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I am preparing for an exam. I found this problem:

I am trying to use the properties then use a pre-transformed function from the table but I couldn't find one.

I saw a solution on the web:

enter image description here

I think he has a mistake since he forgot the \$t\$'s (each \$\tau\$ should be \$t\$).

Is he right or am I missing something?

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  • \$\begingroup\$ Replacing the tau variable with t is just the same as using a dummy variable. You could use the variable g, h, m, or anything in there. It’s just a variable that will disappear when doing the calculus. \$\endgroup\$
    – Leoman12
    Commented Jun 23, 2020 at 13:24

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Tsin(aT+b) cannot be integrated from T=minus infinity, because the integral function contains summed term Tcos(aT+b). It doesn't get eliminated. It's undefined (as a limit value) at T=minus infinite. So, the formula of x(t) doesn't define a function. The starting point is impossible and the rest is meaningless.

It's possible that by applying different Fourier transform rules one gets a formula which contains delta function or its higher powers. I haven't tried it. It seems to need something like a derivative of the delta function which is two gears beyond classical analysis. The delta function alone becomes from the point frequency. Its derivative becomes from multiplying the sine by T in Tsin(aT+B). It's presented in Fourier transform collections like this:

enter image description here

The formula is taken from this website http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html

Another formula for the symbol game can be found from the same collection. It's this:

enter image description here

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  • \$\begingroup\$ i guess the question trying to illustrate the property of integration in fourier transform .. thus you it will be converted to frequncy domain then divide by w but i am having problem with the fact he removed the t in the second step \$\endgroup\$
    – Mhd Ghd
    Commented Jun 23, 2020 at 8:03
  • \$\begingroup\$ thank you @user287001 it seems to be a problem during the solution i am reviewing however thanks a lot \$\endgroup\$
    – Mhd Ghd
    Commented Jun 23, 2020 at 19:37

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