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# sm9_7

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```Problem 9.7 Compute the post-detection SNR in the lower channel for Example 9.2 and
compare to the upper channel.

Solution

The SNR of lower channel is, from Eq. (9.59)

2 2
FM
post 3
0
3 ( /
SNR
2
c fA k P
N W
= 2)

where we have assumed that half the power is in the lower channel. Using the
approximation to Carson\u2019s Rule ( ) kHz20022 212/1 =\u2248+= PkDPkB ffT , that is,
this expression becomes 2 2 / 4f Tk P B=

( )22FM
post
0
3
FM
pre
/ 23SNR
2 2
3SNR
8
Tc
T
T
BA
N B W
B
W
=
\u239b \u239e= \u239c \u239f\u239d \u23a0

With a pre-detection SNR of 12 dB, we determine the post-detection SNR as follows

3
FM FM
post pre
12/10 3
3
3 200SNR SNR
8 19
10 0.375 (10.53)
6.94 10
~ 38.4 dB
\u239b \u239e= \u239c \u239f\u239d \u23a0
= × ×
= ×

(The answer in the text for the lower channel is off by factor 0.5 or 3 dB.) For the upper
channel, Example 9.2 indicates this result should be scaled by 2/52 and
3
FM FM
post pre
3 200 2SNR SNR
8 19 52
~ 24.3 dB
\u239b \u239e= \u239c \u239f\u239d \u23a0 \ufffd

So the upper channel is 10log10(52/2) \u2248 14.1 dB worse than lower channel.

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