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Given the causal system with transfer function

$$ H(z) = \frac{z^{-1} + \frac{1}{2} z^{-2}}{1-\frac{3}{5} z^{-1} + \frac{2}{25} z^{-2}} $$

I can calculate the impulse response to be:

$$ h[n] = \frac{25}{4} \delta[n] - \frac{35}{2} \left(\frac{1}{5}\right)^n u[n] + \frac{45}{4} \left(\frac{2}{5}\right)^n u[n] $$

What is the constant coefficient difference equation relating input and output representing this system?

If I split out the three terms of the impulse function, I can calculate separate difference equations for each term separately, but I'm having trouble combining them back together. Maybe that is not the way to go?

Here is my work in that direction:

\begin{align*} y[n] &= h[n] * x[n] = \sum\limits_{k=-\infty}^\infty x[k] h[n-k] = \sum\limits_{k=-\infty}^\infty x[n-k] h[k] \\ y_1[n] &= \frac{25}{4} x[n] \\ y_2[n] &= \sum\limits_{k=-\infty}^\infty h_2[k] x[n-k] \\ y_2[n] &= \frac{-35}{2} \sum\limits_{k=0}^\infty \left(\frac{1}{5}\right)^k x[n-k] \\ y_2[n] &= \frac{-35}{2} x[n] + \frac{-35}{2} \sum\limits_{k=1}^\infty \left(\frac{1}{5}\right)^k x[n-k] \\ y_2[n] &= \frac{-35}{2} x[n] + \frac{-35}{2} \sum\limits_{k=0}^\infty \left(\frac{1}{5}\right)^{k+1} x[n-(k+1)] \\ y_2[n] &= \frac{-35}{2} x[n] + \frac{1}{5} \cdot \frac{-35}{2} \sum\limits_{k=0}^\infty \left(\frac{1}{5}\right)^k x[(n-1)-k] \\ y_2[n] &= \frac{-35}{2} x[n] + \frac{1}{5} y_2[n-1] \\ y_3[n] &= \frac{45}{4} x[n] + \frac{2}{5} y_3[n-1] \\ y[n] &= \frac{25}{4} x[n] + \frac{-35}{2} x[n] + \frac{1}{5} y_2[n-1] + \frac{45}{4} x[n] + \frac{2}{5} y_3[n-1] \\ y[n] &= \frac{1}{5} y_2[n-1] + \frac{2}{5} y_3[n-1] \\ y[n-1] &= \frac{25}{4} x[n-1] + y_2[n-1] + y_3[n-1] \\ \end{align*}

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We know that the unit impulse response is the transfer function, hence we can write:

\$\dfrac{Y(z)}{X(z)}= H(z) = \dfrac{z^{-1} + \frac{1}{2} z^{-2}}{1-\frac{3}{5} z^{-1} + \frac{2}{25} z^{-2}}\$

where \$X(z)\$ and \$Y(z)\$ are the input and output signals, respectively.

Cross-multiplying:

\$Y(z) (1-\frac{3}{5} z^{-1}+\frac{2}{25} z^{-2}) = X(z) (z^{-1}+\frac{1}{2} z^{-2})\$

Inverse z-transforming and re-arranging:

\$y(n)= x(n-1)+\frac{1}{2} x(n-2) + \frac{3}{5} y(n-1)-\frac{2}{25}y(n-2)\$

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  • \$\begingroup\$ wow, that was way easier than what I was doing. Thank you! \$\endgroup\$ – clay Sep 16 '15 at 23:26

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