(New page: ==Example 1== Compute the Fourier Transform of <math>x(t)=e^{-t}u(t)</math>. <math>X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt</math> <math>=\int_{-\infty}^{\infty}e^{-t}u(t)e^{...) |
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| + | [[Category:problem solving]] | ||
| + | [[Category:ECe301]] | ||
| + | [[Category:signals and system]] | ||
| + | [[Category:Fourier transform]] | ||
| + | |||
| + | =Some Practice Exam Problems for signals and systems ([[ECE301]])= | ||
| + | |||
==Example 1== | ==Example 1== | ||
Compute the Fourier Transform of <math>x(t)=e^{-t}u(t)</math>. | Compute the Fourier Transform of <math>x(t)=e^{-t}u(t)</math>. | ||
| − | <math> | + | <math>\chi(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt</math> |
| + | |||
<math>=\int_{-\infty}^{\infty}e^{-t}u(t)e^{-j\omega t}dt</math> | <math>=\int_{-\infty}^{\infty}e^{-t}u(t)e^{-j\omega t}dt</math> | ||
| + | |||
<math>=\int_{0}^{\infty}e^{-t}e^{-j\omega t}dt</math> | <math>=\int_{0}^{\infty}e^{-t}e^{-j\omega t}dt</math> | ||
| + | |||
<math>=\int_{0}^{\infty}e^{-(1+j\omega )t}dt</math> | <math>=\int_{0}^{\infty}e^{-(1+j\omega )t}dt</math> | ||
| + | |||
<math>=[\frac {e^{-(1+j\omega )t}}{-(1+j\omega)}]|_0^\infty</math> | <math>=[\frac {e^{-(1+j\omega )t}}{-(1+j\omega)}]|_0^\infty</math> | ||
| − | <math> | + | |
| − | <math>=0 | + | <math>=\frac {e^{-(1+j\omega )\infty}}{-(1+j\omega)}-\frac {e^{-(1+j\omega )0}}{-(1+j\omega)}</math> |
| − | <math>=\frac {1}{ | + | |
| + | <math>=0+\frac {1}{(1+j\omega)}</math> | ||
| + | |||
| + | <math>=\frac {1}{1+j\omega}</math> | ||
| + | |||
==Example 2== | ==Example 2== | ||
| + | The impulse response of an LTI system is <math>h(t)=e^{-2t}u(t)+u(t+2)-u(t-2)</math>. | ||
| + | What is the Frequency response <math>H(j\omega)</math> of the system? | ||
| + | |||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
| + | |||
| + | Using the previous example and the time shifting property, | ||
| + | |||
| + | <math>H(j\omega)=\frac {1}{2+j\omega}+\frac {2sin(2\omega)}{\omega}</math> | ||
| + | |||
| + | ==Example 3== | ||
| + | What is the Fourier Transform of the signal <math>x(t)=e^{j\omega _0t}</math>? | ||
| + | |||
| + | To solve this look at the the inverse Fourier transform, but the inverse transform of what? | ||
| + | |||
| + | Take <math>\chi(\omega)=2\pi\delta(\omega-\omega _0)</math> | ||
| + | |||
| + | <math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\chi (\omega)e^{j\omega t}d\omega</math> | ||
| + | <math>=\frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\delta(\omega-\omega _0)e^{j\omega t}d\omega</math> | ||
| + | <math>=\int_{-\infty}^{\infty}\delta(\omega-\omega _0)e^{j\omega t}d\omega</math> | ||
| + | |||
| + | by sifting property, | ||
| + | |||
| + | <math>\int_{-\infty}^{\infty}\delta(\omega-\omega _0)e^{j\omega t}d\omega=e^{j\omega t}|_{\omega=\omega _0}</math> | ||
| + | |||
| + | <math>x(t)=e^{j\omega _0 t}</math> | ||
| + | |||
| + | Thus, the fourier transform of <math>x(t)=e^{j\omega _0t}</math> is <math>\chi(\omega)=2\pi\delta(\omega-\omega _0)</math>. | ||
| + | ==Example 4== | ||
| + | Show that the Fourier transform of <math>x(t)=cos(2\pi t)</math> is <math>\chi (\omega)=\pi\delta(\omega+2\pi)+\pi\delta(\omega-2\pi)</math>. | ||
| + | |||
| + | <math>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</math> | ||
| + | |||
| + | <math>=\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</math> | ||
| + | |||
| + | <math>=\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</math> | ||
| + | |||
| + | <math>=\frac{1}{2}[e^{j2\pi t}+e^{-j2\pi t}]=cos(2\pi)</math> | ||
| + | |||
| + | ==Example 5== | ||
| + | The input of an LTI system is <math>x(t)=1</math>. The unit impulse response of the system is <math>h(t)=\delta(t-3)</math>. What is the Fourier Transform of the response y(t) to the system? | ||
| + | |||
| + | <math>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))</math> | ||
| + | |||
| + | Since <math>\frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\delta(\omega)e^{j\omega t}d\omega=e^{j0t}=1</math>, | ||
| + | |||
| + | <math>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)</math> | ||
| + | |||
| + | Since <math>\delta(\omega)=1</math> only when <math>\omega=0</math>, | ||
| + | |||
| + | <math>Y(\omega)=2\pi\delta(\omega)e^{-3j0}=2\pi\delta(\omega)</math> | ||
| + | ==Example 6== | ||
| + | Assume <math>|\alpha|<1</math>. Compute the Fourier Transform of <math>x[n]=(\alpha)^nu[n]</math>. | ||
| + | |||
| + | <math>\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}}</math> | ||
Latest revision as of 15:48, 23 April 2013
Contents
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}} $
