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CSP401

Test #10

In the adaptive channel equalisation problem shown below, ( )H z is the transferfunction of the channel (assumed unknown), ( )s n is a known signal, ( )v n

represents additive channel noise independent of ( )s n , and ( )nW z is the adaptive

equaliser.

Suppose ( )s n and ( )v n are real with respective autocorrelation functions

( )= - 12( ) 4 m

sr m and

== =

1, 0

( ) 0.2, 1

0, otherwise

v

m

r m m

and ( )n

W z is a 3-tap FIR filter updated by the real LMS algorithm

m+ = +1 ( ) ( )n n e n nw w x

(a) Determine the step size m for a misadjustment of 0.01. (4 Marks)

(b) Determine the modal time constants tk , = 0, 1, 2k , and the average time

constant of the learning curve tmse,av . (4 Marks)

(c) Suppose ( )nW z has been lengthened to 6 taps, but m is kept the same as

in Part (a). Determine the misadjustment and the average time constant of

the learning curve tmse,av of the lengthened adaptive filter. (4 Marks)

Delay

on

= -( ) ( )od n s n n

( )s n ( )x n ( )e n

( )v n

( )d n-

= +1

( ) 1 0.8H z z ( )

n

W z

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Solution

We first observe that, for the real LMS algorithm

( )

2

tr xm R

1k

k

tm l

=

and mse,av1

4

pt

+

To solve the problems, all we need to find, therefore, is ( )xr m . Now

{ }[ ][ ]{ }

2

( ) ( ) ( )

( ) 0.8 ( 1 ) ( ) ( ) 0.8 ( 1) ( )

( ) 0.8 ( 1) 0.8 ( 1) 0.8 ( ) ( )

1.64 ( ) 0.8 ( 1) 0.8 ( 1) ( )

x

s s s s v

s s s v

r m x n m x n

s n m s n m v n m s n s n v n

r m r m r m r m r m

r m r m r m r m

= +

= + + - + + + + - +

= + - + + + +

= + - + + +

Hence

( ) ( )1 12 2(0) 1.64 4 0.8 4 0.8 4 1 4.36xr = + - + - + =

( ) ( )

21 1

2 2(1) 1.64 4 0.8 4 0.8 4 0.2 0.92xr = - + + - + =

( ) ( ) ( )2 31 1 1

2 2 2(2) 1.64 4 0.8 4 0.8 4 0.36xr = - + - + - = -

and

4.36 0.92 0.36

0.92 4.36 0.92

0.36 0.92 4.36

x

- = -

R

(a)( )2 2 2 0.01 0.001529

tr 3 (0) 3 4.36x xrm = = =

R

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(b) The eigenvalues of xR can be found using the formulae in pp. 25 of the Supplement

to the final Exam. Thus1

2 22 (0.92) ( 0.36)0.607467

3Q

+ -= =

{ }2Re 0.92 ( 0.36) 0.304704R = - = -

( )31 0.3047040.607467cos 2.269947 radq - -= =

( )2.269947 2 01 032 0.607467 cos 4.36 5.493469pl l+ = + =

( )2.269947 2 12 232 0.607467 cos 4.36 2.866531pl l+ = + =

( )2.269947 2 23 132 0.607467 cos 4.36 4.72pl l+ = + =

The modal time constants are given, accordingly, by

00

1 1119.05 samples

0.001529 5.4935t

m l= = =

11

1 1138.56 samples

0.001529 4.72t

m l= = =

22

1 1228.15 samples

0.001529 2.8665t

m l= = =

and the average time constant of the learning curve is given by

mse,av

1 375 samples

4 4 0.01

pt

+ = =

(c) ( )0.001529

tr 6 (0) 6 4.36 0.022 2 2

x xrm m

= = =R

mse,av

1 675 samples

4 4 0.02

pt

+ = =

1 Note that the formulae in the Supplement give the eigenvalues as 1 2 3, andl l l . In the solutions, the

eigenvalues were re-ordered and re-numbered to conform to the notation in our theory