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Old 05-03-2012, 05:19 AM
de1337ed de1337ed is offline
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Default Fourier Transform a shifted step function?

I sort of understand how to obtain the Fourier transform of a regular step function (0 when t<0, and A when t>=0), but how do you obtain the fourier transform if the function has been shifted right on the t-axis by T? I've been stuck on this for a while.

What I have as an equation is the following:

\int{Au(t+T)*e^{-jwt}dt} from -∞ to ∞.

I then turn this into
\int{A*e^{-jwt}dt} from T to ∞

I don't know how to proceed from here.
Thank you.

Last edited by de1337ed; 05-03-2012 at 05:24 AM.
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Old 05-04-2012, 01:43 AM
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count_volta count_volta is offline
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First of all, this is useful for you.

So the final result will just be the F transform of the unit step multiplied by e^-jωT

Then as far as doing the problem, take the A out of the integral and take the integral of the exponential. I'm sure you know how to do that. Integral of exponential is an exponential. Remember to do the substitution rule aka u = -jwt and du = -jw.

You get:

-A/jω [ e^-jωt] evaluated at t = infinity and t = T

I end up with A/jw * e^-jwT

I'm not sure where the delta function comes from, but I didn't get one.
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Last edited by count_volta; 05-04-2012 at 01:54 AM.
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Old 05-04-2012, 02:30 AM
t_n_k t_n_k is offline
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Originally Posted by count_volta View Post
First of all, this is useful for you.

So the final result will just be the F transform of the unit step multiplied by e^-jωT

I'm not sure where the delta function comes from, but I didn't get one.
The first part about using the well-known shifting theorem is logical. Your derivation of the Fourier transform of the un-shifted step (Heaviside) function needs a little more careful thought. I suggest you Google "Fourier Transform of the Heaviside Function" to gain some further insights - particularly as to the origin of the delta function term.
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