confused about complex signal

Hi all,

I have a signal a+jb generated inside DSP. Now, I want to feed the signal
to a quadrature modulator, upconvert it to 100MHz and measure it with a
spectrum analyzer or scope. 

My question is, the signal a+jb is complex, the quadrature upconversion is
also complex, exp(j*2*pi*100e6). Will I see the complex signal,
(a+jb)*exp(j*2*pi*100e6) on the spectrum? 

Also, for the complex signal generated from signal generator, GMSK,
CDMA....etc. They have energy at both negative frequency and positive
frequency, or only on one side?


Many Thanks,
cfy30

cfy30 wrote:
> Hi all,
> 
> I have a signal a+jb generated inside DSP. Now, I want to feed the signal
> to a quadrature modulator, upconvert it to 100MHz and measure it with a
> spectrum analyzer or scope. 
> 
> My question is, the signal a+jb is complex, the quadrature upconversion is
> also complex, exp(j*2*pi*100e6). Will I see the complex signal,
> (a+jb)*exp(j*2*pi*100e6) on the spectrum? 

Probably not.  IC quadrature modulators generally have only 
one output and this is 'real,'  but if you had the need, you 
could construct a quadrature modulator to have a complex 
output.  This modulator circuit would necessarily have two 
distinct output connections, one 'real' and one 'imaginary.'

You could not view this complex signal on an analog spectrum 
analyser as they have only a single 'real' input.  If you 
had a spectrum analyser that had a complex input (e.g. a 
FFT-based analyser) then it would be capable of displaying 
both positive-frequency components and negative- frequency 
components, with 0Hz in the middle.  You would then see your 
complex signal appearing on one side or other of the 0Hz point.


> 
> Also, for the complex signal generated from signal generator, GMSK,
> CDMA....etc. They have energy at both negative frequency and positive
> frequency, or only on one side?
> 
> 
> Many Thanks,
> cfy30

Each of the in-phase (I) and quadrature-phase (Q) signals 
taken by itself has both positive-frequency and 
negative-frequency components. When these two signals are 
handled together as a complex IQ signal then either the 
positive-frequency components or the negative-frequency 
components cancel out, so you see energy only on one side of 
the 0Hz point.

Regards,
John

On Dec 7, 6:46&#4294967295;pm, John Monro <johnmo...@optusnet.removethis.com.au>
wrote:

> You could not view this complex signal on an analog spectrum
> analyser as they have only a single 'real' input. &#4294967295;If you
> had a spectrum analyser that had a complex input (e.g. a
> FFT-based analyser) then it would be capable of displaying
> both positive-frequency components and negative- frequency
> components, with 0Hz in the middle. &#4294967295;You would then see your
> complex signal appearing on one side or other of the 0Hz point.
>

> Each of the in-phase (I) and quadrature-phase (Q) signals
> taken by itself has both positive-frequency and
> negative-frequency components. When these two signals are
> handled together as a complex IQ signal then either the
> positive-frequency components or the negative-frequency
> components cancel out, so you see energy only on one side of
> the 0Hz point.

Is the cancellation guaranteed for all possible choices of a and b
asked about by the OP or only for some choices of a and b.
Put another way, given an arbitrary complex signal f(t), is it always
true that either F(w) = 0 for w < 0, or F(w) = 0 for w > 0?  Or
is it just that for a complex signal, the conjugate symmetry
constraint F(-w) = F*(w) does not hold at all frequencies w the
way it should for real-valued signals?

Hi John,

I found something interesting. I thought the function of modulator is to
do exp(j*2*pi*fc) but what I found from the data sheet is, the modulator
simply take the I and Q data or real and imaginery data generated, multiply
the I data with cos(2*pi*fc*t) and Q data with -sin(2*pi*fc*t), them sum
them together at the output. 

If my understanding is correct, then signal at the modulator output is
always real, i.e., there are energy at both negative and positive
frequencies. Still trying to understand this more.


