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− | = Continuous-time Fourier transform of a complex exponential = | + | [[Category:Fourier transform]] |
− | What is the Fourier transform of <math>x(t)= e^{j \pi t}</math>? Justify your answer. | + | |
+ | <center><font size= 4> | ||
+ | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
+ | </font size> | ||
+ | |||
+ | Topic: Continuous-time Fourier transform of a complex exponential | ||
+ | |||
+ | </center> | ||
+ | ---- | ||
+ | ==Question== | ||
+ | What is the Fourier transform of | ||
+ | |||
+ | <math>x(t)= e^{j \pi t}</math>? | ||
+ | |||
+ | Justify your answer. | ||
---- | ---- | ||
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= e^{j\pi t} </math> | = e^{j\pi t} </math> | ||
+ | using the fact that <math>\delta (t-T)f(t) = \delta (t-T)f(T)</math> | ||
+ | :<span style="color:green">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 </span> | ||
===Answer 2=== | ===Answer 2=== | ||
− | + | <math> x(t) = \int_{-\infty}^{\infty} X(f)e^{j2\pi ft} df | |
+ | </math> | ||
+ | |||
+ | In order for the following to be true, | ||
+ | <math> x(t)= e^{j \pi t} </math> | ||
+ | |||
+ | <math> X(f) = \delta(f - \frac{1}{2}) </math> | ||
+ | |||
+ | because | ||
+ | |||
+ | <math> x(t) = \int_{-\infty}^{\infty} \delta(f - \frac{1}{2})e^{j2\pi ft} df = e^{j \pi t} | ||
+ | </math> with careful inspection. | ||
+ | |||
+ | |||
===Answer 3=== | ===Answer 3=== | ||
− | + | ||
+ | <math>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})</math> | ||
+ | |||
+ | ===Answer 4=== | ||
+ | |||
+ | <math>\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} </math> | ||
+ | |||
+ | :<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=== | ||
+ | Using the inverse fourier transform definition, | ||
+ | |||
+ | <math>\, x(t)=e^{j \pi t}= \int_{-\infty}^{\infty}\mathcal{X}(f)e^{j2\pi f t} d f\,</math> | ||
+ | |||
+ | and the sifting property, we can see that an <math>X(f)</math> that works is | ||
+ | |||
+ | <math> \delta (f-\frac{1}{2}) = X(f)</math> | ||
+ | |||
+ | ===Answer 6=== | ||
+ | <math> | ||
+ | \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} </math> | ||
+ | |||
+ | ===Answer 7=== | ||
+ | From the inverse Fourier Transform Definition: | ||
+ | |||
+ | <math>\, x(t)=e^{j \pi t}= \int_{-\infty}^{\infty}\mathcal{X}(f)e^{j2\pi f t} d f\,</math> | ||
+ | |||
+ | After inspection, we can see that need to pluck out only the portion of | ||
+ | <math> e^{j 2\pi f t} </math> where f = <math> 1/2 </math> | ||
+ | |||
+ | The sifting property will sift that portion out if a <math> \delta (f-\frac{1}{2}) </math> is used as X(f), so this is the FT of <math> e^{j \pi t} </math> | ||
+ | |||
+ | ===Answer 8=== | ||
+ | |||
+ | <math>\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} </math> | ||
+ | |||
---- | ---- | ||
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] |
Latest revision as of 09:47, 11 November 2013
Practice Question on "Digital Signal Processing"
Topic: Continuous-time Fourier transform of a complex exponential
Contents
Question
What is the Fourier transform of
$ x(t)= e^{j \pi t} $?
Justify your answer.
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} $