Question

A PAM telemetry system involves the multiplexing of four input signals s(t), i = 1, 2, 3, 4. Two of the signals s1(t) and s2(t) have bandwidths of 80 Hz each, whereas the remaining two signals s3(t) and s4(t) have bandwidths of 1 kHz each. The signals s3(t) and s4(t) are each sampled at the rate of 2400 samples per second. This sampling rate is divided by 2ᴿ (i.e., an integer power of 2) in order to derive the sampling rate for s1(t) and s2(t).

Find the maximum value of R.

Answer 1

The maximum value of R for dividing the sampling rate of 2400 samples per second by 2^R to find the sampling rate for signals s1(t) and s2(t) is 3, based on the Nyquist Theorem sampling criteria.

Step-by-step explanation:

A PAM (Pulse Amplitude Modulation) telemetry system involves multiplexing of four input signals s(t), i = 1, 2, 3, 4. Two of these signals, s1(t) and s2(t), have bandwidths of 80 Hz each, while s3(t) and s4(t) have bandwidths of 1 kHz each. The sampling frequencies for s3(t) and s4(t) are 2400 samples per second, which will be used to derive the sampling rate for s1(t) and s2(t) by dividing it by 2R, where R is an integer power of 2.

To find the maximum value of R, we start by considering the Nyquist Theorem, which states that a signal should be sampled at least twice its highest frequency to avoid aliasing. Therefore, since s1(t) and s2(t) have a maximum frequency of 80 Hz, they should be sampled at minimum 2×80 Hz = 160 samples per second.

Now, we must find the largest R such that 2400 / 2R ≥ 160. The values of R that satisfy this inequality are 0, 1, 2, 3, and 4. The maximum value of R that meets the criteria without going below the needed sampling rate is R = 4, because 2400 / 24 = 2400 / 16 = 150, which is less than the required 160 samples per second. Therefore, the maximum value of R is 3, since 2400 / 23 = 2400 / 8 = 300, which is greater than 160 samples per second.