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Solving Linear Constant-Coefficient Difference Equations

Module by: Richard Baraniuk

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

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  • 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
    Figure 1 (fig1.png)
  • 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
    Figure 2 (fig2.png)
  • 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
    Figure 3 (fig3.png)
  • 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
    Figure 4 (fig4.png)
    If we desire a causal system, then c=0 c 0 and yn=anun y n a n u n
    Figure 5
    Figure 5 (fig5.png)
    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
    Figure 6 (fig6.png)

Notes

  1. Solution 1 was causal and stable.
  2. Solution 2 was anticausal and unstable.
In general, linearity, time-invariance, and causality of a system implemented as a DE will depend on the auxilliary conditions.

Fact:

If we assume that the system is initially at rest ("zero initial conditions"), then it will be LINEAR, TIME-INVARIANT, and CAUSAL.

Note:

Setting input x= δ 0 x δ 0 impulse and setting initial conditions all =0 0 and solving for yp y p yields yp=h y p h as the impulse response of this LSI system.

Example 2: Frequency Response of a "wire"

Figure 7
Figure 7 (fig7.png)
Impulse Response:
Figure 8
Figure 8 (fig8.png)
so Frequency Response: H=𝔽H δ 0 =1N∑n=0N−1hnⅇ-2πNkn=1N∑n=0N−1 δ 0 nⅇ-2πNkn H 𝔽 H δ 0 1 N n N 1 0 h n 2 N k n 1 N n N 1 0 δ 0 n 2 N k n δ 0 n= 1ifn=00otherwise δ 0 n 1 n 0 0 =1N 1 N Flat
Figure 9
Figure 9 (fig9.png)

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