Question

mod
en
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1. For a FM signal v(t) = 25 cos (15 * 108t + 10 sin 1550t), calculate ---
Modulation index
Fd
Maximum frequency deviation
a. 10, 3000.1 Hz
Fro. b. 20, 1550.9Hz
c. 10,2465.9Hz
d. 10, 2000.0Hz​

Answer 1

Answer:

what is this question

Answer 2

Answer:

The modulation index and maximum frequency of deviation is 10 and 2465.9 Hz respectively.i.e.option (c).

Explanation:

The expression for FM signal is given by,

$V(t)=Acos(\omega_c t+m_fsin\omega_m t)$            (1)

Where,

V=value of the signal

A=amplitude of the signal

$\omega_c$=angular frequency of carrier signal

$\omega_m$=angular frequency of message signal

$m_f$=modulation index

And the waveform of the signal given in question is,

$V(t)=25cos(15\times 10^8 t+10sin1550t)$     (2)

By comparing equation (2) with equation (1) the modulation index is given as,

$m_f=10$         (3)

And the maximum frequency deviation is given as,

$\Delta f=m_f\times f_m$       (4)

Now,

$f_m=\frac{\omega_f}{2\pi}=\frac{1550}{2\pi}=246.59Hz$     (5)

By substituting the values in equation (4) we get;

$\Delta f=10\times 246.59=2465.9Hz$

Hence, the modulation index and maximum frequency of deviation is 10 and 2465.9 Hz respectively.i.e.option (c).