Re: meaning of 'j' term in Fourier, Laplace transform
- From: "euler.shannon@xxxxxxxxx" <euler.shannon@xxxxxxxxx>
- Date: Wed, 3 Dec 2008 19:52:07 -0800 (PST)
On Dec 4, 12:23 am, "commengr" <communications_engin...@xxxxxxxxx>
wrote:
Hello, this thing has been bothering me for some time and I have not been
able to find a reasonable answer t this. But can you please me what the
term 'j' stands for in the Fourier transform when we multiply our signal
(be it in time or frequency domain) by an imaginary/complex exponential
function. My question is not based on the idea if quadrature signals.
Secondly, why cant we use the Laplace and Z-transform instead of Fourier
transform in OFDM, Frequency Domain Equalization etc
Dear...
hope now you got what j means....
why do we use j? cos we need it to represent the concise time or
frequency's two dimensional behaviors. Generally Fourier series or
transform give you "orthonormal projection matrix" which will divide
your signal to independent vectors with each component having two
dimentions.
Laplace, Z and Fourier share some fundamental concepts like
orthonormal projection but i think FFT is the main idea why we use in
OFDM because many previous researches have been done to optimize the
calculation of Fourier Transform ( reducing computational complexity
while maintaining targeted accuracy )
with regards,
.
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