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On Dec 23, 10:12*am, "Dave" wrote:
"Keith Dysart" wrote in message ... On Dec 22, 8:43 am, Denny wrote: Nice graphic, Cecil.. But the thread has drifted beyond recognition.. Part of the original dispute across a couple of threads as I remember it, was the contention that there is no energy contained within the reflected wave and therefore no energy contained within the standing wave, i.e. a mere artifact... I'd suggest this is a mischaracterization of the contention. I have seen no disagreement with the notion that the line contains energy. Assertions about a lack of energy in reflected waves is not inconsistent with the line containing energy. I simply wanted to point out that the standing wave on a line does contain energy and it is a childishly simple exercise to prove it, therefore the reflected wave must contain energy... Prepare yourself to rethink this connection. As far as the questioner, where does the energy go between the standing wave peaks - oy vey.... If it is a real question - as opposed to a rhetorical device which I hope was the intent - It was not rhetorical, but an educational question that followed from the claim. With the claim that a lit flourescent bulb demonstrates the presence of energy, it is entirely reasonable to question what a dark lamp means and the original post did not suggest this understanding. then the profound ignorance There is no need to descend to the level of insult commonly used by some of the more prolific posters. of basic physics is vastly beyond the limited space I have to go over it... See ANY introductory level, physics textbook for details... -------------------------------------------------------------- Let us consider a transmission line.... There IS a voltage and current distribution on this line. For the moment attempt to forget standing waves, travelling waves, forward waves, reflected waves, .... *Just that: There IS a voltage and current distribution on this line. These distributions can be expressed as functions of distance along the line and time: *V(x,t) *I(x,t) These are the instantaneous real voltage and current at a particular location (x) and time (t). They can be measured with a voltmeter and ammeter, though this gets more challenging at higher frequencies. Now we know from basic electricity that Power is Volts times Amps, so we have: *P(x,t) = V(x,t) * I(x,t) P(x,t) is the instantaneous power at any point and time on the line. Power being the rate of energy flow, P(x,t) is the instantaneous energy flow at that point and time on the line. If you disagree with any of the above please read no further and post any objections now. Good! Agreement. So let's consider the specific example of sinusoidal signal applied to a transmission line that is open at the end. After settling, there is a voltage and current distribution on this line, but how can we describe it? Now some of you are immediately thinking "standing wave", and you'd be right. Its an excellent description, but we need to look at the details. So V(x,t) = A cos(x) cos(wt) where w is radians/second and x is measured in degrees back from the open end. Consider t=0. The spatial voltage distribution is a sinusoid with a maximum at the open end. As time advances, this spatial sinusoid drops in amplitude until the voltage everywhere on the line is 0, then the amplitude heads towards minus max. Noting that the zero crossings are always in the same place and the shape is sinusoidal leading to the name "standing wave". From a time perspective, every point on the line has a sinusoidal voltage, but the amplitude changes with position. The peaks and zero crossings occur at the same time everywhere, thus the claim that there is no phase shift as one moves down the line. The current is also a sinusoid, but shifted 90 degrees from the voltage sinusoid, thus there is a current zero where-ever there is a voltage maximum. Now power is really interesting. Recall that P(x,t) = V(x,t) * I(x,t) At certain values of t, the voltage everywhere on the line is 0, so at these times, no energy is flowing anywhere on the line. Similarly for current. And at certain positions on the line (n*180+90) the voltage is always 0, so the power is always 0 at these points. No energy is ever flowing at these points. Similarly for current at points (n*180). This is where your argument falls apart. *you can not apply superposition to power as it is a non-linear relationship with the voltage and current. *this is where lots of the arguments on this group begin and get stuck forever. the argument you state in the above paragraph is a contradiction on the most basic level... I am not sure what lead you to think I am superposing power. The only powers I compute are from the actual (in other words, total) voltage and current present at one point on the line. take a step back and consider this: V(x,t)=Z0*I(x,t) In my use, V(x,t) and I(x,t) are the actual voltage and current and current at a point on the line and are only related by Z0 in very exceptional circumstances, an example being when the line is terminated in Z0. which also must be true at every point on the line for the forward and reflected waves, each taken separately. *so you can write equations like: Vf=Z0*If and Vr=Z0*Ir (i'll leave