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[[Category:ECE438Fall2011Boutin]] | [[Category:ECE438Fall2011Boutin]] | ||
[[Category:problem solving]] | [[Category:problem solving]] | ||
− | = Continuous-time Fourier transform computation (in terms of frequency f in hertz)= | + | [[Category:Fourier transform]] |
+ | |||
+ | <center><font size= 4> | ||
+ | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
+ | </font size> | ||
+ | |||
+ | Topic: Continuous-time Fourier transform computation (in terms of frequency f in hertz) | ||
+ | |||
+ | </center> | ||
+ | ---- | ||
+ | ==Question== | ||
Compute the Continuous-time Fourier transform of the two following functions: | Compute the Continuous-time Fourier transform of the two following functions: | ||
Line 67: | Line 77: | ||
<math> y(t)= \frac{ \sin ( \pi t )}{\pi t} | <math> y(t)= \frac{ \sin ( \pi t )}{\pi t} | ||
= \frac{e^{-j\pi t} - e^{j\pi t}}{\pi t} </math> | = \frac{e^{-j\pi t} - e^{j\pi t}}{\pi t} </math> | ||
+ | |||
+ | it should be | ||
+ | <math> y(t)= \frac{ \sin ( \pi t )}{\pi t} | ||
+ | = \frac{e^{j\pi t} - e^{-j\pi t}}{j2\pi t} </math> | ||
For the above equation to be true, | For the above equation to be true, | ||
Line 72: | Line 86: | ||
<math> Y(f) = \frac{\delta(f - \frac{1}{2})}{\pi t} - \frac{\delta(f + \frac{1}{2})}{\pi t} | <math> Y(f) = \frac{\delta(f - \frac{1}{2})}{\pi t} - \frac{\delta(f + \frac{1}{2})}{\pi t} | ||
</math> | </math> | ||
+ | |||
+ | :<span style="color:red">TA's comments: The first part is correct. For the second part, the denominator j2pi*t is not taken into account, which leads to the wrong answer. The CTFT should only have variable f. See answer 9 for reference.</span> | ||
===Answer 3=== | ===Answer 3=== | ||
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= \frac{e^{j\pi t} - e^{-j\pi t}}{j2\pi t} </math> | = \frac{e^{j\pi t} - e^{-j\pi t}}{j2\pi t} </math> | ||
− | Do Fourier transform | + | Do Fourier transform of rect(t), |
<math> \begin{align}\mathcal{F}[\text{rect}(t)] = \int_{-\infty}^{\infty} \text{rect}(t)e^{-j2\pi ft} dt | <math> \begin{align}\mathcal{F}[\text{rect}(t)] = \int_{-\infty}^{\infty} \text{rect}(t)e^{-j2\pi ft} dt | ||
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\\ = \frac{e^{-j2\pi ft}}{-j2\pi f}\vert \end{align}</math> evaluate from -1/2 to 1/2, | \\ = \frac{e^{-j2\pi ft}}{-j2\pi f}\vert \end{align}</math> evaluate from -1/2 to 1/2, | ||
− | <math>\begin{align} | + | <math>\begin{align} Y(f)=\frac{e^{j\pi f}-e^{-j\pi f}}{j2\pi f} |
− | + | =\frac{sin(\pi f)}{\pi f} | |
+ | \end{align} </math> | ||
+ | |||
+ | use duality, <math> Y(f)=rect(x)</math> | ||
+ | |||
+ | :<span style="color:red">TA's comments: From duality, you get Y(f)=rect(-t). Since rect(t) is an even function, rect(t)=rect(-t)=Y(f). </span> | ||
+ | |||
+ | ===Answer 5=== | ||
+ | Since the function is periodic we can integrate over one period: | ||
+ | :<math>F[rect(t)] = \frac{1}{1} \int_{-\frac{1}{2}}^{\frac{1}{2}}\ (1) e^{j2\pi (1)t} df</math> | ||
+ | :<math> = \frac{1}{-j2\pi} e^{-j2\pi t} \Bigg|_{-\frac{1}{2}}^{\frac{1}{2}} = \frac{-1}{j2\pi k } (e^{-j2\pi f\frac{1}{2}} - e^{j2\pi f\frac{1}{2}}) </math> | ||
