Revision as of 19:12, 12 November 2011

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) } $?

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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.

--Xiao1 23:12, 12 November 2011 (UTC)

Answer 3

Write it here.

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