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EE313 Linear Systems and Signals
Fall 2010
Initial conversion of content to PowerPoint
by Dr. Wade C. Schwartzkopf
Prof. Brian L. Evans
Dept. of Electrical and Computer Engineering
The University of Texas at Austin
Difference Equations and Stability
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Input
Response
Impulse
n
x
n
h
n
y
s
=
Example: SecondOrder Equation
•
y
[
n
+2]  0.6
y
[
n
+1]  0.16
y
[
n
] = 5
x
[
n
+2] with
y
[1] = 0 and
y
[2] = 6.25 and
x
[
n
] = 4

n
u[
n
]
•
Zeroinput response
Characteristic polynomial
γ
2
 0.6
γ
 0.16 = (
γ
+
0.2) (
γ
 0.8)
Characteristic equation
(
γ
+ 0.2)
(
γ
 0.8) = 0
Characteristic roots
γ
1
= 0.2
and
γ
2
= 0.8
Solution
y
0
[
n
] =
C
1
(0.2)
n
+
10  3
Example: Impulse Response
•
h
[
n
+2]  0.6
h
[
n
+1]  0.16
h
[
n
] = 5
δ
[
n
+2]
with
h
[1] =
h
[2] = 0 because of causality
•
General form of impulse response
h
[
n
] = (
b
N
/
a
N
)
δ
[
n
] +
y
0
[
n
]
u
[
n
]
•
Since
a
N
= 0.16
and
b
N
= 0
,
h
[
n
] =
y
0
[
n
]
u
[
n
] = [
C
1
(0.2)
n
+
C
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This note was uploaded on 02/16/2011 for the course ECE 313 taught by Professor Evans during the Spring '11 term at University of Texas at Austin.
 Spring '11
 EVANS

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