Amplitude Modulation

Q. The equation of amplitude modulated wave is given by s(t) = 40[1+0.8cos(2π ×10^3t)]cos(4π×10^5t). Find the carrier power, the total sideband power and bandwidth of the signal. The value of resistor given is 30Ω.

Solution: Here,

We are given:

$$s(t)=40[1+0.8\cos (2\pi \cdot {10}^{3}t)]\cos (4\pi \cdot {10}^{5}t)$$

And:

• Load resistance $R = 30\, \Omega$
  

Step 1: Identify the standard AM form

Standard AM wave:

$$s(t)=A_c[1+\mu \cos (2\pi f_{m}t)]\cos (2\pi f_{c}t)$$

From the given equation:

• $A_c = 40$

• $\mu =0.8$

• $f_m={10}^3=1\text{kHz}$

• $f_c=\frac{4\pi \cdot {10}^5}{2\pi }=2\times {10}^5=200\text{kHz}$

a) Carrier Power $P_c$

$$P_c=\frac{A_c^2}{2R}=\frac{{40}^2}{2\cdot 30}=\frac{1600}{60}=\boxed{{\displaystyle 26.67\text{W}}}$$

b) Total Sideband Power $P_{SB}$

$$P_{SB} = P_c \cdot \frac{\mu^2}{2} = 26.67 \cdot \frac{0.8^2}{2} = 26.67 \cdot \frac{0.64}{2} = 26.67 \cdot 0.32 = \boxed{8.53\, \text{W}}$$

c) Bandwidth $BW$

For standard AM:

$$BW=2f_m=2\cdot 1\text{kHz}=\boxed{{\displaystyle 2\text{kHz}}}$$