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Tutorial 2: Amplitude Modulation Solutions: M T T S T T A +M T 0t A A A A A A A A

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10 views5 pages

Tutorial 2: Amplitude Modulation Solutions: M T T S T T A +M T 0t A A A A A A A A

Uploaded by

istiaq8888
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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ETM2046 Analog & Digital Communications

Tutorial 2: Amplitude Modulation


Solutions

1. (a)(b) m(t) = 25cos(21000t) and sc(t) = 75cos(2150000t)


c
AM modulated signal is A [1+mcos(21000t)]cos(2150000t)
m c max c m min c m
A = 25 V and A = 75 V, A = A + A = 100 V, A = A  A = 50 V

Modulation index,

s(t)

100

50

t
50

100
0.5Ac = 37.5V
|S(f)| (AM)
0.25mAc = 6.25V 0.25mAc = 6.25V

fcfm fc fc+fm 0 fcfm fc fc+fm f

c c m c m
f = 150 kHz, f + f = 151 kHz and f – f = 149 kHz

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ETM2046 Analog & Digital Communications

(c) For DSB-SC = Full AM – Carrier

|S(f)| (DSB-SC)
0.5mAc = 12.5V

fcfm fc fc+fm 0 fcfm fc fc+fm


f

2. Let the audio signal be bandlimited to f

f f
m 0 m
m
-f f 2f
m Passband signal’s spectrum: BW = 2fm (+ve freq)
Baseband signal’s spectrum: BW = f (+ve freq)
Bandpass signal BW = 2fm
6 x 2fm = 90 kHz
fm = 7.5 kHz

Total BW

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ETM2046 Analog & Digital Communications

f
m
2f

3.

Given m = 0.85

Ac = 148.45

4. (i)

SDSB(f)
M ( 0)
2
Spectrum of SDSB

f
-fc 0 fc

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ETM2046 Analog & Digital Communications

V(f)
M ( 0)
2

f
Spectrum of v(t) -fm fm
-2fc 0 2fc

M (0) R(f)

Spectrum of r(t) 2

f
m cut c m -fm 0 fm
(ii) f < f < 2f - f

5.

4
ETM2046 Analog & Digital Communications

The power of the information is only 106.2 W in AM compared to 800 W in SSB

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