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Hub intro
This hub covers analog and digital communication systems, modulation schemes, noise analysis, and information theory. Work through the sets in order to build from foundational definitions up to GATE-level problem solving.
Difficulty levels
BeginnerIntermediateAdvanced
Practice sets
Beginner: Definitions and Direct Formulas
BeginnerDefine amplitude modulation (AM) and write the expression for an AM signal.
Answer: AM is a modulation technique where the amplitude of the carrier wave varies in proportion to the instantaneous value of the message signal. The AM signal is: s(t) = A_c [1 + k_a m(t)] cos(2πf_c t), where A_c is the carrier amplitude, k_a is the amplitude sensitivity, and m(t) is the message signal.
In AM, the carrier frequency and phase remain constant. Only the envelope changes. k_a m(t) is the normalised message term. When k_a * max|m(t)| = 1, the modulation index μ = 1 (100% modulation). Values above 1 cause overmodulation and envelope distortion.
State the Nyquist sampling theorem and give the minimum sampling rate for a 4 kHz audio signal.
Answer: The Nyquist theorem states that a bandlimited signal with maximum frequency f_m must be sampled at a rate f_s >= 2 f_m to allow perfect reconstruction. For f_m = 4 kHz, the minimum sampling rate is f_s = 8 kHz.
Sampling below 2 f_m causes aliasing: higher-frequency components fold back into the baseband and corrupt the reconstructed signal. The factor 2 f_m is the Nyquist rate. In practice, anti-aliasing filters are applied before sampling and f_s is set slightly above this limit.
What is the bandwidth of an FM signal according to Carson's rule?
Answer: Carson's rule gives BW_FM = 2(Δf + f_m), where Δf is the maximum frequency deviation and f_m is the maximum message frequency.
Carson's rule approximates that about 98% of the total FM signal power lies within this bandwidth. The modulation index for FM is β = Δf / f_m. Narrowband FM (β << 1) has bandwidth close to 2 f_m, similar to AM. Wideband FM (β >> 1) has bandwidth approximately 2Δf.
Define signal-to-noise ratio (SNR) and express it in decibels.
Answer: SNR is the ratio of signal power S to noise power N: SNR = S/N. In decibels: SNR_dB = 10 log_10(S/N).
SNR quantifies how much stronger the signal is compared to background noise. A higher SNR means cleaner reception. For example, SNR = 100 corresponds to SNR_dB = 20 dB. SNR is measured at a specific point in the system, such as at the receiver input or output.
Intermediate: Multi-step Problems
IntermediateAn AM signal has carrier power 100 W and modulation index μ = 0.6. Calculate the total power and the power in each sideband.
Answer: Total power P_T = P_c (1 + μ²/2) = 100 (1 + 0.36/2) = 100 × 1.18 = 118 W. Each sideband power = P_c μ²/4 = 100 × 0.09 = 9 W.
The AM signal power splits into three components: the carrier at power P_c, the upper sideband at P_c μ²/4, and the lower sideband at P_c μ²/4. Total sideband power = P_c μ²/2 = 18 W. The carrier carries no information; only the sidebands do. Efficiency = (P_c μ²/2) / P_T = 18/118 ≈ 15.3%.
A PCM system uses 8-bit quantisation for a signal with bandwidth 3.4 kHz. Find the minimum bit rate and the bandwidth required for transmission.
Answer: Sampling rate f_s = 2 × 3400 = 6800 samples/s. Bits per sample = 8. Bit rate R_b = 6800 × 8 = 54,400 bps = 54.4 kbps. Minimum transmission bandwidth B_T = R_b / 2 = 27.2 kHz.
Each sample is encoded into n = 8 bits, giving 2^8 = 256 quantisation levels. The bit rate is the product of the sampling rate and bits per sample. Nyquist's criterion for baseband transmission requires a channel bandwidth of at least R_b / 2 to support the bit rate without inter-symbol interference.
The capacity of a noisy channel (AWGN) is C = B log_2(1 + S/N). A channel has bandwidth B = 4 MHz and SNR = 15. Find the channel capacity.
Answer: C = 4 × 10^6 × log_2(1 + 15) = 4 × 10^6 × log_2(16) = 4 × 10^6 × 4 = 16 Mbps.
This is the Shannon-Hartley theorem. log_2(16) = 4 because 2^4 = 16. The result, 16 Mbps, is the theoretical maximum error-free data rate for this channel. No practical coding scheme can exceed this limit. Doubling the bandwidth doubles C, but doubling SNR increases C by only log_2(1 + 2 SNR) / log_2(1 + SNR), which is a diminishing return at high SNR.
Derive the figure of merit (SNR_o / SNR_c) for DSB-SC demodulated using a coherent detector.
