This implies that
1
(
)
X
jw
is zero for


2000
. When
( )
y t
is passed through a lowpass fiter
with cutoff frequency
2000
,the output will clearly be zero .Therefore
( )
y t
=0.
8.4
Consider the signal
3
( )
( )sin(400
)
2sin (400
)
y t
g t
t
t
2
3
sin(200
)sin (400
)
2sin (400
)
sin(200
)
1
cos(800
) / 2
2sin(400
)
1
cos(800
) / 2
(1/ 2)sin(200
)
(1/ 4) sin(1000
)
sin(600
)
sin(400
)
(1/ 2) sin(1200
)
sin(400
)
t
t
t
t
t
t
t
t
t
t
t
t
t
If this signal is passed through a lowpass filter with cutoff frequency 400
,then the output
will be
1
sin(200
)
y
t
.
8.5
The signal
( )
x t
is as shown in Figure S8.5
t
t
Envelop
of
( )
t
A
( )
x t
0
t
1000
c
1000
c
0
c
c
( (
))
c
X
j
( (
))
c
X
j
(
)
Y j
Figure S8.2
134
Figure S8.5
The envelope of the signal
( )
t
is as shown in Figure S8.5.Clearly,is we want to use asynchronous
demodulation to recover the signal
( )
x t
,we need to ensure that A is greater than the height
h
of the highest
sidelobe (see Figure S8.5).Let us now determine the height of the highest sidelobe.The first zerocrossing of
the signal
( )
x t
occurs at time
o
t
such that
0
0
1000
,
1/1000.
t
t
Similarly , the second zero
–
crossing happens at time
1
t
such that
1
1
1000
2 ,
2/1000.
t
t
The highest sidelobe occurs at time
0
1
(
)/ 2
t
t
.that is ,at time
2
3/ 2000
t
. At this time, the amplitude of
( )
x t
is
2
sin(3
/ 2)
2000
(
)
3/ 2000
3
x t
Therefore , A should at least be
2000
3
.The modulation index corresponding to the smallest permissible
value of A is
.
.
Max value
m
Min possible
of
value
( )
x t
of
1
A
=
1000
3
2000/3
2
8.6
Let us denote the Fourier transform of
sin(
)/(
)
c
t
by
(
)
H jw
. This will be rectangular pulse which
is nonzero only in the range


c
. Taking The Fourier transform of the first equation given in the
problem ,we have
(
)
( )cos(
)
( )cos(
)
(
)
( )cos(
)
1
(
)
(1/ 2)
( (
))
( (
))
1
(
)
c
c
c
c
c
G j
FT
x t
t
FT
x t
t
H
jw
FT
x t
t
H
jw
X
j
X
j
H
jw
(
)
G j
is as shown in Figure S8.6
The Fourier transform of
( )cos(
)
c
g t
t
is also shown in Figure S8.6.Clearly,if we want to recover
( )
x t
from
( )cos(
)
c
g t
t
,then we have to pass
( )cos(
)
c
g t
t
through an ideal lowpass filter with gain
4 and cutoff frequency
M
.Therefore ,A=4.
8.7 In Figure S8.7 ,we show
(
)
X
j
,
(
)
G j
,and
(
)
Q j
.We also show a polt of The Fourier transform
of
( )cos(
)
c
g t
t
,then we need to ensure that (1)
0
2
c
, and (2) an ideal lowpass filter with
passband gain of 2 and a cutoff frequency of
c
is used to filter
0
( )cos(
)
g t
t
.
c
1/2
(
)
Q j
0
1/2
c
c
0
c
M
0
1/2
1/2
M
(
)
X
j
1
(
)
G j
2
c
Figure S8.6
M
c
2
c
c
( )cos
c
FT
g t
t
1/4
1/4
1/4
M
c
0
c
(
)
X
j
(
)
G j
1
1/2
1/2
0
M
0
135
8.8
(a) From Figure S8.8,it is clear that
(
)
Y j
is conjugatesymmetric .Therefore,
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 Fall '13
 Answers , Periodic function, LTI system theory, Input/output