Thanks,
cfy30


>cfy30 wrote:
>> Hi all,
>> 
>> I have a signal a+jb generated inside DSP. Now, I want to feed the
signal
>> to a quadrature modulator, upconvert it to 100MHz and measure it with
a
>> spectrum analyzer or scope. 
>> 
>> My question is, the signal a+jb is complex, the quadrature upconversion
is
>> also complex, exp(j*2*pi*100e6). Will I see the complex signal,
>> (a+jb)*exp(j*2*pi*100e6) on the spectrum? 
>
>Probably not.  IC quadrature modulators generally have only 
>one output and this is 'real,'  but if you had the need, you 
>could construct a quadrature modulator to have a complex 
>output.  This modulator circuit would necessarily have two 
>distinct output connections, one 'real' and one 'imaginary.'
>
>You could not view this complex signal on an analog spectrum 
>analyser as they have only a single 'real' input.  If you 
>had a spectrum analyser that had a complex input (e.g. a 
>FFT-based analyser) then it would be capable of displaying 
>both positive-frequency components and negative- frequency 
>components, with 0Hz in the middle.  You would then see your 
>complex signal appearing on one side or other of the 0Hz point.
>
>
>> 
>> Also, for the complex signal generated from signal generator, GMSK,
>> CDMA....etc. They have energy at both negative frequency and positive
>> frequency, or only on one side?
>> 
>> 
>> Many Thanks,
>> cfy30
>
>Each of the in-phase (I) and quadrature-phase (Q) signals 
>taken by itself has both positive-frequency and 
>negative-frequency components. When these two signals are 
>handled together as a complex IQ signal then either the 
>positive-frequency components or the negative-frequency 
>components cancel out, so you see energy only on one side of 
>the 0Hz point.
>
>Regards,
>John
>

On Dec 8, 1:10&#4294967295;pm, "dvsarw...@yahoo.com" <dvsarw...@gmail.com> wrote:
> On Dec 7, 6:46&#4294967295;pm, John Monro <johnmo...@optusnet.removethis.com.au>
> wrote:
>
> > You could not view this complex signal on an analog spectrum
> > analyser as they have only a single 'real' input. &#4294967295;If you
> > had a spectrum analyser that had a complex input (e.g. a
> > FFT-based analyser) then it would be capable of displaying
> > both positive-frequency components and negative- frequency
> > components, with 0Hz in the middle. &#4294967295;You would then see your
> > complex signal appearing on one side or other of the 0Hz point.
>
> > Each of the in-phase (I) and quadrature-phase (Q) signals
> > taken by itself has both positive-frequency and
> > negative-frequency components. When these two signals are
> > handled together as a complex IQ signal then either the
> > positive-frequency components or the negative-frequency
> > components cancel out, so you see energy only on one side of
> > the 0Hz point.
>
> Is the cancellation guaranteed for all possible choices of a and b
> asked about by the OP or only for some choices of a and b.
> Put another way, given an arbitrary complex signal f(t), is it always
> true that either F(w) = 0 for w < 0, or F(w) = 0 for w > 0? &#4294967295;Or
> is it just that for a complex signal, the conjugate symmetry
> constraint F(-w) = F*(w) does not hold at all frequencies w the
> way it should for real-valued signals?

Hi,
The content of the 'a' and 'b' component effects the spectrum either
side of 100MHz point in the up-converted signal, and has no effect on
the cancellation that I mentioned.

The OP was asking whether the converted signal had positive and
negative frequency components. This will depend on the details of the
design of the complex modulator that I referred to.  If such a device
were required I assumed it would be designed to deliver a pure
analytic signal and as a result either the positive-frequency
components or the negative-frequency components would be cancelled
out.

This cancellation is does not depend on the characteristics of signals
'a' and 'b' which may be components of an analytic signal or
alternatively may be unrelated.

Sorry for the delayed reply.  My Internet Provider (OPTUS) has not
been able to provide Newsgroups for the last few days.

