meaning of 'j' term in Fourier, Laplace transform

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

On Wed, 03 Dec 2008 09:23:50 -0600, "commengr"
<communications_engineer@yahoo.com> 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. 

j*j = -1 or j is the complex number with unit magnitude and real part
equal to zero. This is a really interesting question for a
"communications engineer" or I'm missing something.

Muzaffer Kal

DSPIA INC.
ASIC/FPGA Design Services
http://www.dspia.com

commengr 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. 

Holy shit! What qualifies you as "engineer"?

Mathematicians use the letter 'i' to represent sqrt(-1). To avoid 
confusion with the symbol for current, electrical engineers use the 
letter 'j' for the same purpose.

> Secondly, why cant we use the Laplace and Z-transform instead of Fourier
> transform in OFDM, Frequency Domain Equalization etc

You can't use the Hilbert transform either. They are different tools.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

commengr 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

I see no fundamental reason why you can't use the Laplace a Z transforms in 
some appropriate system context.  OFDM, FDE, etc. are system applications 
while those transforms are tools used in system analysis be they analog, 
digital/sampled data, etc.

So, I'll bet we can easily find an OFDM application using analog equipment 
for which a Laplace transform might be useful on some of the parts  [BTW: 
Laplace Transform is for the solution of Ordinary Differential Equations 
with Constant Coefficients].  And a FDE using sampled data for which a Z 
transform might be useful in some of the parts [BTW: Z transform is for the 
solution of Ordinary Difference Equations with Constant Coefficients].

Fred 

On Dec 3, 10:23&#4294967295;am, "commengr" <communications_engin...@yahoo.com>
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

Go find a copy of Paul Nahin's book "Dr. Euler's Fabulous Formula."

The whole book is about exp(i*theta) being cos(theta)+i*sin(theta).

I think you will find many answers in there.

As Fred stated Laplace transforms are mostly useful for turning a
calculus problem into an algebra problem to hopefully make the problem
simpler to solve and analyze. The Z transform is similar but instead
of calculus problems it starts with difference equations.

Clay

On Dec 4, 12:23&#4294967295;am, "commengr" <communications_engin...@yahoo.com>
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,

>On Dec 4, 12:23=A0am, "commengr" <communications_engin...@yahoo.com>
>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,
>

You are the only one who has answered my question SENSIBILY. Ofcourse I
know what 'j' or 'i' for the matter, means. My question was that what does
it mean as far as signal processing is concerned i.e. how does it effect
our practical signal processing rather than just mathematics. 

>On Dec 4, 12:23=A0am, "commengr" <communications_engin...@yahoo.com>
>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,
>



=========================================================================


I was hoping someone would answer my question SENSIBILY. Ofcourse I know
what 'j' or 'i' for the matter, means. My question was that what does it
mean as far as signal processing is concerned i.e. how does it effect our
practical signal processing rather than just mathematics. 

Thanks to the SENSIBLE people who answered my question .... others who
tried to explain it as iota ... imaginary ... are maybe just trying to
increase their posts on this group. Pathetic!

==========================================================================

On Dec 5, 5:22&#4294967295;am, "commengr" <communications_engin...@yahoo.com>
wrote:
> >On Dec 4, 12:23=A0am, "commengr" <communications_engin...@yahoo.com>
> >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,
>
> =========================================================================
>
> I was hoping someone would answer my question SENSIBILY. Ofcourse I know
> what 'j' or 'i' for the matter, means. My question was that what does it
> mean as far as signal processing is concerned i.e. how does it effect our
> practical signal processing rather than just mathematics.
>
> Thanks to the SENSIBLE people who answered my question .... others who
> tried to explain it as iota ... imaginary ... are maybe just trying to
> increase their posts on this group. Pathetic!
>
> ==========================================================================- Hide quoted text -
>
> - Show quoted text -

You are like the guy who begs for a meal and then complains that it is
not good enough for you to eat.

On Dec 5, 9:37�am, buleg...@columbus.rr.com wrote:
> On Dec 5, 5:22&#4294967295;am, "commengr" <communications_engin...@yahoo.com>
> wrote:
>
>
>
>
>
> > >On Dec 4, 12:23=A0am, "commengr" <communications_engin...@yahoo.com>
> > >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,
>
> > =========================================================================
>
> > I was hoping someone would answer my question SENSIBILY. Ofcourse I know
> > what 'j' or 'i' for the matter, means. My question was that what does it
> > mean as far as signal processing is concerned i.e. how does it effect our
> > practical signal processing rather than just mathematics.
>
> > Thanks to the SENSIBLE people who answered my question .... others who
> > tried to explain it as iota ... imaginary ... are maybe just trying to
> > increase their posts on this group. Pathetic!
>
> > ==========================================================================- Hide quoted text -
>
> > - Show quoted text -
>
> You are like the guy who begs for a meal and then complains that it is
> not good enough for you to eat.- Hide quoted text -
>
> - Show quoted text -


It is interesting that what the OP thinks is his SENSIBLE answer did
not really answer his original question.

It also interesting that the OP thinks that people here are trying to
"increase their posts" to the group.  There is certainly a lot to be
gained by doing that. NOT!!!! Maybe that is what the OP does.
"Pathetic"?

Dirk