S-72.227 Digital Communication Systems
Spring 2002 Tutorial#7, 15.04.2002
E 7.1
Binary PAM is used to transmit information over an unequlaized linear
filter channel. When a=1 is transmitted, the noise free output of the
demodulator is
( m  1)
0.3
0.9

xm  
0.3
0
( m  0)
( m  1)
( otherwise )
a) Design a three-tap zero-forcing equalizer so that the
output is
(m  0)
1
qm  
(m  1)
0
b) Determine qm for m=2, 3 by convolving the impulse
response of the equalizer with the channel response.
ANSWER:
a) We denote T as the transmitted bit duration. The equivalent
discrete-time impulse response of the channel is
ht  
m  
 x  t  mT   0.3 t  T   0.9 t   0.3 t  T 
m  
m
Now, if we denote the coefficients of the FIR equalizer by {cn},
then the equalized signal is, qm 
n  
c h
n  
n mn
0.9 0.3 0  c1  0
We can express qm in matrix form as: 0.3 0.9 0.3 *  c0   1

    
 0 0.3 0.9 c1  0
Solving this matrix, we can find the coefficients of the zero forcing
c1   0.4762
equalizer as:  c0    1.4286 
  

c1   0.4762
Course Assistant: Muhammad Imadur Rahman, Phone: 09-4514905, e-mail: [email protected]
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b) The values of qm for m=2, 3 are given by
n  
c h
qm 
n  
so, q2 
n  
n  1
n  
n  1
 cn h2n 
n  
c h
q2 
q3 
n mn
n  
n  2 n

c h
n  2 n
n  1
n  1
n  
n  1
c h
n 3 n
n  
n  1
n  
n  1
 cn h3n 
q 3 
n 2 n
n  1
n  
 cn h3n 
c h
 c1h3  c0 h2  c1h1  0  0  c1h1  0.4762 * 0.3  0.1429
 c1h1  c0 h2  c1h3  c1h1  0  0  0.4762 * 0.3  0.1429
 c1h4  c0 h3  c1h2  0  0  0  0
c h
n 3 n
 c1h2  c0 h3  c1h4  0  0  0  0
E 7.2
Repeat problem (E 7.1) using the MMSE as the criterion for optimizing
the tap coefficients. Assume that the noise power spectral density is
0.1W/Hz.
Answer:
The MMSE for the equalizer having 2k+1 taps, denoted by J(k), is
k
J k   E I k  Iˆk  E I k   c j k  j
2
2
j  k
Minimization of J(k) with respect to {cj}, or equivalently forcing the error,
 k  I k  Iˆk , to be orthogonal to signal samples  *j l ; l  k , yields,
k
c
j  k
j
l j  l
here, l  k ,...,1,0,1,..., k
 l  j  1
l j  xl  j  N 0 lj
where
o
and  
f
*
l
0
otherwise
 L  l  0
otherwise
For our case, l=-1,0,1
With X ( z )  0.3z  0.9  0.3z 1   f 0  f1 z 1  f 0  f1 z 
We obtain the parameters f0 and f1 as,
f 0   0.7854
 0.1146
and
f1   0.1146
 0.7854
The parameters f0 and f1 should have the same sign since f0f1=0.3
Course Assistant: Muhammad Imadur Rahman, Phone: 09-4514905, e-mail: [email protected]
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To have a stable inverse system 1/F*(z-1), we select f0 and f1 in such a
way that zero of the system F*(z-1)= f0+f1z is inside the unit circle. Thus
we choose, f 0  0.1146
and
f1  0.7854
Therefore the desired system for the equalisers coefficient is
 0.7854 
0.3
0.0  c1 
0.9  0.1
 0.3
  c    0.1146 
0
.
9

