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Am trying to compute the Fourier Transform of a function using the properties of the Fourier Transform once and checking my answer using direct integration. My problem is that am not getting the same for both ways.

This is the problem

Sorry for not being able to type the equations as I am not familiar on how to do that, It would be great if somebody can tell me how. Your help is appreciated. Thanks.

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  • $\begingroup$ yes, you can see from the transformation in (1) that I made it two parts. That is, I split g(2t-1) into two parts. First I used the the time shift property for g(t-1), then I used the time scaling property for g(2t). Do you know why using properties for Fourier transform is not matching the direct Integration method? Thanks. $\endgroup$ – Raykh May 28 '17 at 0:21
  • $\begingroup$ I should have included u(t), I apologize I already assumed it. And for the properties part I would still get e^(-jw) $\endgroup$ – Raykh May 28 '17 at 3:16
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Your first solution using the properties of the Fourier transform is correct. Your second solution is wrong, because you forgot to include the unit step function. Your function $g(t)$ should be defined by

$$g(t)=e^{-t}u(t)\tag{1}$$

which gives for $g(2t-1)$

$$g(2t-1)=e^{-(2t-1)}u(2t-1)=e^{-(2t-1)}u\left(t-\frac12\right)\tag{2}$$

Consequently, the Fourier integral is given by

$$\mathcal{F}\{g(2t-1)\}=\int_{\frac12}^{\infty}e^{-(2t-1)}e^{-j\omega t}dt\tag{3}$$

I'm sure you can prove now that the result of $(3)$ equals the result you got from using the properties of the Fourier transform.

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  • $\begingroup$ can you complete the solution please because am still not getting the same answer. Am getting (e^(2-jw/2))/(2+jw) $\endgroup$ – Raykh May 28 '17 at 16:50
  • $\begingroup$ @Raykh: It's exactly like the integral you already solved, just with $1/2$ as the lower integration limit instead of $0$. So in your last expression before evaluating the limits just replace $t$ by $1/2$ and you got it. $\endgroup$ – Matt L. May 28 '17 at 21:15
  • $\begingroup$ That is what I exactly did and I ended up with a different answer as I stated in my previous comment. $\endgroup$ – Raykh May 29 '17 at 3:46
  • $\begingroup$ @Raykh: There must be a very basic mistake in your calculations. Please show us your steps and we can take a look. Probably while writing things down you'll find the error yourself. $\endgroup$ – Matt L. May 29 '17 at 7:13

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