"Randy Yates" <yates@ieee.org> wrote in message news:567ce618.0307080350.43c072e6@posting.google.com...> "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in messagenews:<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 thatit> > > > is centered around 0 radians/second. > > > > > > I should also add that you throw away the signal spectrum from -2*pi*Bto> > 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 Fsprovides> > > 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, thereare T> > Fs unknows, and T Fs equations are needed to solve for them. Samplingthe> > real signal at the appropriate number of points works, sampling thesignal> > and its derivative at half that points works, and sampling theappropriate> > complex function at half the number of points works. The sample datamust> > be linearly independent which restricts it a little bit. > > > > (Sampling the value, first, and second derivative at one third thenumber> > 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

# What is quadrature sampling

Started by ●July 7, 2003

Reply by ●July 12, 20032003-07-12

Reply by ●July 14, 20032003-07-14

> > 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.) > > -- glenHello 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

Reply by ●July 14, 20032003-07-14

"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, thereare T> > Fs unknows, and T Fs equations are needed to solve for them. Samplingthe> > real signal at the appropriate number of points works, sampling thesignal> > and its derivative at half that points works, and sampling theappropriate> > complex function at half the number of points works. The sample datamust> > be linearly independent which restricts it a little bit. > > > > (Sampling the value, first, and second derivative at one third thenumber> > 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

Reply by ●July 14, 20032003-07-14

"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

Reply by ●July 15, 20032003-07-15

"Randy Yates" <yates@ieee.org> wrote in message news:567ce618.0307140625.13982357@posting.google.com...> "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in messagenews:<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 thesignal> > > > and its derivative at half that points works, and sampling theappropriate> > > > complex function at half the number of points works. The sampledata must> > > > be linearly independent which restricts it a little bit. > > > > > > > > (Sampling the value, first, and second derivative at one third thenumber> > > > of points works, too, and doesn't have any connection to complexmath.)> > > 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? Ifso,> this seems completely redundant to me, Glen. If you have N samples, youcan> construct N*Ts second's worth of the signal. Matters not a whit whetherthose> samples are real or complex - in both cases you can construct N*Ts andonly> 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 acomponent.> > 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 valuesat> > 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 complexvalues"? A> sample rate is a rate, not a set of complex values. Perhaps it would bebetter> to use equations? > > Glen, it may seem like I'm being overly pedantic, but at this stage itseems> 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

Reply by ●July 15, 20032003-07-15

"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.

Reply by ●July 15, 20032003-07-15

"Randy Yates" <yates@ieee.org> wrote in message news:567ce618.0307151020.19b41f0b@posting.google.com...> "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in messagenews:<o7KQa.63319$Ph3.6529@sccrnsc04>...> > > Proof by assertion? > > > > Maybe. I complex number is an ordered pair of two real numbers. Howmuch> > 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