# Some Practice Exam Problems for signals and systems (ECE301)

## Example 1

Compute the Fourier Transform of $x(t)=e^{-t}u(t)$.

$\chi(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt$

$=\int_{-\infty}^{\infty}e^{-t}u(t)e^{-j\omega t}dt$

$=\int_{0}^{\infty}e^{-t}e^{-j\omega t}dt$

$=\int_{0}^{\infty}e^{-(1+j\omega )t}dt$

$=[\frac {e^{-(1+j\omega )t}}{-(1+j\omega)}]|_0^\infty$

$=\frac {e^{-(1+j\omega )\infty}}{-(1+j\omega)}-\frac {e^{-(1+j\omega )0}}{-(1+j\omega)}$

$=0+\frac {1}{(1+j\omega)}$

$=\frac {1}{1+j\omega}$

## Example 2

The impulse response of an LTI system is $h(t)=e^{-2t}u(t)+u(t+2)-u(t-2)$. What is the Frequency response $H(j\omega)$ of the system?

\begin{align} H(j\omega) &= H(\omega) \\ &=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt \\ &=\int_{-\infty}^{\infty}(e^{-2t}u(t)+u(t+2)-u(t-2))e^{-j\omega t}dt \\ &=\int_{-\infty}^{\infty}e^{-2t}u(t)e^{-j\omega t}dt+\int_{-\infty}^{\infty}u(t+2)e^{-j\omega t}dt-\int_{-\infty}^{\infty}u(t-2)e^{-j\omega t}dt \end{align}

Using the previous example and the time shifting property,

$H(j\omega)=\frac {1}{2+j\omega}+\frac {2sin(2\omega)}{\omega}$

## Example 3

What is the Fourier Transform of the signal $x(t)=e^{j\omega _0t}$?

To solve this look at the the inverse Fourier transform, but the inverse transform of what?

Take $\chi(\omega)=2\pi\delta(\omega-\omega _0)$

$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\chi (\omega)e^{j\omega t}d\omega$ $=\frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\delta(\omega-\omega _0)e^{j\omega t}d\omega$ $=\int_{-\infty}^{\infty}\delta(\omega-\omega _0)e^{j\omega t}d\omega$

by sifting property,

$\int_{-\infty}^{\infty}\delta(\omega-\omega _0)e^{j\omega t}d\omega=e^{j\omega t}|_{\omega=\omega _0}$

$x(t)=e^{j\omega _0 t}$

Thus, the fourier transform of $x(t)=e^{j\omega _0t}$ is $\chi(\omega)=2\pi\delta(\omega-\omega _0)$.

## Example 4

Show that the Fourier transform of $x(t)=cos(2\pi t)$ is $\chi (\omega)=\pi\delta(\omega+2\pi)+\pi\delta(\omega-2\pi)$.

$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\chi (\omega)e^{j\omega t}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}[\pi\delta(\omega+2\pi)+\pi\delta(\omega-2\pi)]e^{j\omega t}d\omega$

$=\frac{1}{2\pi}\int_{-\infty}^{\infty}\pi\delta(\omega+2\pi)e^{j\omega t}d\omega+\frac{1}{2\pi}\int_{-\infty}^{\infty}\pi\delta(\omega-2\pi)e^{j\omega t}d\omega$

$=\frac{1}{2}\int_{-\infty}^{\infty}\delta(\omega+2\pi)e^{j\omega t}d\omega+\frac{1}{2}\int_{-\infty}^{\infty}\pi\delta(\omega-2\pi)e^{j\omega t}d\omega$

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

## Example 5

The input of an LTI system is $x(t)=1$. The unit impulse response of the system is $h(t)=\delta(t-3)$. What is the Fourier Transform of the response y(t) to the system?

$Y(\omega)=F.T.(x(t)\star h(t))=F.T.(x(t))F.T.(h(t))=F.T.(1)F.T.(\delta(t-3))$

Since $\frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\delta(\omega)e^{j\omega t}d\omega=e^{j0t}=1$,

$Y(\omega)=2\pi\delta(\omega)F.T.(\delta(t-3))=2\pi\delta(\omega)e^{-3j\omega}F.T.(\delta(t))=2\pi\delta(\omega)e^{-3j\omega}(1)$

Since $\delta(\omega)=1$ only when $\omega=0$,

$Y(\omega)=2\pi\delta(\omega)e^{-3j0}=2\pi\delta(\omega)$

## Example 6

Assume $|\alpha|<1$. Compute the Fourier Transform of $x[n]=(\alpha)^nu[n]$.

$\chi(\omega)=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}=\sum_{n=-\infty}^{\infty}(\alpha)^nu[n]e^{-j\omega n}=\sum_{n=0}^{\infty}(\alpha)^ne^{-j\omega n}=\sum_{n=0}^{\infty}(\alpha e^{-j\omega})^n=\frac{1}{1-\alpha e^{-j\omega}}$

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