Communication Systems

A Comprehensive Course — From Foundations to Competitive Examinations

Prof. Mithun Mondal BITS Pilani, Hyderabad Campus EEE — Revision Notes GATE · University Exams

Part 1

Elements of a Communication System

Definition

A communication system conveys information from a source to a destination through a channel. Its basic blocks are: source → transmitter → channel → receiver → destination, with noise added in the channel and in the receiver front end.

Modulation is the process of varying a parameter (amplitude, frequency or phase) of a high-frequency carrier in accordance with a message signal \(m(t)\).

Why Modulate?

Practical antenna size: an antenna must be a sizeable fraction of a wavelength (\(\lambda/4\)); at 3 kHz, \(\lambda = 100\) km.
Frequency-division multiplexing of many signals on one channel.
Shifting the signal into a band where the channel and the hardware perform well.
Trading bandwidth for noise immunity (FM, spread spectrum).

Band	Frequency	Typical use
MF	300 kHz – 3 MHz	AM broadcasting
HF	3 – 30 MHz	Short-wave, sky-wave links
VHF	30 – 300 MHz	FM broadcasting, TV
UHF	300 MHz – 3 GHz	Mobile, Wi-Fi, GPS
SHF	3 – 30 GHz	Satellite, radar, microwave links

Part 2

Amplitude Modulation (AM)

\[ s(t) = A_c\left[1 + \mu\cos(2\pi f_mt)\right]\cos(2\pi f_ct), \qquad \mu = \frac{A_m}{A_c} = \frac{A_\text{max} - A_\text{min}}{A_\text{max} + A_\text{min}} \]

The spectrum has a carrier at \(f_c\) and two side-bands at \(f_c \pm f_m\). Bandwidth \(B = 2f_m\) (or \(2W\) for a message of bandwidth \(W\)).

Power relations — Most of the power is in the carrier

\[ P_t = P_c\left(1 + \frac{\mu^2}{2}\right), \qquad \eta = \frac{\mu^2}{2 + \mu^2}, \qquad I_t = I_c\sqrt{1 + \frac{\mu^2}{2}} \]

Maximum efficiency (at \(\mu = 1\)) is only 33.3 %. For several tones, \(\mu_t = \sqrt{\mu_1^2 + \mu_2^2 + \cdots}\).

\(\mu > 1\) causes over-modulation: envelope distortion that an envelope detector cannot undo.
Generation: square-law modulator, switching modulator, collector (high-level) modulation.
Detection: envelope detector, with time constant satisfying \(\frac{1}{f_c} \ll RC \ll \frac{1}{W}\); to avoid diagonal clipping, \(RC \le \frac{\sqrt{1 - \mu^2}}{2\pi f_m\mu}\).

Problem 1

A carrier of 10 kW is amplitude-modulated to a depth of 80 %. Find the total power and the efficiency.

Solution: \(P_t = 10(1 + 0.64/2) = 13.2\) kW; \(\eta = 0.64/2.64 = 24.2\,\%\).

Problem 2

The same 10 kW carrier is modulated simultaneously by two tones with \(\mu_1 = 0.6\) and \(\mu_2 = 0.8\). Find the total power.

Solution: \(\mu_t = \sqrt{0.36 + 0.64} = 1.0\); \(P_t = 10(1 + 0.5) = 15\) kW.

Part 3

DSB-SC, SSB and VSB

Scheme	Signal	Bandwidth	Detection
DSB-SC	\(m(t)\cos\omega_ct\)	\(2W\)	Coherent (Costas loop)
SSB	\(m(t)\cos\omega_ct \mp \hat m(t)\sin\omega_ct\)	\(W\)	Coherent
VSB	One side-band + vestige of the other	\(W + f_v\)	Coherent, or envelope with carrier
Conventional AM	\([A_c + m(t)]\cos\omega_ct\)	\(2W\)	Envelope detector

\(\hat m(t)\) is the Hilbert transform of \(m(t)\): every frequency component phase-shifted by −90°.
DSB-SC is generated by a balanced or ring modulator; SSB by the filter method or the phase-shift method.
Coherent detection with a local-oscillator phase error \(\phi\) scales the DSB-SC output by \(\cos\phi\) — it vanishes at \(\phi = 90^\circ\) (quadrature null effect). A frequency error \(\Delta\omega\) produces a beat \(\cos(\Delta\omega t)\).
VSB filter must satisfy \(H(f - f_c) + H(f + f_c) = \) constant over the message band; used for analogue TV video.

