GATE 2026 IN Q.27: solution and answer

GATE 2026 · INQ.27 · 1 mark · MCQ

An analog speech signal contains signals from 200 Hz to 2400 Hz. It is sampled at 6 kHz and quantized with 512 levels for pulse code modulation (PCM). The bit rate will be _____ kbps.

(A)
16
(B)
48
(C)
19.2
(D)
54

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Answer and solution

ANSWER

The PCM bit rate is $R_{b} = 9 \times 6 kHz = 54 kbps$ , which is option (D).

Official answer: DSource: Official answer key, GATE 2026 IN (IIT Guwahati)

The sampling rate is given directly as $f_{s} = 6 kHz$ . It is above the Nyquist rate $2 \times 2400 = 4.8 kHz$ , so the stated rate is used as it is.
Bits per sample for $L$ quantization levels: $n = lo g_{2} L = lo g_{2} 512 = 9$ bits.
The PCM bit rate is the number of bits per sample times the number of samples per second: $R_{b} = n f_{s}$ .
$R_{b} = 9 \times 6000 = 54000 bits/s = 54 kbps$
This matches option (D).

KEY IDEA

The PCM bit rate is the sampling rate times the bits per sample, $R_{b} = f_{s} lo g_{2} L$ . Use the sampling rate the question gives, not the Nyquist rate.

WATCH OUT FOR

Using the Nyquist rate $2 f_{m} = 4.8 kHz$ instead of the stated 6 kHz gives $9 \times 4.8 = 43.2$ kbps, which is not an option. Counting 512 levels as 512 bits per sample, or as 8 bits, is also wrong. The bits per sample are $lo g_{2} 512 = 9$ .

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GATE 2026 IN Q.27: solution and answer

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