FM numerical: 10 kHz baseband, 100 MHz carrier, 75 kHz deviation

Given: $f_m = 10$ kHz (highest baseband frequency), $f_c = 100$ MHz, $\Delta f = 75$ kHz.

i) Carrier frequency swing and modulation index

The instantaneous frequency varies between

$$f_{max} = 100\ \text{MHz} + 75\ \text{kHz} = 100.075\ \text{MHz}, \qquad f_{min} = 99.925\ \text{MHz}$$ $$\text{Carrier swing} = 2\Delta f = 2 \times 75 = 150\ \text{kHz}$$

Modulation index (deviation ratio, since $f_m$ is the highest message frequency):

$$\beta = \frac{\Delta f}{f_m} = \frac{75}{10} = 7.5$$

ii) Practical bandwidth

By Carson's rule (contains about 98% of the FM power):

$$B_T = 2(\Delta f + f_m) = 2(75 + 10) = 170\ \text{kHz}$$

Check with Bessel functions: with $\beta = 7.5$, Carson's rule keeps sidebands up to $n = \beta + 1 \approx 8$; the power in these is $J_0^2 + 2\sum_{n=1}^{8}J_n^2(7.5) = 0.981$, i.e. 98.1% of the total.

If the stricter 1% rule is used (keep sidebands with $|J_n(\beta)| > 0.01$), $|J_{11}(7.5)| = 0.015$ and $|J_{12}(7.5)| = 0.005$, so $n_{max} = 11$ and

$$B = 2 n_{max} f_m = 2 \times 11 \times 10 = 220\ \text{kHz}$$

Answer: swing = 150 kHz, $\beta = 7.5$, practical (Carson) bandwidth = 170 kHz (220 kHz by the 1% rule).