0
.
1
0
.
3



 0 


 0



0.3
0.9  0.1  c1 
0


Solving this system, we obtain
c-1=0.8596
c0=0.0886
c1=-0.0266
E 7.3
A time-dispersive channel having an impulse response h(t) is used to
transmit four-phase PSK at a rate R=1/T symbols/s. The equivalent
discrete-time channel is shown in Figure 7.1. The sequence k  is white
noise sequence having zero mean and variance  2  N o .
a) What is the sampled autocorrelation function sequence xk 
for this channel , where {xk}defined by

xk 
 h (t )h(t  kT )dt
*

b) The minimum MSE performance of a linear equalizer and a
decision feedback equalize having an infinite number of
taps depends on the folded spectrum of the channel
1
T
2n
H ( 
)

T
n  

2
where H() is the Fourier transform of h(t).
Determine the folded spectrum.
c) Use your answer in (b) to express the minimum MSE of a
linear equalizer in terms of the folded spectrum.
d) Repeat (c) for an infinite-tap decision feedback equalizer.
Course Assistant: Muhammad Imadur Rahman, Phone: 09-4514905, e-mail: [email protected]
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z-1
{Ik}

0.8

-0.6
+
+
{yk}
Figure 7.1
{k}
Answer:
a)
We denote the sampled autocorrelation function as X ( z )
From the figure 7.1, F  z   0.8  0.6 z 1 ,
  

thus X  z   F  z F * z 1  0.8  0.6 z 1 0.8  0.6 z   1  0.48 z 1  0.48 z
From this values of X ( z ), we can find that x0  1, x1  0.48, x1  0.48
2
b)
1 
2n
H ( 
)  X e jT   1  0.48e  jT  0.48e jT

T n
T
0.48e  jT  0.48e jT
)
2
 1  2 * 0.48 * cos T  1  0.96 * cos T
 1  2(
c) For the linear equalizer based on the mean-square error criterion
we have

J min
T

2

T
N0



 X e

j T
T
T
d 
2
 N0
T



T
N0
d
1  0.96 * cos T  N 0
Changing the var iable  to  , J min
1

2

N0
 1  0.96 * cos  N

d
0


1
Noting that
2
Therefore, J min

1
N0
1
1
0.96
*
*
d ; a 
2 1  N 0  1  a * cos
1 N0
 1  a * cos d 
1
; a 2  1
1 a
N0
N0
1

*

*
2
1 N0
1 N0
1 a

2
1
 0.96 

1  
1

N
0 

2

N0
1  N 0 2  0.962
Course Assistant: Muhammad Imadur Rahman, Phone: 09-4514905, e-mail: [email protected]
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d) For decision feedback equalizer,
2N0
J min 

2
2
1 N0 
1  N 0  1  N 0   4 f 0 f1
Note that for N 0  1, J min 
2N0
1  1  0.96 2
This is contras with linear equalizer ,
where we have, J min 
N0
1  0.96 2
2N0
1  N 0 2  0.96 2
 1.56 N 0
 3.57 N 0
Course Assistant: Muhammad Imadur Rahman, Phone: 09-4514905, e-mail: [email protected]
5
Homework-7
Deadline 29 April 2002 at 10.00
Homework return box is located at Otakaari 5, 2nd floor, near
the E-wing. You can also return the answers to the assistant
just before the class.
The real and imaginary parts of the filter coefficients {cn i } as well as
the signal phase  at the output of the linear equalizer can be adjusted
iteratively towards their optimum values using the steepest descent
algorithm,
Recn i   Recn (i )  
Imcn i   Imcn (i )  
 i  1   (i )  
 ek
 ek
2
 ek
2
Recn 
Imcn 
2

where the error signal is defined as
 M

ek  zk  bˆk    cm rk m  exp( j )  bˆk
 m M

Derive the results given in the handouts by setting this expression of
ek into the iteration equations shown above.
For definition of variables, please refer to lecture handouts.
Course Assistant: Muhammad Imadur Rahman, Phone: 09-4514905, e-mail: [email protected]
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