How to get envelope from AM signal without phase shift
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From: robert bristow-johnson on 31 Mar 2010 16:17 On Mar 31, 2:52 pm, glen herrmannsfeldt <g...(a)ugcs.caltech.edu> wrote: > > Consider this case: A charge is moving along the Z-axis with > position (0,0,vt). That is, velocity v going through the origin at t=0. > For an observer along the X-axis, at what time is the potential > (or field) maximum observed? Two choices: t=0, or t=x/c > (x being the position of the observer on the X axis.) > > As a hint, note that the Lorentz transformation was not derived > to fit special relativity, but to fit Maxwell's equations. > (Maybe that is why it is named after Lorentz and not Einstein.) to simplify things, the thought experiment i prefer is instead of a single moving charge moving on the Z-axis, what one should consider is an infinite line of uniform charge (also having a non-zero lineal mass density) moving along the Z-axis. actually, it should be two parallel infinite lines of uniform charge (parallel to the Z-axis) of known spacing moving together in the Z direction. for the observer moving along with the two parallel lines of charge, there is no motion relative to that observer and the problem is simple electrostatics and, knowing the distance between the two lines, the repulsive acceleration (sideways) of the two lines can be determined purely from electrostatics. now there's a "stationary" observer that watches the two lines of charge (and the "moving" observer) whiz by him and also notices that, due to time dilation, the moving observer's clock is ticking more slowly, so the outward acceleration of the moving lines of charge appears to be slower than what is observed if they are not moving (as the first observer sees). the rate of outward acceleration of the moving lines of charge, when considering *only* electrostatics together with special relativity is exactly the same outward acceleration of the same two moving lines of charge when considering both static and magnetic forces in a classical context (no relativistic effect). that thought experiment, first introduced to me by a physics prof (who now is at Analog Devices) in the 70s, was sufficient to convince me that the electromagnetic interaction (in the classical context) is none other than just the sole electrostatic interaction with relativistic effects applied. i s'pose the same can be done with lines of mass and the static Newtonian gravitational interaction, also applying special relativity and what you'll get out would be consistent with GEM (gravito-electro- magnetism) where Maxwell's and Lorentz equations have mass (or mass density) replacing charge (or charge density) and the Coulomb constant, 1/(4*pi*epsilon_0), is replaced with -G (the minus sign is because like-signed charges repel while like-signed masses attract). for some reason (that i don't really get), the gravito-magnetic force has an extra factor of 2 tossed into it (at least that's what the lit seems to say). but, i think either classical EM or GEM can both be sorta understood by considering the simple Coulomb or Newtonian static model with relativistic effects. that's why we know that the magnetic interaction is really just a consequence of the static interaction and not a separate fundamental interaction. r b-j
From: Clay on 31 Mar 2010 18:03 On Mar 31, 4:17 pm, robert bristow-johnson <r...(a)audioimagination.com> wrote: > On Mar 31, 2:52 pm, glen herrmannsfeldt <g...(a)ugcs.caltech.edu> wrote: > > > > > Consider this case: A charge is moving along the Z-axis with > > position (0,0,vt). That is, velocity v going through the origin at t=0. > > For an observer along the X-axis, at what time is the potential > > (or field) maximum observed? Two choices: t=0, or t=x/c > > (x being the position of the observer on the X axis.) > > > As a hint, note that the Lorentz transformation was not derived > > to fit special relativity, but to fit Maxwell's equations. > > (Maybe that is why it is named after Lorentz and not Einstein.) > > to simplify things, the thought experiment i prefer is instead of a > single moving charge moving on the Z-axis, what one should consider is > an infinite line of uniform charge (also having a non-zero lineal mass > density) moving along the Z-axis. actually, it should be two parallel > infinite lines of uniform charge (parallel to the Z-axis) of known > spacing moving together in the Z direction. > > for the observer moving along with the two parallel lines of charge, > there is no motion relative to that observer and the problem is simple > electrostatics and, knowing the distance between the two lines, the > repulsive acceleration (sideways) of the two lines can be determined > purely from electrostatics. > > now there's a "stationary" observer that watches the two lines of > charge (and the "moving" observer) whiz by him and also notices that, > due to time dilation, the moving observer's clock is ticking more > slowly, so the outward acceleration of the moving lines of charge > appears to be slower than what is observed if they are not moving (as > the first observer sees). > > the rate of outward acceleration of the moving lines of charge, when > considering *only* electrostatics together with special relativity is > exactly the same outward acceleration of the same two moving lines of > charge when considering both static and magnetic forces in a classical > context (no relativistic effect). > > that thought experiment, first introduced to me by a physics prof (who > now is at Analog Devices) in the 70s, was sufficient to convince me > that the electromagnetic interaction (in the classical context) is > none other than just the sole electrostatic interaction with > relativistic effects applied. > > i s'pose the same can be done with lines of mass and the static > Newtonian gravitational interaction, also applying special relativity > and what you'll get out would be consistent with GEM (gravito-electro- > magnetism) where Maxwell's and Lorentz equations have mass (or mass > density) replacing charge (or charge density) and the Coulomb > constant, 1/(4*pi*epsilon_0), is replaced with -G (the minus sign is > because like-signed charges repel while like-signed masses attract). > for some reason (that i don't really get), the gravito-magnetic force > has an extra factor of 2 tossed into it (at least that's what the lit > seems to say). > > but, i think either classical EM or GEM can both be sorta understood > by considering the simple Coulomb or Newtonian