- #1
Saz1
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ω
The function f(x) is defined by:
f(x) = e^2ax when x ≤ 0
0 when x > 0
Show that, for a > 0, its Fourier transform may be written:
Fourier transform = 1 / (2a - iω)
Fourier transform = ∫f(x)e^iωt dx
(the integral is taken over minus infinity to infinity).
I think I have done the 'hard' part correct:
Fourier transform = ∫e^(2ax) × e^iωt
= ∫e^(2a-iω)x
= e^(2a-iω)x / (2a-iω)x
but how would I go about getting it in the form required? I am guessing it related to the fact that the integral is taken over minus infinity to infinity - as x tends to infinity the Fourier transform tends to zero.. as x tends to minus infinity.. ?? What happens to the exponent?
Also, as a side note, is what I have done so far correct? Thanks a lot :)
Homework Statement
The function f(x) is defined by:
f(x) = e^2ax when x ≤ 0
0 when x > 0
Show that, for a > 0, its Fourier transform may be written:
Fourier transform = 1 / (2a - iω)
Homework Equations
Fourier transform = ∫f(x)e^iωt dx
(the integral is taken over minus infinity to infinity).
The Attempt at a Solution
I think I have done the 'hard' part correct:
Fourier transform = ∫e^(2ax) × e^iωt
= ∫e^(2a-iω)x
= e^(2a-iω)x / (2a-iω)x
but how would I go about getting it in the form required? I am guessing it related to the fact that the integral is taken over minus infinity to infinity - as x tends to infinity the Fourier transform tends to zero.. as x tends to minus infinity.. ?? What happens to the exponent?
Also, as a side note, is what I have done so far correct? Thanks a lot :)