Question

Communication system

Problems 

1. A 25 MHz carrier is modulated by a 400 Hz audio sine wave If the carrier voltage 4 V and the maximum deviation is 10 KHz, write the equation of this modulated wave for FM. If the modulating frequency is now changed to 2 KHz, all else remaining constant, write the new equation for FM

2. 2. In an FM system, when the audio frequency (AF) is 500 Hz and the AF voltage is 2.4 V, the deviation is 4.8 KHz. If the AF voltage is now increased to 7.2 V, what is the new deviation? Find the modulation index in each case Middle East College

3. 3. In the above problem if the AF voltage raised to 10V while the AF is dropped to 200 Hz, what if the deviation? Find the modulation index.

4. 4. a) Find the carrier and modulating frequencies, the modulation index, and the maximum deviation of the FM wave represented by the voltage equation v 12 sin (6 108 t+5 sin 1250t)

   
   

Answer 1

\(1 .\)

Given \(: \mathrm{f}_{\mathrm{s}}=400 \mathrm{~Hz}, \mathrm{f}_{\mathrm{c}}=25 \mathrm{MHz}, \Delta \mathrm{f}=10 \mathrm{kHz}\)

Solution:

We know that, voltage equation of FM ia given by:

\(\mathrm{e}=\mathrm{E}_{c} \cos \left(\omega_{\mathrm{c}} \mathrm{t}+\mathrm{m}_{\mathrm{f}} \sin \omega_{\mathrm{s}} \mathrm{t}\right)\)

Where,

\(E_{c}=4 V\)

\(\omega_{c}=2 \pi f_{c}\)

\(=2 \pi \times 25 \times 10^{6}\)

\(=1.57 \times 10^{8} \mathrm{rad} / \mathrm{s}\)

And,

\(\omega_{s}=2 \pi f_{s}\)

\(=2 \pi \times 400\)

\(=2512 \mathrm{rad} / \mathrm{s}\)

\(m_{f}=\frac{\Delta f}{f_{s}}\)

\(m_{f}=\frac{\left(10 * 10^{3}\right) H z}{400 H z}=25\)

Now, putting all values in the equation, we get:

\(\mathrm{e}=4 \cos \left[\left(1.57 \times 10^{8}\right) \mathrm{t}+25 \sin 2512 \mathrm{t}\right]\)

Now, if modulating frequency \(\left(f_{s}\right)\) is changed to \(2 \mathrm{kHz}\), then:

\(\omega_{s}=2 \pi f_{s}\)

\(\omega_{s}=2 \pi *\left(2 * 10^{3} H z\right)\)

\(=1.256 \times 10^{4} \mathrm{rad} / \mathrm{s}\)

\(m_{f}=\frac{\Delta f}{f_{s}}\)

\(m_{f}=\frac{10 k H z}{2 k H z}=5\)

Hence, the new equation is given by:

\(\mathrm{e}=4 \cos \left[\left(1.57 \times 10^{8}\right) \mathrm{t}+5 \sin 12560 \mathrm{t}\right]\)