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

enter image description here

My question is if there is either a property that eliminates the need to evaluate the integral or if there is some simplification of the transform so it's not such a jumble.

I see that i made a mistake with integrating the unit step, and my answer should not have that extra $$t \cdot u(t-1)$$ and $$t \cdot u(t)$$ After the help i got on the DSP exchange and here i ended up with $$ x_1(F)=\frac{e^{2a+j2\pi\cdot F}}{a+j2\pi\cdot F}$$ 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\$ What is \$e^(\$? Also, please rotate your image. \$\endgroup\$ – user17592 Apr 6 '13 at 19:26
  • \$\begingroup\$ chattypics.com/files/droidUpload_qdg2kdu2di_yp4yyclp6d.jpg That is supposed to be e to the quantity -a(t-1), \$\endgroup\$ – retroredeye Apr 6 '13 at 19:32
  • \$\begingroup\$ Okay.. I edited it in for you this time. What exactly is your question? \$\endgroup\$ – user17592 Apr 6 '13 at 19:35
  • \$\begingroup\$ Thanks for fixing that up for me, i don't really get why that didn't work, but my question is if there is either a property that eliminates the need to evaluate the integral or if there is some simplification of the transform so it's not such a jumble. \$\endgroup\$ – retroredeye Apr 6 '13 at 19:40
  • \$\begingroup\$ I edited the question in. You can edit your questions yourself as well. \$\endgroup\$ – user17592 Apr 6 '13 at 19:41
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You have a definite integral with endpoints. Think about how the properties of \$ u(t) \$ let you reduce the integration boundaries.

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  • \$\begingroup\$ I see that you can instead integrate from 1 to infinity on the first integral, and 0 to infinity on the second, but i don't see what that would change in the answer. \$\endgroup\$ – retroredeye Apr 6 '13 at 19:51
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First, you should be able to find the transform for the time domain signal \$f(t) = e^{-at}u(t)\$ in your Fourier transform tables.

For your first signal, a table of Fourier transforms like this one will tell you that if the transform of \$f(t)\$ is \$F(\omega)\$, then the transform of \$f(t-t_0)\$ is \$F(\omega)e^{-j\omega{}t_0}\$.

For your second signal, realize that you can write \$e^{-a(t-1)}\$ as \$e^{a}e^{-at}\$, meaning this signal is just a constant multiplied by the basic decaying exponential signal.

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