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A narrowband FM is approximately given by:

$ x(t)= A[\cos w_ct - k_f a(t)\sin w_ct] $

where

$ a(t)= \int _{ -\infty } ^t m(\alpha) d\alpha$

How does frequency multiplying $x(t)$ by a desired multiplier result in WBFM?

Please show the mathematical steps.

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  • $\begingroup$ what is "frequency multiplying $x(t)$"? $\endgroup$ Apr 20 '20 at 16:34
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An FM signal modulated by $m(t)$ is $x(t)=A_c\cos(\omega_c t+(k_m\int m(\alpha)d\alpha) t)$. This can be expanded as $$ x(t)=A_c\cos(\omega_c t)sin(\theta_m t)-A_c\sin(\omega_c t)\sin(\theta_m t) $$ , where $\theta_m = k_m\int m(\alpha)d\alpha$.

For NBFM, $k_m << 1$, so $\theta_m(t) << 1$, so we can approximate $$ x(t) \approx A_c\cos(\omega_ct)-A_c\theta_m(t)\sin(\omega_ct) $$ For generating WBFM, frequency multiplier are used to generate NBFM at $n\omega_c$ and $n\Delta f$ (frequency deviation). For example, for 8x multiplication, the signal is passed through square law device 3 times, with each stage having a BPF centered around 2 times center frequency after squaring. For squaring, $$ x(t)^2 = A_c^2\frac{1}{2}\Big[1+\cos(2\omega_c t)\Big]+A_c^2\theta_m^2(t)\frac{1}{2}\Big[1-\cos(2\omega_ct)\Big] -A_c^2\theta_m(t)\sin(2\omega_ct) $$ After BPF with center frequency around $2\omega_c$, $$ \tilde{x_2}(t) = \frac{A_c^2}{2}(1-\theta_m^2(t))\cos(2\omega_ct)-A_c^2\theta_m(t)\sin(2\omega_ct)\\ \approx \frac{A_c^2}{2}\cos(2\omega_ct)-\frac{A_c^2}{2}\Big(2\theta_m(t)\Big)\sin(2\omega_ct) $$ You can see that not only carrier frequency has doubled but modulation sensitivity has also doubled, resulting in doubling of frequency deviation $\Delta f$.

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