ECE 308 -8
Solution of Linear Constant-Coefficient
Difference Equations
Z. Aliyazicioglu
Electrical and Computer Engineering Department
Cal Poly Pomona
Solution of Linear Constant-Coefficient Difference
Equations
Two methods
Direct method
Indirect Method (z-transform)
Direct solution Method:
The total solution is the sum of two parts
Part 1 homogeneous solution
Part 2 particular solution
The Homogeneous solution
Assuming that the input . Since , this gives us the zero-input
response of the system
N
∑a
k =0
k
y (n − k ) = 0
ECE 308-8 2
1
Solution of Linear Constant-Coefficient Difference
Equations
The solution is the form of an exponential
yh ( n ) = λ n
substitute this in the previous equation.
λ n + a1λ n−1 + a2 λ n−2 + ... + aN −1λ n−( N −1) + aN λ n− N = 0
λ n− N ( λ N + a1λ N −1 + a2λ N −2 + ... + aN −1λ + aN ) = 0
This is called characteristic polynomial of the system. It has N
roots and denotes by λ1 , λ2 ,..., λN
The roots can be real or complex or some roots are identical.
ECE 308-8 3
Solution of Linear Constant-Coefficient Difference
Equations
Let assume that roots are real and not identical, the solution
becomes
yh (n) = C1λ1n + C2 λ2n + ... + C N λNn
The coefficients Ci , i = 1,2,..., N are determined from the initial
conditions.
If there are two identical roots, the solution becomes
yh (n) = C1λ1n + C1nλ1n + C2λ2n + ... + CN λNn
ECE 308-8 4
2
Solution of Linear Constant-Coefficient Difference
Equations
Example:
Find the zero-input response for the second-order difference
equation
y ( n) − 3 y ( n − 1) − 4 y ( n − 2) = 0
The homogeneous solution form yh ( n) = λ n
λ n − 3λ n−1 − 4λ n−2 = 0
λ n−2 (λ 2 − 3λ − 4) = 0
λ1 = −1, λ2 = 4
The homogenous solution is
yh (n) = C1λ1n + C2 λ2n
= C1 (−1) n + C2 4n
ECE 308-8 5
Solution of Linear Constant-Coefficient Difference
Equations
The Particular solution:
Causal system is the output is depends only on present and past
input signal
Input Signal x ( n )
Particular Solution
y p ( n)
A (constant)
K
AM n
KM n
An M
An n M
K 0 n M + K1n M −1 + ... + K M
 A cos ω 0 n 


 A sin ω 0 n 
K1 cos ω 0 n + K 2 sin ω 0 n
An ( K 0 n M + K1n M −1 + ... + K M )
ECE 308-8 6
3
Solution of Linear Constant-Coefficient Difference
Equations
Example:
Find the particular solution for for
y ( n) =
5
1
y (n − 1) − y (n − 2) + x(n)
x ( n) = 2n , n ≥ 0
6
6
The particular solution form
,
y p ( n) = K 2 n u ( n)
K 2 n u ( n) =
y p ( n) = K 2n , n ≥ 0
5
1
K 2n−1 u (n − 1) − K 2n−2 u (n − 2) + 2n u (n)
6
6
Evaluate the equation for n ≥ 2
5
1
4K = 2K − K + 4
6
6
K=
8
5
Therefore, the particular solution is
8
y p ( n) = 2 n , n ≥ 0
5
ECE 308-8 7
Solution of Linear Constant-Coefficient Difference
Equations
The total solution of the difference equation
y (n) = yh (n) + y p (n)
ECE 308-8 8
4