# LTI system, Fourier transform, unit step function, sine wave

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.

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 $\cos(\omega_0t)=\frac12(e^{j\omega_0 t}+e^{-j\omega_0 t})$.

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

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 $\cos(\omega_0t)$