COMPGROUPS.NET | Search | Post Question | Groups | Stream | About | Register

### What is quadrature sampling

• Email
• Follow

```Hi all,
What is quadrature sampling?It says that sampling frequency be greater
than signal BW.Does it not violate Nyquist criterion?What are its
applications?
Regards.
Raghavendra
```
 0
Reply raghurash (10) 7/7/2003 12:36:25 PM

See related articles to this posting

```In article <becb8p\$3khtp\$1@ID-82263.news.dfncis.de>, bhaskart@deja.com
says...
> "Raghavendra" <raghurash@rediffmail.com> wrote in message
> > Hi all,
> > What is quadrature sampling?It says that sampling frequency be greater
>
> As I understand the term, it means that the sampling freq is exactly 4 times
> the IF frequency. Usually the BW of the signal is smaller than fs/2.
> By picking this sampling freq, it allows you to perform quadrature mixing to
> extract I,Q in a very easy fashion (multiply the signal with only a series
> of 1s, -1s and zeros).
>
> Cheers
>
> > than signal BW.Does it not violate Nyquist criterion?What are its
> > applications?
> > Regards.
> > Raghavendra
>
>
>

heres an angle:

quad sampling exploits the mathematical property that comparing a test
signal (the information carrying signal) to a reference sin wave can be
done independently to comparing it to a reference cosine wave.  ie you
can extract 2 bits of information from a single symbols width of signal
by comparing it to 2 different reference points (sin and cos) at the
receiving end.  So for example, you quite literally extract symbols at
(eg) 1200 baud, but you end up with a bit rate of 2400 bps.

At the transmitting end, you start with a pair of bits (pull them out fo
the incoming stream in pairs), apply one to the sin reference, apply one
to the cos reference, then you add the two signals for transmission.
What you end up with after this addition is just a single sinusoid.  Its
strange to imagine that 2 bits can be extracted from it.  At the
receiving end, to extract the I channel bit, the signal is compared to
the cos reference wave, then to extract the Q channel bit, the same wave
is compared to the sin reference wave.  The curious part is that the 2
data bits can vary independently so that their combination for
transmission will not affect extraction of the 'other' bit at the
receiving end.

Spiro

--

http://www.mobilecomms.com.au
http://www.nexiondata.com
```
 0
Reply Spiro 7/7/2003 11:39:03 PM

```"Spiro" <HOUSTON_we_have_a_problem@127.0.0.1> wrote in message
news:MPG.1974a224e288cf76989691@61.9.128.12...
> In article <becb8p\$3khtp\$1@ID-82263.news.dfncis.de>, bhaskart@deja.com
> says...
> > "Raghavendra" <raghurash@rediffmail.com> wrote in message
> > > Hi all,
> > > What is quadrature sampling?It says that sampling frequency be greater
> >
> > As I understand the term, it means that the sampling freq is exactly 4
times
> > the IF frequency. Usually the BW of the signal is smaller than fs/2.
> > By picking this sampling freq, it allows you to perform quadrature
mixing to
> > extract I,Q in a very easy fashion (multiply the signal with only a
series
> > of 1s, -1s and zeros).

(snip)

> heres an angle:
>
> quad sampling exploits the mathematical property that comparing a test
> signal (the information carrying signal) to a reference sin wave can be
> done independently to comparing it to a reference cosine wave.  ie you
> can extract 2 bits of information from a single symbols width of signal
> by comparing it to 2 different reference points (sin and cos) at the
> receiving end.  So for example, you quite literally extract symbols at
> (eg) 1200 baud, but you end up with a bit rate of 2400 bps.
>
> At the transmitting end, you start with a pair of bits (pull them out fo
> the incoming stream in pairs), apply one to the sin reference, apply one
> to the cos reference, then you add the two signals for transmission.
> What you end up with after this addition is just a single sinusoid.  Its
> strange to imagine that 2 bits can be extracted from it.  At the
> receiving end, to extract the I channel bit, the signal is compared to
> the cos reference wave, then to extract the Q channel bit, the same wave
> is compared to the sin reference wave.  The curious part is that the 2
> data bits can vary independently so that their combination for
> transmission will not affect extraction of the 'other' bit at the
> receiving end.

It sounds pretty much the same as extracting both the signal and its
derivative, which you can also do at half the Nyquist rate.  As long as you
extract the appropriate amount of information, Nyquist doesn't really care
how you do it.

-- glen

