Inverse Z transform of $\frac {-81}{z^4(z-3)(z-1)}$

Question

The question given like this:

Find the convolution of $3^nu(-n+3)*u(n-2)$

Attempt:

I tried to solve it through z transform method

Let $Y(z)=X1(z)X2(z)$

Now $Z$ transform of $x1(n)=3^nu(-n+3)$ will be $\frac{-81z^{-3}}{z-3}$

Z transform of $u(n-2)=\frac{1}{z(z-1)}$

now $Y(z)=\frac {-81}{z^4(z-3)(z-1)}$ I tried to take inverse z transform of this function by residue theorem like this

$Residue at z=3,will be \frac{-3^{n-1}}{2}$.

$Residue at z=1,will be \frac{-81}{2}$

$Residue at z=0,will be -40$

BUT the solution given like this $$ y(n)= \begin{cases} \frac{81}{2},n>=5\\ \frac{3^{n-1}}{2},n<5\\ \end{cases} $$ Now What mistake i am doing,any other method other than Residue throem will be helpful. Thanks

Answer 1

I will use the "engineering" approach to find the inverse $Z$ transform by performing a partial fraction expansion and considering each term separately.

A critical point you have to keep in mind is the region of convergence (ROC) of the $Z$ transform. In particular, since the ROC is $|z|>1$ and $|z|<3$ for $X_1(z)$ and $X_2(z)$, repsectively, it follows that the ROC for $Y(z)$ must be $1<|z|<3$.

Performing a partial fraction expansion gives

$$ Y(z)=\frac{81}{2}\frac{z^{-1}}{1-z^{-1}} +\left(- \frac{1}{2}\frac{z^{-1}}{1-3 z^{-1}}\right)+(-40 z^{-1}-39 z^{-2}-36z^{-3}-27 z^{-4}), 1<|z|<3. $$

Now, by the linearity of the (inverse) Z transform, it follows that $y[n]$ is the sum of three sequences corresponding to the inverse $Z$ transform of each of the three terms of $Y(z)$. These inverse transforms can be easily found using standard tables such as the one provided by wikipedia.

For example, for the second term, it holds

$$ \mathcal{Z}^{-1}\left\{ - \frac{1}{2}\frac{z^{-1}}{1-3 z^{-1}}\right\}=\frac{1}{2}3^{n-1} u[-n], $$

where the transform pair #13 of the table was used.

Remark: Pay attention to the ROC for selecting the right transform formula from the table. #12 has the exact same Z expression as #13, however, its ROC is not compatible with the ROC of $Y(z)$.

It follows that

$$ y[n] = \frac{81}{2}u[n-1]+\frac{1}{2}3^{n-1} u[-n]-40 \delta[n-1]-39 \delta[n-2]-36\delta[n-3]-27 \delta[n-4]. $$

By inspection, it follows that $y[n]=81/2, n\geq 5$ and $y[n]=3^{n-1}/2,n\leq 0$. Interestingly, evaluating the above expression for $n=1,2,3,4$ shows that $y[n]=3^{n-1}/2$ also for these values of $n$.