2
$\begingroup$

I have the following equation: $$X(e^{jw}) = \frac{e^{-jw} - \frac{1}{5}}{1-\frac{1}{5}e^{-jw}}$$ Applying the fourier transform, I have simplified it to: $$e^{-jw}\sum_{0}^{\infty}(\frac{1}{5})^ne^{-jwn} - \frac{1}{5}\sum_{n=0}^{\infty}(\frac{1}{5})^{n}e^{-jwn}$$

However, I'm not sure how to further simplify it. I would potentially like to use the unit function substitution for the transform. However, I'm not sure how to handle the exponential in the first summation.

$\endgroup$
1
$\begingroup$

For the first term you just have to use index substitution to obtain

$$\begin{align}e^{-j\omega}\sum_{n=0}^{\infty}\left(\frac15\right)^ne^{-j\omega n}&=\sum_{n=0}^{\infty}\left(\frac15\right)^ne^{-j\omega (n+1)}\\&=\sum_{n=1}^{\infty}\left(\frac15\right)^{n-1}e^{-j\omega n}\\&=\sum_{n=0}^{\infty}\left(\frac15\right)^{n-1}u[n-1]e^{-j\omega n}\end{align}$$

where $u[n]$ is the unit step.

$\endgroup$
1
$\begingroup$

Put $$ y = e^{\, - j\,\omega } $$ then apply partial fractions $$ \frac{{y - 1/5}} {{1 - y/5}} = - \frac{{5y - 1}} {{y - 5}} = - \frac{{5y - 25 + 24}} {{y - 5}} = - 5 - \frac{{24}} {{y - 5}} $$

$\endgroup$

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.