why in fourier transform we have to multiple with e^{-j\omegat}

Status
Not open for further replies.
O

Osawa_Odessa

Banned
Joined
Dec 31, 2012
Messages
168
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,298
Activity points
0
Hello everybody,
I would like to know why in fourier transform we have to multiple with \[e^{-j\omega\ t}\]?:smile:
 

The short answer would be: that is the definition of the Fourier transform.

The longer answer is below:
You can think of the Fourier transform as a sort of conversion between two bases (from time to frequency). Imagine that you are resolving a vector (in 2D space) into its X and Y components, you would be doing a dot product operation with the x and y unit vectors and obtain the coefficient values. The Fourier transformation can be thought to be something similar. Think of your transform operation as converting from a time basis to frequency basis. You are resolving your function into components expressed in the basis of frequency vectors.
 
Hello jeeudr,
I like the opinion but it still seem not clear to me. Could you explain in more detail or reommend me some sites or material that I should read?
 

Re: why in fourier transform we have to multiple with e^{-j\omega t}

From my understanding (that means it may be wrong).
The Fourier transform is a transformation method which objectively transform a function in normal basis into a polar basis.
This transformation resulting in representation of the signal from time domain to be frequency domain function.
The exponential term exp(j*w*t) represents the polar mapping of the function in the transformation process.
 

Re: why in fourier transform we have to multiple with e^{-j\omega t}

The sinusoidal time function that a phasor represents is obtained by multiplying the phasor by e^ jwt
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…