```On Dec 14, 7:25&#2013266080;pm, communications_engin...@yahoo.com wrote:
> On Dec 6, 12:22&#2013266080;am, Darol Klawetter <darol.klawet...@l-3com.com>
> wrote:
>
>
>
> > On Dec 5, 4:18 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,
>
> > > 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.
>
> > You could find a previous comp.dsp topic to be helpful. It's titled
> > "Complex Number Tutorial" and began in January of this year.
>
> > Darol Klawetter
>
> You're right but it was pretty rude the way I got answered at the
> beginning ... But I also must accept I should have phrased my question
> better ....
>
> THANKS TO ALL THOSE WHO ANSWERED MY QUESTION :)

You're right but it was pretty rude the way I got answered at the
beginning ... But I also must accept I should have phrased my question
better ....

THANKS TO ALL THOSE WHO ANSWERED MY QUESTION :)
```
```On Dec 6, 12:22&#2013266080;am, Darol Klawetter <darol.klawet...@l-3com.com>
wrote:
> On Dec 5, 4:18 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,
>
> > 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.
>
> You could find a previous comp.dsp topic to be helpful. It's titled
> "Complex Number Tutorial" and began in January of this year.
>
> Darol Klawetter

You're right but it was pretty rude the way I got answered at the
beginning ... But I also must accept I should have phrased my question
better ....

THANKS TO ALL THOSE WHO ANSWERED MY QUESTION :)
```
```On Dec 5, 4:18 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,
>
> 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.

You could find a previous comp.dsp topic to be helpful. It's titled
"Complex Number Tutorial" and began in January of this year.

Darol Klawetter
```
```commengr wrote:
>> On Dec 4, 12:23=A0am, "commengr" <communications_engin...@yahoo.com>
>> wrote:
> 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!

What's "sensible"?  Are you thanking ME?  If so, then you're welcome.

Be nice.  Be polite.  Don't question motives.  People here are trying to

```
```commengr wrote:

...

> =========================================================================
>
>
> I was hoping someone would answer my question SENSIBILY. Of course I know
> what 'j' or 'i' for the matter, means.

You asked, "... can you please me what the term 'j' stands for in the
Fourier transform ..." How come, now, "of course"?

>          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.

Then that is the question you should have asked.

> 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!

It is unwise to bite a hand that tries to feed you. Goodbye.

> ==========================================================================

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```On Fri, 05 Dec 2008 04:18:07 -0600, "commengr"
<communications_engineer@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,
>>
>
>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.

If you knew what it meant then you asked the wrong question.  Usenet
is not a good interface for mind reading, people can only go by what
you actually write down.

I still don't know what information you're really looking for.
Consider clarifying your question better and you'll get better
responses.

Garbage in, garbage out.

Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org

Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
```
```On Dec 5, 9:37&#2013266080;am, buleg...@columbus.rr.com wrote:
> On Dec 5, 5:22&#2013266080;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
```
```On Dec 5, 5:22&#2013266080;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 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 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.
```