22-Elec-A3 Signals and Communications · December 2013

Question 2 of 6: Recovering the Message from an Angle-Modulated Signal

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Recovering the Message from an Angle-Modulated Signal (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. \(\varphi_{EM}(t) = 10\cos(12{,}000\,t)\) on \(|t|\le 1\); carrier \(\omega_c = 10{,}000\) rad/s; \(k_p = 1000\) rad/unit; \(k_f = 1000\) rad·s\(^{-1}\)/unit.

Find. The modulating signal \(m(t)\) on \(|t|\le 1\), first under the phase-modulation law and then under the frequency-modulation law.

Approach. Extract the total instantaneous angle \(\theta(t)\) from the given waveform, subtract the carrier term \(\omega_c t\) to isolate the modulation-dependent part, then invert whichever generation law applies — a direct proportionality for PM, an integral relation for FM.

1. Read the instantaneous angle off the given waveform. Writing the signal as \(\varphi_{EM}(t) = A\cos\theta(t)\) with \(A = 10\), $$\theta(t) = 12{,}000\,t \qquad (|t| \le 1)$$ The instantaneous frequency is the derivative of this angle, \(\omega_i(t) = \dot{\theta}(t) = 12{,}000\) rad/s — a constant, offset from the carrier by \(\Delta\omega = 12{,}000 - 10{,}000 = 2000\) rad/s. Every subsequent result flows from that constant 2000 rad/s offset; the two parts differ only in how each modulation law attributes it to \(m(t)\).
2. Invert the PM law (part a). For phase modulation the generated signal is \(A\cos[\omega_c t + k_p m(t)]\), so matching angles gives $$\omega_c t + k_p m(t) = 12{,}000\,t \;\Longrightarrow\; m(t) = \frac{(12{,}000-10{,}000)t}{k_p} = \frac{2000\,t}{1000}$$ $$\boxed{\,m(t) = 2t, \qquad |t| \le 1\,}$$ A linear ramp in the message is what produces a constant frequency offset in PM, because PM responds to the message amplitude but the instantaneous frequency responds to its derivative.
3. Invert the FM law (part b). For frequency modulation the instantaneous frequency itself is driven by the message, \(\omega_i(t) = \omega_c + k_f m(t)\). Equating to the measured 12,000 rad/s, $$10{,}000 + 1000\,m(t) = 12{,}000 \;\Longrightarrow\; \boxed{\,m(t) = 2, \qquad |t| \le 1\,}$$ Equivalently, from the angle form \(\omega_c t + k_f\int_{0}^{t} m(\alpha)\,d\alpha = 12{,}000t\) we get \(\int_0^t m = 2t\), and differentiating returns the constant \(m(t)=2\).
4. Cross-check both answers. Substituting each result back into its own generation law must reproduce the given 12,000 rad/s: for PM, \(\omega_i = \omega_c + k_p\,\dot{m} = 10{,}000 + 1000(2) = 12{,}000\ \checkmark\); for FM, \(\omega_i = \omega_c + k_f m = 10{,}000 + 1000(2) = 12{,}000\ \checkmark\).

The two answers are numerically different functions that produce an identical transmitted waveform, which is the pedagogical point of the question: a received angle-modulated signal cannot be demodulated correctly unless the receiver knows which modulation law the transmitter used.