Note: SSB saves half the bandwidth and, compared with full AM at \(\mu = 1\), 83.3 % of the transmitted power.

Part 4

Angle Modulation — FM and PM

\[ \text{PM: } s(t) = A_c\cos[\omega_ct + k_pm(t)], \qquad \text{FM: } s(t) = A_c\cos\left[\omega_ct + 2\pi k_f\int m(\tau)\,d\tau\right] \]

For a single tone, \(s(t) = A_c\cos(\omega_ct + \beta\sin\omega_mt)\) with modulation index \(\beta = \Delta f/f_m\) (FM). The spectrum consists of the carrier and infinitely many side-bands at \(f_c \pm nf_m\) with amplitudes \(A_cJ_n(\beta)\).

Carson's rule — 98 % of the power lies within

\[ B_T \approx 2(\Delta f + f_m) = 2f_m(\beta + 1) \]

Narrow-band FM (\(\beta \ll 1\)): \(B_T \approx 2f_m\), spectrum like AM but side-band phases differ.
Total power is constant: \(P = A_c^2/2\), independent of \(\beta\).
Carrier vanishes when \(J_0(\beta) = 0\): \(\beta = 2.405,\ 5.52,\ 8.65\) — used to measure \(\Delta f\).
Frequency multiplication by \(n\) multiplies both \(f_c\) and \(\Delta f\) (and \(\beta\)) by \(n\); mixing changes \(f_c\) only.
Generation: direct (VCO, varactor) or indirect (Armstrong: NBFM + multipliers). Detection: slope / balanced discriminator, Foster–Seeley, ratio detector, PLL.
Pre-emphasis / de-emphasis (75 µs in commercial FM) improves SNR at high audio frequencies because FM output noise PSD rises as \(f^2\).

Problem 3

Commercial FM: \(\Delta f = 75\) kHz, \(f_m = 15\) kHz. Find \(\beta\) and the Carson bandwidth.

Solution: \(\beta = 5\); \(B_T = 2(75 + 15) = 180\) kHz (channels are allotted 200 kHz).

Part 5

Radio Receivers

The superheterodyne receiver mixes the incoming signal with a local oscillator to a fixed intermediate frequency where most gain and selectivity are obtained.

\[ f_\text{LO} = f_s + f_\text{IF}, \qquad f_\text{image} = f_s + 2f_\text{IF}, \qquad \alpha = \sqrt{1 + Q^2\rho^2},\ \ \rho = \frac{f_\text{im}}{f_s} - \frac{f_s}{f_\text{im}} \]

Standard IFs: 455 kHz (AM), 10.7 MHz (FM).
A high IF improves image rejection; a low IF improves selectivity and adjacent-channel rejection — hence double conversion.
Receiver figures of merit: sensitivity, selectivity, fidelity, image-frequency rejection ratio.

Problem 4

An AM receiver with IF 455 kHz is tuned to 1000 kHz. Find the LO frequency and the image frequency.

Solution: \(f_\text{LO} = 1455\) kHz; \(f_\text{image} = 1000 + 910 = 1910\) kHz.

Part 6

Noise

\[ P_n = kTB, \qquad v_{n,\text{rms}} = \sqrt{4kTRB}, \qquad kT_0 = -174~\text{dBm/Hz at } 290~\text{K} \]

Sources: thermal (Johnson), shot \(\big(i_n^2 = 2qI B\big)\), flicker (\(1/f\)), partition, atmospheric and cosmic noise.
White noise has PSD \(N_0/2\) (two-sided); after an ideal filter of bandwidth \(B\), noise power is \(N_0B\).
Noise figure \(F = \dfrac{(S/N)_\text{in}}{(S/N)_\text{out}}\); equivalent noise temperature \(T_e = (F - 1)T_0\).

Friis Formula

\[ F = F_1 + \frac{F_2 - 1}{G_1} + \frac{F_3 - 1}{G_1G_2} + \cdots \] The first stage dominates — use a low-noise, high-gain front end (LNA).

Problem 5

An LNA (NF 3 dB, gain 20 dB) precedes a mixer of NF 10 dB. Find the overall noise figure.

