Question

Prob. 3 An FM modulator output signal is given by Prm 5 cos(2nft + 2nk, m(t)dr). where k, = 30 H2/V. Assume that the message signal is the rectangular pulse m(1) = 8rect «- 2) (a) Sketch the phase deviation in radians. (b) Sketch the frequency deviation in Hz. (c) Determine the peak frequency deviation in Hz. (d) Is this NBFM or WBFM?

Answer 1

Although, In this question, we need to sketch the output waveform, let's discuss the concept first for better understanding.

An FM modulator O/P is given, Kf is given, which is frequency deviation sensitivity.

Now, FM modulation, Modulation index = phase deviation(Kp) and Frequency Deviation $\Delta$ f = Kf*Am =  30*8 = 240 Hz, where Am = 8V

And Modulation index = $\Delta$ f / fm, = 240*4 = 960 where fm = frequency of modulating wave/ message signal = 1/Time periode= 1/4 Hz

Therefore, Phase deviation = 960 radians

Note: In the case of FM modulator phase deviation will be zero.

(d) Ans: As Modulation index >> 1, Therefore, it is WBFM or Wide Band frequency Modulation

Note: Though practically the frequency of rectangular message signal is infinite, but we will take the frequency as (1/ pulse width of rectangular pulse) using a Low Pass filter, using a cut-off frequency of (1/T). As 99% of Energy lies between this bandwidth.

Therefore, Peak frequency deviation for this modulated wave will be = fc $\pm$ 240 Hz

redding modulok (e - 240 Hz PHEE DEVIHTI ON zero