# Impulse response from difference equation without partial fractions

I have a system with input $x[n]$ and output $y[n]$, described by the difference equation

$$y[n] - 0.3y[n-1] + 0.1 y[n-2] = 3x[n] - x[n -1]$$

and I am trying to find the frequency and impulse responses. The frequency response I found was

$$H(e^{jw}) = \frac{3-e^{-jw}}{1 - 0.3e^{-jw} + 0.1e^{-j2w}}$$

but I am having trouble converting this to $h[n]$ because the denominator can't be factored using partial fractions.

Also, I tried using the recursion method to find $h[n]$ but I am unsure whether the system is causal or not, so I don't know where to start for the recursion. Could anyone give me some tips on what to do next?

## 1 Answer

First of all you can apply partial fraction expansion method to get the inverse DTFT to find the impulse resposne h[n] for this system. You have to assume casual ot anti-causal system.

$$H(e^{jw}) = \frac{3-e^{-jw}}{1 - 0.3e^{-jw} + 0.1e^{-j2w}} = \frac{3-e^{-jw}}{ (1 - \alpha e^{-jw})(1 - \alpha^* e^{-jw})}$$

$$H(e^{jw})= \frac{A}{1 - \alpha e^{-jw}} + \frac{B}{1 - \alpha^* e^{-jw}}$$

Where $\alpha = 0.1500 + j0.2784$ for this particular case. Hence assuming causality and using inverse DTFT look-up tables, deduce as:

$$\boxed{ h[n] = A \alpha^n u[n] + B (\alpha^*)^n u[n] }$$

Time domain recursion is also possible, requiring only two auxiliary values. For example assuming causality one can use recursion from the LCCDE to obtain $y[n]$ for all $n \geq 0$ from the given LCCDE $$y[n] - 0.3y[n-1] + 0.1 y[n-2] = 3x[n] - x[n -1]$$

by first rearranging it for $y[n]$ as $$y[n] = 0.3y[n-1] - 0.1 y[n-2] + 3x[n] - x[n -1]$$

Then the recursion follows for $n \geq 0$ assuming that $x[n]=\delta[n]$ \begin{align} y &= 0.3y[-1] - 0.1 y[-2] + 3x - x[-1] \\ y &= 0.3y - 0.1 y[-1] + 3x - x \\ y &= 0.3y - 0.1 y + 3x - x \\ y[n] &= 0.3y[n-1] - 0.1 y[n-2] \\ \end{align}

Note that in computing $y$ you need $y[-1]$ and $y[-2]$ as auxiliary (initial) conditions. Also note that $x[-1]=0$ and $x=1$ and $x[n]=0$ for all remaining $n>0$.

A similar analysis fallows for the anti-causal case.