+ | :<math> = \frac{sin(\pi f)}{\pi f}</math> | ||
+ | |||
+ | To find the FT of the sinc function, simply use duality from the first solution: | ||
+ | |||
+ | <math> F[sinc(t)] = rect(-f)</math> | ||
+ | |||
+ | and since <math> rect</math> is even, | ||
+ | |||
+ | <math> | ||
+ | rect(-f)= \text{rect}(f) = \left\{ | ||
+ | \begin{array}{ll} | ||
+ | 1, & \text{ if } |f|<\frac{1}{2}\\ | ||
+ | 0, & \text{ else} | ||
+ | \end{array} | ||
+ | \right. | ||
+ | </math> | ||
+ | |||
+ | :<span style="color:red">TA's comments: It doesn't matter whether it's periodic or not. Just integrate over the real line, which is equivalent to integrating from -0.5 to 0.5 in this problem. </span> | ||
+ | |||
+ | |||
+ | ===Answer 6=== | ||
+ | |||
+ | <math> | ||
+ | \begin{align} | ||
+ | X(f)&=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dx \\ | ||
+ | &=\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dx \\ | ||
+ | &=\frac{e^{-j2\pi ft}}{-j2\pi f} \big| _{-1/2}^{1/2} \\ | ||
+ | &=\frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} \\ | ||
+ | &=\frac{sin(\pi f)}{\pi f} | ||
+ | \end{align} | ||
+ | </math> | ||
+ | |||
+ | duality: | ||
+ | |||
+ | <math>f(t) \Leftrightarrow F(-\omega)</math> | ||
+ | |||
+ | <math>F(t) \Leftrightarrow 2\pi f(-\omega)</math> | ||
+ | |||
+ | so | ||
+ | |||
+ | <math>rect(t) \Leftrightarrow sinc(\omega/2) </math> | ||
+ | |||
+ | <math>2 \pi sinc(t) \Leftrightarrow 2 \pi rect(-\omega/2\pi) = rect(f)</math> | ||
+ | |||
+ | :<span style="color:red">TA's comments: You can use the duality for continuous time signal directly. </span> | ||
+ | |||
+ | ===Answer 7=== | ||
+ | Start with x(t) = rect(t) | ||
+ | |||
+ | <math> X(f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt </math> | ||
+ | |||
+ | <math> = \int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt </math> | ||
+ | |||
+ | <math> = \frac{e^{-j2\pi ft}}{-j2\pi f} \Bigg| _{-1/2}^{1/2} </math> | ||
+ | |||
+ | <math> = \frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} </math> | ||
+ | |||
+ | <math> = \frac{ \sin (\pi f)}{\pi f} </math> | ||
+ | |||
+ | Now the FT of the sinc function is much easier. Use Duality to show: | ||
+ | |||
+ | <math> F{(sinc(t))} = rect(-f) = rect(f) </math> | ||
+ | |||
+ | ===Answer 8=== | ||
+ | <math> X(f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt </math> | ||
+ | |||
+ | Since x(t) only exists from -1/2<t<1/2, | ||
+ | <math> = \int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt </math> | ||
+ | |||
+ | <math> = \frac{e^{-j2\pi ft}}{-j2\pi f} \Bigg| _{-1/2}^{1/2} </math> | ||
+ | |||
+ | <math> = \frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} </math> | ||
+ | |||
+ | <math> = \frac{ \sin (\pi f)}{\pi f} </math> | ||
+ | |||
+ | :<span style="color:red">TA's comments: You should consider the case where f=0 separately to make it more rigorous. </span> | ||
+ | |||
+ | ===Answer 9=== | ||
+ | <math> X(f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt | ||
+ | </math> | ||
+ | |||
+ | <math> = \int_{-\infty}^{\infty} rect(t)e^{-j2\pi ft} dt | ||
+ | </math> | ||
+ | |||
+ | <math> = \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt | ||
