Difference between revisions of "Obtain CSFT complex exponential" - Rhea

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.

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

$x[n] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{j \pi (ax +by) }e^{-2j \pi (ux +vy) }dxdy$ 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

Back to ECE438 Fall 2011 Prof. Boutin

@@ Line 4: / Line 4: @@
 [[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>?
@@ Line 14: / Line 22: @@
 ----
 ===Answer 1===
-Write it here.
+<math> x[n] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{j \pi (ax +by) }e^{-2j \pi (ux +vy) }dxdy
+</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>
+<math> = \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}
+</math>
+<math> = {\infty}
+</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===
-Write it here
+Claim that <math>CSFT  \{ e^{j \pi (ax +by) }\} = \delta(u-a,v-b) =\delta(u-a)\delta(v-b)</math>
+Proof:
+<math>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</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>
+:<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>
+then <math>F(u,v)=G(u)H(v),G(u) = CTFT(f(x)),H(v) = CTFT(h(y))</math>
+by CTFT pairs,  <math>G(u) = /delta(u-a),H(v) = /delta(v-b)</math>
+which shows <math>CSFT  \{ e^{j \pi (ax +by) }\} = \delta(u-a)\delta(v-b)</math>,
+as the same above.
+--[[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===
-Write it here.
+<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]]