Question 1

Q.1 (a) What is meant by Quantization? Explain the need of quantization of si…

Accepted Answer

Quantization is the process of mapping a continuous-amplitude sample to the nearest of a finite set of discrete amplitude levels, necessary to represent an analog signal with a finite number of bits for digital transmission/storage; this introduces quantization error (noise), whose mean-square value for a uniform quantizer with step size Δ is Δ²/12; for the given 3cos(2π×500t) signal with 10-bit PCM, the resulting signal-to-quantization-noise ratio is approximately 61.96 dB.

Question 2

Q.1 OR (a) With neat diagram, explain the adaptive delta modulation and demod…

Accepted Answer

Adaptive Delta Modulation (ADM) improves on basic delta modulation by dynamically varying the step size based on recent bit patterns, expanding the step size during periods of rapid signal change (reducing slope-overload distortion) and shrinking it during slowly-varying periods (reducing granular noise); calculating the PCM and DM signal-to-noise ratios for Pe=10⁻⁶ and n=8 gives PCM SNR ≈ 51.92 dB, substantially exceeding typical DM performance at a comparable bit rate, since PCM's SQNR grows with 2^(2n) while DM's is fundamentally more limited by its 1-bit-per-sample structure.

Question 3

Q.2 (a) Give the classification hierarchy of line encoding techniques. Discus…

Accepted Answer

Line encoding techniques are classified into unipolar, polar, and bipolar formats, each further divided into NRZ and RZ variants; unipolar NRZ is simplest but has poor DC content and no error detection, bipolar RZ improves synchronization and reduces DC bias, and bipolar AMI-RZ (alternate mark inversion) additionally provides simple error-detection capability and eliminates DC content entirely; for the quaternary PAM output waveform, the input binary sequence 0010110111 is grouped into dibits and mapped to one of four discrete amplitude levels per symbol.

Question 4

Q.2 OR (a) A matched filter has the frequency response H(f) = (1-e^(-j2πft))/…

Accepted Answer

For the given matched filter frequency response H(f)=(1-e^(-j2πft))/(j2πf), the impulse response h(t) is a rectangular pulse of unit amplitude and duration T (found via inverse Fourier transform), matched to a rectangular pulse signal of the same duration and amplitude; Inter-Symbol Interference (ISI) arises when the transmission channel's bandwidth limitation causes pulse spreading such that adjacent symbol pulses overlap and interfere at the receiver's sampling instants, degrading bit-error-rate performance and requiring pulse shaping (e.g., raised-cosine filtering) satisfying Nyquist's zero-ISI criterion to control.

Question 5

Q.3 (a) Draw the signal constellation of a binary FSK modulation scheme. Draw…

Accepted Answer

The signal constellation of binary FSK consists of two orthogonal signal points at distance √(2Eb) apart along two perpendicular axes (representing the two orthogonal frequency basis functions), rather than being distinguished purely by phase/amplitude as in PSK/ASK; coherent binary FSK is generated using two separate oscillators (or a single VCO) switched by the data and detected using two correlators/matched filters (one per frequency) followed by a comparator; for the given BPSK system (1 Mbps bipolar NRZ data, 100 MHz carrier), the symbol rate equals the bit rate at 1 Msps (since BPSK carries 1 bit/symbol), and the null-to-null bandwidth requirement is 2 MHz.

Question 6

Q.3 OR (a) Derive the relation for the average probability of symbol error fo…

Accepted Answer

The average symbol error probability for coherent M-ary PSK is approximately Pe ≈ 2Q(√(2Es/N0)·sin(π/M)) for M≥4, derived from the minimum Euclidean distance between adjacent constellation points in the M-ary PSK signal space; coherent detection requires the receiver to precisely reconstruct and synchronize with the transmitted carrier's phase reference (achieving optimal performance but requiring complex carrier recovery circuitry), while non-coherent detection avoids this requirement (simpler receiver) at the cost of a higher error probability for the same signal-to-noise ratio; QPSK's signal constellation comprises four equally-spaced points at 45°, 135°, 225°, 315° around a circle of radius √Es.

Question 7

Q.4 (a) State and explain Shannon-Hartley theorem on channel capacity and its…

Accepted Answer

The Shannon-Hartley theorem states that the maximum error-free channel capacity C (bits/s) of a channel with bandwidth B and signal-to-noise ratio S/N is C=B·log2(1+S/N), implying capacity can be traded between bandwidth and SNR, and that a channel with infinite bandwidth still has finite capacity limited by the noise floor; for the given telegraph source (dot=0.2s, dash=0.6s, P(dot)=2/3, P(dash)=1/3, 0.2s inter-symbol gap), the calculated information rate is approximately 1.72 bits/second.

Question 8

Q.4 OR (a) Describe trade-off between SNR and bandwidth of a channel capacity…

Accepted Answer

For a channel with capacity C held constant, bandwidth B and required signal-to-noise ratio S/N can be traded off according to C=B·log2(1+S/N), so reducing bandwidth requires an exponentially increasing S/N to maintain the same capacity, while increasing bandwidth allows the required S/N to be reduced (with diminishing returns as B grows large); for the given DMS with 5 symbols, the Shannon-Fano code achieves an average codeword length of 2.25 bits/symbol (efficiency ≈95.5%) while the Huffman code achieves 2.2 bits/symbol (efficiency ≈97.7%), confirming Huffman coding is at least as efficient as Shannon-Fano for the same source.

Question 9

Q.5 (a) What are Hamming codes? How many errors can be detected and corrected…

Accepted Answer

Hamming codes are a class of linear block codes with minimum Hamming distance 3, capable of detecting up to 2 errors and correcting exactly 1 error per code word, constructed such that each syndrome (from the parity-check matrix) uniquely identifies the single-bit error position; for the given parity-check matrix H (a (6,3) code), the generator matrix is derived as G=[I3|P], the codeword beginning 101 is found to be 101011, and the received word 110110 is decoded (via syndrome calculation identifying an error in bit position 2) to the corrected codeword 100110, giving decoded message 100.

Question 10

Q.5 OR (a) Discuss the error detecting and error correcting capabilities of c…

Accepted Answer

Convolutional codes provide continuous, memory-based error correction using a shift register encoder and Viterbi (maximum-likelihood sequence) decoding, with error correction capability governed by the code's free distance (the minimum Hamming distance between any two distinct encoded output sequences), generally providing effective error correction over the whole data stream rather than block-by-block, well-suited to channels with bursty or continuously-distributed errors; for the given (7,4) cyclic code with generator polynomial g(x)=1+x+x³, the systematic generator matrix is derived by polynomial division, giving the code word 1010001 for the message d=(1,0,1,0).

Digital Communication