--- title: "Working principle of PCM" subject: "Communication Systems" chapter: 4 key: "semester_6/communication_systems:pcm-working-principle" source: https://www.bctengineer.com/q/semester_6/communication_systems/pcm-working-principle site: BCT Engineering --- # Working principle of PCM Communication Systems · Chapter 4: Pulse Modulation - Asked in: 2073 Shrawan (CS II), 5 marks - Video: [PCM - PULSE CODE MODULATION - Block diagram, Waveform, Advantages, Disadvantages, Applications](https://www.youtube.com/watch?v=ojPJeEnY5oo) — Electronics Subjectified - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/pcm-working-principle - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4 ## Question Explain working principle of PCM with necessary figures and equations. ## Short answer PCM converts an analog signal to digital by anti-alias filtering, sampling at fs ≥ 2W, quantizing to L levels, and encoding each level to n = log₂ L bits. Bit rate Rb = n fs, minimum bandwidth BT = Rb/2. At the receiver, decoding and a reconstruction LPF recover the signal; regenerative repeaters prevent noise accumulation. ## Answer **Pulse Code Modulation (PCM)** converts an analog signal into a sequence of binary code words by sampling, quantizing and encoding; the bits are then sent as pulses. ```text Transmitter m(t)->[LPF]->[Sampler]->[Quantizer]->[Encoder]->line anti- fs>=2W L levels n bits alias Channel: --[Regenerative repeaters]--> Receiver -->[Regen]->[Decoder]->[Hold]->[Recon LPF]->m(t) ``` **Transmitter** 1. **Anti-aliasing LPF** limits the message to $W$ Hz. 2. **Sampling** at $f_s \ge 2W$ (flat-top, sample-and-hold), e.g. speech 3.4 kHz at 8 kHz. 3. **Quantization** rounds each sample to one of $L$ levels. For uniform steps: $$ \Delta = \frac{2m_p}{L}, \qquad N_q = \frac{\Delta^2}{12} $$ Non-uniform (µ/A-law) quantization is used for speech. 4. **Encoding** maps each level to an $n$-bit code word, $n = \log_2 L$. Bit rate and minimum bandwidth: $$ R_b = nf_s, \qquad B_T = \frac{R_b}{2} = nW\ (\text{at } f_s = 2W) $$ **Channel** - **Regenerative repeaters** placed along the line re-time, re-shape and re-decide the pulses, so noise does not accumulate (major advantage of PCM). **Receiver** 1. **Regeneration** of clean pulses. 2. **Decoding**: each code word is converted back to a quantized PAM sample. 3. **Reconstruction LPF** (cut-off $W$) recovers the analog signal, with only quantization noise. **Performance** $$ \text{SQNR} = 1.5L^2 \;\Rightarrow\; \text{SQNR}_{dB} = 1.76 + 6.02n\ \text{dB} $$ (full-load sinusoid). Each extra bit adds 6 dB but increases bandwidth. **Example:** telephony with $f_s = 8$ kHz, $n = 8$: $R_b = 64$ kbps, SQNR ≈ 49.9 dB. **Merits:** high noise immunity, regenerative repeaters, easy encryption and TDM, uniform digital format. **Demerits:** larger bandwidth, need for synchronization, quantization noise. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which sequence correctly describes the PCM transmitter? A. Quantizer, sampler, encoder, anti-aliasing LPF B. Anti-aliasing LPF, sampler, quantizer, encoder C. Sampler, anti-aliasing LPF, encoder, quantizer D. Encoder, quantizer, sampler, LPF Answer: B. The message is band-limited first, then sampled at fs >= 2W, quantized to L levels, and finally encoded into n-bit code words. (Key point: PCM steps) 2. Why does PCM tolerate noise on a long link better than analog systems? A. Regenerative repeaters re-time, re-shape and re-decide pulses so noise does not accumulate B. Quantization noise is zero in PCM C. PCM uses a smaller bandwidth than analog D. Amplifiers boost the noise together with the signal Answer: A. Repeaters decide each pulse afresh and retransmit a clean one, so channel noise is removed at each hop. (Key point: Regenerative repeaters) 3. For n bits per sample and sampling at fs = 2W, the bit rate and minimum bandwidth are: A. Rb = n/fs, BT = nW B. Rb = nfs, BT = 2nW C. Rb = nfs, BT = Rb/2 = nW D. Rb = fs/n, BT = W Answer: C. Rb = n fs, and the minimum bandwidth is Rb/2 = nW. (Key point: Bit rate and bandwidth) 4. In PCM, adding one extra bit per sample changes SQNR and bandwidth how? A. SQNR rises by about 6 dB and bandwidth also increases B. SQNR rises by 3 dB and bandwidth is unchanged C. SQNR rises by 6 dB with no bandwidth change D. SQNR is unchanged but bandwidth increases Answer: A. SQNR(dB) = 1.76 + 6.02n, so each bit gives 6 dB, but more bits per sample need a higher bit rate and bandwidth. 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