1
$\begingroup$

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

$$ne^{-an}u[n]$$

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

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $b=e^{-a}$.

$\endgroup$

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.