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# sm9_17

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```Problem 9.17. The signal ( ) cos(400 )m t t\u3c0= is transmitted via FM. There is an ideal
band-pass filter passing 100 \u2264 |f| \u2264 300 at the discriminator output. Calculate the post-
detection SNR given that kf = 1 kHz per volt, and the pre-detection SNR is 500. Use
Carson\u2019s rule to estimate the pre-detection bandwidth.

Solution
We begin by estimating the Carson\u2019s rule bandwidth
( )
( )
Hz2400
200)1(10002
2
=
+=
+= mfT fAkB

We are given that the pre-detection SNR is 500. From Section 9.7 this implies

2400
1
2
500
2
0
2
0
2
N
A
BN
ASNR
c
T
cFM
pre
=
=

Re-arranging this equation, we obtain

Hz102.1
2
6
0
2
×=
N
Ac

The nuance in this problem is that the post-detection filter is not ideal with unity gain
from 0 to W and zero for higher frequencies. Consequently, we must re-evaluate the post-
detection noise using Eq. (9.58)

[ ]
7
2
0
33
2
0
300
100
2
100
300
2
2
0
106.2
3
2
100300
3
2
powernoisedetection-postAvg.
×=
\u2212=
\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1 += \u222b\u222b\u2212
\u2212
c
c
c
A
N
A
N
dffdff
A
N

The post-detection SNR then becomes

Continued on next slide
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
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Problem 9.17 continued
( )
( )( )
8.69230
106.2
5.01000102.13
106.22
3
106.22
3
SNR
7
2
6
7
2
0
2
7
0
22
FM
post
=
××=
×\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b=
×=
Pk
N
A
N
PkA
fc
fc

where we have used the fact that kf = 1000 Hz/V and P = 0.5 watts. In decibels, the post-
detection SNR is 48.4 dB.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.```