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By: Richard BaraniukAs a part of collection:"Signals and Systems"
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FFT convolution DFT signal processing system CTFT DTFT Fourier Technology

Solving Linear Constant-Coefficient Difference Equations

Module by: Richard Baraniuk

Summary: A module concerning the concepts involved in solving linear constant-coefficient difference equations.

Step 1 - Given the input xn x n , find a solution to ∑k=0N a k y p n-k=∑k=0M b k xn-k k N 0 a k y p n k k M 0 b k x n k
Note: Just any old solution will do!
y p n y p n - particular solution.
Solve the homogeneous equation ∑k=0N a k y h n-k=0 k N 0 a k y h n k 0 for y h n y h n - homogeneous solution.
Complete solution given by yn= y p n+ y h n y n y p n y h n

Solving The Homogeneous Equation

What does it mean?

Figure 1
Clearly y h n y h n depends on the INITIAL CONDITIONS of the system T T.
- Linearity
- Time-Invariance
- Causality
will each depend on these conditions.
In this course, we will emphasize the simplest case, when T T is "initially at rest" with "zero initial conditions." → we will get LTI and causal solutions. (although possibly at the expense of stability

Example 1

Solve yn-ayn-1=xn

y n a y n 1 x n

where |a|<1

a 1

for xn=δn

x n δ n

Step 1: Particular Solution - Assume n≥0 n 0 and "zero initial conditions" y p 0=δ0→1+a y p -1→0=1 y p 0 δ 0 1 a y p -1 0 1 y p 1=δ1→0+a y p 0→1=a y p 1 δ 1 0 a y p 0 1 a y p 2=δ2→0+a y p 1→a=a2 y p 2 δ 2 0 a y p 1 a a 2
y p n=an y p n a n (1)
where n≥0 n 0

Figure 2
Step 2: Homogeneous Solution - If xn=0 x n 0 , then y h n-a y h n-1=0 y h n a y h n 1 0 y h n=a y h n-1 y h n a y h n 1 A solution is given by
y h n=can y h n c a n (2)
for all n n.

Figure 3
Step 3: Reconcile - yn= y n p+ y h n y n y n p y h n =anun+can a n u n c a n How to pick c c? Need auxilliary conditions.

Figure 4
If we desire a causal system, then c=0 c 0 and yn=anun y n a n u n

Figure 5
If we desire an anticausal system then choose c=-1 c -1 , so yn=-anu-n y n a n u n This does not assume "system initially at rest!"

Figure 6