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Low freq "analog" of Nyquist? ( possibly naive question )

Started by Richard Owlett July 2, 2003
"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
news:by1Pa.2506$Jk5.1594965@feed2.centurytel.net...
>
(snip)
> Maybe I've come around a little: What got me going on this was when Glen > said: > > "Consider a signal, f, sampled over time T, from t=0 to t=T. Assume that > f(0)=f(T)=0 for now. All the components must be sine with periods that
are
> multiples of 2T. If it is known to have a maximum frequency component <
Fn
> then the number of possible frequency components is 2 T Fn. A system with
2
> T Fn unknowns needs 2 T Fn equations, so 2 T Fn sampling points. 2 T Fn > sampling points uniformly distributed over time T are 2 Fn apart." > > This seems to have a couple of errors - which may have contributed to my > misunderstanding: > > It should say: frequencies that are integer multiples of (1/T) instead of > periods that are multiples of 2T. The frequencies need to get larger in > integer multiples, not the periods.
I corrected that somewhere along the way. Though it is 1/(2T), as sin(pi)=0. Also, in terms of waves on a cable, the wave travels one length, is reflected and inverted, travels to the other end, and is again reflected and inverted.
> It should say sampling points (in time) that are 1/(2Fn) apart. Time is > divided into seconds not Hz. > > Then, I missed that Glen was talking about sampling in time - which is
only
> introduced at the end. I was focused on "must be sine with periods". In > the mean time, I was talking about "sampled over time T" as in "taking a > temporal epoch of length T as a sample".
(snip)
> Now, since Glen introduced it, assume that f(0)=f(T)=0. > What does this do? > I can't tell that it does anything in particular.
It forces only integer multiples of the fundamental. Well, first it is the form of the solutions to a large number of physics problems, such as modes of a violin string, air column inside a tube, or voltage on a coaxial cable. I was once trying to explain Nyquist sampling to some students. After realizing that the more traditional explanations weren't so obvious (I even had a copy of Nyquists paper), I realized that they all had done such problems in physics, and that the solutions had the required restrictions. Consider what happens without such restriction. Say I have a signal defined for t=0 to t=pi. sin(t) is one obvious basis function, but why can't sin(1.2 t) also be a basis function? sin(t) and sin(1.2 t) are linearly independent, though not orthogonal (over 0 to pi). Violin strings make musical notes because the allowed vibrational modes are restricted by tying the ends of the strings. It doesn't unduly restrict the arbitrary function over the desired range, yet it does restrict the allowed basis functions to describe that solution. Once done, only T Fs basis functions are needed to describe a band limited signal.
> I can see if we say something different: > for *all* f(t) such that f(0)=f(T)=0 then: > f(t)=sum over n [bn*sin(n*pi*t/T) + an*cos(n*pi*t/T)] > an=0 for n even > and > (sum over n of an for n odd)=0 > which may be interesting but I'm not tuned in.... > At least this isn't what I'd call "all components must be sine" because > there are functions with nonzero elements that are cosine. Perhaps that > wasn't what was meant. Still doesn't get my attention. > > Either of these temporal functions are known to have infinite spectral > extent, so they *can't* have a maximum frequency component < Fn and we
can't
> say "the number of possible frequency components is 2 T Fn". That would
be
> the case only if there were temporal sampling and a periodic spectrum - > which isn't a point we've reached and can't reach without changing the > definition of the original function. That doesn't mean that such
functions
> don't exist, it only means that we can't get there starting from an > *arbitrary* time-limited function.
I learned Fourier series a long time before Fourier transforms, and was very confused about Fourier transforms for a while. Finally someone explained that Fourier transform was the limit of a periodic function as the period goes to infinity. Is it still periodic when the period is infinite? You can take lim T --> infinity for the time limited function, or limit Fn --> infinity for the band limited function. Both require an infinite number of samples, so aren't very practical, but you can see the connection between continuous and discrete functions.
> It bothered me to think that a continous spectrum could be expressed as a > discrete spectrum. But, with time limiting, and after considering that
the
> temporal record could be made periodic with no loss of information, I can > see how this is OK in an analytical context. > > Solutions to differential equations don't have to be introduced to take
this
> analytical trip do they? That continues to elude me. Saying that sin(t)
is
> the solution to a particular differential equation is interesting but not > compelling - why should it be when discussing an arbitrary f(t)?
Consider analog filter design using complex impedance. You can use Z=iwL, Z=-i/(wC) and design all kinds of filters without knowing that they are solutions to differential equations. Mortgage lenders can use a table of payments for different interest rates to make loans, without understanding exponential functions. Both come from differential equations, though. Consider that one could expand an arbitrary function in terms of triangle waves of appropriate frequency and phase. Triangle waves make a fine set of basis functions. Not being solutions to a simple differential equation they don't have the useful properties of sin() and cos(). The property of band limited functions doesn't make sense with triangle basis functions. Resonant systems, which follow the solution to differential equations, have modes where the band limiting make sense. With all the emphasis on sin() and cos() solutions, it may seem that they are the only useful functions. Consider the rotationally symmetric modes of a circular drum head. They will be related to solutions of J0(w T)=0 where T is a constant (depending on the drum diameter, mass per unit area, and tension in the drum head) and J0 is a Bessel function, a solution to a different differential equation. The modes will not be frequencies that are integer multiples of a fundamental, but they will be solutions to differential equations. -- glen
"Clay S. Turner" <physicsNOOOOSPPPPAMMMM@bellsouth.net> wrote in message
news:Vc7Pa.2844$zJ6.2343@fe02.atl2.webusenet.com...

