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# sm9_3

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Problem 9.3. For the same received signal power, compare the post-detection SNRs of
DSB-SC with coherent detection and envelope detection with ka = 0.2 and 0.4. Assume
the average message power is P = 1.
Solution
From Eq. (9.23), the post-detection SNR of DSB-SC with received power 2
2PA
c
DSB
is
WN
PA
SNR c
DSB
DSB
post
0
2
2
=
From Eq. (9.30), the post-detection SNR of AM with received power
()
Pk
A
a
c
AM
2
2
1
2+ is
WN
PkA
SNR ac
AM
AM
post
0
22
2
=
So, by equating the transmit powers for DSB-Sc and AM, we obtain
()
22
2
22
2
1
22
221
DSB AM
cc
a
AM DSB
cc
a
AP A kP
AA
P
kP
=+
⇒= +
Substituting this result into the expression for the post-detection SNR of AM,
22
AM DS B
p
ost post
2
0
SNR SNR
21
DSB
ca
a
AP k P
NW k P
⎛⎞
=
=∆
⎜⎟
+
⎝⎠
Where the factor is
Pk
Pk
a
a
2
2
1+
=
With ka= 0.2 and P= 1, the AM SNR is a factor
(
)
04.
04.1
2. 2
== less.
With ka = 0.4 and P = 1, the AM SNR is a factor
(
)
14.0
16.1
16.
16.1
4. 2
=
+
= less.
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