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Math Help - Causal LTI system: homogeneous-, impulse- and step-response

  1. #1
    Member courteous's Avatar
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    Causal LTI system: homogeneous-, impulse- and step-response

    1. The problem
    A causal LTI system is described by the difference equation y[n] - 5y[n-1] + 6y[n-2] = 2x[n-1].

    Determine the
    (a) homogeneous ( x[n]=0 for all n),
    (b) impulse,
    (c) step
    response of the system.

    2. Relevant equations and definitions
    2.1 Causal are systems for which the output y[n_0] depends only on the input samples x[n], for n \leq n_0

    2.2 For the (a) part of the problem:
    \sum_{k=0}^{N} a_k y_h[n-k] = 0 \text{    ................... (2.96)}
    Equation (2.96) is called the homogeneous difference equation and y_h[n] the homogeneous solution. The sequence y_h[n] is in fact a member of a family of solutions of the form
    y_h[n] = \sum_{m=1}^{N} A_m z_m^{n} \text{    ................... (2.97)}
    where the coefficients A_m can be chosen to satisfy a set of auxiliary conditions on y[n].

    3. The attempt at a solution
    First of all, there is no additional information to be gained from knowing that the LTI system is also "causal", right? That is, being causal ( y[n_0] only depends on x[n], n \leq n_0) is already in the given equation, correct?

    (a) homogeneous response, y_h[n]
    The official solution is given in the form of Eq. (2.97): y_h[n] = A_1(2)^n + A_2(3)^n

    I have no idea how one should arrive at this solution. Neither writing it out for particular n's (and hoping to glimpse a pattern), nor writing the system in the general form of \sum_{k=0}^N a_k y[n-k] = \sum_{m=0}^M b_m x[n-m] and then finding the relevant coefficients (N=2: a_0=1, a_1=-5, a_2=6; M=1: b_0=0, b_1=2), gives way toward the solution.

    (b) impulse response, h[n]
    I do get the official solution ( h[n] = 2(3^n - 2^n)u[n]), but I'm exploiting the Z-transform, which is introduced only after this chapter ... so there must be other way without Z-transform:

    H(z) = \frac{Y(z)}{X(z)} = \frac{2z^{-1}}{1 - 5z^{-1} + 6z^{-2}} = 2\left(\frac{1}{1-3z^{-1}} - \frac{1}{1-2z^{-1}}\right) \Rightarrow h[n] = 2(3^n - 2^n)u[n]

    (c) step response, s[n]
    As in the (a) part, I'm completely stuck here. The solution should be s[n] = \left( -8(2)^{n-1} + 9(3)^{n-1} + 1 \right) u[n], which smells of Z-transform, but I'm also interested if/how could one do without it?

    All I can think of for (c) is using the result from (b) (the impulse response), but I get stuck:
    s[n] = \sum_{k=-\infty}^{\infty} x[n-k] h[k]= \sum_{k=-\infty}^{\infty} u[n-k] 2(3^k - 2^k)u[k] = 2 \left( \sum_{k=-\infty}^{\infty} 3^k u[k] u[n-k]  -  \sum_{k=-\infty}^{\infty} 2^k u[k] u[n-k] \right) = \text{?}
    Last edited by courteous; September 30th 2012 at 04:57 AM.
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  2. #2
    Member courteous's Avatar
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    Re: Causal LTI system: homogeneous-, impulse- and step-response

    Bumping in hope of an answer.
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  3. #3
    Member courteous's Avatar
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    Re: Causal LTI system: homogeneous-, impulse- and step-response

    Think I've found the solution to (a), but I'm still stuck with (c).

    (a) Since every complex number can be represented as exponential A e^{st}, all that is needed for a homogeneous response, is to solve the characteristic polynomial s^2 - 5s + 6 = (s-3)(s-2) = 0. Hence, y_h is of the form A_1 (2)^n + A_2(3)^n.

    (c) Can somebody help me out? Is there an easy way to finding the step response?
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  4. #4
    Member courteous's Avatar
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    Re: Causal LTI system: homogeneous-, impulse- and step-response

    (c) The solution (the step response) in t-domain via evaluating the convolution sum:
    s[n] = \sum_{k=-\infty}^{\infty} x[n-k] h[k]= \sum_{k=-\infty}^{\infty} u[n-k] 2(3^k - 2^k)u[k] = 2 \sum_{k=0}^{n} (3^k - 2^k) = 2( \frac{1-3^{n+1}}{1 - 3} - \frac{1-2^{n+1}}{1-2} ) = 3^{n+1} - 2^{n+2} + 1

    Perhaps simpler to go directly in the z-domain: S(z)=H(z)U(z)=...

    Thanks for the help.
    Last edited by courteous; October 14th 2012 at 11:19 PM.
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