0
$\begingroup$

I just finished learning about Fourier Transforms and don't understand this signal:

$$x(t) = \cos(\omega t)u(t) $$

This is a cosine wave but only where $\omega$ is positive. My question is what can I do to figure out the output when the impulse response $h(t) = e^{-2\pi t}u(t)$ ?

I first try to break $x(t)$ up using Euler to get:

$$x(t) = \frac{1}{2} \left[e^{j2\pi t} + e^{-j2\pi t}\right]$$

then get FT of each component and use convolution property to get:

$$Y(j\omega) = H(j\omega)X(j\omega)$$

but kept getting an incorrect answer.

$\endgroup$
1
$\begingroup$

Some errors in your question were already pointed out in gsmafra's answer. Here I'll give you some more hints.

The input signal is a cosine that is switched on at $t=0$. There's no real shortcut here, you just have to compute the convolution integral:

$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau\tag{1}$$

Since the input signal starts at $t=0$, and since the system is causal (i.e., $h(t)=0$ for $t<0$), the integral (1) can be written as

$$y(t)=u(t)\int_{0}^{t}\cos(\omega_0\tau)e^{-2\pi(t-\tau)}d\tau= e^{-2\pi t}u(t)\int_{0}^{t}\cos(\omega_0\tau)e^{2\pi\tau}d\tau\tag{2}$$

This integral can be easily computed using $\cos(\omega_0t)=\frac12(e^{j\omega_0 t}+e^{-j\omega_0 t})$.

$\endgroup$
  • $\begingroup$ this helped a lot, I got it figured out now, thank you! $\endgroup$ – Bryan Ehlers May 9 '15 at 15:53
2
$\begingroup$

I can see three problems here:

1) $x(t)$ is a cosine wave but only where $t$ is positive, not $\omega$

2) You must keep the step function when applying Euler's identity, and I can't see the cosine frequency in your second expression for $x(t)$ either. Always make sure your identity is right, with WolframAlpha for example

3) Do not mix up your $\omega$ that is a parameter to $\cos(\omega t)$ and the $\omega$ variable of the frequency domain in $X(j\omega)$. Use different notations, like $\cos(\omega_0t)$

$\endgroup$

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.