sm10_27

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Problem 10.27. A BPSK signal is applied to a matched-filter receiver that lacks perfect 
phase synchronization with the transmitter. Specifically, it is supplied with a local carrier 
whose phase differs from that of the carrier used in the transmitter by φ radians. 
(a) Determine the effect of the phase error φ on the average probability of error of 
this receiver. 
(b) As a check on the formula derived in part (a), show that when the phase error is 
zero the formula reduces to the same form as in Eq. (10.44). 
 
Solution 
(a) With BPSK, assume the transmitted signal is (10.36): 
 
0
( ) ( ) cos(2 )
N
c k c
k
s t A b h t kT f tπ
=
= −∑ , 
where bk = +1 for a 1 and bk = -1 for a 0, h(t) is the rectangular pulse 
/ 2rect t T
T
−⎛ ⎞⎜ ⎟⎝ ⎠ . 
 
The received signal is 
 
0
( ) ( ) ( )
( ) cos(2 ) ( ) cos(2 ) ( ) sin(2 )
N
c k c I c Q c
k
x t s t n t
A b h t kT f t n t f t n t f tπ π
=
= +
= − + −∑ π 
 
The receiver matched filter is the integrate-and-dump filter. The output for the kth symbol 
after down-conversion with phase error φ and match filtering is: 
 
[ ]
( 1)
( 1) ( 1)
( 1) ( 1)
( ) cos(2 )
( ) cos(2 )cos(2 ) ( )sin(2 )cos(2 )
1 1[ ( )][cos cos(4 )] ( )[sin(4 ) sin( )]
2 2
kT
k ck T
kT kT
c k I c c Q c ck T k T
kT kT
c k I c Q ck T k T
Y x t f t dt
A b n t f t f t dt n t f t f t dt
A b n t f t dt n t f t dt
π φ
π π φ π π φ
φ π φ π φ φ
−
− −
− −
= +
= + + − +
= + + + − + +
≅
∫
∫ ∫
∫ ∫ −
( 1) ( 1)
1 1cos ( ) cos ( )sin
2 2 2
cos
2
kT kT
c k I Qk T k T
c k k
T A b n t dt n t dt
T A b N
φ φ φ
φ
− −+ +
= +
∫ ∫
 
where we define 
( 1) ( 1)
1 1( ) cos ( )sin
2 2
kT kT
k I Qk T k T
N n t dt n t dtφ φ− −= +∫ ∫ 
 
 
Continued on next slide 
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Problem 10.27 continued 
 
 
The random variable Nk has zero mean and variance 
 
2 20 0
20
var[ ] cos sin
4 4
4
k
N T N TN
N T
φ φ
σ
= +
= =
 
 
Let cos
2 c
T Aµ φ= . Then the probability of bit error Pe is 
 
2 20
2 20
[ 1] [ 0 | 1] [ 1] [ 0 | 1]
1 1 ( ) 1 1 ( )exp exp
2 22 2
e k k k k k kP b Y b b Y b
y ydy dy
Q
µ µ
σ σπσ πσ
µ
σ
+∞
−∞
= = < = + = − > = −
⎧ ⎫ ⎧− += − + − ⎫⎨ ⎬ ⎨⎩ ⎭ ⎩
⎛ ⎞= ⎜ ⎟⎝ ⎠
∫ ∫
P P P P
⎬⎭
 
 
with 
2
0
1, cos ,
2 2 2
c
b c
A T TE Aµ φ σ= = = N T , we have 
0
2 cosb
e
EP Q
N
φ⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠
 
 
(b) When the phase error φ=0, 
0
2 b
e
EP Q
N
⎛ ⎞= ⎜⎜⎝ ⎠⎟⎟
, as the same as Eq. (10.44). 
 
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.