Numerical Problems 2

In the previous chapter, we have discussed the parameters used in Angle 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 Frequency Modulation.

Problem 1

A sinusoidal modulating waveform of ampptude 5 V and a frequency of 2 KHz is appped to FM generator, which has a frequency sensitivity of 40 Hz/volt. Calculate the frequency deviation, modulation index, and bandwidth.

Solution

Given, the ampptude of modulating signal, $A_m=5V$

Frequency of modulating signal, $f_m=2 KHz$

Frequency sensitivity, $k_f=40 Hz/volt$

We know the formula for Frequency deviation as

$$Delta f=k_f A_m$$

Substitute $k_f$ and $A_m$ values in the above formula.

$$Delta f=40 imes 5=200Hz$$

Therefore, frequency deviation, $Delta f$ is $200Hz$

The formula for modulation index is

$$eta = frac{Delta f}{f_m}$$

Substitute $Delta f$ and $f_m$ values in the above formula.

$$eta=frac{200}{2 imes 1000}=0.1$$

Here, the value of modulation index, $eta$ is 0.1, which is less than one. Hence, it is Narrow Band FM.

The formula for Bandwidth of Narrow Band FM is the same as that of AM wave.

$$BW=2f_m$$

Substitute $f_m$ value in the above formula.

$$BW=2 imes 2K=4KHz$$

Therefore, the bandwidth of Narrow Band FM wave is $4 KHz$.

Problem 2

An FM wave is given by $sleft ( t ight )=20 cosleft ( 8 pi imes10^6t+9 sinleft ( 2 pi imes 10^3 t ight ) ight )$. Calculate the frequency deviation, bandwidth, and power of FM wave.

Solution

Given, the equation of an FM wave as

$$sleft ( t ight )=20 cosleft ( 8 pi imes10^6t+9 sinleft ( 2 pi imes 10^3 t ight ) ight )$$

We know the standard equation of an FM wave as

$$sleft ( t ight )=A_c cosleft ( 2 pi f_ct + eta sin left ( 2 pi f_mt ight ) ight )$$

We will get the following values by comparing the above two equations.

Ampptude of the carrier signal, $A_c=20V$

Frequency of the carrier signal, $f_c=4 imes 10^6 Hz=4 MHz$

Frequency of the message signal, $f_m=1 imes 10^3 Hz = 1KHz$

Modulation index, $eta=9$

Here, the value of modulation index is greater than one. Hence, it is Wide Band FM.

We know the formula for modulation index as

$$eta=frac {Delta f}{f_m}$$

Rearrange the above equation as follows.

$$Delta=eta f_m$$

Substitute $eta$ and $f_m$ values in the above equation.

$$Delta=9 imes 1K =9 KHz$$

Therefore, frequency deviation, $Delta f$ is $9 KHz$.

The formula for Bandwidth of Wide Band FM wave is

$$BW=2left ( eta +1 ight )f_m$$

Substitute $eta$ and $f_m$ values in the above formula.

$$BW=2left ( 9 +1 ight )1K=20KHz$$

Therefore, the bandwidth of Wide Band FM wave is $20 KHz$

Formula for power of FM wave is

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

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

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

Therefore, the power of FM wave is $200$ watts.