Principles of Communication Engineering

Answer

Correct Option: (iii) Bit error rate

Explanation

BER = Bit Error Rate = bits in error / total bits transmitted, measure of link quality vs $E_b/N_0$.

Not binary error range, bit energy ratio ($E_b$), or bandwidth efficiency ($\eta$).

Final Answer

Bit error rate → Option (iii).

Answer

Correct Option: (ii) $4$

Explanation

QPSK (Quadrature PSK): $M=4$, phases $0°,90°,180°,270°$ or $45°,135°,225°,315°$, 2 bits/symbol ($00,01,10,11$).

Final Answer

$4$ phase states → Option (ii).

Answer

Correct Option: Depends on phrasing — typically $E_b/N_0$ (SNR) determines BER

Explanation

BER error probability $P_e=Q(\sqrt{2E_b/N_0})$ for BPSK, depends on $E_b/N_0$ (energy per bit to noise density), which relates to PSD $N_0/2$ and bandwidth efficiency. Higher $E_b/N_0$ → lower BER.

If options: power spectral density, bandwidth efficiency, SNR — SNR/$E_b/N_0$ is primary.

Final Answer

$E_b/N_0$ (SNR) → power spectral density related.

Answer

PSK Phase Shift Keying encodes bits in carrier phase, amplitude constant.

Principle: $s(t)=A\cos(\omega_c t+\theta(t))$, BPSK $\theta=0°$ for 0, $180°$ for 1. Generation: bits → NRZ ±1 → multiply carrier. Coherent detection correlates with reference.

Advantages over ASK:

	ASK	PSK
Amplitude	Varies → fading sensitive	Constant → robust
Power	Wasteful	Constant envelope efficient
BER	$Q(\sqrt{E_b/N_0})$	$Q(\sqrt{2E_b/N_0})$ 3dB better
Noise	Amplitude noise errors	Phase noise only

Final: PSK 3dB better, constant power, preferred for noisy links.

Answer

AWGN BER: BPSK coherent $P_e=Q(\sqrt{2E_b/N_0})$, distance $2\sqrt{E_b}$. BFSK orthogonal $P_e=Q(\sqrt{E_b/N_0})$, non-coherent $0.5e^{-E_b/2N_0}$. BPSK 3dB better than FSK. Derivation threshold 0 decision.

Answer

Digital Communication System: Source→Source Encoder (compression, Huffman)→Channel Encoder (error control, Hamming/conv, adds parity)→Modulator (ASK/PSK/FSK/QAM, digital→analog)→Channel (AWGN)→Demodulator→Channel Decoder→Source Decoder→Destination. Metrics $R_b$, BER, $E_b/N_0$. Source removes redundancy, channel adds controlled redundancy.

Answer

Analog vs Digital: Analog continuous, accumulates noise, low BW, simple, FDM, hard encryption. Digital discrete samples+quantized, regenerative threshold, high immunity, higher BW (PCM $n f_s$), complex codec, TDM efficient, easy encryption. Digital dominates modern due to robustness despite BW cost.

Answer

Nomenclature: Source/destination, message $m(t)$, bits/symbols, $R_b$ bps, baud $R_s=R_b/\log_2M$, bandwidth $B$ Hz, channel AWGN $N_0$, modulation ASK/PSK/QAM, BER $P_e$, SNR $E_b/N_0$, constellation I-Q, encoding source vs channel.

Answer

PSD $S(f)$: Power per Hz $P=\int S(f)df$, via autocorrelation $R(\tau)$. White $N_0/2$ flat, BPSK sinc. $\eta=R_b/B$ bps/Hz, Nyquist $\le2\log_2M$, Shannon $\le\log_2(1+SNR)$. PSD guides interference/filter, $\eta$ measures efficiency. Trade higher $\eta$ needs more power.