Generate FM (50 MHz, β = 1) from NBFM

Idea: a frequency multiplier multiplies both the carrier and the deviation (so it multiplies $\beta$); a mixer shifts only the carrier and leaves the deviation unchanged.

Given NBFM: $f_{c1} = 1$ MHz, $\beta_1 = 0.1$, $f_m = 10$ kHz

$$\Delta f_1 = \beta_1 f_m = 0.1 \times 10\ \text{kHz} = 1\ \text{kHz}$$

Required FM: $f_c = 50$ MHz, $\beta = 1$

$$\Delta f = \beta f_m = 1 \times 10\ \text{kHz} = 10\ \text{kHz}$$

Step 1: Multiplication factor. The mixer does not change $\beta$, so the multiplier must give

$$n = \frac{\beta}{\beta_1} = \frac{1}{0.1} = 10$$

After a $\times 10$ multiplier:

$$f_{c2} = 10 \times 1\ \text{MHz} = 10\ \text{MHz}, \qquad \Delta f_2 = 10 \times 1\ \text{kHz} = 10\ \text{kHz}$$

Step 2: Frequency translation. The carrier is now 10 MHz but 50 MHz is needed. Mix with a local oscillator and select the sum frequency with a band-pass filter:

$$f_{LO} + 10\ \text{MHz} = 50\ \text{MHz} \;\Rightarrow\; f_{LO} = 40\ \text{MHz}$$

(Choosing the difference, $f_{LO} = 60$ MHz, also works.) The deviation stays 10 kHz.

NBFM      +-----+ 10 MHz  +-------+  +-----+ FM out
1 MHz --->| x10 |-------->| Mixer |->| BPF |------->
df=1kHz   +-----+ df=10kHz+-------+  |50MHz| 50 MHz
b=0.1                         ^      +-----+ df=10kHz
                         LO 40 MHz           b = 1

Check: $\beta = \Delta f / f_m = 10\ \text{kHz}/10\ \text{kHz} = 1$, $f_c = 50$ MHz.

The order can also be reversed: mix the 1 MHz NBFM with a 4 MHz LO to get 5 MHz, then multiply by 10 to get 50 MHz with $\Delta f = 10$ kHz.

Answer: frequency multiplier $n = 10$ followed by a mixer with $f_{LO} = 40$ MHz (sum taken) gives $f_c = 50$ MHz, $\Delta f = 10$ kHz, $\beta = 1$.