$

Answer: $u(t) = \frac12[1 + \text{sgn}(t)]$; the unit step has infinite energy and finite average power $P = 0.5$ W (normalised), so it is a power signal.

Noise, interference and distortion all degrade a received signal, but they differ in origin and in how they can be reduced.

• Noise: unwanted, random electrical signals that add to the wanted signal. Sources: thermal (Johnson) noise in resistors, shot noise in devices, atmospheric, cosmic and man-made noise. Example: hiss in a radio, snow on a TV screen.
• Interference: contamination by unwanted signals from other man-made transmitters or systems, usually having a definite structure and frequency. Example: a neighbouring radio station heard in the background, crosstalk between telephone lines.
• Distortion: change in the shape of the signal caused by the imperfect response of the system itself (non-flat amplitude, non-linear phase, non-linearity). Example: a square pulse becoming rounded after a band-limited channel.

Key point: distortion can in principle be undone by an inverse system (equalizer), because it is a deterministic function of the signal; noise cannot be removed completely, only reduced, because it is random.

A received signal can be written as $r(t) = f\{s(t)\} + i(t) + n(t)$: the channel distorts the signal $s(t)$, unwanted signals from other sources add interference $i(t)$, and random noise $n(t)$ is added. These three impairments differ as follows.

Distortion

Alteration of the waveform shape by the system or channel itself, so that the output is not $K\,s(t - t_d)$.

• Linear distortion: amplitude distortion (non-flat $|H(f)|$) and phase/delay distortion (non-linear phase), e.g. ISI in data channels, multipath.
• Non-linear distortion: caused by non-linear devices (amplifier saturation), producing harmonics and intermodulation.
• It exists only while the signal is present and is deterministic, so it can be corrected by equalization or by improving linearity.

Noise

Random, unpredictable unwanted electrical energy from natural sources.

• External: atmospheric (lightning), solar and cosmic, industrial (ignition, motors).
• Internal: thermal noise $P_n = kTB$, shot noise, partition and flicker noise.
• Always present, even with no signal. Cannot be removed fully; reduced by limiting bandwidth, low-noise amplifiers, higher transmit power, wideband modulation (FM) and error-correcting codes.

Interference

Unwanted signals from other communication sources that fall in the receiver band.

• Examples: co-channel and adjacent-channel interference, crosstalk between cables, image-frequency interference in superheterodyne receivers, power-line hum.
• Usually has a definite form and frequency, so it can be reduced by good filtering, shielding, frequency planning, directional antennas.

Modulation

Modulation is the process by which some characteristic of a high-frequency carrier $c(t) = A_c\cos(2\pi f_c t + \phi)$, namely its amplitude, frequency or phase, is varied in accordance with the instantaneous amplitude of the message signal $m(t)$. The results are amplitude modulation (AM), frequency modulation (FM) and phase modulation (PM) respectively. For example, in AM, $s(t) = A_c[1 + k_a m(t)]\cos 2\pi f_c t$.

Modulation shifts the message spectrum from baseband to a band around $f_c$, which allows practical antenna sizes, multiplexing of many signals, and better noise performance.

Functional block diagram of an analog communication system

+--------+   +----------+   +-------------+
| Source |-->|  Input   |-->| Transmitter |
|        |   |transducer|   | mod + amp   |
+--------+   +----------+   +------+------+
                                   |
                     Noise   +-----v-----+
                     ------->|  Channel  |
                             +-----+-----+
                                   |
+--------+   +----------+   +------v------+
| Desti- |<--|  Output  |<--|  Receiver   |
| nation |   |transducer|   | amp + demod |
+--------+   +----------+   +-------------+

Blocks

1. Information source: generates the message, e.g. speech, music, picture. Its output is usually non-electrical.
2. Input transducer: converts the message into an electrical baseband signal $m(t)$, e.g. microphone, camera, temperature sensor.
3. Transmitter: processes $m(t)$ for efficient transmission. It band-limits and amplifies the message, modulates it on a carrier generated by an oscillator, then amplifies the modulated signal (power amplifier) and couples it to the antenna or cable.
4. Channel: the medium between transmitter and receiver: wire pair, coaxial cable, optical fibre or free space. It attenuates the signal and introduces distortion; noise and interference are added mainly here.
5. Receiver: selects the wanted signal from the antenna, amplifies it (RF amplifier), converts it to an intermediate frequency (mixer + local oscillator in a superheterodyne receiver), filters and amplifies it, and demodulates it to recover $m(t)$, followed by audio/video amplification.
6. Output transducer: converts the recovered electrical signal back into the original form, e.g. loudspeaker, display.
7. Destination: the listener, viewer or machine that uses the message.

