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I am stuck on a problem about finding the complete solution of a Discrete Linear System by using the parametrization $y_{c}(n) = y_{h}(n) + y_{p}(n)$ where $y_{h}(n)$ and $y_{p}(n)$ are, respectively, the homogenous and particular solutions.

The System, in this case, is given by the following difference equation:

$y(n) = y(n-2) + x(n) + x(n-2)$ $\{y(-2) = 1; y(-1) = -1\}$

The input sequence is $x(n) = n^{2}u(n-1)$.

I had no problems finding the homogenous solution. It was pretty straightforward. The next step would be to find the particular solution and then calculate the constants that appear on the homogenous solution. I chose a model for $y_{p}(n)$ by looking at a table, which, in this case, yielded: $y_{p}(n) = (K_{1}n^{2} + K_{2}n + K_{3})u(n-1)$

The problem was that after plugging this model on the system's difference equation and doing some algebra, the constant $K_{3}$ canceled out and I arrived at the following expression:

$-4K_{1} + 4nK_{1} + 2K_{2} = 2n^{2} - 4n + 4$

That was the first example where a constant got canceled and where I found no expressions of $K_{n}$ multiplying $n^{2}$, in order to create a linear system of equations and find the constants. My book hasn't yet talked about such cases, and I don't know how to proceed. Am I doing things right? How can I proceed in order to solve this problem? Thanks in advance.

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  • $\begingroup$ "homogeneous and particular solutions" have meaning with continuous differential equations, but i am not sure of their meaning in a discrete-time context. i suppose discrete-time LTI systems have a "natural response" and "forced response" or "transient response" and "steady-state response", so i guess we can attach meaning. $\endgroup$ – robert bristow-johnson Mar 18 '14 at 2:46
  • $\begingroup$ @robertbristow-johnson: You're correct, there is an analog for discrete-time systems. The homogeneous and particular solution method can be used to calculate the impulse response for an IIR filter, for example. I'd have to dig out a textbook to recall how to do it, though, as I've never really used it in practice (it's usually a lot easier to just inverse-$z$-transform the transfer function). $\endgroup$ – Jason R Mar 18 '14 at 16:44
  • $\begingroup$ as far as i can tell, the particular solution has nothing to do with the impulse response. the impulse response is the response to a Kronecker impulse which can be modeled as a zero input (because $\delta[n]=0$ for all $n \ne 0$) with an initial state that might be non-zero. that is what the homogeneous solution is (a.k.a. "natural response" or "transient response"). for the particular solution, you have to specify the particular driving function. $\endgroup$ – robert bristow-johnson Mar 18 '14 at 23:59
  • $\begingroup$ I think you're right. Instead of impulse response, I think I should have said "response for a specified input signal." $\endgroup$ – Jason R Mar 19 '14 at 2:22

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