Answer: For DSB-SC with coherent detection, SNR_o / SNR_c = 1, meaning the output SNR equals the channel SNR. DSB-SC achieves the same figure of merit as baseband transmission.
SNR_c is the carrier-to-noise ratio at the receiver input computed with carrier power S_T and noise in bandwidth B_T. After coherent demodulation and low-pass filtering, the signal power is S_T/2 and the noise power is N_0 f_m. Dividing gives SNR_o = S_T / (2 N_0 f_m), which equals SNR_c. Compare with standard AM where figure of merit = μ² / (2 + μ²), always less than 1.
Advanced: GATE-style Questions
AdvancedA message signal m(t) = 2 cos(2π × 1000t) frequency-modulates a carrier. The frequency sensitivity k_f = 5000 Hz/V. Determine β, classify the FM type, and find the bandwidth using the Bessel function table given that significant sidebands exist for β = 10.
Answer: Δf = k_f × A_m = 5000 × 2 = 10,000 Hz = 10 kHz. f_m = 1000 Hz. β = Δf / f_m = 10,000 / 1000 = 10. This is wideband FM (β >> 1). From Bessel tables, for β = 10, there are approximately 14 significant sideband pairs. BW = 2 × 14 × f_m = 2 × 14 × 1000 = 28 kHz.
The trap here is to use Carson's rule without checking whether Bessel analysis gives a different answer for large β. Carson's rule gives BW = 2(10 + 1) kHz = 22 kHz, which underestimates the true bandwidth for large β. GATE problems sometimes supply a Bessel table and expect the student to count significant coefficients (those above 1% of unmodulated carrier amplitude). For β = 10, the 14th-order Bessel coefficient J_14(10) ≈ 0.014, still significant, hence 14 pairs.
A binary symmetric channel (BSC) has crossover probability p = 0.1. Find the channel capacity.
Answer: C = 1 + p log_2(p) + (1-p) log_2(1-p) = 1 + 0.1 log_2(0.1) + 0.9 log_2(0.9). log_2(0.1) ≈ -3.322, log_2(0.9) ≈ -0.152. C = 1 + 0.1(-3.322) + 0.9(-0.152) = 1 - 0.3322 - 0.1368 = 0.531 bits/use.
The BSC capacity formula uses binary entropy: C = 1 - H(p), where H(p) = -p log_2(p) - (1-p) log_2(1-p). A common trap is to confuse this formula with the AWGN Shannon-Hartley theorem or to use log base 10 instead of log base 2. C = 0 when p = 0.5 (completely random channel) and C = 1 when p = 0 or p = 1 (perfectly deterministic).
In a QPSK system, each symbol carries 2 bits. The symbol rate is 10 Msymbols/s. The E_b/N_0 required for BER = 10^-5 is 9.6 dB. If the signal power is 1 mW and the noise PSD is N_0/2, find the required noise PSD N_0.
Answer: Bit rate R_b = 2 × 10 × 10^6 = 20 Mbps. E_b/N_0 = 9.6 dB = 10^(9.6/10) ≈ 9.12. E_b = P / R_b = 10^-3 / (20 × 10^6) = 5 × 10^-11 J/bit. N_0 = E_b / 9.12 = 5 × 10^-11 / 9.12 ≈ 5.48 × 10^-12 W/Hz.
E_b is the energy per bit. In QPSK, bits per symbol = 2, so the bit rate is twice the symbol rate. The trap is to use symbol energy E_s instead of bit energy E_b. E_s = 2 E_b in QPSK. Also, N_0/2 is the two-sided PSD; BER formulas for QPSK use E_b/N_0 with one-sided N_0. Always confirm which convention the formula uses before substituting.
Lab exercises
- Simulate AM modulation and demodulation in MATLAB and measure the modulation index from the envelope: lab-am-modulation-matlab
- Generate a PCM encoded bitstream for a sinusoidal input and reconstruct the signal, plotting quantisation error: lab-pcm-quantisation
- Plot BER vs E_b/N_0 curves for BPSK and QPSK using Monte Carlo simulation in Python: lab-ber-simulation
Revision checklist
- Can you derive the power efficiency formula for standard AM and compute it for μ = 0.5 without looking at notes?
- Can you state Carson's rule and explain when it overestimates or underestimates FM bandwidth?
- Can you apply the Shannon-Hartley theorem to find the minimum SNR needed to achieve a target data rate over a given bandwidth?
- Can you write the BSC capacity formula and evaluate it for p = 0, p = 0.5, and p = 1?
- Can you explain the difference between coherent and non-coherent detection and state which gives better SNR performance?
- Can you convert between E_b/N_0 in dB and linear, and relate it to BER for BPSK?