Numerical Problems 1

In the previous chapter, we have discussed the parameters used in Amplitude Modulation. Each parameter has its own formula. By using those formulas, we can find the respective parameter values. In this chapter, let us solve a few problems based on the concept of amplitude modulation.

Problem 1

A modulating signal $m\left ( t \right )=10 \cos \left ( 2\pi \times 10^3 t\right )$ is amplitude modulated with a carrier signal $c\left ( t \right )=50 \cos \left ( 2\pi \times 10^5 t\right )$. Find the modulation index, the carrier power, and the power required for transmitting AM wave.

Solution

Given, the equation of modulating signal as

$$m\left ( t \right )=10\cos \left ( 2\pi \times 10^3 t\right )$$

We know the standard equation of modulating signal as

$$m\left ( t \right )=A_m\cos\left ( 2\pi f_mt \right )$$

By comparing the above two equations, we will get

Amplitude of modulating signal as $A_m=10 volts$

and Frequency of modulating signal as $$f_m=10^3 Hz=1 KHz$$

Given, the equation of carrier signal is

$$c\left ( t \right )=50\cos \left ( 2\pi \times 10^5t \right )$$

The standard equation of carrier signal is

$$c\left ( t \right )=A_c\cos\left ( 2\pi f_ct \right )$$

By comparing these two equations, we will get

Amplitude of carrier signal as $A_c=50volts$

and Frequency of carrier signal as $f_c=10^5 Hz=100 KHz$

We know the formula for modulation index as

$$\mu =\frac{A_m}{A_c}$$

Substitute, $A_m$ and $A_c$ values in the above formula.

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

Therefore, the value of modulation index is 0.2 and percentage of modulation is 20%.

The formula for Carrier power, $P_c=$ is

$$P_c=\frac{{A_{c}}^{2}}{2R}$$

Assume $R=1\Omega$ and substitute $A_c$ value in the above formula.

$$P_c=\frac{\left ( 50 \right )^2}{2\left ( 1 \right )}=1250W$$

Therefore, the Carrier power, $P_c$ is 1250 watts.

We know the formula for power required for transmitting AM wave is

$$\Rightarrow P_t=P_c\left ( 1+\frac{\mu ^2}{2} \right )$$

Substitute $P_c$ and $\mu$ values in the above formula.

$$P_t=1250\left ( 1+\frac{\left ( 0.2 \right )^2}{2} \right )=1275W$$

Therefore, the power required for transmitting AM wave is 1275 watts.

Problem 2

The equation of amplitude wave is given by $s\left ( t \right ) = 20\left [ 1 + 0.8 \cos \left ( 2\pi \times 10^3t \right ) \right ]\cos \left ( 4\pi \times 10^5t \right )$. Find the carrier power, the total sideband power, and the band width of AM wave.

Solution

Given, the equation of Amplitude modulated wave is

$$s\left ( t \right )=20\left [ 1+0.8 \cos\left ( 2\pi \times 10^3t \right ) \right ]\cos \left ( 4\pi \times 10^5t \right )$$

Re-write the above equation as

$$s\left ( t \right )=20\left [ 1+0.8 \cos\left ( 2\pi \times 10^3t \right ) \right ]\cos \left ( 2\pi \times 2 \times 10^5t \right )$$

We know the equation of Amplitude modulated wave is

$$s\left ( t \right )=A_c\left [ 1+\mu \cos\left ( 2\pi f_mt \right ) \right ]\cos\left ( 2 \pi f_ct \right )$$

By comparing the above two equations, we will get

Amplitude of carrier signal as $A_c=20 volts$

Modulation index as $\mu=0.8$

Frequency of modulating signal as $f_m=10^3Hz=1 KHz$

Frequency of carrier signal as $f_c=2\times 10^5Hz=200KHz$

The formula for Carrier power, $P_c$is

$$P_c=\frac{{A_{e}}^{2}}{2R}$$

Assume $R=1\Omega$ and substitute $A_c$ value in the above formula.

$$P_c=\frac{\left ( 20 \right )^2}{2\left ( 1 \right )}=200W$$

Therefore, the Carrier power, $P_c$ is 200watts.

We know the formula for total side band power is

$$P_{SB}=\frac{P_c\mu^2}{2}$$

Substitute $P_c$ and $\mu$ values in the above formula.

$$P_{SB}=\frac{200\times \left ( 0.8 \right )^2}{2}=64W$$

Therefore, the total side band power is 64 watts.

We know the formula for bandwidth of AM wave is

$$BW=2f_m$$

Substitute $f_m$ value in the above formula.

$$BW=2\left ( 1K \right )=2 KHz$$

Therefore, the bandwidth of AM wave is 2 KHz.