Fourier Transform

WBahn

Joined Mar 31, 2012
30,062
Look at a table of Fourier transforms and you will see that what your prof told you is wrong (or, more likely, that you have made an error in transcribing from the slides to your post).

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.
 

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arman19940326

Joined Jul 31, 2014
43
Just look at this picture.the solution seems right to me.As you can see the inverse Fourier transform of the signal (1/(jw+2j)) yields to (e^(-2jt) u(t)) in time domain. here my problem begins...
 

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arman19940326

Joined Jul 31, 2014
43
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.

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).
 

WBahn

Joined Mar 31, 2012
30,062
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).
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.
 

Thread Starter

arman19940326

Joined Jul 31, 2014
43
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.

Yes,I understand what you meant.but for convenience I assume the integration over "1R" is identical to integration from -∞ to +∞. Excuse me if I didn't mention my assumptions.
 

Thread Starter

arman19940326

Joined Jul 31, 2014
43
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.

View attachment 76597
View attachment 76598

Yeah,I have seen this notation before but for the course I have this semester, we do not use this alternative definition.
Well in your worked example the function is real valued in time domain but what I want is the Fourier transform of a complex valued function in time domain as I wrote in my starting question.
The main question is this:What is the Fourier transform of the signal (e^(-2jt) u(t))? it definitely can't be obtained from the usual definition of the Fourier transform (as I showed it in post #1)....I think maybe there is an alternative way to define it as we define the Fourier transform of periodic signals.
 

studiot

Joined Nov 9, 2007
4,998
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?
 

Thread Starter

arman19940326

Joined Jul 31, 2014
43
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?
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)
 
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