Analog Communica...3

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(t)=10cos(2π×103t)$m\left(t\right)=10\cos \left(2\pi \times {10}^{3}t\right)$ is amplitude modulated with a carrier signal c(t)=50cos(2π×105t)$c\left(t\right)=50\cos \left(2\pi \times {10}^{5}t\right)$. Find the modulation index, the carrier power, and the power required for transmitting AM wave.

Solution

Given, the equation of modulating signal as

m(t)=10cos(2π×103t)$$m\left(t\right)=10\cos \left(2\pi \times {10}^{3}t\right)$$

We know the standard equation of modulating signal as

m(t)=Amcos(2πfmt)$$m\left(t\right)=A_m\cos \left(2\pi f_{m}t\right)$$

By comparing the above two equations, we will get

Amplitude of modulating signal as Am=10volts$A_m=10volts$

and Frequency of modulating signal as

fm=103Hz=1KHz$$f_m={10}^{3}Hz=1KHz$$

Given, the equation of carrier signal is

c(t)=50cos(2π×105t)$$c\left(t\right)=50\cos \left(2\pi \times {10}^{5}t\right)$$

The standard equation of carrier signal is

c(t)=Accos(2πfct)$$c\left(t\right)=A_c\cos \left(2\pi f_{c}t\right)$$

By comparing these two equations, we will get

Amplitude of carrier signal as Ac=50volts$A_c=50volts$

and Frequency of carrier signal as fc=105Hz=100KHz$f_c={10}^{5}Hz=100KHz$

We know the formula for modulation index as

μ=AmAc$$\mu =\frac{A_m}{A_c}$$

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

μ=1050=0.2$$\mu =\frac{10}{50}=0.2$$

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

The formula for Carrier power, Pc=$P_c=$ is

Pc=Ac22R$$P_c=\frac{{A_c}^2}{2R}$$

Assume R=1Ω$R=1\mathrm{\Omega }$ and substitute Ac$A_c$ value in the above formula.

Pc=(50)22(1)=1250W$$P_c=\frac{{\left(50\right)}^2}{2\left(1\right)}=1250W$$

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

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

⇒Pt=Pc(1+μ22)$$\Rightarrow P_t=P_c\left(1+\frac{{\mu }^2}{2}\right)$$

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

Pt=1250(1+(0.2)22)=1275W$$P_t=1250\left(1+\frac{{\left(0.2\right)}^2}{2}\right)=1275W$$

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

Problem 2

The equation of amplitude wave is given by s(t)=20[1+0.8cos(2π×103t)]cos(4π×105t)$s\left(t\right)=20\left[1+0.8\cos \left(2\pi \times {10}^{3}t\right)\right]\cos \left(4\pi \times {10}^{5}t\right)$. 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(t)=20[1+0.8cos(2π×103t)]cos(4π×105t)$$s\left(t\right)=20\left[1+0.8\cos \left(2\pi \times {10}^{3}t\right)\right]\cos \left(4\pi \times {10}^{5}t\right)$$

Re-write the above equation as

s(t)=20[1+0.8cos(2π×103t)]cos(2π×2×105t)$$s\left(t\right)=20\left[1+0.8\cos \left(2\pi \times {10}^{3}t\right)\right]\cos \left(2\pi \times 2\times {10}^{5}t\right)$$

We know the equation of Amplitude modulated wave is

s(t)=Ac[1+μcos(2πfmt)]cos(2πfct)$$s\left(t\right)=A_c\left[1+\mu \cos \left(2\pi f_{m}t\right)\right]\cos \left(2\pi f_{c}t\right)$$

By comparing the above two equations, we will get

Amplitude of carrier signal as Ac=20volts$A_c=20volts$

Modulation index as μ=0.8$\mu =0.8$

Frequency of modulating signal as fm=103Hz=1KHz$f_m={10}^{3}Hz=1KHz$

Frequency of carrier signal as fc=2×105Hz=200KHz$f_c=2\times {10}^{5}Hz=200KHz$

The formula for Carrier power, Pc$P_c$is

Pc=Ae22R$$P_c=\frac{{A_e}^2}{2R}$$

Assume R=1Ω$R=1\mathrm{\Omega }$ and substitute Ac$A_c$ value in the above formula.

Pc=(20)22(1)=200W$$P_c=\frac{{\left(20\right)}^2}{2\left(1\right)}=200W$$

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

We know the formula for total side band power is

PSB=Pcμ22$$P_{SB}=\frac{P_c{\mu }^2}{2}$$

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

PSB=200×(0.8)22=64W$$P_{SB}=\frac{200\times {\left(0.8\right)}^2}{2}=64W$$

Therefore, the total side band power is 64 watts.

We know the formula for bandwidth of AM wave is

BW=2fm$$BW=2f_m$$

Substitute fm$f_m$ value in the above formula.

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

Therefore, the bandwidth of AM wave is 2 KHz.