```
 0
Reply Glen 7/8/2003 12:00:07 AM

```On Tue, 08 Jul 2003 00:00:07 GMT, "Glen Herrmannsfeldt"
<gah@ugcs.caltech.edu> wrote:

>
>"Spiro" <HOUSTON_we_have_a_problem@127.0.0.1> wrote in message
>news:MPG.1974a224e288cf76989691@61.9.128.12...
>> In article <becb8p\$3khtp\$1@ID-82263.news.dfncis.de>, bhaskart@deja.com
>> says...
>> > "Raghavendra" <raghurash@rediffmail.com> wrote in message
>> > > Hi all,
>> > > What is quadrature sampling?It says that sampling frequency be greater
>> >
>> > As I understand the term, it means that the sampling freq is exactly 4
>times
>> > the IF frequency. Usually the BW of the signal is smaller than fs/2.
>> > By picking this sampling freq, it allows you to perform quadrature
>mixing to
>> > extract I,Q in a very easy fashion (multiply the signal with only a
>series
>> > of 1s, -1s and zeros).
>
>(snip)
>
>> heres an angle:
>>
>> quad sampling exploits the mathematical property that comparing a test
>> signal (the information carrying signal) to a reference sin wave can be
>> done independently to comparing it to a reference cosine wave.  ie you
>> can extract 2 bits of information from a single symbols width of signal
>> by comparing it to 2 different reference points (sin and cos) at the
>> receiving end.  So for example, you quite literally extract symbols at
>> (eg) 1200 baud, but you end up with a bit rate of 2400 bps.
>>
>> At the transmitting end, you start with a pair of bits (pull them out fo
>> the incoming stream in pairs), apply one to the sin reference, apply one
>> to the cos reference, then you add the two signals for transmission.
>> What you end up with after this addition is just a single sinusoid.  Its
>> strange to imagine that 2 bits can be extracted from it.  At the
>> receiving end, to extract the I channel bit, the signal is compared to
>> the cos reference wave, then to extract the Q channel bit, the same wave
>> is compared to the sin reference wave.  The curious part is that the 2
>> data bits can vary independently so that their combination for
>> transmission will not affect extraction of the 'other' bit at the
>> receiving end.
>
>It sounds pretty much the same as extracting both the signal and its
>derivative, which you can also do at half the Nyquist rate.  As long as you
>extract the appropriate amount of information, Nyquist doesn't really care
>how you do it.
>
>-- glen

Yes, and lately I've seen Julius and a few folks use the term
"innovation rate" in papers rather than sample rate to clarify this
distinction (at least I think that's the intent).   This touches on
the real or complex sample debate that we have here occasionally as
well...

Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be Intel's opinions.
http://www.ericjacobsen.org
```
 0
Reply eric 7/8/2003 12:41:50 AM

```A couple of extra points:

Randy Yates wrote:
> [...]
> In quadrature demodulation, you take the signal spectrum from 0 to
> 2*pi*B (or, more generally, from 2*pi*(fc - B/2) to 2*pi*(fc + B/2),
> i.e., some bandwidth around a carrier fc) and shift it down so that it
> is centered around 0 radians/second.

I should also add that you throw away the signal spectrum from -2*pi*B to 0.

I must inform you that you've been lied to all your life with this
Nyquist criterion thing. It says that sampling at a rate of Fs provides
a bandwidth of Fs/2. This is a lie! Sampling at a rate of Fs provides
a bandwidth of Fs, that is, from -Fs/2 to +Fs/2! The problem is that
real signals contain no useful information from -Fs/2 to 0, but that's
not the sampling process's fault! It does indeed provide you with a
bandwidth of Fs, and a quadrature demodulator is a device that allows
you to make use of this full bandwidth.
--
%  Randy Yates                  % "...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO
```
 0
Reply Randy 7/8/2003 2:08:35 AM

```"Randy Yates" <yates@ieee.org> wrote in message
news:3F0A2847.BA306D6C@ieee.org...
> A couple of extra points:
>
> Randy Yates wrote:
> > [...]
> > In quadrature demodulation, you take the signal spectrum from 0 to
> > 2*pi*B (or, more generally, from 2*pi*(fc - B/2) to 2*pi*(fc + B/2),
> > i.e., some bandwidth around a carrier fc) and shift it down so that it
> > is centered around 0 radians/second.
>
> I should also add that you throw away the signal spectrum from -2*pi*B to
0.
>
> I must inform you that you've been lied to all your life with this
> Nyquist criterion thing. It says that sampling at a rate of Fs provides
> a bandwidth of Fs/2. This is a lie! Sampling at a rate of Fs provides
> a bandwidth of Fs, that is, from -Fs/2 to +Fs/2! The problem is that
> real signals contain no useful information from -Fs/2 to 0, but that's
> not the sampling process's fault! It does indeed provide you with a
> bandwidth of Fs, and a quadrature demodulator is a device that allows
> you to make use of this full bandwidth.

Well, you can do it that way.

The way I like to think of it is that, for a signal of length T, there are T
Fs unknows, and T Fs equations are needed to solve for them.  Sampling the
real signal at the appropriate number of points works, sampling the signal
and its derivative at half that points works, and sampling the appropriate
complex function at half the number of points works.  The sample data must
be linearly independent which restricts it a little bit.

(Sampling the value, first, and second derivative at one third the number
of points works, too, and doesn't have any connection to complex math.)

-- glen

