| (5 intermediate revisions by 3 users not shown) | |||
| Line 4: | Line 4: | ||
[[Category:continuous-space Fourier transform]] | [[Category:continuous-space Fourier transform]] | ||
| − | = Continuous-space Fourier transform of a complex exponential | + | <center><font size= 4> |
| + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
| + | </font size> | ||
| + | |||
| + | Topic: Continuous-space Fourier transform of a complex exponential | ||
| + | |||
| + | </center> | ||
| + | ---- | ||
| + | ==Question== | ||
What is the Continuous-space Fourier transform (CSFT) of <math>f(x,y)= e^{j \pi (ax +by) }</math>? | What is the Continuous-space Fourier transform (CSFT) of <math>f(x,y)= e^{j \pi (ax +by) }</math>? | ||
| Line 15: | Line 23: | ||
===Answer 1=== | ===Answer 1=== | ||
<math> x[n] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{j \pi (ax +by) }e^{-2j \pi (ux +vy) }dxdy | <math> x[n] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{j \pi (ax +by) }e^{-2j \pi (ux +vy) }dxdy | ||
| − | </math> | + | </math> <span style="color:red">Instructor's comment: Are you sure you mean x[n] on the left? -pm </span> |
<math> = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{j \pi x(a-2u) }e^{j \pi y(b - 2v) }dxdy | <math> = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{j \pi x(a-2u) }e^{j \pi y(b - 2v) }dxdy | ||
| Line 27: | Line 35: | ||
</math> | </math> | ||
| + | |||
| + | :<span style="color:red">Instructor's comment: Please, please, please, do not try to integrate functions that are not integrable. This approach does not work. -pm </span> | ||
===Answer 2=== | ===Answer 2=== | ||
| − | Claim that <math>CSFT \{ e^{j \pi (ax +by) }\} = \delta(u-a)\delta(v-b)</math> | + | Claim that <math>CSFT \{ e^{j \pi (ax +by) }\} = \delta(u-a,v-b) =\delta(u-a)\delta(v-b)</math> |
Proof: | Proof: | ||
| Line 36: | Line 46: | ||
<math>=\int_{-\infty}^{\infty}\delta(u-a)e^{2j \pi (ux) }du \int_{-\infty}^{\infty}\delta(v-b)e^{2j \pi (vy) }dv = e^{j \pi (ax)} e^{j \pi (by)} = e^{j \pi (ax +by)}</math> | <math>=\int_{-\infty}^{\infty}\delta(u-a)e^{2j \pi (ux) }du \int_{-\infty}^{\infty}\delta(v-b)e^{2j \pi (vy) }dv = e^{j \pi (ax)} e^{j \pi (by)} = e^{j \pi (ax +by)}</math> | ||
| − | Another way is to show by | + | :<span style="color:green">Instructor's comment: Pretty good! -pm </span> |
| + | Another way is to show by separality, since | ||
<math>f(x,y)=g(x)h(y),g(x) = e^{j \pi (ax)},h(y) = e^{j \pi (by) } </math> | <math>f(x,y)=g(x)h(y),g(x) = e^{j \pi (ax)},h(y) = e^{j \pi (by) } </math> | ||
| Line 49: | Line 60: | ||
--[[User:Xiao1|Xiao1]] 23:12, 12 November 2011 (UTC) | --[[User:Xiao1|Xiao1]] 23:12, 12 November 2011 (UTC) | ||
| + | :<span style="color:red">Instructor's comment: On the exam, you could use the second approach if you were given a table of continuous-time fourier transform in which the Fourier transform of an exponential is given. Otherwise, you would need to justify this Fourier transform pair. -pm </span> | ||
===Answer 3=== | ===Answer 3=== | ||
| − | + | ||
| + | <math> F(u,v) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{j \pi (ax +by) }e^{-2j \pi (ux +vy) }dxdy=\int_{-\infty}^{\infty}e^{j \pi (ax)} e^{-j2 \pi u x}dx \int_{-\infty}^{\infty}e^{j \pi (by)} e^{-j2 \pi v y}dy=\delta(u-a/2) \delta(v-b/2) | ||
| + | </math> | ||
| + | |||
| + | :<span style="color:red">Instructor's comment: How do you justify the last equality? (For example, how do you know the answer is not 7 times what you wrote, or 42 times what you wrote (which both give you infinity and zero for the same values of u and v as the answer you give.) It seems like you just "plugged in" the answer. -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 11:57, 26 November 2013
Practice Question on "Digital Signal Processing"
Topic: Continuous-space Fourier transform of a complex exponential
Question
What is the Continuous-space Fourier transform (CSFT) of $ f(x,y)= e^{j \pi (ax +by) } $?
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
$ x[n] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{j \pi (ax +by) }e^{-2j \pi (ux +vy) }dxdy $ Instructor's comment: Are you sure you mean x[n] on the left? -pm
$ = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{j \pi x(a-2u) }e^{j \pi y(b - 2v) }dxdy $ $ = \frac{1}{j\pi(a-2u)}\frac{1}{j\pi(b-2v)}[e^{j \pi x(a-2u)}e^{j \pi y(b - 2v) }]{-\infty}^{\infty} $ $ = {\infty} $
- Instructor's comment: Please, please, please, do not try to integrate functions that are not integrable. This approach does not work. -pm
Answer 2
Claim that $ CSFT \{ e^{j \pi (ax +by) }\} = \delta(u-a,v-b) =\delta(u-a)\delta(v-b) $
Proof:
$ iCSFT\{\delta(u-a)\delta(v-b)\} = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\delta(u-a)\delta(v-b)e^{2j \pi (ux +vy) }dudv $
$ =\int_{-\infty}^{\infty}\delta(u-a)e^{2j \pi (ux) }du \int_{-\infty}^{\infty}\delta(v-b)e^{2j \pi (vy) }dv = e^{j \pi (ax)} e^{j \pi (by)} = e^{j \pi (ax +by)} $
- Instructor's comment: Pretty good! -pm
Another way is to show by separality, since
$ f(x,y)=g(x)h(y),g(x) = e^{j \pi (ax)},h(y) = e^{j \pi (by) } $
then $ F(u,v)=G(u)H(v),G(u) = CTFT(f(x)),H(v) = CTFT(h(y)) $
by CTFT pairs, $ G(u) = /delta(u-a),H(v) = /delta(v-b) $
which shows $ CSFT \{ e^{j \pi (ax +by) }\} = \delta(u-a)\delta(v-b) $,
as the same above.
--Xiao1 23:12, 12 November 2011 (UTC)
- Instructor's comment: On the exam, you could use the second approach if you were given a table of continuous-time fourier transform in which the Fourier transform of an exponential is given. Otherwise, you would need to justify this Fourier transform pair. -pm
Answer 3
$ F(u,v) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{j \pi (ax +by) }e^{-2j \pi (ux +vy) }dxdy=\int_{-\infty}^{\infty}e^{j \pi (ax)} e^{-j2 \pi u x}dx \int_{-\infty}^{\infty}e^{j \pi (by)} e^{-j2 \pi v y}dy=\delta(u-a/2) \delta(v-b/2) $
- Instructor's comment: How do you justify the last equality? (For example, how do you know the answer is not 7 times what you wrote, or 42 times what you wrote (which both give you infinity and zero for the same values of u and v as the answer you give.) It seems like you just "plugged in" the answer. -pm
