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(Solved): 5. An FM modulator has a frequency deviation constant of 5KHz/V. Assume that a message signal is in ...



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5. An FM modulator has a frequency deviation constant of . Assume that a message signal is input to the modulator as follows: , where is the step function. a) Determine the peak frequency deviation for the FM signal. b) Give a mathematical expression for the FM signal that can be evaluated and plotted if the carrier frequency is . Assume that the average power of the FM signal is 1. c) Give an approximate value for the bandwidth of the FM signal for each of the following three cases: (i) (ii) (iii) d) Suppose an FM signal has the form where is some baseband signal with bandwidth less than . If this signal is input to an demodulator give an expression for the demodulated signal.


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a) The peak frequency deviation for an FM signal can be determined using the formula: ?f = ? × A_max where ?f is the peak frequency deviation, ? is the frequency deviation constant, and A_max is the maximum amplitude of the message signal. In this case, the frequency deviation constant is given as 5 kHz/V. The maximum amplitude of the message signal can be determined by evaluating the given expression m(t) at t = 0, which gives: m(0) = 2u(0) - 4u(-T) + 2u(-2T) = 2 - 4(0) + 2(0) = 2 Therefore, A_max = 2. Substituting the values into the formula, we have: ?f = 5 kHz/V × 2 = 10 kHz So, the peak frequency deviation for the FM signal is 10 kHz. b) The mathematical expression for the FM signal can be given by: s(t) = Acos(2?f_c t + ??m(?)d?) where s(t) is the FM signal, A is the amplitude of the FM signal, f_c is the carrier frequency, ? is the frequency deviation constant, and m(t) is the message signal. In this case, the carrier frequency is given as 100 kHz, and the message signal is m(t) = 2u(t) - 4u(t - T) + 2u(t - 2T). Substituting the values into the expression, we have: s(t) =   
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