```
 0
Reply Glen 7/8/2003 2:35:31 AM

```"Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message news:<T5qOa.2362\$N7.679@sccrnsc03>...
> "Randy Yates" <yates@ieee.org> wrote in message
> news:3F0A2847.BA306D6C@ieee.org...
> > A couple of extra points:
> >
> > Randy Yates wrote:
> > > [...]
> > > In quadrature demodulation, you take the signal spectrum from 0 to
> > > 2*pi*B (or, more generally, from 2*pi*(fc - B/2) to 2*pi*(fc + B/2),
> > > i.e., some bandwidth around a carrier fc) and shift it down so that it
> > > is centered around 0 radians/second.
> >
> > I should also add that you throw away the signal spectrum from -2*pi*B to
>  0.
> >
> > I must inform you that you've been lied to all your life with this
> > Nyquist criterion thing. It says that sampling at a rate of Fs provides
> > a bandwidth of Fs/2. This is a lie! Sampling at a rate of Fs provides
> > a bandwidth of Fs, that is, from -Fs/2 to +Fs/2! The problem is that
> > real signals contain no useful information from -Fs/2 to 0, but that's
> > not the sampling process's fault! It does indeed provide you with a
> > bandwidth of Fs, and a quadrature demodulator is a device that allows
> > you to make use of this full bandwidth.
>
> Well, you can do it that way.
>
> The way I like to think of it is that, for a signal of length T, there are T
> Fs unknows, and T Fs equations are needed to solve for them.  Sampling the
> real signal at the appropriate number of points works, sampling the signal
> and its derivative at half that points works, and sampling the appropriate
> complex function at half the number of points works.  The sample data must
> be linearly independent which restricts it a little bit.
>
>  (Sampling the value, first, and second derivative at one third the number
> of points works, too, and doesn't have any connection to complex math.)

Glen,

I don't follow you. What are the T Fs equations?

I also don't see how the number of equations required to solve
for a certain number of unknowns has a relationship to the
spectrum of a signal.
```
 0
Reply yates 7/8/2003 11:50:18 AM

```Randy Yates wrote:
>
...
>
> I also don't see how the number of equations required to solve
> for a certain number of unknowns has a relationship to the
> spectrum of a signal.

That part is easy. Given one period T of a periodic function bandlimited
so that no frequency higher than f exists in it, then its spectrum can
consist of no more than f*T sines and an equal number of cosines. We
know what they are -- harmonics of sin(2*pi/T) and cos(2*pi/T).
Determining the spectrum consists of solving for 2*f*T amplitudes, so
2*f*T measurements are needed. The signal and its first (2*f*T - 1)th
derivatives at t = zero, for example, suffice. When the measurements are
spread uniformly in time and consist of the requisite number either of
the waveform alone or the waveform and its derivative, a Fourier
transform becomes a convenient way to do the computation.

Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
```
 0
Reply Jerry 7/8/2003 4:51:51 PM

```"Randy Yates" <yates@ieee.org> wrote in message
> "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
news:<T5qOa.2362\$N7.679@sccrnsc03>...
> > "Randy Yates" <yates@ieee.org> wrote in message
> > news:3F0A2847.BA306D6C@ieee.org...
> > > A couple of extra points:
> > >
> > > Randy Yates wrote:
> > > > [...]
> > > > In quadrature demodulation, you take the signal spectrum from 0 to
> > > > 2*pi*B (or, more generally, from 2*pi*(fc - B/2) to 2*pi*(fc + B/2),
> > > > i.e., some bandwidth around a carrier fc) and shift it down so that
it
> > > > is centered around 0 radians/second.
> > >
> > > I should also add that you throw away the signal spectrum from -2*pi*B
to
> >  0.
> > >
> > > I must inform you that you've been lied to all your life with this
> > > Nyquist criterion thing. It says that sampling at a rate of Fs
provides
> > > a bandwidth of Fs/2. This is a lie! Sampling at a rate of Fs provides
> > > a bandwidth of Fs, that is, from -Fs/2 to +Fs/2! The problem is that
> > > real signals contain no useful information from -Fs/2 to 0, but that's
> > > not the sampling process's fault! It does indeed provide you with a
> > > bandwidth of Fs, and a quadrature demodulator is a device that allows
> > > you to make use of this full bandwidth.
> >
> > Well, you can do it that way.
> >
> > The way I like to think of it is that, for a signal of length T, there
are T
> > Fs unknows, and T Fs equations are needed to solve for them.  Sampling
the
> > real signal at the appropriate number of points works, sampling the
signal
> > and its derivative at half that points works, and sampling the
appropriate
> > complex function at half the number of points works.  The sample data
must
> > be linearly independent which restricts it a little bit.
> >
> >  (Sampling the value, first, and second derivative at one third the
number
> > of points works, too, and doesn't have any connection to complex math.)
>
> Glen,
>
> I don't follow you. What are the T Fs equations?
>
> I also don't see how the number of equations required to solve
> for a certain number of unknowns has a relationship to the
> spectrum of a signal.

Sorry for the slow response.  I thought Jerry's was fine, but anyway.  It is
not a relationship with the spectrum, but with the number of sample values
needed to reconstruct a signal.  A complex number is twice as good as a
single real number.

Another way to consider it is to determine the amplitude and phase of a
component.  This is a single complex phasor, or the amplitude of sin() and
cos() terms.  Also, a function value and first derivative will do it.   So,
a sampling rate of Fs/2 complex values or value and derivative are
sufficient, or two real values at a rate of Fs/2.  But two real values at
Fs/2, uniformly spaced, is sampling as Fs.

-- glen