Regards,
John

On Dec 8, 2:44&#4294967295;pm, "cfy30" <cf...@yahoo.com> wrote:
> Hi John,
>
> I found something interesting. I thought the function of modulator is to
> do exp(j*2*pi*fc) but what I found from the data sheet is, the modulator
> simply take the I and Q data or real and imaginery data generated, multiply
> the I data with cos(2*pi*fc*t) and Q data with -sin(2*pi*fc*t), them sum
> them together at the output.
>
> If my understanding is correct, then signal at the modulator output is
> always real, i.e., there are energy at both negative and positive
> frequencies. Still trying to understand this more.
>
> Thanks,
> cfy30
>
>
>
> >cfy30 wrote:
> >> Hi all,
>
> >> I have a signal a+jb generated inside DSP. Now, I want to feed the
> signal
> >> to a quadrature modulator, upconvert it to 100MHz and measure it with
> a
> >> spectrum analyzer or scope.
>
> >> My question is, the signal a+jb is complex, the quadrature upconversion
> is
> >> also complex, exp(j*2*pi*100e6). Will I see the complex signal,
> >> (a+jb)*exp(j*2*pi*100e6) on the spectrum?
>
> >Probably not. &#4294967295;IC quadrature modulators generally have only
> >one output and this is 'real,' &#4294967295;but if you had the need, you
> >could construct a quadrature modulator to have a complex
> >output. &#4294967295;This modulator circuit would necessarily have two
> >distinct output connections, one 'real' and one 'imaginary.'
>
> >You could not view this complex signal on an analog spectrum
> >analyser as they have only a single 'real' input. &#4294967295;If you
> >had a spectrum analyser that had a complex input (e.g. a
> >FFT-based analyser) then it would be capable of displaying
> >both positive-frequency components and negative- frequency
> >components, with 0Hz in the middle. &#4294967295;You would then see your
> >complex signal appearing on one side or other of the 0Hz point.
>
> >> Also, for the complex signal generated from signal generator, GMSK,
> >> CDMA....etc. They have energy at both negative frequency and positive
> >> frequency, or only on one side?
>
> >> Many Thanks,
> >> cfy30
>
> >Each of the in-phase (I) and quadrature-phase (Q) signals
> >taken by itself has both positive-frequency and
> >negative-frequency components. When these two signals are
> >handled together as a complex IQ signal then either the
> >positive-frequency components or the negative-frequency
> >components cancel out, so you see energy only on one side of
> >the 0Hz point.
>
> >Regards,
> >John

Hi cfy30,

This illustration may help:

Take a complex signal containing only positive-frequency components.
Physically this will have a wire ('I')  with the in-phase components
and a wire ('Q') with the quadrature components.  The components on
'Q' is the same as the components on 'I' except that all components on
Q have a phase of -90 degrees relative to I.

Now take an otherwise identical complex signal with only negative-
frequency components.  It is similar to the above except that the
components on its Q wire have a phase of +90 relative to the 'I'
components.

If we now sum the two 'I' signals and we separately sum the two 'Q'
signals we find that the resulting I signal is twice the amplitude and
the resulting Q signal is zero (because the phase difference between
the two Q components is 180 degrees).  What we have now is a 'real'
signal which necessarily contains both positive and negative frequency
components and yet exists on one single wire.

Admittedly it does seem strange at first that while one wire is enough
to carry both positive and negative frequencies, it requires two wires
to carry either the positive or the negative frequencies alone.

Sorry about the delayed reply, but my Internet Provider (OPTUS) is
having trouble providing Newsgroups and it does not seem to be a high
priority for them to fix the problem.  On the other hand,  I CAN
download any number of ring-tones to my OPTUS phone so I suppose one
takes the good with the bad (:=)

