and 3.5 kHz. It is applied as one input to an ideal bal- anced modulator, and the other input is a sinusoid having a frequency of 1.2 MHz. (a) List the frequencies appear- ing at the output of the balanced modulator. (b) Determine the required transmission bandwidth.

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A discrete-spectrum baseband modulating signal has com-
ponents at the following frequencies: 500 Hz, 1.5 kHz,
6-3
Transcribed Image Text:A discrete-spectrum baseband modulating signal has com- ponents at the following frequencies: 500 Hz, 1.5 kHz, 6-3
and 3.5 kHz. It is applied as one input to an ideal bal-
anced modulator, and the other input is a sinusoid having
a frequency of 1.2 MHz. (a) List the frequencies appear-
ing at the output of the balanced modulator. (b) Determine
the required transmission bandwidth.
Transcribed Image Text:and 3.5 kHz. It is applied as one input to an ideal bal- anced modulator, and the other input is a sinusoid having a frequency of 1.2 MHz. (a) List the frequencies appear- ing at the output of the balanced modulator. (b) Determine the required transmission bandwidth.
Expert Solution
Step 1

The modulating signal has a component at the following frequencies: 0.5 kHz, 1.5 kHz and 3.5 kHz.

The carrier frequency is: 1.2 MHz=1200 kHz

Let us consider the modulating signal is given by:

m(t) = E, cos(Wmit) +E, cos(Wm2 t) + Ez cos(Wm3t) ;Ez > Ez > E4 ;

And the carrier frequency:

c(t) = cos(wet)

The balanced modulator is used for generating the Double sideband suppressed carrier(DSB-SC) modulated signal. So that the output of the Balance modulator :

f(t) = m(t)c(t)

f(t) = [E, cos(Wmit) + E, cos(Wm2t) + Ez cos(mat)] cos(wit)

f(t) = E, cos(Wmit) cos(wet) + E, cos(Wm2t) cos(wet) + Ez cos(Wm2 t) cos(wet)

f(t) = 5; [2 cos(Wmat) cos(wÄ…t)] + 52 [2 cos(Wm2t) cos(wÄ…t)] + *> [2 cos(Wm3 t) cos(wet)]

f (t) = [cos(wc + Wma)t + cos(wc – Wma)t} + { [cos(we + Wm2)t + cos(wc – Wm2)t] We + Wm3)t + cos(we - Wm3)t) ......... (1)

Given data:

fm1 = 0.5 x 103 Hz, fm2 = 1.5 x 103 Hz, fm3 = 3.5 x 103 Hz, fe = 1.2 x 106 Hz

Or the frequency in rad/sec

Wmı = 21 x fm1 = 211 X 0.5 x 103rad/secWm2 = 21 x fm2 = 211 x 1.5 x 103rad/sec

Wm3 = 21 x fm2 = 211 x 3.5 x 103rad/sec

w = 211 x fc = 211 X 1.2 x 106rad/sec

Put the all frequency values in the equation (1)

f(t) = (21(1.2 x 106 +0.5 x 103)t + cos(21t(1.2 x 106 – 0.5 x 103)t] + [cos(21(1.2 x 106 + 1.5 x 103)t + cos(21(1.2 x 106 – 1

(a) List the frequency appears at the output of the balanced modulator

The frequency component available in the modulated signal:

The upper sideband frequency component of the first modulated signal:

We +Wmı = 21(1.2 x 106 + 0.5 x 103)rad/sec

21(1.2 x 106 + 0.5 X 103) fe+fm1 = 21

fc+fm1 = 1200 + 0.5 = 1200.5 KHz

The Lower sideband frequency component of the first modulated signal:

we - Wmı = 21(1.2 x 106 – 0.5 x 103)rad/sec

21(1.2 x 106 – 0.5 x 103) fc-fm1 = 211 HZ

fc-fm1 = 1200 - 0.5 = 1199.5 KHz

The upper sideband frequency component of the second modulated signal:

We +Wm2 = 21(1.2 x 106 + 1.5 x 103)rad/sec

21(1.2 x 106 + 1.5 X 103) fe + fm2 = 21

f. + f m2 = 1200 + 1.5 = 1201.5 KHz

The Lower sideband frequency component of the second modulated signal:

Wc-Wm2 = 21(1.2 x 106 – 1.5 X 103) rad/sec

21(1.2 x 106 – 1.5 x 103) fe - fm2 = 21

fc-fm2 = 1200 - 1.5 = 1198.5 KHz

The upper sideband frequency component of the third modulated signal:

We + Wm3 = 21(1.2 x 106 + 3.5 x 103)rad/sec

21(1.2 x 106 + 3.5 103) fe + fm3 = 21

f. + fm3 = 1200 + 3.5 = 1203.5 KHz

The lower sideband frequency component of the third modulated signal:

we - Wm3 = 21(1.2 x 106 – 3.5 x 103)rad/sec

21(1.2 x 106 - 3.5 X 103) fe - fm3 = 21

fc-fm3 = 1200 - 3.5 = 1196.5 KHz

The frequency spectrum is:

f(t) Lower side bands uppermanente E2 fc-fm3 11gcs кн? tofma fitmo 118.5 10.5 к? kН fettmi fettma 1200-5 1205 KH Кң KH3 tetim

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