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- Introduction
- Modulation
- Amplitude Modulation
- Numerical Problems 1
- AM Modulators
- AM Demodulators
- DSBSC Modulation
- DSBSC Modulators
- DSBSC Demodulators
- SSBSC Modulation
- SSBSC Modulators
- SSBSC Demodulator
- VSBSC Modulation
- Angle Modulation
- Numerical Problems 2
- FM Modulators
- FM Demodulators
- Multiplexing
- Noise
- SNR Calculations
- Transmitters
- Receivers
- Sampling
- Pulse Modulation
- Transducers
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- Analog Communication - Discussion
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 amplitude 5 V and a frequency of 2 KHz is applied to FM generator, which has a frequency sensitivity of 40 Hz/volt. Calculate the frequency deviation, modulation index, and bandwidth.
Solution
Given, the amplitude 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.
$$\Delta f=40 \times 5=200Hz$$
Therefore, frequency deviation, $\Delta f$ is $200Hz$
The formula for modulation index is
$$\beta = \frac{\Delta f}{f_m}$$
Substitute $\Delta f$ and $f_m$ values in the above formula.
$$\beta=\frac{200}{2 \times 1000}=0.1$$
Here, the value of modulation index, $\beta$ 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.
$$BW=2 \times 2K=4KHz$$
Therefore, the bandwidth of Narrow Band FM wave is $4 KHz$.
Problem 2
An FM wave is given by $s\left ( t \right )=20 \cos\left ( 8 \pi \times10^6t+9 \sin\left ( 2 \pi \times 10^3 t \right ) \right )$. Calculate the frequency deviation, bandwidth, and power of FM wave.
Solution
Given, the equation of an FM wave as
$$s\left ( t \right )=20 \cos\left ( 8 \pi \times10^6t+9 \sin\left ( 2 \pi \times 10^3 t \right ) \right )$$
We know the standard equation of an FM wave as
$$s\left ( t \right )=A_c \cos\left ( 2 \pi f_ct + \beta \sin \left ( 2 \pi f_mt \right ) \right )$$
We will get the following values by comparing the above two equations.
Amplitude of the carrier signal, $A_c=20V$
Frequency of the carrier signal, $f_c=4 \times 10^6 Hz=4 MHz$
Frequency of the message signal, $f_m=1 \times 10^3 Hz = 1KHz$
Modulation index, $\beta=9$
Here, the value of modulation index is greater than one. Hence, it is Wide Band FM.
We know the formula for modulation index as
$$\beta=\frac {\Delta f}{f_m}$$
Rearrange the above equation as follows.
$$\Delta=\beta f_m$$
Substitute $\beta$ and $f_m$ values in the above equation.
$$\Delta=9 \times 1K =9 KHz$$
Therefore, frequency deviation, $\Delta f$ is $9 KHz$.
The formula for Bandwidth of Wide Band FM wave is
$$BW=2\left ( \beta +1 \right )f_m$$
Substitute $\beta$ and $f_m$ values in the above formula.
$$BW=2\left ( 9 +1 \right )1K=20KHz$$
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, $R=1\Omega$ and substitute $A_c$ value in the above equation.
$$P=\frac{\left ( 20 \right )^2}{2\left ( 1 \right )}=200W$$
Therefore, the power of FM wave is $200$ watts.
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