Question 4 of 10: FM Spectrum — Reading Modulation Index from a Bessel-Line Plot

Question 4: FM Spectrum — Reading Modulation Index from a Bessel-Line Plot (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. $A_c=100$ V; the printed one-sided spectrum $|\Phi_{FM}(f)|$ has 7 lines at 497–503 kHz, magnitudes 1, 5.5, 22, 38.5, 22, 5.5, 1; a table/plot of $J_n(\beta)$ is provided on the paper.

Find. (a) $\Delta f_{peak}$, $f_m$, $f_c$; (b) $K_f$ for $A_m=2.0$ V; (c) the rescaled spectrum for $A_m=4.0$ V, $f_m\to2f_m$.

Approach. The line spacing gives $f_m$ directly and the tallest line gives $f_c$; because the plot shows only $f\ge0$, each printed line height equals $(A_c/2)|J_n(\beta)|$ (half the two-sided coefficient, from splitting $\cos$ into $e^{\pm}$-exponentials) — match these against the provided Bessel table to read off $\beta$.

1. Part (a) — Δf_peak, f_m, f_c. Adjacent lines are spaced $(503-497)/6=1$ kHz apart, so $f_m=\boxed{1\ \text{kHz}}$. The tallest (centre) line sits at $f=500$ kHz, so $f_c=\boxed{500\ \text{kHz}}$. Reading each line as $(A_c/2)|J_n(\beta)|=50|J_n(\beta)|$ for $n=0,1,2,3$ and comparing against $50\,J_n(1.0)$ from the provided Bessel table ($J_0(1)=0.77,J_1(1)=0.44,J_2(1)=0.11,J_3(1)=0.02$) gives $50\times(0.77,0.44,0.11,0.02)=(38.5,22.0,5.5,1.0)$ — an exact match to the four printed amplitudes, so $$\beta=\boxed{1.0}.$$ The peak frequency deviation follows from $\beta=\Delta f_{peak}/f_m$: $$\Delta f_{peak}=\beta f_m=\boxed{1\ \text{kHz}}.$$
2. Part (b) — K_f. For single-tone FM, $\Delta f_{peak}=K_fA_m$, so $\beta=K_fA_m/f_m$. With $A_m=2.0$ V, $$K_f=\frac{\beta f_m}{A_m}=\frac{1\ \text{kHz}}{2.0\ \text{V}}=\boxed{500\ \text{Hz/V}}.$$
3. Part (c) — rescaled spectrum. With $A_m'=4.0$ V (doubled) and $f_m'=2f_m=2$ kHz (doubled), the new peak deviation is $\Delta f_{peak}'=K_fA_m'=500\times4.0=2000$ Hz, so the new modulation index is $$\beta'=\frac{\Delta f_{peak}'}{f_m'}=\frac{2000}{2000}=\boxed{1.0}\ \text{(UNCHANGED)}.$$ Because $\beta$ is the same, the Bessel envelope — and hence every line's amplitude $(A_c/2)J_n(\beta)$ — is identical to part (a). Only the LINE SPACING changes: sidebands now sit at $f_c\pm2,4,6$ kHz instead of $f_c\pm1,2,3$ kHz.