Exponents and Imaginary Numbers

[Natalie89]

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!

 

[╔(σ_σ)╝]

Please, check your brackets out. I doubt you got an exponential function to the power of another exponential function after you integrated.

 

[Natalie89]

Oh I am sorry, I made a typo.

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

 

[╔(σ_σ)╝]

Does the second exponent have a j in it? Please consider using latex your post is a messy and difficult to read.

 

[Natalie89]

Here's an attachment in word! Sorry for the confusion.

 

[╔(σ_σ)╝]

Here's an attachment in word! Sorry for the confusion.

Use the following

$$sin(x) = \frac{(e^{-ix} - e^{ix})}{2i}$$

What you have right now is the same as

$$\frac{Aj e^{jwT}}{w} \left( e^{-jw\frac{t}{2}} -e^{jw \frac{t}{2}} \right)$$

 

[Natalie89]

I am still able to do this without the imaginary number in the denominator?

 

[╔(σ_σ)╝]

I am still able to do this without the imaginary number in the denominator?

Well, $$2j*sin(x) = \left(e^{-jx} - e^{jx} \right)$$
Yes, you can use it without the imaginary number in the demonimator.