```
 0
Reply Glen 7/12/2003 5:39:12 AM

```>
> The way I like to think of it is that, for a signal of length T, there are T
> Fs unknows, and T Fs equations are needed to solve for them.  Sampling the
> real signal at the appropriate number of points works, sampling the signal
> and its derivative at half that points works, and sampling the appropriate
> complex function at half the number of points works.  The sample data must
> be linearly independent which restricts it a little bit.
>
>  (Sampling the value, first, and second derivative at one third the number
> of points works, too, and doesn't have any connection to complex math.)
>
> -- glen

Hello Glen

I have been following this thread and it wasfine up to the point when
you started talking about derivatives. I am sorry, but I haven't a
clue what you are talking about derivatives. Do you mind elaborating
it a little bit?

Do you mean to say that you can sample derivative of the signal at
Fs/2, and the signal itself and Fs/2 and then reconstruct the signal
from these two? How? I am clueless.. May be I know this already, just
need a hint.. :-)

Abhijit
```
 0
Reply abhijit_patait 7/14/2003 5:05:55 AM

```"Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message news:<4aNPa.40408\$N7.5284@sccrnsc03>...
> "Randy Yates" <yates@ieee.org> wrote in message
> > "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
>  news:<T5qOa.2362\$N7.679@sccrnsc03>...
> > > "Randy Yates" <yates@ieee.org> wrote in message
> > > news:3F0A2847.BA306D6C@ieee.org...
> > > > A couple of extra points:
> > > >
> > > > Randy Yates wrote:
> > > > > [...]
> > > > > In quadrature demodulation, you take the signal spectrum from 0 to
> > > > > 2*pi*B (or, more generally, from 2*pi*(fc - B/2) to 2*pi*(fc + B/2),
> > > > > i.e., some bandwidth around a carrier fc) and shift it down so that
>  it
> > > > > is centered around 0 radians/second.
> > > >
> > > > I should also add that you throw away the signal spectrum from -2*pi*B
>  to
>  0.
> > > >
> > > > I must inform you that you've been lied to all your life with this
> > > > Nyquist criterion thing. It says that sampling at a rate of Fs
>  provides
> > > > a bandwidth of Fs/2. This is a lie! Sampling at a rate of Fs provides
> > > > a bandwidth of Fs, that is, from -Fs/2 to +Fs/2! The problem is that
> > > > real signals contain no useful information from -Fs/2 to 0, but that's
> > > > not the sampling process's fault! It does indeed provide you with a
> > > > bandwidth of Fs, and a quadrature demodulator is a device that allows
> > > > you to make use of this full bandwidth.
> > >
> > > Well, you can do it that way.
> > >
> > > The way I like to think of it is that, for a signal of length T, there are T
> > > Fs unknows, and T Fs equations are needed to solve for them.  Sampling the
> > > real signal at the appropriate number of points works, sampling the signal
> > > and its derivative at half that points works, and sampling the appropriate
> > > complex function at half the number of points works.  The sample data must
> > > be linearly independent which restricts it a little bit.
> > >
> > >  (Sampling the value, first, and second derivative at one third the number
> > > of points works, too, and doesn't have any connection to complex math.)
> >
> > Glen,
> >
> > I don't follow you. What are the T Fs equations?
> >
> > I also don't see how the number of equations required to solve
> > for a certain number of unknowns has a relationship to the
> > spectrum of a signal.
>
> Sorry for the slow response.  I thought Jerry's was fine, but anyway.

I'm sure you are fine with it. The question I raise is my own and not
yours.

> It is
> not a relationship with the spectrum, but with the number of sample values
> needed to reconstruct a signal.

"The number of sample values"? Do you mean time-domain sample values? If so,
this seems completely redundant to me, Glen. If you have N samples, you can
construct N*Ts second's worth of the signal. Matters not a whit whether those
samples are real or complex - in both cases you can construct N*Ts and only
N*Ts seconds of the original signal.

> A complex number is twice as good as a
> single real number.

Proof by assertion?

> Another way to consider it is to determine the amplitude and phase of a
> component.

Do you mean a component in the frequency spectrum? OK, let's go with that.

> This is a single complex phasor, or the amplitude of sin() and
> cos() terms.

Yes, r(w)*e~{j*theta(w)} = r(w)*cos(theta(w)) + j*r(w)*sin(theta(w)). Euler
