Question

A device with input x(t) and output y(t) is characterized by y(t) = x 2 (t). An fm signal with frequency deviation of 100 khz and modulating signal bandwidth of 5 khz is applied to this device. Then what is the bandwidth of the output signal?

Answer 1

Answer:

410 kHz

Explanation:

In present case

?f = 100; fm = 5

β = [?f / fm] = [100/5] = 20

FM equation

A cos [wct  + β = sin wmt]

= A cos [wct  + 18 sin wmt]

y(t) = x2 (t) = A2 cos2 [wct  + 20 Sin wmt]

Note : Cos2 q = [1 + Cos2q] / 2

If there is change in frequency the modulation index also changes in same ratio

y(t) = A2 [(1/2) + (1/2) Cos {2wct  + 40Sin wmt}]

y(t) = [(A2/2) + (A2/2) Cos {2wct  + 40Sin wmt}]

After the device,

β(new) = 40 = [?f(new) / fm]

?f(new) = 40 x 5 = 200

By carson's rule

Bandwidth = 2(?f +  fm)

              = 2 (200 + 5)

Bandwidth = 410 kHz