Amplitude Modulation

Q. An audio frequency signal 10sin(1000πt) is used for a single tone amplitude modulation with a carrier of 50sin(2π×10^(5)t). Calculate:

(i) Modulation index

(ii) Bandwidth requirement

(iii) Total power delivered if load = 60Ω.

Solution: Here,

We are given:

• Message signal: $m(t)=10\sin (1000\pi t)$

• Carrier signal: $c(t)=50\sin (2\pi \times {10}^{5}t)$

• Load Resistance: $R = 60\, \Omega$
  

Let’s solve the parts step-by-step.

(i) Modulation Index $\mu$

The modulation index is defined as:

$$\mu =\frac{A_m}{A_{\mathrm{c}}}$$

Where:

• $A_m =$Amplitude of message = 10

• $A_c=\; Amplitude\; of\; carrier\; =\; 50$

$$\mu =\frac{10}{50}=0.2$$

So, Modulation Index = 0.2

(ii) Bandwidth Requirement

For AM (Amplitude Modulation), bandwidth is:

$$BW=2f_{\mathrm{m}}$$

Where:

• $f_m = \frac{1000\pi}{2\pi} = 500 \, \text{Hz}$
  

$$BW=2\times 500=1000\text{Hz}$$

So, Bandwidth = 1000 Hz or 1 kHz

(iii) Total Power Delivered

In AM, Total Power is given by:

$$P_{total}=P_c(1+\frac{{\mu }^2}{2})$$

Where:

• $P_c=\; Carrier\; power\; =$$\frac{A_c^2}{2R}=\frac{{50}^2}{2\times 60}=\frac{2500}{120}=20.83\text{ W}$

• $\mu = 0.2$
  

$$P_{total}=20.83(1+\frac{{0.2}^2}{2})=20.83(1+0.02)=20.83\times 1.02\approx 21.25\text{W}$$

So, Total Power Delivered = 21.25 W