Determine (a) the peak frequency deviation, (b) minimum bandwidth, and (c) baud for a binary FSK signal with a mark frequency of 38 kHz, a space frequency of 40 kHz, and an input bit rate of 4 kbps

The peak frequency deviation, minimum bandwidth, and baud for a binary FSK signal for the given frequencies are respectively;

a) 0.5 kHz

b) 9 kHz

c) 4000

1) The peak frequency deviation is gotten from the formula;

∆f = |f_m - f_s|/f_b

where;

Thus;

∆f = |38 - 40|/4

∆f = 0.5 kHz

2) The minimum bandwidth is given by the formula;

B = 2(∆f + f_b)

B = 2(0.5 + 4)

B = 9 kHz

3) For FSK signal, N = 1, and the baud is gotten from the Equation;

baud = f_b/1

f_b = 4 kbps = 4000 bps

Thus; baud = 4000/1 = 4000

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explanation: