Finding Fourier Transform for x(t): A Math Student's Query

AI Thread Summary
The discussion focuses on finding the Fourier transform of the function x(t) = θ(t-t1) - θ(t-t2), where t2 > t1. The user initially struggles with the assignment due to a lack of exposure to convolution and Fourier transforms in their math class. A participant advises that the integrand is zero outside the interval [t1, t2], simplifying the integral to x(ω) = ∫(t1 to t2) e^(-iωt) dt. The user confirms their understanding of this simplification, indicating they have grasped the concept with the help provided. Overall, the exchange highlights the importance of breaking down functions to facilitate integration in Fourier analysis.
cepheid
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In an assignment, I've been given a function:

x(t) = \theta(t-t_1) - \theta(t-t_2)

Assume t_2 > t_1

and we are asked to find the Fourier transform. So I wrote down:

x(\omega) = \int_{-\infty}^{\infty}{e^{-i\omega t} [\theta(t-t_1) - \theta(t-t_2)] dt}

I know that the function given is the heaviside step function. Its derivative is the dirac delta function, and it is itself the derivative of the ramp function. But I just found this stuff out by looking online. We've learned neither convolution nor Fourier transform in math class, yet somehow this physics prof expects us to do it. Can someone at least point me in the right direction?

Thanks.
 
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Your integrand is zero everywhere except in the interval from t1 to t2 - integrate it directly!
 
Ok, I feel stupid! :smile:

Just to see if I'm understanding you correctly, x(t) can be broken down as follows:

x(t) = 0 - 0 for t < t1

x(t) = 1 - 0 for t1 < t < t2

x(t) = 1 - 1 for t > t2

This is just the first statement in your post.

So the integral reduces to:

x(\omega) = \int_{t_1}^{t_2}{e^{-i\omega t}dt}

Am I right? The help was much appreciated!
 
You got it - WTG!
 
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