# 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?

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] &= 0.3y[-1] - 0.1 y[-2] + 3x[0] - x[-1] \\ y[1] &= 0.3y[0] - 0.1 y[-1] + 3x[1] - x[0] \\ y[2] &= 0.3y[1] - 0.1 y[0] + 3x[2] - x[1] \\ y[n] &= 0.3y[n-1] - 0.1 y[n-2] \\ \end{align}

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

A similar analysis fallows for the anti-causal case.