Two analog signals x 1 (t) and x 2 (t) are to be transmitted over

Two analog signals x 1 (t) and x 2 (t) are to be transmitted over
| Two analog signals x1(t) and x2(t) are to be transmitted over a channel by means of time division multiplexing (TDM). The highest frequency of x1(t) = 4 kHz and that of x2(t) is 4.5 kHz. The minimum value of the sampling rate is:

A. 8 kHz

B. <p>9 kHz</p>

C. 4.5 kHz

D. 4 kHz

Please scroll down to see the correct answer and solution guide.

Right Answer is: B

SOLUTION

Concept:

Two signals when time division multiplexed must be sampled at a Nyquist frequency of the highest frequency signal.

i.e. if x1(t) is limited to fm1 and x2(t) is limited to fm2,

such that fm2 > fm1

then fm2 will decide the Nyquist rate.

Calculation:

Given, Highest frequency of x1(t) = 4 kHz

And highest frequency of x2(t) = 4.5 kHz

So, the minimum sampling frequency is decided by x2(t)

(Sampling Frequency) fs= 2.fm2

= 2 × 4.5 k

= 9 kHz

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