Question 5 of 6: Multitone FM — Message, Power, Peak Deviation, Bandwidth, Demodulator

Question 5: Multitone FM — Message, Power, Peak Deviation, Bandwidth, Demodulator (1/6 of paper)

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. $A_c=5$ V; $\phi(t)=a\cos(2\pi f_1t)+b\cos(2\pi f_2t)$, $a=1$, $b=2$, $f_1=2000$ Hz, $f_2=2f_1=4000$ Hz; frequency deviation constant $k_f=5$ kHz/V.

Find. (a) $m(t)$; (b) $P_{\text{avg}}$ of $s(t)$; (c) peak frequency deviation $\Delta f_{\text{peak}}$; (d) transmission bandwidth $B_T$; (e) a demodulator block diagram.

Approach. Relate the given phase $\phi(t)$ to the message through the FM instantaneous-frequency relation $\dfrac{1}{2\pi}\dfrac{d\phi}{dt}=k_fm(t)$; get power from the constant-envelope property of FM; get the peak deviation from the two tones' individual contributions (Carson's-rule sum, the standard multitone design bound); apply Carson's rule for bandwidth.

1. Part (a) — message signal. The instantaneous frequency deviation is $\Delta f(t)=\tfrac{1}{2\pi}\dfrac{d\phi}{dt}=-af_1\sin(2\pi f_1t)-bf_2\sin(2\pi f_2t)$, and by definition $\Delta f(t)=k_fm(t)$, so $$m(t)=\dfrac{-af_1\sin(2\pi f_1t)-bf_2\sin(2\pi f_2t)}{k_f}=\boxed{-0.4\sin(2\pi\cdot2000t)-1.6\sin(2\pi\cdot4000t)\ \text{V}}$$ (using $af_1/k_f=2000/5000=0.4$ and $bf_2/k_f=8000/5000=1.6$).
2. Part (b) — average power. $s(t)$ is a constant-envelope (angle-modulated) signal, so its average power does not depend on $\phi(t)$ at all: $$P_{\text{avg}}=\dfrac{A_c^2}{2}=\dfrac{25}{2}=\boxed{12.5\text{ W (per }\Omega\text{)}}$$
3. Part (c) — peak frequency deviation. Each tone contributes its own peak deviation, $\Delta f_1=af_1=1\times2000=2000$ Hz and $\Delta f_2=bf_2=2\times4000=8000$ Hz. The standard multitone-FM design value (Carson's rule) takes the peak deviation as the sum of the individual tone deviations — the conservative bound used for bandwidth allocation, since it is never exceeded even though the two sinusoids cannot both peak in the same direction at exactly the same instant: $$\Delta f_{\text{peak}}=\Delta f_1+\Delta f_2=2000+8000=\boxed{10\ \text{kHz}}$$ (A direct calculus maximisation of $|\Delta f(t)|$, done as a cross-check since $f_2=2f_1$ makes it a solvable single-variable problem, gives a true joint maximum of $9.44$ kHz — slightly below the $10$ kHz sum, confirming the sum is a safe, standard upper bound rather than an underestimate.)
4. Part (d) — bandwidth. Carson's rule with the highest message frequency $f_{m,\max}=f_2=4$ kHz: $$B_T=2\big(\Delta f_{\text{peak}}+f_{m,\max}\big)=2(10+4)=\boxed{28\ \text{kHz}}$$