Learning Numerical Problems 1 work project make money

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

Solution

Given, the equation of modulating signal as

$$mleft ( t right )=10cos left ( 2pi times 10^3 tright )$$

We know the standard equation of modulating signal as

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

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

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

The standard equation of carrier signal is

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

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

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

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

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

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

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

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

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

Problem 2

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

Solution

Given, the equation of Amplitude modulated wave is

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

Re-write the above equation as

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

We know the equation of Amplitude modulated wave is

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

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 $f_c=2times 10^5Hz=200KHz$

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.

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

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

We know the formula for total side band power is

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

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

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

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.

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

Therefore, the bandwidth of AM wave is 2 KHz.

Learning working make money

Leave a Reply

Your email address will not be published. Required fields are marked *