Solution: \(F = 2 + (10 - 1)/100 = 2.09 \Rightarrow 3.2\) dB.

Problem 6

Find the thermal noise power in a 1 MHz bandwidth at 290 K.

Solution: \(P_n = 1.38\times10^{-23}(290)(10^6) = 4.0\times10^{-15}\) W \(= -114\) dBm.

Noise Performance of Analogue Systems

System	Figure of merit \((S/N)_o / (S/N)_\text{baseband}\)
DSB-SC, SSB (coherent)	1
AM (envelope, tone)	\(\mu^2/(2 + \mu^2) \le 1/3\)
FM (tone, above threshold)	\(\tfrac32\beta^2\)

FM shows a threshold effect: below a carrier-to-noise ratio of about 10 dB the output SNR collapses. AM envelope detection shows a similar threshold.

Part 7

Sampling and Pulse Modulation

Sampling Theorem

A signal band-limited to \(W\) Hz is completely described by samples taken at \(f_s \ge 2W\) (the Nyquist rate). Sampling below this causes aliasing; an anti-aliasing low-pass filter is used before the sampler.

PAM (amplitude), PWM/PDM (width), PPM (position). Flat-top PAM causes aperture effect, corrected by an equaliser \(\propto 1/\text{sinc}(fT)\).
Band-pass sampling: a signal occupying \(f_L\)–\(f_H\) can be sampled at \(f_s = 2f_H/k\), \(k = \lfloor f_H/B \rfloor\).

Pulse-Code Modulation (PCM)

\[ R_b = nf_s, \qquad L = 2^n, \qquad \Delta = \frac{2m_\text{max}}{L}, \qquad \text{SQNR} = 6.02n + 1.76~\text{dB (full-scale sine)} \]

Each extra bit adds about 6 dB of SQNR and doubles the bandwidth.
Minimum transmission bandwidth \(B_T = R_b/2\) (Nyquist pulses).
Companding — μ-law (μ = 255, North America/Japan) and A-law (A = 87.6, Europe/India) — gives nearly constant SQNR over a wide range of signal levels.
DPCM quantises the prediction error; delta modulation uses one bit per sample. To avoid slope overload with a tone \(A\sin 2\pi f_mt\): \(\Delta f_s \ge 2\pi f_mA\). Too large a step causes granular noise.

Problem 7

Telephone speech (3.4 kHz) is sampled at 8 kHz with 8-bit PCM. Find the bit rate, the minimum bandwidth and the SQNR.

Solution: \(R_b = 64\) kb/s; \(B_T = 32\) kHz; SQNR \(= 6.02(8) + 1.76 = 49.9\) dB.

Multiplexing

FDM separates signals in frequency (guard bands); TDM in time (guard times, frame synchronisation). The E1 system carries 32 × 64 kb/s = 2.048 Mb/s (30 voice channels); T1 carries 24 channels + framing = 1.544 Mb/s.

Part 8

Baseband Transmission

Line codes: unipolar/polar NRZ, RZ, AMI (bipolar), Manchester (split-phase). Manchester has no DC and guarantees transitions, but needs twice the bandwidth.
Intersymbol interference (ISI) is avoided if the pulse satisfies Nyquist's criterion \(p(nT_b) = \delta[n]\).
Raised-cosine pulse with roll-off \(\alpha\): \(B_T = \dfrac{R_b}{2}(1 + \alpha)\).
The eye diagram shows ISI and noise margin: eye opening = noise margin, eye width = timing margin, slope = sensitivity to timing error.

Matched Filter

The filter \(h(t) = s(T - t)\) maximises output SNR at \(t = T\): \(\left(S/N\right)_\text{max} = 2E/N_0\), independent of pulse shape.

