How to generate WBFM from NBFM?

Question

An FM signal is given by

$$ s_{FM}(t) = A_c \cos [\omega_C t +k_f \int _{-\infty}^t m(\alpha)d \alpha] $$

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

Therefore, $ s_{FM}(t) = A_c \cos [\omega_C t +k_fa(t)]$

Now, for NBFM, $k_f$ is very small such that $|k_fa(t) << 1|$

We can then approximate $ s_{FM}(t)$ by

$ s_{FM}(t) \approx A_c[\cos \omega _c t - k_fa(t)\sin w_ct] $

We can generate an approximate NBFM at point (a). At point (b) we get the signal

$ x_b(t) = A_c \cos [n\omega_C t +nk_fa(t)]$. Where n is the multiplying factor.

At point (c) the signal is passed through a mixer and we get the signal

$ x_c(t) = A_c \cos [(n\omega_C - f_{crystal}) t +nk_fa(t)]$

After passing the signal through the frequency multiplier, we finally get the following signal at (d).

$ x_d(t) = A_c \cos [n_1(n\omega_C - f_{crystal}) t +n_1nk_fa(t)]$ , Where $n_1$ is the multiplying factor.

At point (d) since $k_f$ is multiplied by $n_1n$ the overall value of $k_f' (=nn_1k_f)$ increases. Thus, we are able to convert NBFM into WBFM by effectively increasing $k_f$ to $k_f'$.

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I wold like to know if this explanation is correct or not?

Answer 1

Yes, the explanation is correct. In a recent related QA (How to generate WBFM from NBFM), it is shown how $\times 2$ frequency multiplication doubles both the carrier frequency $\omega_c$ as well as frequency deviation $\Delta f$. So for a $\times 64$ is implemented as six stages of $\times 2$ multiplier. Similarly $\times 128$ is implemented as 7 stages of $\times 2$ multiplier. So (b) and (d) will result in $\times 64$ and $\times 128$ $\omega_c$ as well as $\Delta f$ multiplication. For frequency converter stage, it will change only $\omega_c$. Its output is $12.8 \pm 10.9\text{MHz}$. The $23.7\text{MHz}$ is rejected resulting in $1.9\text{MHz}$ carrier but the same $\Delta f = 1.6\text{kHz}$.