Example: FM radio: microphone, FM transmitter at 88–108 MHz, free space, FM receiver with discriminator, loudspeaker.

Analog source vs digital source

Effects of atmospheric noise on analog communication

Atmospheric (static) noise is produced mainly by lightning discharges and other natural electrical disturbances in the atmosphere. It travels to the receiver like radio waves.

• It appears as impulsive crackling or "static" in AM radio and as spots on analog TV pictures.
• Strongest at low and medium frequencies (below about 30 MHz) and decreases at higher frequencies; it is negligible above about 30 MHz.
• Since it is additive, it directly changes the amplitude of the received signal; AM is badly affected because the information is in the amplitude. FM is far less affected because a limiter removes amplitude variations.
• It reduces the SNR and fidelity of the received message and limits the usable range of LF/MF/HF links.
• In an analog system the noise cannot be separated from the signal once added, and repeaters amplify it along with the signal, so it accumulates over the link.
• It is worst in the tropics and in summer/rainy seasons when thunderstorms are frequent.

Advantages of analog over digital communication

1. Less bandwidth: e.g. a voice channel needs about 4 kHz in AM-SSB versus 64 kbit/s (about 32 kHz or more) in PCM.
2. Simpler and cheaper hardware: no ADC/DAC, coding or complex synchronization.
3. No quantization noise: the signal is represented exactly, apart from added noise.
4. Graceful degradation: quality falls slowly as SNR drops, instead of a sudden failure (cliff effect) seen in digital links.
5. Real-time with no processing delay (no coding or buffering latency).
6. Natural signals are already analog, so no conversion is needed.

Impulse response and transfer function

• Impulse response $h(t)$: the output of an LTI system when the input is a unit impulse $\delta(t)$, with zero initial conditions. For any input, $y(t) = x(t) * h(t) = \int x(\lambda)h(t-\lambda)\,d\lambda$.
• Transfer function $H(f)$: the Fourier transform of the impulse response, $H(f) = \int h(t)e^{-j2\pi ft}\,dt = Y(f)/X(f)$. It gives the gain $|H(f)|$ and phase shift $\theta(f)$ the system applies to each frequency: $H(f) = |H(f)|e^{j\theta(f)}$.

Transfer function for distortionless transmission

Transmission is distortionless when the output is an exact replica of the input except for a constant gain $K$ and a constant delay $t_d$:

$$y(t) = K\,x(t - t_d)$$

Take the Fourier transform of both sides, using the time-shift property $x(t - t_d) \leftrightarrow X(f)e^{-j2\pi f t_d}$:

$$Y(f) = K\,X(f)\,e^{-j2\pi f t_d}$$

Therefore

$$H(f) = \frac{Y(f)}{X(f)} = K\,e^{-j2\pi f t_d}$$

and the impulse response is $h(t) = K\,\delta(t - t_d)$.

Comparing with $H(f) = |H(f)|e^{j\theta(f)}$, the two conditions are:

1. Amplitude response constant: $|H(f)| = K$ for all frequencies in the signal band.
2. Phase response linear in frequency: $\theta(f) = -2\pi f t_d$, i.e. constant time delay $t_d = -\frac{1}{2\pi}\frac{d\theta}{df}$ for every component. (A phase of $-2\pi f t_d \pm n\pi$ is also acceptable; $\pm\pi$ only inverts the sign.)

