sm9_23

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Problem 9.23. In a receiver using coherent detection, the sinusoidal wave generated by 
the local oscillator suffers from a phase error θ(t) with respect to the carrier wave 
cos(2 )cf tπ . Assuming that θ(t) is a zero-mean Gaussian process of variance 2θσ and that 
most of the time the maximum value of θ(t) is small compared to unity, find the mean-
square error of the receiver output for DSB-SC modulation. The mean-square error is 
defined as the expected value of the squared difference between the receiver output and 
message signal component of a synchronous receiver output. 
 
Solution
Based on the solution of Problem 9.22, we have the DSB-SC demodulator output is 
 
[ ] [ ] [ ]1 1 1( ) ( ) cos ( ) ( ) cos ( ) ( ) sin ( )
2 2 2c I Q
y t A m t t n t t n t tθ θ θ= + + 
 
Recall from Section 9. that the output of a synchronous receiver is 
 
1 1( ) ( )
2 2c I
A m t n t+ 
 
The mean-square error (MSE) is defined by 
 
21MSE ( ) ( )
2 c
y t A m t
⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
E 
 
Substituting the above expression for y(t), the mean-square error is 
 
( ) ( ) ( )
( ) ( ) ( )
2
2
22 2 2 2
1 1 1MSE ( ) cos ( ) 1 ( ) cos ( ) ( )sin ( )
2 2 2
1 1( ) cos ( ) 1 ( )cos ( ) ( )sin ( )
4 4 4
c I Q
c
I Q
A m t t n t t n t t
A m t t n t t n t t
θ θ θ
θ θ
⎡ ⎤⎡ ⎤⎡ ⎤= − + +⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦
⎡ ⎤ 2 θ⎡ ⎤ ⎡⎡ ⎤= − + +⎣ ⎦ ⎤⎣ ⎦ ⎣⎢ ⎥⎣ ⎦
E
E E E ⎦
 
 
where we have used the independence of m(t), nI(t), nQ(t), and θ(t) and the fact that [ ]( ) ( ) 0I Qn t n t⎡ ⎤= ⎣ ⎦E E = to eliminate the cross terms. 
 
 
 
 
 
Continued on next slide 
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Problem 9.23 continued 
( )( ) ( )
( )( ) ( )
( )( )
2
22 2 2 2
2
2 2 2
0 0
2
2 0
1 1 2SE ( ) 1 cos ( ) ( ) cos ( ) ( ) sin ( ( )
4 4 4
1 11 cos ( ) cos ( ) sin ( ( )
4 4 4
1 cos ( )
4 2
c
I Q
c
c
A m t t n t t n t t
A P t N W t N W t
A P N Wt
θ θ θ
θ θ θ
θ
⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= − + +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦
⎡ ⎤ ⎡ ⎤ ⎡ ⎤= − + +⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦
⎡ ⎤= − +⎢ ⎥⎣ ⎦
E E E E E E
E E E
E
M
 
 
where we have used the equivalences of E[m2(t)] = P, and [ ] [ ] WNtntn QI 022 2)()( == EE . 
The last line uses the fact that cos2(θ(t))+sin2(θ(t)) = 1. If we now use the relation that 
1-cos A = 2sin2(A/2), this expression becomes 
 
2 4 0( )MSE sin
2 2c
N WtA P θ⎡ ⎤⎛ ⎞= +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦E 
 
Since the maximum value of θ(t) << 1, sin(θ(t)) ≈ θ(t) and we have 
 
4
2 0
2 4 0
( )MSE
2 2
3
16 2
c
c
N WtA P
N WA P θ
θ
σ
⎡ ⎤⎛ ⎞≈ +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
= +
E
 
 
where we have used the fact that if θ is a zero-mean Gaussian random variable then 
 [ ] [ ]( ) 4224 33 θσθθ == EE 
 
The mean square error is therefore 2 4 0
3 1
16 2c
A P Nθσ + W . 
 
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.