+ | </math> | ||
+ | |||
+ | <math> = \frac{e^{-j2\pi ft}}{-j2\pi f} \Bigg|_{-\frac{1}{2}}^{\frac{1}{2}} | ||
+ | </math> | ||
+ | |||
+ | <math> = \frac{1}{\pi f} \frac{e^{-j\pi f} -e^{j\pi f}}{2j} | ||
+ | </math> | ||
+ | |||
+ | <math> = \frac{\sin\pi f}{\pi f} = sinc(\pi f) | ||
+ | </math> | ||
+ | |||
+ | By duality: | ||
+ | |||
+ | <math> F[y(t)]\rightarrow rect(-f) | ||
+ | </math> | ||
+ | |||
+ | Since rect() is an even function, | ||
+ | |||
+ | <math> | ||
+ | rect(-f) = rect(f) | ||
+ | </math> | ||
+ | :<span style="color:green">Instructor's comments: This answer is almost completely correct. The only issue is that if f=0, then your integration is incorrect. You should really consider the case f=0 separately. -pm </span> | ||
---- | ---- | ||
[[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 09:45, 11 November 2013
Practice Question on "Digital Signal Processing"
Topic: Continuous-time Fourier transform computation (in terms of frequency f in hertz)
Contents
Question
Compute the Continuous-time Fourier transform of the two following functions:
$ x(t)= \text{rect}(t) = \left\{ \begin{array}{ll} 1, & \text{ if } |t|<\frac{1}{2}\\ 0, & \text{ else} \end{array} \right. $
$ y(t)= \frac{ \sin ( \pi t )}{\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
Fourier Transform of rect(t):
$ X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dx =\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dx =\frac{e^{-j2\pi ft}}{-j2\pi f} $ from t=-1/2 to t=1/2
$ =\frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} =\frac{sin(\pi f)}{\pi f} $
- Instructor's comments: Technically, you should look at the case f=0 separately, because your solution involves a division by f. -pm
Fourier Transform of $ \frac{sin(\pi t)}{\pi t} $:
Guess: $ X(f)=rect(t) $
Proof:
$ x(t)=\int_{-\infty}^{\infty} X(f)e^{j2\pi ft} df =\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{j2\pi ft} df =\frac{e^{j2\pi ft}}{j2\pi t} $ from f=-1/2 to f=1/2
$ =\frac{e^{j\pi t}-e^{j\pi t}}{j2\pi t} =\frac{sin(\pi t)}{\pi t} $
- Instructor's comments: Guessing the answer and proving it using the inverse Fourier transform is a good trick. One could also obtain this Fourier transform using the duality property and your previous answer. -pm
Answer 2
$ X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt =\int_{-\infty}^{\infty} rect(t)e^{-j2\pi ft} dt =\int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt $
$ = -\frac{e^{-j2\pi ft}}{-j2\pi f} $ integrating from -0.5 to +0.5. $ = \frac{e^{-j\pi f} - e^{j\pi f}}{-j2\pi f} = \frac{sin(\pi f)}{\pi f} $
For y(t), we know that
$ y(t) = \int_{-\infty}^{\infty} Y(f)e^{j2\pi ft} df $
$ y(t)= \frac{ \sin ( \pi t )}{\pi t} = \frac{e^{-j\pi t} - e^{j\pi t}}{\pi t} $
it should be $ y(t)= \frac{ \sin ( \pi t )}{\pi t} = \frac{e^{j\pi t} - e^{-j\pi t}}{j2\pi t} $
For the above equation to be true,
$ Y(f) = \frac{\delta(f - \frac{1}{2})}{\pi t} - \frac{\delta(f + \frac{1}{2})}{\pi t} $
- TA's comments: The first part is correct. For the second part, the denominator j2pi*t is not taken into account, which leads to the wrong answer. The CTFT should only have variable f. See answer 9 for reference.
