I need to find the Fourier transform of the following signal:


The answers start by using the rule of the basic signal:

$$a^nu[n] \rightarrow \frac{1}{1-ae^{-j\omega}} $$

and then uses the property of differentiation in the frequency domain. However I think there's something wrong with the way the rule is used. The answers assume that:

$$e^{-an}u[n] = a^nu[n] \rightarrow \frac{1}{1-ae^{-j\omega}} $$

Is there something wrong or am I missing something?

Here's the answer for reference:

The answer


You're right, the first Fourier transform correspondence in your reference is wrong. It should be

$$\mathcal{F}\{e^{-\alpha n}u[n]\}=\frac{1}{1-e^{-\alpha}e^{-j\omega}}\tag{1}$$

You just need to substitute $a=e^{-\alpha}$ and use the formulas you know.

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Let's find the DTFT of $x[n]=e^{-an}u[n]$ by definition:

$$X(e^{j\omega})=\sum\limits_{n=-\infty}^\infty x[n]e^{-j\omega n}=\sum\limits_{n=-\infty}^\infty e^{-an}u[n]e^{j\omega n}=\sum\limits_{n=0}^\infty e^{-an -j\omega n}=\sum\limits_{n=0}^\infty \left(e^{-a-j\omega}\right)^n$$

This is a famous series and its result is

$$X(e^{j\omega})=\sum\limits_{n=0}^\infty \left(e^{-a-j\omega}\right)^n=\frac{1}{1-e^{-a-j\omega}}=\frac{1}{1-e^{-a}\cdot e^{-j\omega}}$$

Another way of seeing it without doing this by definition is by using the property you said $$b^nu[n] \rightarrow \frac{1}{1-be^{-j\omega}}$$ considering that $b=e^{-a}$.

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