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Difference equation

  1. Apr 27, 2006 #1
    This is not quite a homework question, but I hope close enough that it can be posted here. I'm going through a signal processing book on my free time, doing all the problems and so on, and I've come across a problem which I'm not too sure about how to solve. Here it is--

    Consider the sequence:

    h: Z -> C (integers to complex)

    h(n) = [C1*(p1^n) + C2*(p2^n) + . . . + CM*(pM^n)]u(n) (0),

    where u(n) = 1 for n >= 0, 0 else.

    without using Z transforms, show that h(n) satisfies the difference equation for n >= M:

    h(n) + a1*h(n-1) + a2*h(n-2) + . . . + aM*h(n-M) = 0 (1),

    where {1,a1,a2,...,aM} are the coefficients of the polynomial whose roots are (complex) numbers {p1,p2,...,pM}, that is,

    1 + a1*(z^-1) + a2*(z^-2) + ... + aM*(z^-M) =


    Note: the Ci are arbitrary and the restriction n >= M is necessary.

    By the substitution of (0) into (1) one gets an equation in ai,Ci,pi, but I don't quite see how to show that the coefficients ai are what the problem says they are. I'm thinking about matrix multiplication, determinants and characteristic polynomials but my memory of that stuff is fading and I sold back all my college books which talk specifically about those. Do any of you have any ideas, or know how to do this type of problem? Help would be appreciated. Thanks!
  2. jcsd
  3. Apr 28, 2006 #2
    You don't really have to manually substitute all of (0) into (1) al at once. Note that (1) is a homogenous difference equation, i.e. it is set equal to zero. Thus, sort of like a homogenous differential equation, any g(n) that is a solution, so would any C*g(n). And if f(n) and g(n) are solutions, then any C1*f(n) + C2*g(n) is also a solution. So if you can first just let p be any root of the polynomial, and show p^n is a solution to (1) from there you can easily show that show that (0) is the general solution.
  4. Apr 28, 2006 #3
    Ok, that makes some sense.

    I've worked out explicitely the cases for M = 2, M = 3-- the restriction for n >= M now makes intuitive sense. The full difference equation for all n is inhomogeneous with M shifted unit delta terms as well. For example, for M = 3, the eqn is inhomogeneous with b0*delta(n)+b1*delta(n-1) + b2*delta(n-2) showing up on the right... these terms will vanish and the equation will be homogeneous when n >= 3.
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