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}} $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva