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I have to find the Fourier transform of this function.

enter image description here

(- sign exists BEFORE modulus). I thought of using this standard result.

enter image description here

If I split this fourier transform into this will it be correct to use this standard result. enter image description here

Otherwise do I have to intergrate this by hand - I thought of using the convolution theoram but firstly I cannot work out the Fourier transform of t^2 (when I try using the fourier transfer equation I get 0) and I also cannot find that in any standard signal processing books.

I have also sketched this function in the time domain. enter image description here Please can you advise me on this question.

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  • \$\begingroup\$ Have you checked with WolframAlpha? Oh, and the function in your title doesn't match the function in your question. \$\endgroup\$ – Roger Rowland Feb 20 '16 at 9:09
  • \$\begingroup\$ I am very sorry, it was just a typing accident. I will the title change it immediately. \$\endgroup\$ – Saavin Feb 20 '16 at 9:13
  • \$\begingroup\$ I changed the title \$\endgroup\$ – Saavin Feb 20 '16 at 9:14
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    \$\begingroup\$ I know it's related but wouldn't he get a quicker answer on math.stackexchange.com \$\endgroup\$ – Rob Feb 20 '16 at 13:41
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Your representation of the negative branch of the function is incorrect.Try putting $$t=-\infty $$ to see that your function diverges as opposed to the definition.The correct integral is as stated: $$X(\omega)=\int_{0}^\infty t^2 exp(-j\omega t-t) dt +\int_{-\infty}^0 t^2 exp(-j\omega t+t)dt$$ (In most mathematical problems you might find $$\frac{1}{(\sqrt{2\pi})}$$ multiplied to the integral,so do adjust that for yourself.)

I think you are talking about trying to evaluate it from your available results. In that case try writing $$f(t)=t^2e^{-t}u(t)+t^2e^tu(-t)$$ You can use the results you have stated along with the time scaling property for the required integral.

I think this is a homework problem,so I leave the final values for you to figure out.

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  • \$\begingroup\$ Thank you sir you have really helped me. I very much understand your explanation \$\endgroup\$ – Saavin Feb 20 '16 at 19:12
  • \$\begingroup\$ Can you please tell me if you mean use the time scaling property for U(t) in the standard result (t^2)*(e^t)*U(-t). Also how to do that? \$\endgroup\$ – Saavin Feb 20 '16 at 23:31
  • \$\begingroup\$ do you mean to use u(-t)=1-u(t) \$\endgroup\$ – Saavin Feb 20 '16 at 23:36
  • \$\begingroup\$ @Saavin, Isn't the second term in the addition a time reversed signal? What is the fourier transform of $$f(-t)$$? After all $$t^2 e^tu(-t) =(-t)^2e^{-(-t)}u(-t)$$ which is just the first term reversed in time.You need to only calculate the first one,for this you can use a scaling of -1 for time and use the property. \$\endgroup\$ – ObnoxiousPerson1 Feb 21 '16 at 4:29
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The key of your problem is how to define the absolute value. I think you have to solve that integral, please look at the picture and let me know if it answers your question, regards:

enter image description here

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  • \$\begingroup\$ i meant to say that the - sign exists BEFORE modulus \$\endgroup\$ – Saavin Feb 20 '16 at 11:30

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