told us that a long time ago.

> Also, a function value and first derivative will do it.

Now you've lost me.

> So,
> a sampling rate of Fs/2 complex values or value and derivative are
> sufficient, or two real values at a rate of Fs/2.  But two real values at
> Fs/2, uniformly spaced, is sampling as Fs.

I'm sorry Glen - I'm still lost. It seems to me the sentences above don't
even make much grammatical sense. "a sampling rate of Fs/2 complex values"? A
sample rate is a rate, not a set of complex values. Perhaps it would be better
to use equations?

Glen, it may seem like I'm being overly pedantic, but at this stage it seems
that it's important for me not to let things slide by if understanding is
to take place.

--Randy
```
 0
Reply yates 7/14/2003 6:25:16 PM

```"Randy Yates" <yates@ieee.org> wrote in message
> "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
news:<4aNPa.40408\$N7.5284@sccrnsc03>...

(snip)

> > > > The way I like to think of it is that, for a signal of length T,
there are T
> > > > Fs unknows, and T Fs equations are needed to solve for them.
Sampling the
> > > > real signal at the appropriate number of points works, sampling the
signal
> > > > and its derivative at half that points works, and sampling the
appropriate
> > > > complex function at half the number of points works.  The sample
data must
> > > > be linearly independent which restricts it a little bit.
> > > >
> > > >  (Sampling the value, first, and second derivative at one third the
number
> > > > of points works, too, and doesn't have any connection to complex
math.)

> > > I don't follow you. What are the T Fs equations?
> > >
> > > I also don't see how the number of equations required to solve
> > > for a certain number of unknowns has a relationship to the
> > > spectrum of a signal.

(snip)

> > It is not a relationship with the spectrum, but with the number
> > of sample values needed to reconstruct a signal.
>
> "The number of sample values"? Do you mean time-domain sample values? If
so,
> this seems completely redundant to me, Glen. If you have N samples, you
can
> construct N*Ts second's worth of the signal. Matters not a whit whether
those
> samples are real or complex - in both cases you can construct N*Ts and
only
> N*Ts seconds of the original signal.
>
> > A complex number is twice as good as a single real number.
>
> Proof by assertion?

Maybe.  I complex number is an ordered pair of two real numbers.   How much
proof do you need?

> > Another way to consider it is to determine the amplitude and phase of a
component.
>
> Do you mean a component in the frequency spectrum? OK, let's go with that.
>
> > This is a single complex phasor, or the amplitude of sin() and
> > cos() terms.
>
> Yes, r(w)*e~{j*theta(w)} = r(w)*cos(theta(w)) + j*r(w)*sin(theta(w)).
> Euler told us that a long time ago.
>
> > Also, a function value and first derivative will do it.
>
> Now you've lost me.
>
> > So,
> > a sampling rate of Fs/2 complex values or value and derivative are
> > sufficient, or two real values at a rate of Fs/2.  But two real values
at
> > Fs/2, uniformly spaced, is sampling as Fs.
>
> I'm sorry Glen - I'm still lost. It seems to me the sentences above don't
> even make much grammatical sense. "a sampling rate of Fs/2 complex
values"? A
> sample rate is a rate, not a set of complex values. Perhaps it would be
better
> to use equations?
>
> Glen, it may seem like I'm being overly pedantic, but at this stage it
seems
> that it's important for me not to let things slide by if understanding is
> to take place.

OK, again without being overly pedantic what do you believe about sampling
rate?  Nyquist's paper didn't have anything to do with sampling of analog
signals.  The paper was on putting telegraph pulses through a band limited
analog channel.  He wanted to know how close he could put the pulses
together.

So, do you believe that a signal band limited at 2 Fn can be sampled, and
the samples be used to reconstruct the original signal?  Why do you believe
that?

Some years ago I was trying to figure out how to explain that to some
people.  The arguments I was coming up with weren't very convincing.  Then I
remembered that they had done vibrational mode problems, such as modes of a
string or air column.  When I considered the number of modes for a string of
a given length that were below a given frequency I wasn't surprised to find
that it was T Fn.     It helped me understand why sampled signals can be
reconstructed.

-- glen