Regards,
John

On Dec 13, 7:47�pm, johnmo...@optusnet.com.au wrote:
> On Dec 8, 2:44&#4294967295;pm, "cfy30" <cf...@yahoo.com> wrote:
>
>
>
> > Hi John,
>
> > I found something interesting. I thought the function of modulator is to
> > do exp(j*2*pi*fc) but what I found from the data sheet is, the modulator
> > simply take the I and Q data or real and imaginery data generated, multiply
> > the I data with cos(2*pi*fc*t) and Q data with -sin(2*pi*fc*t), them sum
> > them together at the output.
>
> > If my understanding is correct, then signal at the modulator output is
> > always real, i.e., there are energy at both negative and positive
> > frequencies. Still trying to understand this more.
>
> > Thanks,
> > cfy30
>
> > >cfy30 wrote:
> > >> Hi all,
>
> > >> I have a signal a+jb generated inside DSP. Now, I want to feed the
> > signal
> > >> to a quadrature modulator, upconvert it to 100MHz and measure it with
> > a
> > >> spectrum analyzer or scope.
>
> > >> My question is, the signal a+jb is complex, the quadrature upconversion
> > is
> > >> also complex, exp(j*2*pi*100e6). Will I see the complex signal,
> > >> (a+jb)*exp(j*2*pi*100e6) on the spectrum?
>
> > >Probably not. &#4294967295;IC quadrature modulators generally have only
> > >one output and this is 'real,' &#4294967295;but if you had the need, you
> > >could construct a quadrature modulator to have a complex
> > >output. &#4294967295;This modulator circuit would necessarily have two
> > >distinct output connections, one 'real' and one 'imaginary.'
>
> > >You could not view this complex signal on an analog spectrum
> > >analyser as they have only a single 'real' input. &#4294967295;If you
> > >had a spectrum analyser that had a complex input (e.g. a
> > >FFT-based analyser) then it would be capable of displaying
> > >both positive-frequency components and negative- frequency
> > >components, with 0Hz in the middle. &#4294967295;You would then see your
> > >complex signal appearing on one side or other of the 0Hz point.
>
> > >> Also, for the complex signal generated from signal generator, GMSK,
> > >> CDMA....etc. They have energy at both negative frequency and positive
> > >> frequency, or only on one side?
>
> > >> Many Thanks,
> > >> cfy30
>
> > >Each of the in-phase (I) and quadrature-phase (Q) signals
> > >taken by itself has both positive-frequency and
> > >negative-frequency components. When these two signals are
> > >handled together as a complex IQ signal then either the
> > >positive-frequency components or the negative-frequency
> > >components cancel out, so you see energy only on one side of
> > >the 0Hz point.
>
> > >Regards,
> > >John
>
> Hi cfy30,
>
> This illustration may help:
>
> Take a complex signal containing only positive-frequency components.
> Physically this will have a wire ('I') &#4294967295;with the in-phase components
> and a wire ('Q') with the quadrature components. &#4294967295;The components on
> 'Q' is the same as the components on 'I' except that all components on
> Q have a phase of -90 degrees relative to I.
>
> Now take an otherwise identical complex signal with only negative-
> frequency components. &#4294967295;It is similar to the above except that the
> components on its Q wire have a phase of +90 relative to the 'I'
> components.
>
> If we now sum the two 'I' signals and we separately sum the two 'Q'
> signals we find that the resulting I signal is twice the amplitude and
> the resulting Q signal is zero (because the phase difference between
> the two Q components is 180 degrees). &#4294967295;What we have now is a 'real'
> signal which necessarily contains both positive and negative frequency
> components and yet exists on one single wire.
>
> Admittedly it does seem strange at first that while one wire is enough
> to carry both positive and negative frequencies, it requires two wires
> to carry either the positive or the negative frequencies alone.
>
> Sorry about the delayed reply, but my Internet Provider (OPTUS) is
> having trouble providing Newsgroups and it does not seem to be a high
> priority for them to fix the problem. &#4294967295;On the other hand, &#4294967295;I CAN
> download any number of ring-tones to my OPTUS phone so I suppose one
> takes the good with the bad (:=)
>
> Regards,
> John

BUT do negative frequency means the orientation of the sinusoid (for
e.g) right? Correct me if I'm wrong ... Also what is the practical
significance of using an imaginary component when we are dealing with
a quadrature composite signal. If I may also ask does 'j' represent
orthonormal signals ?

