Connexions

You are here: Home » Content » Solving Linear Constant-Coefficient Difference Equations
Content Actions
Lenses

What is a lens?

Lenses

A lens is a custom view of Connexions content. You can think of it as a fancy kind of list that will let you see Connexions through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to Connexions materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual Connexions member, a community, or a respected organization.

This content is ...
Affiliated with (?)
This content is either by members of the organizations listed or about topics related to the organizations listed. Click each link to see a list of all content affiliated with the organization.
  • This module is included inLens: Rice University OpenCourseWare
    By: OpenCourseWare ConsortiumAs a part of collection:"Signals and Systems"

    Click the "Rice University OCW" link to see all content affiliated with them.

    Rice University OCW
Also in these lenses
  • This module is included inLens: richb's DSP resources
    By: Richard BaraniukAs a part of collection:"Signals and Systems"

    Comments:

    "My introduction to signal processing course at Rice University."

    Click the "richb's DSP" link to see all content selected in this lens.

    richb's DSP
Tags

(?)

These tags come from the endorsement, affiliation, and other lenses that include this content.

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?
    fig1.png
    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 n0 n 0 and "zero initial conditions" y p 0=δ01+a y p -10=1 y p 0 δ 0 1 a y p -1 0 1 y p 1=δ10+a y p 01=a y p 1 δ 1 0 a y p 0 1 a y p 2=δ20+a y p 1a=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 n0 n 0
    fig2.png
    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.
    fig3.png
    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.
    fig4.png
    Figure 4
    If we desire a causal system, then c=0 c 0 and yn=anun y n a n u n
    fig5.png
    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!"
    fig6.png
    Figure 6
    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" 
fig7.png
Figure 7
Impulse Response:
fig8.png
Figure 8
so Frequency Response: H=𝔽H δ 0 =1Nn=0N-1hn-2πNkn=1Nn=0N-1 δ 0 n-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
fig9.png
Figure 9

Comments, questions, feedback, criticisms?

Send feedback