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ECE 460: Communication & Information Theory
Midterm Exam Solution รฑ Problem 2
Willy W Huaracha
April 1, 2002
AM Modulation
Throughout this problem assume that the message signal m t ( ) is given as
m t ( ) = sin(2 ฯ f tm ). The carrier frequency fc is much higher than fm.
(a) Assume m t ( ) is modulated using conventional DSB AM. Sketch the resulting
time domain signal.
x t ( ) = [ A + m t ( ) .cos(2 ] ฯ f tc )
Envelope Carrier
Figure 1
x t ( ) = [ A +cos(2 ฯ f tm ) .cos(2] ฯ f tc )
Then we simplify it by using the identity [ ]
1 cos( )cos( ) cos( ) cos( ) 2
x y = x โ y + x + y
x t ( ) = A .cos(2 ฯ f tc ) +cos(2ฯ f tm ).cos(2 ฯ f tc )
[ ]
( ) .cos(2 ) cos(2 ( ) ) cos(2 ( ) ) 2
x t = A ฯ f t c + ฯ fc โ fm t + ฯ f (^) c + fm t
(b) Compute and sketch the Fourier transform of the modulated signal
Taking the Fourier transform of x ( ) t results in the following equation:
[ ] [ ]
[ ]
c c c m c m
c m c m
A
X f f f f f f f f f f f
f f f f f f
Below is the Fourier transform plot of the modulated signal.
X ( f )
2
A 2
A
1 4
1 4
1 4
1 4
โ +( fc f (^) m ) โ f (^) c โ โ( fc f (^) m ) ( fc โ fm ) fc ( fc + fm )
Figure 2
(c) Repeat parts (a) and (b) if DSB-SC AM is used.
Taking the same steps as for part (a) we determine the modulated signal x ( ) t
using a DSB-SC AM.
x t ( ) = m t ( ).cos(2 ฯ f tc ) =cos(2 ฯ f tm ).cos(2ฯ f tc )
[ ]
( ) cos(2 ( ) ) cos(2 ( ) ) 2
x t = ฯ fc โ f (^) m t + ฯ fc + f (^) mt
Envelope Carrier
Figure 3
First, we simplify (^) x 1 (^) ( ) t by using the identity (^) [ ]
cos( )cos( ) cos( ) cos( ) 2
x y = x โ y + x + y , and
take its Fourier transform as follows.
x 1 ( ) t = m t ( ).cos(2 ฯ f tm ) = cos(2ฯ f tm ) cos(2 โ
ฯ f tm )
2 1
( ) cos (2 ) (1 cos(4 ) 2
x t = ฯ f t m = + ฯ f tm
[ ]
X f = ฮด f โ f (^) m + ฮด f + fm
X ( f )
1 4
1 4
โ 2 fm โ f^ m f (^) m 2 fm
Figure 6
(e) Compute the outputs of the lowpass filters, y 1 (^) ( ) t and y 2 (^) ( ) t , as well as their
Fourier transforms Y 1 (^) ( f )and Y 2 (^) ( f ).
When the signals x 1 (^) ( ) t and x 2 (^) ( ) t are passed through the lowpass filters, the filter
cuts off all-higher frequencies signals and the following signals result:
1
Y f = ฮด f
1
y t =
Y 2 (^) ( f ) = 0
y 2 (^) ( ) t = 0
(f) Compute the output signal s t ( ) and its Fourier transform S ( f ).
First we determine from the block diagram that s 2 (^) ( ) t = z 1 (^) ( ) t โ z 2 ( ) t , which results
as follows.
1 1
( ) ( ) cos(2 ( ) ) cos(2 ( ) ) 2
z t = y t โ
ฯ f c โ f (^) m t = โ
ฯ fc โ f (^) mt
z 2 ( ) t = y 2 ( ) sin(2 t โ
ฯ ( f c โ fm ) ) t = 0
( ) cos(2 ( ) ) 2
s t = โ
ฯ f c โ f m t
Ultimately we determine we the Fourier transform of s ( t )
[ ]
S f = ฮด f โ f c โ f m + ฮด f + fc โ fm
(g) How would you describe what this system does?
This system is a LSB-SSB AM Modulator.
