ECE 301
Detailed Solutions  Set 6:
Analog Equalizers
1
62.
A channel transfer function having no interference is given by
C s
s
s
62 832
A signal
x t
is sent through this channel and the received signal
r t
is given by
r t
c t
x t
.
The received signal
r t
is then input to an inverse filter equalizer which was designed for the
channel
C s
. The output of this equalizer is the signal
y t
. Compute the simplest math form for
y t
as a function of
r t
.
DETAILED
SOLUTION:
From the Notes we know that
y t
, the output of the equalizer, is given by
y t
h
E
t
r t
a
The
s
domain relation corresponding to
a
is given by
Y s
H
E
s R s
b
The function
H
E
s
is the transfer function of the inverse filter equalizer. Using
C s
given in the problem
statement we have
H
E
s
1
C s
s
62 832
s
1
62 832
s
c
Substitute
c
into
b
to see the
s
domain expression for
Y s
:
Y s
1
62 832
s
R s
R s
62 832
1
s
R s
d
Taking the inverse Laplace transform of both sides of
d
will give the desired answer. To do this we need
the relation between dividing by
s
in the
s
domain and integration in the timedomain. For a general signal
g t
this relation is as follows:
If
g t
G s
e
then
t
τ
0
g
τ
d
τ
1
s
G s
f
Applying
f
to the specific
s
domain function
R s
in
d
then gives the desired answer:
y t
r t
62 832
t
τ
0
r
τ
d
τ
g
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ECE 301
Detailed Solutions  Set 6:
Analog Equalizers
2
63.
A channel transfer function is
C s
s
50 . A signal
x t
is sent through this channel and
the received signal
r t
is given by
r t
c t
x t
. The received signal
r t
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 Spring '08
 Hoozturk
 Laplace, inverse ﬁlter equalizer

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