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ISRO (VSSC) Technical Assistant Electronics 2019 Official Paper

Option 1 : 90 Mbps

ST 1: Building Materials

1159

20 Questions
20 Marks
20 Mins

__Concept__:

The bandwidth of a PCM system for an encoded signal sampled at a frequency of fs is given by:

Bit rate = n fs

fs = Sampling frequency

n = number of bits used for encoding.

n is related to the number of quantization levels (L) as:

L = 2^{n}

or n = log_{2 }L

__Calculation__:

For a minimum bit rate, the modulating signal must be sampled at the minimum sampling rate, i.e. at the Nyquist Rate.

fs = 2f_{m}

fm = Maximum frequency present at the modulating signal.

∴ For the given bandlimited signal with a maximum frequency of 4.5 MHz, the sampling frequency will be:

fs = 2 × 4.5 M = 9 MHz

With L = 1024, the number of bits will be:

n = log_{2} 1024 = log_{2} 2^{10}

n = 10 bits

Now, the required minimum bit rate will be:

R_{b} = n f_{s} = 10 × 9 Mbps

Rb = 90 Mbps