Question 2 of 6: Recovering the Modulating Signal from an Angle-Modulated Phase Function

Question 2: Recovering the Modulating Signal from an Angle-Modulated Phase Function (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The total instantaneous phase of the angle-modulated carrier, $\theta_i(t)=\varphi_{EM}(t)=10\cos(12{,}000t)$ rad, valid for $|t|\le1$; carrier angular frequency $\omega_c=10{,}000\ \text{rad/s}$; deviation constants $k_p=1000\ \text{rad/unit}$ (PM) and $k_f=1000\ \text{rad/s per unit}$ (FM).

Find. (a) $m(t)$ assuming $\theta_i(t)$ is a PM phase function; (b) $m(t)$ assuming instead it is an FM phase function.

Approach. Use the two standard definitions of instantaneous phase for angle modulation, $\theta_i(t)=\omega_c t+k_p\,m(t)$ for PM and $\theta_i(t)=\omega_c t+k_f\int_0^t m(\tau)\,d\tau$ for FM (equivalently $\dot\theta_i(t)=\omega_c+k_f\,m(t)$), and solve each algebraically for $m(t)$ using the one $\theta_i(t)$ given.

1. Part (a) — PM. By definition $\theta_i(t)=\omega_c t+k_p\,m(t)$, so $$m(t)=\frac{\theta_i(t)-\omega_c t}{k_p}=\frac{10\cos(12{,}000t)-10{,}000t}{1000}.$$ $$m(t)=\boxed{0.01\cos(12{,}000t)-10t},\qquad |t|\le1.$$
2. Part (b) — FM. For FM the modulating signal sets the instantaneous frequency deviation, so differentiate $\theta_i(t)$ first: $$\dot\theta_i(t)=\frac{d}{dt}\big[10\cos(12{,}000t)\big]=-120{,}000\sin(12{,}000t).$$ Then $\dot\theta_i(t)=\omega_c+k_f\,m(t)$ gives $$m(t)=\frac{\dot\theta_i(t)-\omega_c}{k_f}=\frac{-120{,}000\sin(12{,}000t)-10{,}000}{1000}.$$ $$m(t)=\boxed{-120\sin(12{,}000t)-10},\qquad |t|\le1.$$