Question

You are given an analog-to-digital converter (ADC) which is used to convert an input analog signal to a digital signal. The highest frequency present in the input analog signal is 20 Hz. The ADC samples the input analog signal at 1.25 times the minimum required sampling rate. (Note; this ADC “over-samples” the analog input signal; the sampling rate of this ADC is higher than the minimum required sampling rate. The ADC samples at 1.25 times the minimum required sampling rate). Each sample is then quantized (“measured") by comparing the amplitude of each sample to a standard set of 65,536 levels. Determine the total number of bits generated in a four-minute interval. (20 points)

Answer 1

Solution:

Highest frequency content in the input analog signal = 20Hz

As per the sampling theorem, the minimum sampling frequency required is twice the frequency of the signal = 2* 20 = 40 Hz.

But, ADC samples the 1.25 times the minimum sampling rate = 1.25 * 40 = 50Hz.

One second, the number of samples being taken = 50.

each sample is quantized to no of levels = 65536.

The number of bits used to code one sample value = 2n <= 65536

where n is the number of bits.

Here n =16 bits.

So, each sample is coded by using 16 bits.

So, for every one second, 50 samples are being sampled.

So, the number of bits being coded in one second = 50 * 16 bits = 800 bits.

In Four minutes interval (4*60sec), the number of bits generated = 4*60*800bits = 192,000 bits.