GATE 2021 EC – Question 62
A message signal having peak-to-peak value of 2 V, root mean square value of 0.1 V and bandwidth of 5 kHz is sampled and fed to a pulse code modulation (PCM) system that uses a uniform quantizer. The PCM output is transmitted over a channel that can support a maximum transmission rate of 50 kbps. Assuming that the quantization error is uniformly distributed, the maximum signal to quantization noise ratio that can be obtained by the PCM system (rounded off to two decimal places) is ________.
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Correct answer: 30.57 to 30.87
Explanation
The sampling rate is at least $2\times5=10$ kHz, so a 50 kbps channel allows $\frac{50}{10}=5$ bits per sample, i.e. $L=32$ levels over the $2$ V range. The step size is $\Delta=\frac{2}{32}=0.0625$ V and the quantization noise power is $\frac{\Delta^2}{12}=3.255\times10^{-4}\ \text{V}^2$. The signal power is $0.1^2=0.01\ \text{V}^2$, so $\text{SQNR}=\frac{0.01}{3.255\times10^{-4}}=30.72$.