sm9_3
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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
2 PAc
DSB
 is 
WN
PASNR c
DSB
DSB
post
0
2
2
= 
 
From Eq. (9.30), the post-detection SNR of AM with received power ( )PkA acAM 22 12 + is 
WN
PkASNR ac
AM
AM
post
0
22
2
= 
 
So, by equating the transmit powers for DSB-Sc and AM, we obtain 
 
( )2 2 2
2 2
2
1
2 2
2 2 1
DSB AM
c c
a
AM DSB
c c
a
A P A k P
A A P
k P
= +
\u21d2 = +
 
 
Substituting this result into the expression for the post-detection SNR of AM, 
 
2 2
AM DSB
post post2
0
SNR SNR
2 1
DSB
c a
a
A P k P
N W k P
\u239b \u239e= = \u2206\u239c \u239f+\u239d \u23a0
 
 
 
Where the factor \u2206 is 
 
Pk
Pk
a
a
2
2
1+=\u2206 
 
 With ka= 0.2 and P= 1, the AM SNR is a factor 
( ) 04.
04.1
2. 2 ==\u2206 less. 
 
 With ka = 0.4 and P = 1, the AM SNR is a factor 
( ) 14.0
16.1
16.
16.1
4. 2 \u2248=+=\u2206 less. 
 
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