0
$\begingroup$

Find the inverse Z transform:

enter image description here

I have done the solution but my answer does not match with the one given in the textbook. What I may have done wrong?

enter image description here

$\endgroup$
1
$\begingroup$

HINT: Your partial fraction expansion is wrong; you should always check it by recombining the factors. The correct expansion is

$$X(z)=\frac{z^{-1}-\frac12}{\left(1-\frac12 z^{-1}\right)^2}=\frac{\frac32}{\left(1-\frac12 z^{-1}\right)^2}-\frac{2}{\left(1-\frac12 z^{-1}\right)}\tag{1}$$

You should try to derive this result yourself.

BUT: note that you don't even need partial fractions here because if you know (or look up) the $\mathcal{Z}$-transform correspondence

$$\frac{az^{-1}}{\left(1-a z^{-1}\right)^2}\Longleftrightarrow na^nu[n]\tag{2}$$

then you can directly apply it by writing $X(z)$ as

$$X(z)=2\frac{\frac12z^{-1}}{{\left(1-\frac12 z^{-1}\right)^2}}-z\frac{\frac12z^{-1}}{{\left(1-\frac12 z^{-1}\right)^2}}\tag{3}$$

Finally, note that the given result is wrong. It should be

$$x[n]=2n\left(\frac12\right)^nu[n]-(n+1)\left(\frac12\right)^{n+1}u[n+1]\tag{4}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.