ECE 565 Homework 2
Do homework problems in Phillips Text for next class:
2-4. Find the z-transform of the number sequence generated by sampling e(t) at T seconds
beginning at t = 0.
2(1   5s )
where L{e(t)} = E(s) =
s( s  2)
T = 1second
E1(s) =
2
1
1
 
s( s  2) s s  2
inverse Laplace Tranfom: e1(t) = (1--2 t)u(t)  e(kT) = (1--2 kT)u(kT)
definition of z-transform:
E1(z) = (1 + z -1 +z -2 + z -3 + ….) – (1 -  -2Tz -1 +  - 4Tz-2 +  -6Tz -3 + …)
E1(z) =
1
1 z
1

1
1
2T
z 1
grouping terms:
z[( z   2T )  ( z  1)]
z
z
E1(z) =


z  1 z   2T
( z  1)( z   2T )
E1(z) =
z (1   2T )
z   2T ( z  1)
delay property gives: E(z) = E1(z) – z -5 E1(z)
substitute & simplifiy
E(z) =
E(z) =
z(1   2T )  z 4 (1   2T )
( z   2T )( z  1)
(1   2 )( z 5  1)
z 4 ( z   2 )( z  1)


T 1
z 4 ( z 5  1)(1   2 )
( z   2 )( z  1)
0.8764( z 5  1)
z 4 ( z  1)( z  0.1353)
2-5.
a. for E(z) =
e(k) =
0.1
, find e(0), e(10), and e(10) using inversion method
z ( z  0.9)
1
k 1
of E(z)z k-1 ]
 [residues
 E ( z ) z dz 
k-1
2 
at pole sof E(z)z
for a simple pole at z = a: residue z=a = [ z  a ] E( z )z k 1
for a pole of order m at z = a: residue z=a =
e(k) = resicdues
k 1
z a
1
d m1
[ z  a]m E ( z ) z k 1
( m  1)! dz m1
k 2
0.1z
0.1z

z( z  0.9 ) ( z  0.9 )
for k = 0 : e(0) = resicdues
residue|z=0 =
z ( z  0.9 )
d
0.1
dz ( z  0.9 )
residue|z=0.9 =
thus e(0) =
0.1
2
0.1
2
z
0.1
0.9
2

z 0.9

0.1
0.9 2
for k = 1 : e(1) = resicdues
residue|z=0 =
0.1
z  0.9

z 0
0.1
( z  0.9 )
0.1
0.9 2
0
0.1
z( z  0.9 )

z 0
0.1
0.9
0.1
0.1

z z 0.9 0.9
0.1 0.1

0
thus e(1) =
0.9 0.9
residue|z=0.9 =
for k = 10 : e(10) = resicdues
residue|z=0.9 = 0.1z 8
z 0.9
thus e(10) = 0.1(0.9)8
0.1z 8
( z  0.9 )
 0.1( 0.9 )8

2
z 0
0.1
0.9 2
z a
(2-5) - continued
b. check e(0) using initial value property
e(0) = lim E( z )
z 
e(0) = lim
z 
0.1
0
z( z  0.9 )
c. check values in (a) using partial fractions expansion
A
A
E( z )
0.1
B
 2
 21  2 
z
z
z  0.9
z ( z  0.9 ) z
0.1
z  0.9
A1 =
z 0
d 0.1
dz z  0.9
A2 =
B=

0.1
E( z )  
z
2
z
2

z 0
0.1
( z  0.9 )

2
z 0

1
0.81

0.1234 z 0.1234 z

z
z  0.9
z 0.9
0.1111z
1
9
0.1
0.81
e(k) = -0.1111 (k-1) – 0.1234(k) + 0.1234 (0.9)k
e(0) = 0 – 0.1234 + 0.1234
e(1) = -0.1111 + 0.1234 (0.9)k = 0
e(10) = 0.1234 (0.9)10 = 0.1 (0.9)8
(2-5) - continued
d. find e(k) for k = 0,1,2,3,4 for
Z{e(k)} =
expand into power series: E(z) =
1.98z
z  ...
5
1.98z
( z  0.9z  0.9)( z  0.8)( z 2  1.2z  0.27)
2
 1.98z 4  (...) z 5  (...) z 6  ...
e(0) = e(1) = e(2) = e(3) = 0, e(4) = 1.98
e. find e(t) that when sampled at T = 0.1 results in a transform of E(z) = 2z/(z-0.8)
2z
2z
E(z) =

z  0.8 z   aT
-aT

= 0.8,
T = 0.1
 a = -ln(0.8)/0.1 = 2.231
e(t) = 2 -2.231tu(t)
f. repeat for (e) for E(z) = 2z/(z+0.8)

2z
2z


2
(   aT z 1 ) k
E(z) =

z  0.8 z  (  aT )
k 0

-aT
 j = -0.8, ( j = -1)
aT= 2.231
(-0.8)k changes sign at each sample  sinusoid with period = 2T
e(t) = 2at cos(ws/2) =- 2 -2.231 cos(10)
g. what is the effect of chainging the sign on a real pole?
e. e(k) = (0.8)k ,  sampled decaying exponential
f. e(k) = (-0.8)k, sign alternates  sampled decaying sinusoid
2-7 a thru c
for the sequence {e(k)}
z
E(z) =
( z  1) 2
a. apply the final value theorm (FVT)
lim e( k )  lim ( z  1) E ( z )
k 
z 1
lim e( k )
k 
z ( z  1)
( z  1) 2
0
z 1
b. check reuslts by finding the inverse z-transform
E( z ) 
z
( z  1) 2
residue z  - 1 
1 dz k
1! dz
 kz k 1
z  1
z  1
 k ( 1 k 1 )
- increases without bound
- FVT only works if bounds exists!
c. repeat a & b for E(z) = z/(z-1)2
apply the final value theorm
lim e( k )  lim ( z  1) E ( z )
k 
lim e( k )
k 
z 1
z ( z  1)
( z  1)2
z 1
z
lim e( k ) ( z  1)
k 

z 1
check by inverse z-transform
Z-1{E(z)} = e(k) = k  unbounded (z-transform table)