Question

5. (15 pts) Consider the baseband signal m(t) and determine the instantaneous frequency range min (fmax) of the PM modulated signal. Assume kp-10π and the carrier frequency is c100 MHz. -1

Answer 1

In phase modulation system: Assume, cos = AC cos 2 pi Fc(t) (Because carrier signal before modulation) After modulation: S_pm = Ac cos[2 pi F)c + phi(t)] where, phi(t) = K_p m(t) SP_m = A_C cos[2 pi F_ct + k_pm(t)] \r\n Instantaneous phase theta_i = [2 pi f_c t + K_p m(t)] Instantaneous Frequency f_i = 1/2 pi middot d theta_i(t)/dt = F_c + Kp/2 pi d/dt m(t) d/dt m(t) \r\n Slope of m(t) = 1 - (-1)/10^-4 = 2 times 10^4 F_i = f_c Kp/2 pi middot d/dt m(t) where, F_c = 100 MHz Kp = 10 pi F_i Vertical-bar _max = F_c + Kp/2 pi [d/dt m(t)]_max = 100 MHz + 10 pi/2 pi times 2 times 10^4 = 100 MHz + 10^5 Hz = 100 MHz + 0.1 MHz Therefore F_i_max = 100.1 MHz \r\n F_i Vertical-bar _min = F_c + Kp/2 pi [d/dt m(t)]_min = 100 MHz + 10 pi/2 pi times [-2 times 10^-4]