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() b u+
() =
 y 1 b u 1 1 ( ) a y− 1 ( ) 0
0
0

( ) b u+
0
for k = 1 ,  = bu 1 1 () a b u− 1 0 () 0
0
 = () (b+ −
 bu 1 1 a b ) ( )u 0
0
1 0
Similarly,
y()2 = b u() (2 + b − a b u) ( )1 − a b( − a b u) ( )0
0 1 10 1 1 10
y( )3 = b u( ) (3 + b − ab u) ( )2 − a b( − a b u) ( )1 + a b( − ab u) ( )0
2
0 1 10 1 1 10 1 1 1 0
or, more generally, if:
y(k) = w(0)u(k) + w(1)u(k – 1) + … + w(k)u(0)
then,

w()0 = b
0
−
, , ,3 …
wi() =− a ) ( b − a b ) for i = 12
i 1
(
1 1 1 0
In-Class Exercises
Pb. 2.17 Using the convolution-summation technique, find the closed form
solution for:

y k() = uk() − 1 uk( − 1 ) + 1 y k( − 1 )
3 2

 uk() = 0 for negative
k
and the input function given by: 
 uk() = 1 otherwise
Compare your analytical answer with the numerical solution.

Pb. 2.18 Show that the resultant weight functions for two systems are,
respectively:

w(k) = w (k) + w (k) if connected in parallel
1
2
k
wk() = ∑ w i w k i( ) 1 ( − ) if connected in cascade
2
i=0

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