# How to generate WBFM from NBFM?

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 $$k_f'$$.

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

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 $$\Delta f = 1.6\text{kHz}$$.