Amplitude and phase response

 |H(f)|
    |
  K +-------------------------
    |
    +-------------------------> f
   -B            0           B

 theta(f)
    |\
    | \   slope = -2*pi*td
 ---+--\----------------------> f
    |   \
    |    \

The amplitude response is a horizontal line at height $K$ and the phase response is a straight line through the origin with negative slope $-2\pi t_d$. In practice these need to hold only over the bandwidth $-B \le f \le B$ of the signal. If $|H(f)|$ varies, amplitude distortion results; if $\theta(f)$ is not linear, phase (delay) distortion results.

Parseval's theorem for power signals

For a periodic power signal $x(t)$ with period $T_0$ and complex Fourier series coefficients $c_n$, the average power computed in the time domain equals the sum of powers of its harmonics:

$$P = \frac{1}{T_0}\int_{T_0} |x(t)|^2\,dt = \sum_{n=-\infty}^{\infty} |c_n|^2$$

In general (any power signal), $P = \int_{-\infty}^{\infty} S_x(f)\,df$, where $S_x(f)$ is the power spectral density. So power can be found either in the time domain or in the frequency domain.

Proof: $S_y(f) = |H(f)|^2 S_x(f)$

Let a power signal $x(t)$ be applied to an LTI system with impulse response $h(t)$ (real) and transfer function $H(f)$. The output is

$$y(t) = \int_{-\infty}^{\infty} h(\alpha)\,x(t-\alpha)\,d\alpha$$

Step 1: autocorrelation of the output. With $R_x(\tau) = \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2} x(t)x(t-\tau)\,dt$,

$$\begin{aligned} R_y(\tau) &= \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2} y(t)\,y(t-\tau)\,dt \\ &= \int\!\!\int h(\alpha)h(\beta)\left[\lim_{T\to\infty}\frac1T\int x(t-\alpha)x(t-\tau-\beta)\,dt\right]d\alpha\,d\beta \\ &= \int\!\!\int h(\alpha)h(\beta)\,R_x(\tau - \alpha + \beta)\,d\alpha\,d\beta \end{aligned}$$

(the bracket is $R_x$ at the lag $(t-\alpha) - (t-\tau-\beta) = \tau - \alpha + \beta$).

Step 2: take the Fourier transform with respect to $\tau$. By the Wiener–Khinchine theorem $S_y(f) = \mathcal{F}\{R_y(\tau)\}$:

$$\begin{aligned} S_y(f) &= \int\!\!\int h(\alpha)h(\beta)\left[\int R_x(\tau-\alpha+\beta)e^{-j2\pi f\tau}d\tau\right]d\alpha\,d\beta \end{aligned}$$

Put $\lambda = \tau - \alpha + \beta$, so $\tau = \lambda + \alpha - \beta$:

$$\begin{aligned} S_y(f) &= \int h(\alpha)e^{-j2\pi f\alpha}d\alpha \int h(\beta)e^{+j2\pi f\beta}d\beta \int R_x(\lambda)e^{-j2\pi f\lambda}d\lambda \\ &= H(f)\,H^*(f)\,S_x(f) \end{aligned}$$

since for real $h(t)$, $\int h(\beta)e^{j2\pi f\beta}d\beta = H(-f) = H^*(f)$.

Result:

$$S_y(f) = |H(f)|^2\,S_x(f)$$

The output PSD equals the input PSD multiplied by the squared amplitude response; the phase response has no effect on the PSD. The output power is then $P_y = \int |H(f)|^2 S_x(f)\,df$.

Example: white noise of PSD $N_0/2$ through an ideal LPF of bandwidth $B$ and gain 1 gives $S_y = N_0/2$ for $|f| < B$ and output power $N_0B$.

Block diagram of a communication system

A communication system carries information from a source to a destination through a channel.