On Dec 14, 9:12�am, communications_engin...@yahoo.com wrote:
>
> BUT do negative frequency means the orientation of the sinusoid (for
> e.g) right? Correct me if I'm wrong ... Also what is the practical
> significance of using an imaginary component when we are dealing with
> a quadrature composite signal. If I may also ask does 'j' represent
> orthonormal signals ?- Hide quoted text -
>
> - Show quoted text -

Negative frequency is really only relevent in the complex domain.  A
sinusoidal signal is comprised of two phasors: A negative frequency
phasor, and a positive frequency phasor.  Thes two complex
exponentials add up to make a real sinusoid.

I have built some flash programs that illustrate how phasors add up to
make sinusoids:

http://www.fourier-series.com/fourierseries2/complex_tutorial.html

You want to work with complex numbers because you don;t really
understand how sinusoidal motion (as a solution to a differential
equation) makes sense, until you can see the complex exponentials.

On Dec 15, 1:12�am, communications_engin...@yahoo.com wrote:
> On Dec 13, 7:47�pm, johnmo...@optusnet.com.au wrote:
>
>
>
> > On Dec 8, 2:44&#4294967295;pm, "cfy30" <cf...@yahoo.com> wrote:
>
> > > Hi John,
>
> > > I found something interesting. I thought the function of modulator is to
> > > do exp(j*2*pi*fc) but what I found from the data sheet is, the modulator
> > > simply take the I and Q data or real and imaginery data generated, multiply
> > > the I data with cos(2*pi*fc*t) and Q data with -sin(2*pi*fc*t), them sum
> > > them together at the output.
>
> > > If my understanding is correct, then signal at the modulator output is
> > > always real, i.e., there are energy at both negative and positive
> > > frequencies. Still trying to understand this more.
>
> > > Thanks,
> > > cfy30
>
> > > >cfy30 wrote:
> > > >> Hi all,
>
> > > >> I have a signal a+jb generated inside DSP. Now, I want to feed the
> > > signal
> > > >> to a quadrature modulator, upconvert it to 100MHz and measure it with
> > > a
> > > >> spectrum analyzer or scope.
>
> > > >> My question is, the signal a+jb is complex, the quadrature upconversion
> > > is
> > > >> also complex, exp(j*2*pi*100e6). Will I see the complex signal,
> > > >> (a+jb)*exp(j*2*pi*100e6) on the spectrum?
>
> > > >Probably not. &#4294967295;IC quadrature modulators generally have only
> > > >one output and this is 'real,' &#4294967295;but if you had the need, you
> > > >could construct a quadrature modulator to have a complex
> > > >output. &#4294967295;This modulator circuit would necessarily have two
> > > >distinct output connections, one 'real' and one 'imaginary.'
>
> > > >You could not view this complex signal on an analog spectrum
> > > >analyser as they have only a single 'real' input. &#4294967295;If you
> > > >had a spectrum analyser that had a complex input (e.g. a
> > > >FFT-based analyser) then it would be capable of displaying
> > > >both positive-frequency components and negative- frequency
> > > >components, with 0Hz in the middle. &#4294967295;You would then see your
> > > >complex signal appearing on one side or other of the 0Hz point.
>
> > > >> Also, for the complex signal generated from signal generator, GMSK,
> > > >> CDMA....etc. They have energy at both negative frequency and positive
> > > >> frequency, or only on one side?
>
> > > >> Many Thanks,
> > > >> cfy30
>
> > > >Each of the in-phase (I) and quadrature-phase (Q) signals
> > > >taken by itself has both positive-frequency and
> > > >negative-frequency components. When these two signals are
> > > >handled together as a complex IQ signal then either the
> > > >positive-frequency components or the negative-frequency
> > > >components cancel out, so you see energy only on one side of
> > > >the 0Hz point.
>
> > > >Regards,
> > > >John
>
> > Hi cfy30,
>
> > This illustration may help:
>
> > Take a complex signal containing only positive-frequency components.
> > Physically this will have a wire ('I') &#4294967295;with the in-phase components
> > and a wire ('Q') with the quadrature components. &#4294967295;The components on
> > 'Q' is the same as the components on 'I' except that all components on
> > Q have a phase of -90 degrees relative to I.
>
> > Now take an otherwise identical complex signal with only negative-
> > frequency components. &#4294967295;It is similar to the above except that the
> > components on its Q wire have a phase of +90 relative to the 'I'
> > components.
>
> > If we now sum the two 'I' signals and we separately sum the two 'Q'
> > signals we find that the resulting I signal is twice the amplitude and
> > the resulting Q signal is zero (because the phase difference between
> > the two Q components is 180 degrees). &#4294967295;What we have now is a 'real'
> > signal which necessarily contains both positive and negative frequency
> > components and yet exists on one single wire.
>
> > Admittedly it does seem strange at first that while one wire is enough
> > to carry both positive and negative frequencies, it requires two wires
> > to carry either the positive or the negative frequencies alone.
>
> > Sorry about the delayed reply, but my Internet Provider (OPTUS) is
> > having trouble providing Newsgroups and it does not seem to be a high
> > priority for them to fix the problem. &#4294967295;On the other hand, &#4294967295;I CAN
> > download any number of ring-tones to my OPTUS phone so I suppose one
> > takes the good with the bad (:=)
>
> > Regards,
> > John
>
> BUT do negative frequency means the orientation of the sinusoid (for
> e.g) right? Correct me if I'm wrong ...

