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
