GATE 2019 EC – Question 56
A voice signal $m(t)$ is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal $f(t)=A(1+m(t))\cos2\pi f_ct$, where $f_c=600$ kHz. The AM signal $f(t)$ is to be digitized and archived. This is done by first sampling $f(t)$ at 1.2 times the Nyquist frequency, and then quantizing each sample using a 256-level quantizer. Finally, each quantized sample is binary coded using $K$ bits, where $K$ is the minimum number of bits required for the encoding. The rate, in Megabits per second (rounded off to 2 decimal places), of the resulting stream of coded bits is ________ Mbps.
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Correct answer: 11.75 to 11.87
Explanation
The highest frequency of the AM signal is $600+15=615$ kHz, so the Nyquist rate is $1230$ kHz and the sampling rate is $1.2\times1230=1476$ kHz. A 256-level quantizer needs $K=8$ bits, so the rate is $1.476\times8=11.81$ Mbps.