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= Practice Question on the Properties of the Continuous-time Fourier Transform =
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= [[:Category:Problem_solving|Practice Question]] on the Properties of the Continuous-time Fourier Transform =
 
Let x(t) be a continuous time signal with Fourier transform <math class="inline">{\mathcal X} (\omega) </math>.
 
Let x(t) be a continuous time signal with Fourier transform <math class="inline">{\mathcal X} (\omega) </math>.
  
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</math>
 
</math>
 
=== Answer 2  ===
 
=== Answer 2  ===
Write it here.
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<math>
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\begin{align}
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{\mathcal F} \left( x(3t+7) \right) &= \int_{-\infty}^\infty x(3t+7) e^{-j\omega t} dt \\
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&= \int_{-\infty}^\infty x(u) e^{-j\omega \frac{(u-7)}{3}} \frac{du}{3}, \text{ (letting }u=3t+7), \\
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&= \frac{1}{3} \int_{-\infty}^\infty x(u)e^{-\frac{j\omega u}{3}}e^{\frac{7j\omega}{3}} du\\
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&= \frac{e^{\frac{7j\omega}{3}}}{3}\int_{-\infty}^\infty x(u)e^{-\frac{j\omega u}{3}} du \\
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&= \frac{e^{\frac{7j\omega}{3}}}{3} \mathcal{X}(\omega)
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\end{align}
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</math>
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--[[User:Ekhall|Ekhall]] 11:45, 2 March 2011 (UTC)
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:<span style="color:red">Instructor's note: You should replace <math class="inline">\mathcal{X}(\omega)</math> by <math class="inline">\mathcal{X}(\frac{\omega}{3})</math>. </span>
 
=== Answer 3  ===
 
=== Answer 3  ===
 
Write it here.
 
Write it here.
 
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[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]
 
[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]

Latest revision as of 10:27, 11 November 2011


Practice Question on the Properties of the Continuous-time Fourier Transform

Let x(t) be a continuous time signal with Fourier transform $ {\mathcal X} (\omega) $.

We have seen the time-shifting property of the Fourier transform:

$ {\mathcal F} \left( x(t-t_0) \right) = e^{-j \omega t_0} {\mathcal X} (\omega) $

which gives us an expression for the Fourier transform of $ y(t)=x(t-t_0) $ in terms of $ {\mathcal X} (\omega) $. Note that, to prove this property, one can proceed as follows:

$ \begin{align} {\mathcal F} \left( x(t-t_0) \right) &= \int_{-\infty}^\infty x(t-t_0) e^{-j\omega t} dt \\ &= \int_{-\infty}^\infty x(u) e^{-j\omega (u+t_0)} du, \text{ (letting }u=t-t_0), \\ &= e^{-j\omega t_0} \int_{-\infty}^\infty x(u) e^{-j\omega u} du \\ &= e^{-j\omega t_0} {\mathcal X} (\omega). \end{align} $

Using a similar approach as above, derive an expression for the Fourier transform of y(t)=x(3t+7) in terms of $ {\mathcal X} (\omega) $.


Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!


Answer 1

Instructor's hint: You should start like this

$ \begin{align} {\mathcal F} \left( x(3t+7) \right) &= \int_{-\infty}^\infty x(3t+7) e^{-j\omega t} dt \\ &= \int_{-\infty}^\infty x(u) e^{-j\omega \frac{(u-7)}{3}} \frac{du}{3}, \text{ (letting }u=3t+7), \\ &= ... \end{align} $

Answer 2

$ \begin{align} {\mathcal F} \left( x(3t+7) \right) &= \int_{-\infty}^\infty x(3t+7) e^{-j\omega t} dt \\ &= \int_{-\infty}^\infty x(u) e^{-j\omega \frac{(u-7)}{3}} \frac{du}{3}, \text{ (letting }u=3t+7), \\ &= \frac{1}{3} \int_{-\infty}^\infty x(u)e^{-\frac{j\omega u}{3}}e^{\frac{7j\omega}{3}} du\\ &= \frac{e^{\frac{7j\omega}{3}}}{3}\int_{-\infty}^\infty x(u)e^{-\frac{j\omega u}{3}} du \\ &= \frac{e^{\frac{7j\omega}{3}}}{3} \mathcal{X}(\omega) \end{align} $ --Ekhall 11:45, 2 March 2011 (UTC)

Instructor's note: You should replace $ \mathcal{X}(\omega) $ by $ \mathcal{X}(\frac{\omega}{3}) $.

Answer 3

Write it here.


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