Lite Explanation – Why This Topic is Important for ECET
PCM and Delta Modulation are core topics of Digital Communication. They are asked in ECET 2026 because:
- They test both conceptual clarity (quantization, coding, sampling) and numerical skills (bit rate, SNR, quantization error).
- PCM is the foundation of all digital telephony, audio, and video systems.
- Delta Modulation is a simpler method of analog-to-digital conversion → ECET often asks comparison questions.
By mastering PCM & DM, you get high-scoring opportunities with direct formula-based and conceptual MCQs.
📘 Concept Notes
1. Pulse Code Modulation (PCM)
- PCM is a method to digitize an analog signal.
- Steps:
- Sampling – Convert continuous signal into discrete time.
- Quantization – Approximate each sample to the nearest level.
- Encoding – Represent quantized levels into binary code.
Quantization Error (Qe):
where 
= step size.
Signal-to-Quantization-Noise Ratio (SQNR):
where 
= number of bits per sample.
Example: If 8-bit PCM is used, SQNR ≈ 49.9 dB.
2. Delta Modulation (DM)
- Instead of encoding absolute sample values, DM encodes the difference between successive samples.
- Only 1 bit per sample (up or down).
- Advantages: Simpler, less bandwidth.
- Disadvantages:
- Slope overload (if step size too small).
- Granular noise (if step size too large).
Step Size Relation:
where 
= slope, 
= sampling interval.
3. Difference Between PCM & DM
| Feature | PCM | Delta Modulation |
|---|---|---|
| Bits per sample | n bits (multi-bit) | 1 bit only |
| Quality | High, less noise | Lower, prone to noise |
| Complexity | High (quantizer + encoder) | Simple |
| Bandwidth | Higher | Lower |
⚙️ Formulas
- Quantization Error:
SQNR for PCM:
Bit Rate:
where 
= sampling frequency.
Nyquist Rate:
where 
= maximum signal frequency.
Step Size in DM:
🔟 10 MCQs
Q1. In PCM, quantization error is:
a) ±Δ
b) ±Δ/2
c) Δ
d) 2Δ
Q2. For 8-bit PCM, the SQNR is approximately:
a) 24 dB
b) 36 dB
c) 50 dB
d) 60 dB
Q3. A voice signal of 4 kHz is sampled at Nyquist rate and encoded with 8 bits per sample. Find the bit rate.
Q4. Which modulation technique uses 1 bit per sample?
a) PCM
b) DPCM
c) Delta Modulation
d) AM
Q5. In PCM, increasing number of quantization levels leads to:
a) Higher error
b) Lower SQNR
c) Higher SQNR
d) No effect
Q6. If Δ is too small in Delta Modulation, the system suffers from:
a) Granular noise
b) Slope overload
c) Aliasing
d) Quantization error
Q7. The minimum sampling frequency for a 5 kHz signal is:
a) 2.5 kHz
b) 5 kHz
c) 10 kHz
d) 20 kHz
Q8. In PCM, the bit rate is given by:
a) n / fs
b) fs / n
c) n × fs
d) n + fs
Q9. Delta Modulation requires:
a) Quantizer + Encoder
b) Only 1-bit quantizer
c) No sampling
d) Frequency modulation
Q10. Which of the following is a disadvantage of PCM compared to DM?
a) High SQNR
b) High bandwidth requirement
c) Less accuracy
d) Simpler hardware
✅ Answer Key
| Q No | Answer |
|---|---|
| Q1 | b |
| Q2 | c |
| Q3 | 64 kbps |
| Q4 | c |
| Q5 | c |
| Q6 | b |
| Q7 | c |
| Q8 | c |
| Q9 | b |
| Q10 | b |
🧠 Explanations
- Q1: Quantization error lies within ±Δ/2 → (b).
- Q2: SQNR = 6.02n + 1.76 = 6.02×8 + 1.76 = 49.9 dB → ≈ 50 dB → (c).
- Q3: Nyquist rate = 2×4k = 8 kHz. Bit rate = 8×8k = 64 kbps → (64 kbps).
- Q4: DM encodes with 1 bit per sample → (c).
- Q5: More levels → smaller quantization step → higher SQNR → (c).
- Q6: If Δ is too small, input slope > Δ → slope overload → (b).
- Q7: Nyquist sampling → 2×5k = 10 kHz → (c).
- Q8: PCM bit rate = n×fs → (c).
- Q9: DM uses a 1-bit quantizer → (b).
- Q10: PCM needs more bandwidth compared to DM → (b).
🎯 Motivation / Why Practice Matters
PCM & DM problems are direct scoring areas in ECET 2026:
- Most questions are formula-based, can be solved in under 1 minute.
- Conceptual clarity avoids silly mistakes (e.g., Nyquist rate confusion).
- Practicing SQNR, bit rate, and error formulas gives a competitive edge in exam speed and accuracy.
📲 CTA
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