```
 0
Reply Glen 7/15/2003 3:00:36 AM

```"Randy Yates" <yates@ieee.org> wrote in message
> "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
news:<o7KQa.63319\$Ph3.6529@sccrnsc04>...
> > > Proof by assertion?
> >
> > Maybe.  I complex number is an ordered pair of two real numbers.   How
much
> > proof do you need?
>
> I buy that. I do not buy that this is the reason that a sample rate of
> Fs provides a bandwidth of Fs for complex sampling.

Yes, you have to be careful how you define the complex numbers.

> > OK, again without being overly pedantic what do you believe
> > about sampling rate?
>
> I believe that the model which describes sampling as multiplication in
> time by an infinite impulse train perfectly describes everything we need
> to know about the subject. From this it is very easy to see using the
> convolution property of the Fourier transform why both a real signal
> and a complex signal both have a bandwidth of Fs (from, e.g., - Fs/2 to
> +Fs/2) and how a real signal simply doesn't utilize that full bandwidth
> because of its symmetry in frequency.

I will think about that one then.  That was the one I was being less
successful at explaining, but that I didn't remember yesterday.

-- glen

```
 0
Reply Glen 7/15/2003 11:15:50 PM

12 Replies
355 Views

Similar Articles

12/9/2013 10:11:31 PM
page loaded in 5500 ms. (0)