Frequency Modulation

Q. A single tone FM is represented by the voltage equation as v(t)=12cos(5x10^8t+5sin1250t). Determine following:

a) Carrier frequency

b) Modulating frequency

c) Modulation index

d) Maximum frequency deviation

Solution: Here,

We are given a single-tone FM signal:

$$v(t)=12\cos (5\times {10}^{8}t+5\sin (1250t))$$

This equation is in the general form of an FM signal:

$$v(t)=A_c\cos (2\pi f_{c}t+\beta \sin (2\pi f_{m}t))$$

Where:

• $f_c$ = carrier frequency

• $f_m$ = modulating frequency

• $\beta$ = modulation index

• $\Delta f = \beta f_m$ = frequency deviation

Step-by-step Extraction from Given Equation

We compare:

$$v(t)=12\cos (5\times {10}^{8}t+5\sin (1250t))$$

with:

$$v(t)=A_c\cos ({\omega }_{c}t+\beta \sin ({\omega }_{m}t))$$

From this, we get:

• $\omega_c = 5 \times 10^8 \Rightarrow f_c = \frac{\omega_c}{2\pi} = \frac{5 \times 10^8}{2\pi} \approx 79.58 \times 10^6 = 79.58 \, \text{MHz}$

• $\omega_m = 1250 \Rightarrow f_m = \frac{1250}{2\pi} \approx 199 \, \text{Hz}$
  

• $\beta = 5$

So, Answers:

a) Carrier Frequency $f_c$

$$f_c=\frac{5\times {10}^8}{2\pi }\approx \boxed{{\displaystyle 79.58\text{MHz}}}$$

b) Modulating Frequency $f_m$

$$f_m=\frac{1250}{2\pi }\approx \boxed{{\displaystyle 199\text{Hz}}}$$

c) Modulation Index $\beta$

$$\boxed{{\displaystyle \mathrm{5}}}$$

d) Maximum Frequency Deviation $\Delta f$

$$\mathrm{\Delta }f=\beta f_m=5\times 199=\boxed{{\displaystyle 995\text{Hz}}}$$