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I'm trying to do a CT Fourier Transform of these two signals $$x_1(t)=e^{−a(t−1)} \cdot u(t−1)$$ and $$x_2(t)=e^{−a(t−1)} \cdot u(t)$$ Where $a$ is any real number, and $u(t)$ is the unit step function and that is multiplication, not convolution.

Picture of my work and answers here.

The answers listed should not contain the extra $$t \cdot u(t-1)$$ and $$t \cdot u(t)$$

Okay, thanks to Dilip i now have my answers to look like $$ x_1(F)=\frac{e^{2a+j2\pi\cdot F}}{a+j2\pi\cdot F}$$ and $$x_2(F) =\frac{e^a}{a+j2\pi\cdot F} $$ That does seem more elegant than what i previously had, I feel like I could possibly manipulate $x_1(F)$ into a sinc function, but I don't think it's necessary.

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  • $\begingroup$ Please check your displayed formulas. Right now, they don't make sense. Also, does that $*$ stand for a convolution or multiplication? Once you have corrected the question, draw a diagram of the signals whose Fourier transforms you seek. Then it should be fairly straightforward via a change of variables such as putting $-j\omega-a(t-1) = u$. $U(t)$ or $U(t-1)$ merely changes the limits on the integral; they DO NOT appear inside the integral at all as you have done in your work that you have linked to. $\endgroup$ – Dilip Sarwate Apr 6 '13 at 19:44
  • $\begingroup$ I understand what the signals would look like, but how does that help? And i don't really get why changing the variables would help give a nicer answer. I do see that i made that mistake when integrating the u(t). $\endgroup$ – retroredeye Apr 6 '13 at 20:03
  • $\begingroup$ U(t-n) essentially sets the limits of integration. The integration starts at n and ends at infinity. Once you get the limits right, the U(t-n) portion of the time domain function is redundant and can be tossed. After that, a change of variables as suggested by Dilip may simplify the integration. Note that once you perform the integration and apply the limits, t will be gone from the result (generally no functions of t remain in a Fourier transform). $\endgroup$ – user2718 Apr 6 '13 at 20:22
  • $\begingroup$ I find easier just integrating from 1 to infinite the exponential(by removing the unit step from integrand),this way you get a simple integral to solve. $\endgroup$ – Daniel Conde Marin Apr 6 '13 at 20:24
  • $\begingroup$ @DanielConde - Exactly :-) $\endgroup$ – user2718 Apr 6 '13 at 20:28

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