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</math> | </math> | ||
===Answer 2=== | ===Answer 2=== | ||
| − | Claim that <math>CSFT{e^{j \pi (ax +by)}</math> | + | Claim that <math>CSFT \{ e^{j \pi (ax +by) }\} = \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> | ||
| + | |||
| + | 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. | ||
| + | |||
===Answer 3=== | ===Answer 3=== | ||
Revision as of 19:12, 12 November 2011
Contents
Continuous-space Fourier transform of a complex exponential (Practice Problem)
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 $
$ = \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} $
Answer 2
Claim that $ CSFT \{ e^{j \pi (ax +by) }\} = \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)} $
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.
Answer 3
Write it here.
