Numerical Problems 1

In the previous chapter, we have discussed the parameters used in Ampptude 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 ampptude modulation.

Problem 1

A modulating signal $mleft ( t ight )=10 cos left ( 2pi imes 10^3 t ight )$ is ampptude modulated with a carrier signal $cleft ( t ight )=50 cos left ( 2pi imes 10^5 t ight )$. Find the modulation index, the carrier power, and the power required for transmitting AM wave.

Solution

Given, the equation of modulating signal as

$$mleft ( t ight )=10cos left ( 2pi imes 10^3 t ight )$$

We know the standard equation of modulating signal as

$$mleft ( t ight )=A_mcosleft ( 2pi f_mt ight )$$

By comparing the above two equations, we will get

Ampptude 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

$$cleft ( t ight )=50cos left ( 2pi imes 10^5t ight )$$

The standard equation of carrier signal is

$$cleft ( t ight )=A_ccosleft ( 2pi f_ct ight )$$

By comparing these two equations, we will get

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

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

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

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

$$Rightarrow P_t=P_cleft ( 1+frac{mu ^2}{2} ight )$$

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

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

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

Problem 2

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

Solution

Given, the equation of Ampptude modulated wave is

$$sleft ( t ight )=20left [ 1+0.8 cosleft ( 2pi imes 10^3t ight ) ight ]cos left ( 4pi imes 10^5t ight )$$

Re-write the above equation as

$$sleft ( t ight )=20left [ 1+0.8 cosleft ( 2pi imes 10^3t ight ) ight ]cos left ( 2pi imes 2 imes 10^5t ight )$$

We know the equation of Ampptude modulated wave is

$$sleft ( t ight )=A_cleft [ 1+mu cosleft ( 2pi f_mt ight ) ight ]cosleft ( 2 pi f_ct ight )$$

By comparing the above two equations, we will get

Ampptude 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 $f_c=2 imes 10^5Hz=200KHz$

The formula for Carrier power, $P_c$is

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

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

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

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

We know the formula for total side band power is

$$P_{SB}=frac{P_cmu^2}{2}$$

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

$$P_{SB}=frac{200 imes left ( 0.8 ight )^2}{2}=64W$$

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.

$$BW=2left ( 1K ight )=2 KHz$$

Therefore, the bandwidth of AM wave is 2 KHz.