sm10_24

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Problem 10.24. Consider binary FSK transmission where 1 2( )f f T− is not an integer. 
(a) What is the mean output of the upper correlator of Fig. 10.12, if a 1 is 
transmitted? What is the mean output of the lower correlator? 
(b) Are the random variables N1 and N2 independent under these conditions? What is 
the variance of N1 – N2? 
(c) Describe the properties of the random variable D of Fig. 10.12 in this case. 
 
 
Solution: 
(a) If a 1 is transmitted, 
)()2cos()( 1 tntfAtr c += π 
 
where n(t) is a narrow band Gaussian noise. The output of the upper correlator is Y1: 
 
∫
∫∫
∫
+≅
+=
=
T
c
TT
c
T
dttftnTA
dttftndttftfA
dttftrY
0 1
0 10 11
0 11
)2cos()(2
2
1
)2cos()(2)2cos()2cos(2
)2cos(2)(
π
πππ
π
 
 
The expected value of Y1 is 1
1[ ]
2 c
Y A=E T , since n(t) has zero mean. 
 
The output of the lower correlator is Y2: 
 
∫∫
∫∫∫
∫∫
∫
+−≅
+−++=
+=
=
TTc
TTcTc
TT
c
T
dttftndttffA
dttftndttffAdttffA
dttftndttftfA
dttftrY
0 20 21
0 20 210 21
0 20 21
0 22
)2cos()(2))(2cos(
2
)2cos()(2))(2cos(
2
))(2cos(
2
)2cos()(2)2cos()2cos(2
)2cos(2)(
ππ
πππ
πππ
π
 
 
 
 
 
 
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 10.24 continued 
 
where the first term of the third line is negligible due to the bandpass assumption. The 
expected value of Y2 is 
( ) ( )
( ) ( )
2 1 20
1 2
1 2
1 2
1 2
[ ] cos(2 ( ) )
2
1 sin 2
22
sin(2 )
2 2
Tc
Tc
c
AY f f t dt
A
0f f tf f
A f f T
f f
π
ππ
ππ
= −
⎡ ⎤= ⋅ −⎣ ⎦−
= −−
∫E
 
 
which clearly differs from the orthogonal case. 
 
(b) The random variables N1 and N2 are given by 
∫= T dttftnN 0 11 )2cos()(2 π 
∫= T dttftnN 0 22 )2cos()(2 π 
Since n(t) is a Gaussian process, both N1 and N2 are Gaussian. To show N1 and N2 are 
correlated consider 
 
[ ]
1 2 1 20 0
1 20 0
0
1 20 0
0
1 20
0
1 2 1 2
[ ] ( ) cos(2 ) ( )cos(2 )
[ ( ) ( )]cos(2 )cos(2 )
( )cos(2 )cos(2 )
2
cos(2 )cos(2 )
2
cos(2 ( ) ) cos(2 ( ) )
4
T T
T T
T T
T
N N n t f t dt n f d
n t n f t f dtd
N t f t f dtd
N f t f t dt
N f f t f f t
π τ π τ τ
τ π π τ
δ τ π π τ τ
π π
π π
⎡ ⎤= ⋅⎢ ⎥⎣ ⎦
=
= −
=
= + + −
∫ ∫
∫ ∫
∫ ∫
∫
E E
E τ
( ) ( )
0
1 2 1 20
0 0
1 2 1 2
0
1 2
sin 2 ( ) sin 2 ( )
4 2 ( ) 2 ( )
sinc(2( ) )
4
T
T T
dt
f f t f f tN
f f f f
N f f T
π π
π π
+ −= ++ −
≅ −
∫
 
 
 
Continued on next slide 
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. 
Problem 10.24 continued 
 
 
where the first term of the second last line is assumed negligible due to the bandpass 
assumption. Since N1 and N2 are correlated, they are not independent. The variance of 
(N1-N2) is 
 
( )
1 2 1 2 1 2
0
0 1 2
var[ ] var[ ] var[ ] 2 [ ]
sinc 2( )
2
N N N N N N
NN f f
− = + −
= − −
E
T
 
 
(c) The random variable D is Gaussian with zero mean and variance var[N1-N2]. 
 
 
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.