I'm not sure what you mean here.
 I take a negative frequency sinusoid to be one in which the imaginary
(Q) has a phase of +90 degrees relative to the real (I) component.

 Also what is the practical significance of using an imaginary
component when we are dealing with
> a quadrature composite signal.

It is useful to regard a real signal as containing equal positive and
negative frequency components.  The imaginary parts cancel out, hence
the signal can exist on one wire.

If you want either the negative frequency or the positive frequency
component alone, then the imaginary component is non-zero and so you
need two wires (I and Q) to contain the signal.

 If I may also ask does 'j' represent
> orthonormal signals ?

In this context the j operator is used to designate the imaginary
component of a complex time-domain signal.
It is also used to designate the imaginary component in the phasor
representation of a signal.

On Dec 15, 4:10�am, johnmo...@optusnet.com.au wrote:
> On Dec 15, 1:12�am, communications_engin...@yahoo.com wrote:
>
>
>
> > On Dec 13, 7:47�pm, johnmo...@optusnet.com.au wrote:
>
> > > On Dec 8, 2:44&#4294967295;pm, "cfy30" <cf...@yahoo.com> wrote:
>
> > > > Hi John,
>
> > > > I found something interesting. I thought the function of modulator is to
> > > > do exp(j*2*pi*fc) but what I found from the data sheet is, the modulator
> > > > simply take the I and Q data or real and imaginery data generated, multiply
> > > > the I data with cos(2*pi*fc*t) and Q data with -sin(2*pi*fc*t), them sum
> > > > them together at the output.
>
> > > > If my understanding is correct, then signal at the modulator output is
> > > > always real, i.e., there are energy at both negative and positive
> > > > frequencies. Still trying to understand this more.
>
> > > > Thanks,
> > > > cfy30
>
> > > > >cfy30 wrote:
> > > > >> Hi all,
>
> > > > >> I have a signal a+jb generated inside DSP. Now, I want to feed the
> > > > signal
> > > > >> to a quadrature modulator, upconvert it to 100MHz and measure it with
> > > > a
> > > > >> spectrum analyzer or scope.
>
> > > > >> My question is, the signal a+jb is complex, the quadrature upconversion
> > > > is
> > > > >> also complex, exp(j*2*pi*100e6). Will I see the complex signal,
> > > > >> (a+jb)*exp(j*2*pi*100e6) on the spectrum?
>
> > > > >Probably not. &#4294967295;IC quadrature modulators generally have only
> > > > >one output and this is 'real,' &#4294967295;but if you had the need, you
> > > > >could construct a quadrature modulator to have a complex
> > > > >output. &#4294967295;This modulator circuit would necessarily have two
> > > > >distinct output connections, one 'real' and one 'imaginary.'
>
> > > > >You could not view this complex signal on an analog spectrum
> > > > >analyser as they have only a single 'real' input. &#4294967295;If you
> > > > >had a spectrum analyser that had a complex input (e.g. a
> > > > >FFT-based analyser) then it would be capable of displaying
> > > > >both positive-frequency components and negative- frequency
> > > > >components, with 0Hz in the middle. &#4294967295;You would then see your
> > > > >complex signal appearing on one side or other of the 0Hz point.
>
> > > > >> Also, for the complex signal generated from signal generator, GMSK,
> > > > >> CDMA....etc. They have energy at both negative frequency and positive
> > > > >> frequency, or only on one side?
