"Randy Yates" <yates@ieee.org> wrote in message
news:567ce618.0307151020.19b41f0b@posting.google.com...
> "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

"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.

> 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.

"Randy Yates" <yates@ieee.org> wrote in message
news:567ce618.0307140625.13982357@posting.google.com...
> "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

"Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message news:<4aNPa.40408$N7.5284@sccrnsc03>...
> "Randy Yates" <yates@ieee.org> wrote in message
> news:567ce618.0307080350.43c072e6@posting.google.com...
> > "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
>  news:<T5qOa.2362$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

"Abhijit Patait" <abhijit_patait@yahoo.com> wrote in message
news:d16f58e8.0307132105.5d85fae1@posting.google.com...
> >
> > 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 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.. :-)

If you are considering the signal as a sum of sin() and cos(), then for a
sample at t=0, f(0) comes from the cos() and f'(0) comes from the sin()
term.

Otherwise, f(t) and f'(t) are popular boundary conditions for many
differential equations.

Mathematically, you need T Fs linearly independent constraints on the
solution.  There are many different ways to get those constraints, uniform
sampling of f(t) at Fs being a popular one.

-- glen

> 
> 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

"Randy Yates" <yates@ieee.org> wrote in message
news:567ce618.0307080350.43c072e6@posting.google.com...
> "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

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.
�����������������������������������������������������������������������

"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.

"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