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

Question

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 $h[n]$.

Thank you for any help you can give me.

Answer 1

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

Answer 2

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

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

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

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. $u[n-1]$