Part 9

Digital Modulation

Scheme	Bandwidth (null-to-null)	Bit-error probability (coherent, AWGN)
BASK (OOK)	\(2R_b\)	\(Q\left(\sqrt{E_b/N_0}\right)\)
BFSK	\(2R_b + \|f_1 - f_2\|\)	\(Q\left(\sqrt{E_b/N_0}\right)\)
BPSK	\(2R_b\)	\(Q\left(\sqrt{2E_b/N_0}\right)\)
QPSK	\(R_b\)	\(Q\left(\sqrt{2E_b/N_0}\right)\) (per bit, Gray)
DPSK	\(2R_b\)	\(\tfrac12 e^{-E_b/N_0}\)
Non-coherent FSK	—	\(\tfrac12 e^{-E_b/2N_0}\)

BPSK is 3 dB better than coherent BFSK/ASK (antipodal vs orthogonal signals).
QPSK sends 2 bits/symbol with the same BER as BPSK but half the bandwidth. MSK is continuous-phase FSK with \(\Delta f = R_b/2\).
\(M\)-ary schemes: bandwidth efficiency \(\log_2M\) bits/s/Hz (Nyquist); M-QAM is preferred to M-PSK for \(M \ge 16\) because its points are farther apart.

Problem 8

Find the BER of coherent BPSK at \(E_b/N_0 = 9.6\) dB.

Solution: \(E_b/N_0 = 9.12\); \(P_e = Q(\sqrt{18.24}) = Q(4.27) \approx 1\times10^{-5}\).

Part 10

Information Theory and Coding

\[ I(x_i) = \log_2\frac{1}{p_i}~\text{bits}, \qquad H(X) = -\sum_i p_i\log_2p_i, \qquad 0 \le H \le \log_2M \]

Information rate \(R = rH\) bits/s for \(r\) symbols/s.
Source coding: Shannon–Fano and Huffman codes approach the entropy; efficiency \(\eta = H/\bar L\).
Mutual information \(I(X;Y) = H(X) - H(X|Y)\); channel capacity \(C = \max I(X;Y)\). For a binary symmetric channel, \(C = 1 - H(p)\).

Shannon–Hartley theorem — The ultimate limit for an AWGN channel

\[ C = B\log_2\left(1 + \frac{S}{N}\right)~\text{bits/s}, \qquad \lim_{B\to\infty} C = 1.44\,\frac{S}{N_0}, \qquad \left(\frac{E_b}{N_0}\right)_\text{min} = -1.59~\text{dB} \]

Channel coding: an \((n,k)\) block code has rate \(k/n\); minimum distance \(d_\text{min}\) detects \(d_\text{min} - 1\) and corrects \(\lfloor (d_\text{min} - 1)/2\rfloor\) errors. Hamming codes: \(n = 2^m - 1\), \(d_\text{min} = 3\). Convolutional codes are decoded by the Viterbi algorithm.

Problem 9

A source emits four symbols with probabilities 1/2, 1/4, 1/8, 1/8. Find its entropy and a Huffman code.

Solution: \(H = 0.5 + 0.5 + 0.375 + 0.375 = 1.75\) bits/symbol. Codewords 0, 10, 110, 111 give \(\bar L = 1.75\) — 100 % efficient.

Problem 10

A telephone channel has \(B = 3.1\) kHz and SNR = 30 dB. Find its capacity.

Solution: \(C = 3100\log_2(1001) = 30.9\) kb/s.

Part 11

Quick Reference

Quantity	Formula
AM total power	\(P_c(1 + \mu^2/2)\)
AM efficiency	\(\mu^2/(2 + \mu^2)\)
FM index	\(\beta = \Delta f/f_m\)
Carson bandwidth	\(2(\Delta f + f_m)\)
Image frequency	\(f_s + 2f_\text{IF}\)
Friis	\(F_1 + (F_2 - 1)/G_1 + \cdots\)
PCM bit rate / SQNR	\(nf_s\); \(6.02n + 1.76\) dB
Raised-cosine bandwidth	\(R_b(1 + \alpha)/2\)
BPSK BER	\(Q(\sqrt{2E_b/N_0})\)
Shannon capacity	\(B\log_2(1 + S/N)\)

Part 12

Summary and Exam Tips

AM wastes power in the carrier but allows the simplest receiver; SSB saves both bandwidth and power.
FM trades bandwidth for SNR: improvement grows as \(\beta^2\) until the threshold is reached.
Digital systems are judged by \(E_b/N_0\) for a target BER and by bits/s/Hz.
Shannon's limit fixes what no modulation or coding scheme can exceed.

Exam tip: keep the side-band power \(P_c\mu^2/4\) per side-band, the image frequency \(f_s + 2f_\text{IF}\), and the 3 dB advantage of BPSK over coherent BFSK at your fingertips — they appear in GATE almost every year.