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Exponents and Imaginary Numbers

  • Thread starter Natalie89
  • Start date
  • #1
29
0
Hello,

I did the integral of a Fourier Transform which resulted in this:

A(je^(-jwe^(To+t/2) - je^-jw(T0-t/2))/(1/w)

Where A is the amplitude, j the imaginary number, and w is omega or 2*pi*f.

My question is, how can this be further simplifier. I am forgetting how to simplify these exponents, and what trigonometric identities would further simplify the solution.

Thank you!
 

Answers and Replies

  • #2
Please, check your brackets out. I doubt you got an exponential function to the power of another exponential function after you integrated.
 
  • #3
29
0
Oh I am sorry, I made a typo.

A(je^(-jw(T0+t/2)-je^(T0-t/2)w))(1/w)
 
  • #4
Does the second exponent have a j in it? Please consider using latex your post is a messy and difficult to read.
 
  • #5
29
0
Here's an attachment in word! Sorry for the confusion.
 

Attachments

  • #6
Here's an attachment in word! Sorry for the confusion.
Use the following

[tex]sin(x) = \frac{(e^{-ix} - e^{ix})}{2i}[/tex]

What you have right now is the same as

[tex] \frac{Aj e^{jwT}}{w} \left( e^{-jw\frac{t}{2}} -e^{jw \frac{t}{2}} \right) [/tex]
 
  • #7
29
0
I am still able to do this without the imaginary number in the denominator?
 
  • #8
I am still able to do this without the imaginary number in the denominator?

Well,


[tex]
2j*sin(x) = \left(e^{-jx} - e^{jx} \right)
[/tex]



Yes, you can use it without the imaginary number in the demonimator.
 

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