# Impulse response of an LTI system given the input and output signals

I have been given the input and output signals of an LTI system as:

$$x[n] = (\frac{1}{2})^nu[n] + 2^nu[-n-1]$$

$$y[n] = 6(\frac{1}{2})^nu[n] - 6(\frac{3}{4})^nu[n]$$

I have found the system function $$H(z)$$ by using $$H(z) = \frac{Y(z)}{X(z)}$$ and using the standard transform tables to get:

$$X(z) = \frac{1}{1-\frac{1}{2} {z^{-1}}} - \frac{1}{1-2z^{-1}} = \frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-2z^{-1})}$$

$$Y(z) = 6*\frac{1}{1-\frac{1}{2} {z^{-1}}} - 6*\frac{1}{1-\frac{3}{4}z^{-1}} = \frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}$$

$$H(z) = \frac{Y(z)}{X(z)} = \frac{\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}}{\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-2z^{-1})}} = \frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}$$

Then this can be further simplified to:

$$H(z) = 1 - \frac{\frac{5}{4}z^{-1}}{1-\frac{3}{4}z^{-1}}$$

However from here I am stuck on how to get the form of the second term back into the time domain as it does not match anything from the standard tables.

I am just wondering if I have missed anything or made a mistake in my calculations, or if there is some way to get it into time domain form from here that I am not seeing, so that i can get the impule response $$h[n]$$.

For a term $$\frac{1}{1-az^{-1}}$$ it can either be $$a^nu[n]$$ with Region of Convergence(ROC) $$|z| \gt |a|$$ OR $$-a^nu[-n-1]$$ with ROC $$|z| \lt |a|$$.

Since your input and output are both stable, your $$h[n]$$ is also stable. If you see its original form of $$\frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}$$, it has pole at $$z=3/4$$, it's ROC is $$|z| \gt 3/4$$ so that the region includes unit circle. Since this region doesn't include $$z=0$$, there will not be negative powers of $$z$$. So your $$h[n]$$ will be of the composed of only negative powers of $$z$$. So from the final form of $$H(z)$$, the first term $$1$$ will translate to $$\delta[n]$$. The second term $$(\frac{3}{4})^nu[n]$$ will be scaled by $$5/4$$ and delayed by 1 sample due to presence of $$z^{-1}$$. So $$h[n] = \delta[n] - (\frac{5}{4})(\frac{3}{4})^{n-1}u[n-1]$$ OR equivalently $$h[n] = (\frac{3}{4})^nu[n] - 2(\frac{3}{4})^{n-1}u[n-1]$$

$$H(z) = \frac{Y(z)}{X(z)} = \frac{\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}}{\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-2z^{-1})}} = \frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}$$

$$H(z)= \frac{1}{1-\frac{3}{4}z^{-1}}-\frac{2z^{-1}}{1-\frac{3}{4}z^{-1}}$$

$$h[n] = (\frac{3}{4})^nu[n] - 2*(\frac{3}{4})^{n-1}u[n-1]$$

Since $$z^{-1}$$(the delay term) is multiplied to the z-transform of $$(\frac{3}{4})^nu[n]$$, the unit step function gets delayed by 1 i.e. $$u[n-1]$$

• In the second term, don't you need to have $\frac{3}{4}^{(n-1)}$ instead of $\frac{3}{4}^n$ ? – jithin Mar 29 '20 at 6:38