6/28/2016
Z Transforms Finding the inverse Z transform
First Way: Use the definition
Consider: = 1 + 2 + 3 + 2 ELEC 407
Since
→ ( − )
= + 2 − 1 + 3 − 2 + 2 − 3
Z Transforms
= {1,2,3,2}
1
2
Z Transforms Finding the inverse Z transform
Z Transforms Finding the inverse Z transform
Second Way: Use long division
Second Way: Use long division
Consider: = . .
.
Consider: = = 0.5 + + 0.75 + ⋯
0.5 + Z-1 +0.75 Z-2+…
Z2-Z+0.5 0.5 Z2 + 0.5 Z
0.5 Z2 - 0.5 Z + 0.25
substract
Z - 0.25
Z - 1.0 + 0.5 Z-1
0.75 - 0.5 Z-1
0.75 – 0.75 Z-1 +0.375 Z-2
=
. .
.
0.5 + +
0.75 + ⋯
So by inspection
ℎ = 0.5δ + δ − 1 + 0.75δ( − 2)+…
This is only useful for the first few terms of the
sequence.
3
4
1
6/28/2016
Z Transforms Finding the inverse Z transform
Z Transforms Finding the inverse Z transform
Third Way: Partial fraction expansion
Case 1: Non Repeated Real Poles
PFE Case 1 (continued)
Step 1 : Divide H(z) by z
()
4
+
+
= =
+
− 0.25 − 1/2 + 1/2
Step 2: Find +
a) Multiply by denominator of +
1
4
+ +
( − 2)
= +
+ 1/2
+ 1/2
1
b) Evaluate at the pole z = ½
' ( )
Consider: = .
= /
+
( /
And so from inspection of the z transform table
1
1
ℎ = + ( ), ++
(− ),
2
2
)
'( )
And all we need is + -./+
5
/
/
=
()
) )
+
( ()
/
0
→ + = = 2
6
Z Transforms Finding the inverse Z transform
Z Transforms Finding the inverse Z transform
Step 3: Find +
a) Multiply by denominator of +
1
+ ( + )
4
2 ++
=
− 1/2
− 1/2
b) Evaluate at z = -1/2 ( which eliminates C1)
−2
= +
→ +
= 2
−1
PFE Case 2: Non Repeated Complex poles
So = And finally
1()
2
Consider: = Step 1: Divide Y(z) by z
4 ()
=
( − 0.5)( + 0.5)( − 5)( + 5)
(
)
= .
+
+2
/
/
,
2( ⁄
) +2(⁄
),
=2
ℎ =
' 4
( .
)( )
7
(
.
(
(
4
+ 6
+ 76
8
2
6/28/2016
Z Transforms Finding the inverse Z transform
Z Transforms Finding the inverse Z transform
Step 2: Find the C’s
4 ( − 0.5)
+ =
|
( − 0.5)( + 0.5)( + 1) 9:.
4(0.5)
=
= 0.8
(0.5 + 0.5)(0.5
+ 1)
Step 2: Find the C’s
4 ( − 5)
+ = |
( − 0.25)( + 5)( − 5) 9:6
4(5)
= = −1.65 = 1.6= 6>/
(5 − 0.25)(5 + 5)
4 ( + 0.5)
+
=
|
( − 0.5)( + 0.5)( + 1) 9:.
4(−0.5)
=
= −0.8
(−0.5 − 0.5)((−0.5)
+1)
+' =
4 ( + 5)
|
( − 0.25)( + 5)( − 5) 9:6
4(−5)
= 1.65 = 1.6= 6>/
=
( −5 − 0.25)(−5 − 5)
9
Z Transforms Finding the inverse Z transform
Step 3: ?()
=
2
6>
Z Transforms Finding the inverse Z transform
6>
+ 1.6= (=
6>,
)
= 0.8(0.5), −0.8 −0.5
1.6(=
,
C@(ABCA)
+ 1.6=
>
6>
=
Has
6>,
Thus:
,
+ 1.6 =
@ ABCA
.JK C@A/ →
K @A/
1.6= 6>/
(=
A
6 ,
)
+
)
G
G
= 0.8(0.5) −0.8 −0.5 + 3.2 cos −
2
2
G
,
,
= 0.8(0.5) −0.8 −0.5 + 3.2sin( )
2
,
.JK C@A/ .JK @A/ +
K @A/
K C@A/
form: ↔ (-),
L
Note:
0.8
0.8
1.6= 1.6= 6 −
+
+
=
>
6>
− 0.5 + 0.5
− = 6 −=
= 0.8(0.5), −0.8 −0.5
10
and:
,
11
.JK @A/ →
K C@A/
1.6= 6>/
(=
A
6 ,
)
12
3
6/28/2016
Z Transforms Finding the inverse Z transform
And 1.6= 6>/
{−=
A
6( ),
A
6
,
−1.65{= + =
− =
A
6( ),
A
6( ),
A
A
@
B
C@( )B
K K 3.2{
}
6
>
3.2sin( )
Z Transforms Finding the inverse Z transform
Case 3 Repeated Roots
}=
Consider = }=
' 4
( .
)()
Step 1 Y(z)/z
()
4 =
( − 0.5)( + 0.5)( − 1)
+
+
+
+'
=
+
+
+
− 0.5 + 0.5 ( − 1)
−1
Step 2
4 ( − 0.5)
+ =
|
=4
( − 0.5)( + 0.5)( − 1)
:.
=
13
14
Z Transforms Finding the inverse Z transform
Z Transforms Finding the inverse Z transform
4 ( + 0.5)
4
+
=
|
=
−
:.
( − 0.5)( + 0.5)( − 1)
9
()
4 |: =
( − 0.5)( + 0.5)( − 1)
+
+
+
+'
=
+
+
+
− 0.5 + 0.5 ( − 1)
−1
()
4(0)
|: =
(−0.5)(0.5)(−1)
16O
4O
4
3 + +'
=
+ 9+
−0.5 0.5 (−1)
−1
+' = −32/9
+ =
' ()() |
.( .)() :
=16/3
How to find C4?
Evaluate Y(z)/Z at z=0 or some other convenient
value of z.
15
16
4
6/28/2016
Z Transforms Finding the inverse Z transform
Z Transforms Finding the inverse Z transform
Some Partial Fraction checks
Step3
4 () =
( − 0.5)( + 0.5)( − 1)
16O − 4O9 4
3
=
+
+
− 0.5 + 0.5 ( − 1)
−32O 9
+
−1
Step 4
= {4
.
,
− 1. Compare to long division
2. Reassemble the fractions and compare
3. Check final result against values found
recursively
4. Check for complex conjugates
5. Check using computer. {matlab residue}
'
J
. B + , T }1(k)
T
17
18
19
5