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Suppose the output $y_h[n]$ of a linear, time-invariant system is described by by the following equation for input $x[n]=0$,

$$\sum_{k=0}^{N}a_k y_h[n-k] = 0$$

My book states the $y_h[n]$ is in fact a member of a family of solution of the form:

$$y_h[n]=\sum_{m=1}^{N}A_mz_m^n$$

where the complex numbers $z$ are the zeros of $A(z)=\sum_{k=0}^Na_kz^{-k}$.


What is the proof that $y_h$ is in fact of the form given?

A simple substitution would result in:

$$\sum_{k=0}^N (a_k \sum_{m=1}^NA_mz_m^n) = 0$$

but I don't see to proof showing a set $\{z_1,z_2,...,z_N\}$ and $\{A_1, A_2,..., A_N\}$ indeed exist for all n. Sorry if this seems like a poor question.

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  • $\begingroup$ Do you have a definition for $A_m$? $\endgroup$ – zabop Mar 26 at 16:51
  • $\begingroup$ The first equation doesn't seem to me like a good description of the output. For example, you could make $a_k=0$ and then the equation is true, but it doesn't tell you anything about $y_h$. Also: the output of an LTI system to input $x[n]=0$ must be $y[n]=0$. $\endgroup$ – MBaz Mar 26 at 16:59
  • $\begingroup$ @zabop $A_m$ are the coefficients for output formula. $\endgroup$ – Shukant Pal Mar 26 at 21:01
  • $\begingroup$ @MBaz You're correct that the output of an LTI system for $x[n]=0$ must be $y[n]=0$. I think that right side of the difference equation $\sum_{m=0}^Mb_kx[n-m]$ disappears when $x[n]=0 $ which allows you to solve for $y_h$ easily. $\endgroup$ – Shukant Pal Mar 26 at 21:08
  • $\begingroup$ This question is related to pg. 40 of Alan V. Oppenheim's book "Discrete-Time Signal Processing". There he described how to solve difference equations. $\endgroup$ – Shukant Pal Mar 26 at 21:09
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If you substitute your second equation back into the first equation, you obtain

$$\begin{align}\sum_{k=0}^{N}a_k\sum_{m=1}^NA_mz_m^{n-k}&=\sum_{m=1}^NA_mz_m^n\sum_{k=0}^Na_kz_m^{-k}\end{align}\tag{1}$$

If $z_m$ are zeros of the polynomial

$$A(z)=\sum_{k=0}^Na_kz^{-k}\tag{2}$$

then Eq. $(1)$ equals zero, as required for the homogenous solution.

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