>
> > > > >> Many Thanks,
> > > > >> cfy30
>
> > > > >Each of the in-phase (I) and quadrature-phase (Q) signals
> > > > >taken by itself has both positive-frequency and
> > > > >negative-frequency components. When these two signals are
> > > > >handled together as a complex IQ signal then either the
> > > > >positive-frequency components or the negative-frequency
> > > > >components cancel out, so you see energy only on one side of
> > > > >the 0Hz point.
>
> > > > >Regards,
> > > > >John
>
> > > Hi cfy30,
>
> > > This illustration may help:
>
> > > Take a complex signal containing only positive-frequency components.
> > > Physically this will have a wire ('I') &#4294967295;with the in-phase components
> > > and a wire ('Q') with the quadrature components. &#4294967295;The components on
> > > 'Q' is the same as the components on 'I' except that all components on
> > > Q have a phase of -90 degrees relative to I.
>
> > > Now take an otherwise identical complex signal with only negative-
> > > frequency components. &#4294967295;It is similar to the above except that the
> > > components on its Q wire have a phase of +90 relative to the 'I'
> > > components.
>
> > > If we now sum the two 'I' signals and we separately sum the two 'Q'
> > > signals we find that the resulting I signal is twice the amplitude and
> > > the resulting Q signal is zero (because the phase difference between
> > > the two Q components is 180 degrees). &#4294967295;What we have now is a 'real'
> > > signal which necessarily contains both positive and negative frequency
> > > components and yet exists on one single wire.
>
> > > Admittedly it does seem strange at first that while one wire is enough
> > > to carry both positive and negative frequencies, it requires two wires
> > > to carry either the positive or the negative frequencies alone.
>
> > > Sorry about the delayed reply, but my Internet Provider (OPTUS) is
> > > having trouble providing Newsgroups and it does not seem to be a high
> > > priority for them to fix the problem. &#4294967295;On the other hand, &#4294967295;I CAN
> > > download any number of ring-tones to my OPTUS phone so I suppose one
> > > takes the good with the bad (:=)
>
> > > Regards,
> > > John
>
> > BUT do negative frequency means the orientation of the sinusoid (for
> > e.g) right? Correct me if I'm wrong ...
>
> I'm not sure what you mean here.
> &#4294967295;I take a negative frequency sinusoid to be one in which the imaginary
> (Q) has a phase of +90 degrees relative to the real (I) component.
>
> &#4294967295;Also what is the practical significance of using an imaginary
> component when we are dealing with
>
> > a quadrature composite signal.
>
> It is useful to regard a real signal as containing equal positive and
> negative frequency components. &#4294967295;The imaginary parts cancel out, hence
> the signal can exist on one wire.
>
> If you want either the negative frequency or the positive frequency
> component alone, then the imaginary component is non-zero and so you
> need two wires (I and Q) to contain the signal.
>
> &#4294967295;If I may also ask does 'j' represent
>
> > orthonormal signals ?
>
> In this context the j operator is used to designate the imaginary
> component of a complex time-domain signal.
> It is also used to designate the imaginary component in the phasor
> representation of a signal.


Thanks.