Principles of Communication Engineering

Answer

Correct Option: (iii) $16$ kHz

Explanation

Nyquist Sampling Theorem: $f_s \ge 2f_m$ to avoid aliasing.

Given $f_m = 8$ kHz → $f_s^{min}=2\times8=16$ kHz.

Sampling below 16 kHz causes aliasing (foldover), reconstruction fails.

Final Answer

$16$ kHz → Option (iii).

Answer

Correct Option: (ii) Higher noise immunity

Explanation

PCM quantizes PAM to binary, regenerative repeaters restore levels via threshold, noise does not accumulate. Analog PAM/AM accumulates noise.

PCM requires more bandwidth ($n f_s$) and complex codec, but SNR greatly higher ($\approx6n$ dB). Not less bandwidth, not simpler.

Final Answer

Higher noise immunity → Option (ii).

Answer

Correct Option: (iii) PPM

Explanation

PAM varies amplitude, PWM varies width, PPM varies position of pulse relative to reference. PPM constant amplitude → better noise immunity vs PAM.

Final Answer

PPM → Option (iii).

Answer

PAM: Sample amplitude ∝ $m(t)$, generation via FET switch, BW $f_s$, noise-sensitive. PPM: Pulse position varies, constant amplitude, better than PAM, needs sync. PCM: PAM→quantize→binary (8-bit), BW $n f_s$ large, high immunity, TDM. Applications: PAM backbone, PPM telemetry, PCM telephony 64kbps, CD. Table BW $f_s$, $f_s$, $n f_s$; noise low, medium, high.

Answer

PAM: Sample via FET switch (natural) or S/H (flat-top). Detection via LPF $f_s/2$ with $\text{sinc}$ equalizer. Spectrum replicated at $nf_s$ weighted $\text{sinc}$. Applications intermediate for PCM, TDM.

Answer

Comparison PAM/PPM/PCM: PAM amplitude BW $f_s$ low noise, PPM position BW $f_s$ medium, PCM binary $n f_s$ high immunity. PAM TDM backbone, PPM optical/control, PCM telephony 64kbps CD. PCM most robust but BW heavy.