+--------+  +--------+  +-----------+
| Source |->| Input  |->|Transmitter|
|        |  | transd.|  |           |
+--------+  +--------+  +-----+-----+
                              |
             Noise,     +-----v-----+
             interf. -->|  Channel  |
                        +-----+-----+
                              |
+--------+  +--------+  +-----v-----+
| Desti- |<-| Output |<-| Receiver  |
| nation |  | transd.|  |           |
+--------+  +--------+  +-----------+

• Information source: produces the message, e.g. voice, text, picture, data.
• Input transducer: converts the message into an electrical signal (microphone, camera, keyboard).
• Transmitter: makes the signal suitable for the channel: filtering, encoding (digital), modulation onto a carrier, and power amplification.
• Channel: the physical path (wire, cable, optical fibre, free space). It attenuates and distorts the signal; noise and interference are added here.
• Receiver: picks up the weak signal, amplifies and filters it, demodulates/decodes it, and reconstructs the message as closely as possible.
• Output transducer: converts the electrical signal back into the original form (loudspeaker, display, printer).
• Destination: the person or machine for whom the message is meant.

Example: in a mobile phone call, the microphone, the phone's transmitter, the radio path, the other phone's receiver and its earpiece form the chain.

Needs for modulation

1. Reasonable antenna height: antenna length $\approx \lambda/4$; a 1 kHz signal would need a 75 km antenna, but at 1 MHz only 75 m.
2. Multiplexing: several signals can share one channel by using different carrier frequencies.
3. Better noise immunity: FM and other wideband methods improve output SNR.
4. Long-distance propagation and placing the signal in the frequency band permitted for the service.
5. Narrowbanding: easier design of antennas, filters and amplifiers.

Auto-correlation function

The auto-correlation function measures the similarity between a signal and a time-shifted copy of itself. For an energy signal $x(t)$:

$$R_x(\tau) = \int_{-\infty}^{\infty} x(t)\,x^*(t-\tau)\,dt$$

where $\tau$ is the time shift (lag).

Properties (energy signal)

1. Value at origin equals energy: $R_x(0) = \int |x(t)|^2dt = E$.
2. Conjugate symmetry: $R_x(\tau) = R_x^*(-\tau)$; for a real signal it is an even function, $R_x(\tau) = R_x(-\tau)$.
3. Maximum at origin: $|R_x(\tau)| \le R_x(0)$ for all $\tau$.
4. Fourier pair with ESD: $R_x(\tau) \leftrightarrow \Psi_x(f) = |X(f)|^2$.
5. $R_x(\tau) \to 0$ as $|\tau| \to \infty$ for energy signals.

Proof: $R_x(\tau)$ and ESD form a Fourier transform pair

Take the Fourier transform of $R_x(\tau)$:

$$\begin{aligned} \mathcal{F}\{R_x(\tau)\} &= \int_{-\infty}^{\infty} R_x(\tau)e^{-j2\pi f\tau}\,d\tau \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} x(t)\,x^*(t-\tau)\,e^{-j2\pi f\tau}\,dt\,d\tau \end{aligned}$$

Change the variable in the inner integral: let $\lambda = t - \tau$, so $\tau = t - \lambda$ and $d\tau = -d\lambda$ (limits swap, cancelling the sign):

$$\begin{aligned} \mathcal{F}\{R_x(\tau)\} &= \int_{-\infty}^{\infty} x(t)\left[\int_{-\infty}^{\infty} x^*(\lambda)e^{-j2\pi f(t-\lambda)}\,d\lambda\right]dt \\ &= \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft}\,dt \cdot \left[\int_{-\infty}^{\infty} x(\lambda)e^{-j2\pi f\lambda}\,d\lambda\right]^* \\ &= X(f)\,X^*(f) = |X(f)|^2 = \Psi_x(f) \end{aligned}$$

Hence

$$R_x(\tau) \;\xleftrightarrow{\ \mathcal{F}\ }\; \Psi_x(f) = |X(f)|^2$$

i.e. the auto-correlation function and the energy spectral density are a Fourier transform pair.

Check: putting $\tau = 0$ in the inverse transform, $R_x(0) = \int \Psi_x(f)\,df = \int |X(f)|^2df = E$, which is Rayleigh's energy theorem.

Example: for $x(t) = e^{-at}u(t)$, $R_x(\tau) = \frac{e^{-a|\tau|}}{2a}$ and $\Psi_x(f) = \frac{1}{a^2 + 4\pi^2f^2}$, which are indeed a Fourier pair.