Can you help me continue the solution?
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What happened to the u(t) in line 4 of your working?
Also surely the FT is X(w) not X(jw) in your notation?
Unless I'm misreading your script.
You generally see the Fourier transform written as a function of just ω while you see the steady-state system transfer function written as a function of jω.Because u(t) is zero for t<0 and equals to 1 in t>0 then the integral in line 3 can be written as the integral in line 4.
I don't understand your second question but my Profs here uses various notation to show the Fourier transform of a signal.some use X(w) and some use X(jw).
What limits of integration are you using? It looks like you have a "1R" at the bottom of the integral sign. If you mean to say that you are integrating over the reals, then expressing it this way isn't really good enough. Integrating from +∞ to -∞ would also qualify as integrating over the reals but would not yield the same answer.
If I now read your post#5 correctly your function to find the FT is the same as the one in my worked example, exccept that I have used the alternative definition of the trnasform pair as in my post#2 of this thread
http://forum.allaboutcircuits.com/threads/table-of-fourier-transforms.104024/
You can compare your notation with mine.
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View attachment 76598
The Fourier transform of the product of the signals in time domain equals to the convolution of the Fourier transforms of those signals (with a coefficient as you can see in the picture I've posted)Does your line 4 conform to this
The product of the transforms equals the transform of the convolution
Or should it be the inverse transform in line 4?
by Jake Hertz
by Jake Hertz
by Jeff Child
by Aaron Carman