Answer 3
$ X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt = \int_{-1/2}^{1/2} e^{-j2\pi ft} dt = \frac{e^-j2\pi ft}{j2 \pi f},where x from-1/2 to 1/2 =sinc(f) $ $ use duality, Y(f)=rect(x) $
Answer 4
Using Euler's equation, we know
$ y(t)= \frac{ \sin ( \pi t )}{\pi t} = \frac{e^{j\pi t} - e^{-j\pi t}}{j2\pi t} $
Do Fourier transform of rect(t),
$ \begin{align}\mathcal{F}[\text{rect}(t)] = \int_{-\infty}^{\infty} \text{rect}(t)e^{-j2\pi ft} dt \\ = \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt \\ = \frac{e^{-j2\pi ft}}{-j2\pi f}\vert \end{align} $ evaluate from -1/2 to 1/2,
$ \begin{align} Y(f)=\frac{e^{j\pi f}-e^{-j\pi f}}{j2\pi f} =\frac{sin(\pi f)}{\pi f} \end{align} $
use duality, $ Y(f)=rect(x) $
- TA's comments: From duality, you get Y(f)=rect(-t). Since rect(t) is an even function, rect(t)=rect(-t)=Y(f).
Answer 5
Since the function is periodic we can integrate over one period:
- $ F[rect(t)] = \frac{1}{1} \int_{-\frac{1}{2}}^{\frac{1}{2}}\ (1) e^{j2\pi (1)t} df $
- $ = \frac{1}{-j2\pi} e^{-j2\pi t} \Bigg|_{-\frac{1}{2}}^{\frac{1}{2}} = \frac{-1}{j2\pi k } (e^{-j2\pi f\frac{1}{2}} - e^{j2\pi f\frac{1}{2}}) $
- $ = \frac{sin(\pi f)}{\pi f} $
To find the FT of the sinc function, simply use duality from the first solution:
$ F[sinc(t)] = rect(-f) $
and since $ rect $ is even,
$ rect(-f)= \text{rect}(f) = \left\{ \begin{array}{ll} 1, & \text{ if } |f|<\frac{1}{2}\\ 0, & \text{ else} \end{array} \right. $
- TA's comments: It doesn't matter whether it's periodic or not. Just integrate over the real line, which is equivalent to integrating from -0.5 to 0.5 in this problem.
Answer 6
$ \begin{align} X(f)&=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dx \\ &=\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dx \\ &=\frac{e^{-j2\pi ft}}{-j2\pi f} \big| _{-1/2}^{1/2} \\ &=\frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} \\ &=\frac{sin(\pi f)}{\pi f} \end{align} $
duality:
$ f(t) \Leftrightarrow F(-\omega) $
$ F(t) \Leftrightarrow 2\pi f(-\omega) $
so
$ rect(t) \Leftrightarrow sinc(\omega/2) $
$ 2 \pi sinc(t) \Leftrightarrow 2 \pi rect(-\omega/2\pi) = rect(f) $
- TA's comments: You can use the duality for continuous time signal directly.
Answer 7
Start with x(t) = rect(t)
$ X(f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt $
$ = \int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt $
$ = \frac{e^{-j2\pi ft}}{-j2\pi f} \Bigg| _{-1/2}^{1/2} $
$ = \frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} $
$ = \frac{ \sin (\pi f)}{\pi f} $
Now the FT of the sinc function is much easier. Use Duality to show:
$ F{(sinc(t))} = rect(-f) = rect(f) $
Answer 8
$ X(f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt $
Since x(t) only exists from -1/2<t<1/2, $ = \int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt $
$ = \frac{e^{-j2\pi ft}}{-j2\pi f} \Bigg| _{-1/2}^{1/2} $
$ = \frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} $
$ = \frac{ \sin (\pi f)}{\pi f} $
- TA's comments: You should consider the case where f=0 separately to make it more rigorous.
Answer 9
$ X(f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt $
$ = \int_{-\infty}^{\infty} rect(t)e^{-j2\pi ft} dt $
$ = \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt $
$ = \frac{e^{-j2\pi ft}}{-j2\pi f} \Bigg|_{-\frac{1}{2}}^{\frac{1}{2}} $
$ = \frac{1}{\pi f} \frac{e^{-j\pi f} -e^{j\pi f}}{2j} $
$ = \frac{\sin\pi f}{\pi f} = sinc(\pi f) $
By duality:
$ F[y(t)]\rightarrow rect(-f) $
Since rect() is an even function,
$ rect(-f) = rect(f) $
- Instructor's comments: This answer is almost completely correct. The only issue is that if f=0, then your integration is incorrect. You should really consider the case f=0 separately. -pm