static model with > relativistic effects. that's why we know that the magnetic > interaction is really just a consequence of the static interaction and > not a separate fundamental interaction. > > r b-j Hello Robert, et al, The quick and easy way is via 4-vectors. Here the scalar electric potential and the magnetic vector potential form a 4-vector. It is interesting to note that the fields themselves do not transform nicely. In fact it is the potentials that affect things and not the fields. This is well demonstrated by the Aharonov-Boehm effect. Yes there are cases where you have non zero potentials with zero fields and can observe the quantum interference being affected by varying the potential! But start with a single stationary charge in one frame of reference (in this frame the potentials are trivial) and then view the charge from a moving frame and using 4-vector calculus you get the new potentials. The curl of the magnetic vector potential will give you the B fields resulting from a single moving charge. A lot of Physics books will start with the Biot-Savart law and work from there avoiding the relativity approach. But it makes it much easier to calculate. Clay
From: Jerry Avins on 31 Mar 2010 18:15 On 3/31/2010 5:46 PM, WWalker wrote: > Jerry, > > I am sorry I do not understand your comment here... > >> Whatever power is >> developed in your resistor comes from the field being warped by the >> receiver, so the equations you use don't apply to the simulation. The probe alters the field that it's probing. > The Poynting vector just addresses the radiative power, but there is also > nonradiative power in the nearfield which also propagates. This is what I > am discussing in my simulation. The power doesn't radiate, but it nevertheless propagates through space. Hmmm. I'd like to understand how. I imagine it's akin to the evanescent wave just outside a surface of total internal reflection. When another interface is put sufficiently close to detect that wave, it is altered. Jerry -- "It does me no injury for my neighbor to say there are 20 gods, or no God. It neither picks my pocket nor breaks my leg." Thomas Jefferson to the Virginia House of Delegates in 1776. ���������������������������������������������������������������������
From: WWalker on 31 Mar 2010 18:42 Hi Clay, The propagation distance in my 500MHz carrier simulation is 10cm. But the distance can be a lot larger for lower carrier frequencies. For example, if the carrier frequency (fc) is 1MHz (typical AM radio) then the optimum propagation distance is 300m (1/6 carrier wavelength) and the envelope will arrive 80ns earlier than a light speed (propagating envelope (0.08/fc). For lower carrier frequencies, even larger distances and larger light speed time differences are possible. In terms of quantum mechanics I think the following might be happening in this system. If a photon is created a t=0 then as it propagates, because of the uncertainty principle, the uncertainty of the velocity of the photon is much larger than c in the nearfield and much less than c in the farfield. Which means the photon can be much faster than light in the nearfield but reduces to the speed of light as it propagates into the farfield. Below is the argument that shows this. Lets calculate the uncertainty of the velocity of a photon that propagates one wavelength after it is created: According to the Heisenberg uncertainty principle, the relation between the uncertainty in Energy (dE) and the uncertainty in time (dt) is: dE*dt >= h. The time for a photon to cross one wavelength distance is: dt = lambda/c. Since dE = h*df and df=dv/lambda then dE*dt=h*dv/c, but dE*dt <= h therefore: dv >= c For smaller distances the uncertianty will be greater and for larger distances the uncertainty will be much smaller. William >On Mar 29, 12:18=A0pm, "WWalker" <william.walker(a)n_o_s_p_a_m.imtek.de> >wrote: >> Jerry, >> >> The speed of light is a corner stone in physics and if it is not a consta= >nt >> then many of our theories in physics will be affected. There may be direc= >t >> practical uses as well, but I just guessing: improving accuracy of high >> speed doppler radar, speeding up communication to spacecraft where time >> delays are problematic, increasing speed of computers when they are >> eventually limited by light speed delays etc. As I said, these are only >> guesses, the main effect would be a change in many of our theories in >> physics, which would eventually lead to new practical uses and >> technologies. >> >> William >> >> >> >> >Eric Jacobsen wrote: >> >> > =A0 ... >> >> >> I think until you can demonstrate something like that the more likely >> >> explanation of bandlimited prediction would be expected to prevail. >> >> >Even allowing the unlikely possibility that the 6-degree phase advance >> >*in the near field* represents a real speed increase, and that the >> >"pulse" in the far field is expected to show no advance at all, What >> >practical use can this have? >> >> >Jerry >> >-- >> >Discovery consists of seeing what everybody has seen, and thinking what >> >nobody has thought. =A0 =A0.. Albert Szent-Gyorgi >> >- Hide quoted text - >> >> - Show quoted text - > >Hello William, > >I suggest you 1st study the EPR paradox and then look up Bell's >theorem and see how it applies to Relativity. You are not going to get >information over any significant distance with superluminal speed. >Sure there is a probability that a particle will travel faster than >light for a short distance (say for example across the nucleus of an >atom about 10^-14 to 10^-15 meters) but when you start to add up all >of the paths in a Feynman diagram, you will see the probability of it >happening across a room is not even likely in a time period of the age >of the Universe. > >Clay > >
From: glen herrmannsfeldt on 31 Mar 2010 19:02
Jerry Avins <jya(a)ieee.org> wrote: (snip) > The power doesn't radiate, but it nevertheless propagates through space. > Hmmm. I'd like to understand how. I imagine it's akin to the evanescent > wave just outside a surface of total internal reflection. When another > interface is put sufficiently close to detect that wave, it is altered. There is an example in Feynman, not long after he introduces the Poynting vector. He has a point charge near a bar magnet, with E cross B circulating but not propagating. Later on, this turns out to be necessary to conserve angular momentum. That is the angular momentum of the electromagnetic field. And still no answer to my multiple choice, only two choice, question. -- glen |