Question 4 of 10: Reading an FM Spectrum with Bessel Coefficients

Question 4: Reading an FM Spectrum with Bessel Coefficients (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. Carrier amplitude $A_c = 100$ V. From Figure 4, spectral lines at 497, 498, 499, 500, 501, 502 and 503 kHz with magnitudes 1, 5.5, 22, 38.5, 22, 5.5 and 1 respectively (shown for $f \ge 0$ only).

Find. $\Delta f_{\text{peak}}$, $f_m$ and $f_c$; then $K_f$ given $A_m = 2.0$ V; then the spectrum after doubling both $A_m$ and $f_m$.

[Figure not reproduced: Figure 4 (redrawn): line spacing gives fm = 1 kHz and the centre line gives fc = 500 kHz. Matching the heights against (Ac/2)|Jn(β)| fixes β = 1. See the official exam paper.]

Approach. The line spacing gives $f_m$ and the centre line gives $f_c$ by inspection; the relative line heights identify $\beta$ through the Bessel table, and $\Delta f = \beta f_m$ then follows.

1. Read $f_m$ and $f_c$ off the figure. The lines are equally spaced by 1 kHz, and in the single-tone FM expansion the components sit at $f_c + nf_m$, so the spacing is the modulating frequency: $f_m = 1$ kHz. The spectrum is symmetric about the tallest line at 500 kHz, which is the $n = 0$ (carrier) term, so $f_c = 500$ kHz.
2. Normalise the line heights to find $\beta$. The one-sided spectrum shows each cosine as a single line of height $A_c|J_n(\beta)|/2$, since a cosine of amplitude $A_cJ_n$ splits half its weight to each of $\pm f$. Dividing the measured heights by $A_c/2 = 50$: $$|J_0|=\frac{38.5}{50}=0.77,\quad |J_1|=\frac{22}{50}=0.44,\quad |J_2|=\frac{5.5}{50}=0.11,\quad |J_3|=\frac{1}{50}=0.02$$
3. Match against the supplied Bessel table. Scanning the table row by row, the entries $J_0 = 0.77$, $J_1 = 0.44$, $J_2 = 0.11$, $J_3 = 0.02$ are precisely the $\beta = 1.0$ row — all four orders agree simultaneously, so the identification is unambiguous: $$\boxed{\beta = 1.0}$$
4. Deduce the peak deviation. By the definition given in the question, $\beta = \Delta f/f_m$, so $$\Delta f_{\text{peak}} = \beta f_m = 1.0\times 1\ \text{kHz} = \boxed{1\ \text{kHz}}$$ Collecting part (a): $\Delta f_{\text{peak}} = 1$ kHz, $f_m = 1$ kHz, $f_c = 500$ kHz.
5. Part (b): modulation sensitivity. For single-tone FM the instantaneous frequency swings by $K_f$ hertz per volt of message, so the peak deviation is $\Delta f_{\text{peak}} = K_f A_m$. Solving for $K_f$ with $A_m = 2.0$ V: $$K_f = \frac{\Delta f_{\text{peak}}}{A_m} = \frac{1000\ \text{Hz}}{2.0\ \text{V}} = \boxed{500\ \text{Hz/V}}$$

Part (c) is the interesting one, because the two changes pull in opposite directions and it is the ratio that governs the shape of the spectrum.

6. Recompute the deviation and the index. Doubling the message amplitude to $A_m = 4.0$ V doubles the deviation, $\Delta f = K_fA_m = 500\times 4.0 = 2$ kHz, while doubling the modulating frequency gives $f_m = 2$ kHz. Therefore $$\beta = \frac{\Delta f}{f_m} = \frac{2\ \text{kHz}}{2\ \text{kHz}} = \boxed{1.0\ \text{(unchanged)}}$$
7. Sketch the consequence. Because $\beta$ is unchanged, every Bessel coefficient is unchanged, so the line heights are exactly as before: 38.3, 22.0, 5.7 and 1.0 (the small differences from the figure's 38.5/22/5.5/1 are just its reading precision). What changes is the spacing, which doubles to 2 kHz, so the lines now sit at 500, $500\pm 2$, $500\pm 4$ and $500\pm 6$ kHz.
8. Quantify the bandwidth growth. Carson's rule gives $B_T = 2(\Delta f + f_m)$, so the bandwidth grows from $2(1+1) = 4$ kHz to $$B_T = 2(2\ \text{kHz} + 2\ \text{kHz}) = \boxed{8\ \text{kHz}}$$ The signal occupies twice the spectrum while carrying the same number of significant sidebands — the classic demonstration that in FM it is the index, not the deviation alone, that sets the shape of the spectrum.