FM expression after change of modulating signal

Given: $f_c = 10^6$ Hz, $A_c = 3$ V, original $A_m = 1$ V, $f_m = 500$ Hz, $\Delta f = 1$ kHz.

Step 1: Frequency sensitivity of the modulator. In FM, the deviation is proportional to the modulating amplitude:

$$k_f = \frac{\Delta f}{A_m} = \frac{1\ \text{kHz}}{1\ \text{V}} = 1\ \text{kHz/V}$$

Step 2: New deviation. New $A_m = 5$ V, new $f_m = 2$ kHz. The deviation depends only on amplitude, not on $f_m$:

$$\Delta f' = k_f A_m' = 1\ \text{kHz/V} \times 5\ \text{V} = 5\ \text{kHz}$$

Step 3: New modulation index.

$$\beta' = \frac{\Delta f'}{f_m'} = \frac{5\ \text{kHz}}{2\ \text{kHz}} = 2.5$$

(The original index was $\beta = 1000/500 = 2$.)

Step 4: New FM waveform. Taking the modulating signal as a cosine, $m(t) = 5\cos(2\pi \times 2000\,t)$:

$$\begin{aligned} s(t) &= A_c \cos\left[2\pi f_c t + \beta' \sin 2\pi f_m' t\right] \\ &= 3\cos\left[2\pi \times 10^6\, t + 2.5 \sin(2\pi \times 2000\, t)\right]\ \text{V} \\ &= 3\cos\left[6.283\times10^6\, t + 2.5\sin(12566\, t)\right]\ \text{V} \end{aligned}$$

(If a sine carrier is used, write $3\sin[\ldots]$ with the same argument.)

Answer: $s(t) = 3\cos[2\pi\times10^6 t + 2.5\sin(4000\pi t)]$ V, with $\Delta f = 5$ kHz and $\beta = 2.5$.