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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.

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2 Answers 2

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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.

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

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