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= Continuous-time Fourier transform of a complex exponential =
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[[Category:Fourier transform]]
What is the Fourier transform of <math>x(t)= e^{j \pi t}</math>? Justify your answer.
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'''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]'''
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Topic:  Continuous-time Fourier transform of a complex exponential
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----
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==Question==
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What is the Fourier transform of  
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<math>x(t)= e^{j \pi t}</math>?  
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Justify your answer.
 
   
 
   
 
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       \\=\delta (f-\frac{1}{2})  \end{align} </math>
 
       \\=\delta (f-\frac{1}{2})  \end{align} </math>
  
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:<span style="color:red">TA's comments: This is an infeasible solution! You cannot integrate a complex exponential over the range from -infinity to infinity. See the first solution for reference. </span>
 
===Answer 5===
 
===Answer 5===
 
Using the inverse fourier transform definition,  
 
Using the inverse fourier transform definition,  

Latest revision as of 09:47, 11 November 2013


Practice Question on "Digital Signal Processing"

Topic: Continuous-time Fourier transform of a complex exponential


Question

What is the Fourier transform of

$ x(t)= e^{j \pi t} $?

Justify your answer.


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

Guess: $ X(f)=\delta (f-\frac{1}{2}) $

Proof:

$ x(t)=\int_{-\infty}^{\infty} X(f)e^{j2\pi ft} df = \int_{-\infty}^{\infty} \delta (f-\frac{1}{2})e^{j2\pi ft} df = \int_{-\infty}^{\infty} \delta (f-\frac{1}{2})e^{j\pi t} df = e^{j\pi t} \int_{-\infty}^{\infty} \delta (f-\frac{1}{2}) df = e^{j\pi t} $

using the fact that $ \delta (t-T)f(t) = \delta (t-T)f(T) $

Instructor's comments: Nice and clear solution! One can also justify the answer using the shifting property directly, which would save a couple of steps.-pm

Answer 2

$ x(t) = \int_{-\infty}^{\infty} X(f)e^{j2\pi ft} df $

In order for the following to be true, $ x(t)= e^{j \pi t} $

$ X(f) = \delta(f - \frac{1}{2}) $

because

$ x(t) = \int_{-\infty}^{\infty} \delta(f - \frac{1}{2})e^{j2\pi ft} df = e^{j \pi t} $ with careful inspection.


Answer 3

$ x(t)=e^{j2\pi 1/2t}=e^{j\omega_0 t},where \omega_0=1/2. F(e^{j\omega_0 t})=2\pi \delta(\omega-\omega_0),also C\delta(Cn)=\delta(n). so, X(f)=\delta (f-\frac{1}{2}) $

Answer 4

$ \begin{align} \mathcal{F}[e^{j\pi t}]=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt \\=\int_{-\infty}^{\infty} e^{j\pi t}e^{-j2\pi ft} dt \\=\int_{-\infty}^{\infty} e^{-j2\pi (f-\frac{1}{2})t} dt \\=\delta (f-\frac{1}{2}) \end{align} $

TA's comments: This is an infeasible solution! You cannot integrate a complex exponential over the range from -infinity to infinity. See the first solution for reference.

Answer 5

Using the inverse fourier transform definition,

$ \, x(t)=e^{j \pi t}= \int_{-\infty}^{\infty}\mathcal{X}(f)e^{j2\pi f t} d f\, $

and the sifting property, we can see that an $ X(f) $ that works is

$ \delta (f-\frac{1}{2}) = X(f) $

Answer 6

$ \begin{align} \mathcal{X}(f)&=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt \\ &=\int_{-\infty}^{\infty} e^{j\pi t}e^{-j2\pi ft} dt \\ &=\delta \left (f-\frac{1}{2} \right) \end{align} $

Answer 7

From the inverse Fourier Transform Definition:

$ \, x(t)=e^{j \pi t}= \int_{-\infty}^{\infty}\mathcal{X}(f)e^{j2\pi f t} d f\, $

After inspection, we can see that need to pluck out only the portion of $ e^{j 2\pi f t} $ where f = $ 1/2 $

The sifting property will sift that portion out if a $ \delta (f-\frac{1}{2}) $ is used as X(f), so this is the FT of $ e^{j \pi t} $

Answer 8

$ \begin{align} \mathcal{F}[e^{j\pi t}]=\int_{-\infty}^{\infty} e^{j\pi t}e^{-j2\pi ft} dt \\=\int_{-\infty}^{\infty} e^{-j2\pi (f-\frac{1}{2})t} dt \\=\delta (f-\frac{1}{2}) \end{align} $


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