(snip)

> Actually reflections off of multiple surfaces are well described by FIR > methods. The simple case is Bragg diffraction with crystals. But wait > there's more, volume holograms are 3-dimensional FIR filters. Not only are > they frequency selective (that's why you can see the image in color), they > also are direction selective. A simple holographic mirror if viewed in
cross
> section is a series of Bragg planes all equally spaced and their extent > covers the thickness of the film's emulsion. Of course they offer some > tricks not easily had in DSP. A standard hologram works by modulating the > amplitude just like the taps all have scale factors. If you bleach the > hologram, it then will modulate the phase of the signal. Imagine a signal > that speeds up and slows down over and over as it traverses the filter.
The
> advantage is conservation of energy. The disadvantage, as with angle > modulation, is harmonic generation which results in veiling glare.
Maybe my favorite lab class experiment was in an optics lab, which goes something like this: Take a two dimensional transparent object, send a plane wave from a laser through it (appropriate lenses are needed) then send it through a lens with a focal length of f, and put the image plane of a television camera at the focal point of the lens. The 'picture' you get is the two dimensional Fourier transform (well, the magnitude, anyway) of the object. With additional lenses you can generate the transform, filter it (by blocking parts of the transform plane) then put it through another lens and invert the transform. For example, you can put an opaque sheet with a hole in it in the tranform plane, which is a low pass filter in real space. The high pass filter is a little harder, as blocking the DC component removes much of the optical power. It can be done, though. But I do like the analogy between holograms and FIR filters. -- glen
Glen Herrmannsfeldt wrote:
>
...
> > Maybe my favorite lab class experiment was in an optics lab, which goes > something like this: Take a two dimensional transparent object, send a > plane wave from a laser through it (appropriate lenses are needed) then send > it through a lens with a focal length of f, and put the image plane of a > television camera at the focal point of the lens. The 'picture' you get is > the two dimensional Fourier transform (well, the magnitude, anyway) of the > object. With additional lenses you can generate the transform, filter it > (by blocking parts of the transform plane) then put it through another lens > and invert the transform. > > For example, you can put an opaque sheet with a hole in it in the tranform > plane, which is a low pass filter in real space. The high pass filter is a > little harder, as blocking the DC component removes much of the optical > power. It can be done, though. > > But I do like the analogy between holograms and FIR filters. > > -- glen
Glen, There's a simpler experiment that doesn't need a laser, and that some of us can do at home. It relies on the observation that the image of a distant light source formed by a typical 10x microscope objective is on or very near the back surface of the rear doublet. Put a small obstruction -- a tiny stick-on or a touch of india ink -- in the center of the rear element, and adjust the substage mirror so that the image of a distant -- a meter is usually plenty -- bare filament is obstructed by it. Effectively, this blocks the "carrier" in the same way that your high-pass filter blocks "DC". The sidebands that get past the obstruction (they really are off to the side!) carry all the information about the image, but it comes out darkfield without the carrier. Jerry P.S. If you don't mind permanently altering the objective, you can dissolve away the quarter-wave coating where the obstruction had been. Then with the same illumination scheme as above, the quadrature shift of the carrier converts PM to AM, yielding phase contrast. That's good for examining your bleached holograms. -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Thanks to Glen and Rune and Jerry and others for sticking to the discussion!
I'm going to ponder one of the implications in a new thread.

Fred