off the (x,t) for now) I would suggest not leaving off (x,t). I understand the short-hand but it leads many to start thinking in terms of average or RMS. Keeping (x,t) re-inforces the idea that the measurement is at one point and one time. and by superposition you can also do: V=Vf+Vr I=If+Ir Which work just fine if you do fancy graphs and animations to create the illusion of 'standing' waves along the line. *And as long as you keep the two equations separate everything is simple. BUT, lets look at the a place where the standing current wave is always zero and the voltage standing wave is of course a maximum. *if you plug those into ohm's law to find the impedance at that point you get: Z=V/I *(where V is large, and I is zero) *so you get an infinite impedance. does this surprise anyone? *it shouldn't, this is what the smith chart shows you should happen every half wave along the line. *the result of superimposing the waves results in changes in the measured impedance as you move along the line. *note this is NOT a change in Z0, that is forever a constant and property of the physical line independent of the waves imposed on it. So what does this really mean? *on the surface it would tend to support the assertion that there is no power flow at the points where the current is zero, and the impedance measures is infinite. *BUT there is a catch. P=VI * *is a basic representation for instantaneous power at a point given voltage and current, again (x,t) left off but that doesn't matter. but also you have to keep the relationship: V=Z0*I Except that V(x,t) and I(x,t) are not, in general, related by Z0. so you can rewrite the P equation 2 different ways: P=V^2/Z0 = I^2*Z0 now, obviously these have to hold for any wave that exists in the line. *so for the forward wave they hold up just fine, and for the reflected wave they hold up just fine.. BUT if you look at the 'standing' wave they fall apart.. i.e. for the spot where the current is zero and the voltage is a maximum you get: P=V^2/Z0 which is a large number AND P=I^2*Z0 which is a small number you can't have it both ways at one point at the same time! now, what is really happening? Given: 1. you have 2 waves. No. The two wave view is merely an alternate set of expressions which, when summed (i.e. using superposition), provide the actual voltage and current on the line. These alternate expressions are obtained by algebraic maniupulation of the more fundamental descriptive equations. Just as in basic ciruit theory, the partial results, which are eventually superposed to arrive at the final results, have no particular meaning. For sure, they are not what is "really happening". None-the-less, this "two wave" model is extremely useful. 2. these 2 waves are traveling in opposite directions. 3. each of these waves obeys ohms law. 4. the voltage and/or current of these waves obeys the superposition principle. 5. you only need to look at voltage OR current since they are linearly related to each other at every point in each wave. If you dispute any of the above, do not pass go, do not collect 200$, go back to school and take fields and waves 101 all over again. Tsk. Tsk. Insults. Unbecoming. Lets think voltage waves for now, this is an arbitrary decision as noted above. *and lets consider a lossless line with a short or open end so there is 100% reflection. So, as these two waves travel along the line they periodically go in and out of phase with each other, there are animations that show this very nicely. lets think about the time where the two waves completely cancel each other so the voltage along the line is zero... did the 2 waves dissappear? *no, because obviously an instant later they are back out of phase and the voltage doesn't cancel on the line. *is the power zero?? no, since energy can neither be created nor destroyed, it didn't just dissappear, so neither can the flow of energy that is called power. *and since we know that when this voltage minimum occurs there is a current maximum in the superimposed waves the power represented by that superposition would contradict the power represented by the voltage minimum... they can't both be correct at the same time! So what do you take away from this? 1. Standing waves have no physical significance, they do not represent power or energy, they do not obey ohms law, they are ONLY a result of superposition of the voltage and/or current waves in the line. 2. can you measure standing waves? *Yes, of course. *that is how they got their name, you could measure them and they didn't seem to move on the line. but this is only because simple measurement tools can't distinguish the forward and reflected components that make them up. 3. if you want to talk about power and energy you MUST use the individual traveling waves. now what does that mean for this lossless line that has 100% reflection?? 1. the reflected wave magnitude is equal to the forward wave just traveling in the opposite direction. (voltage and/or current) 2. the power in the forward wave is equal to the power in the reflected wave but traveling in the opposite direction. 3. there is energy stored in the line equal to the sum of the integrated power in the two waves each taken separately. *DO NOT sum them first then integrate, that is NOT a legal operation since as has been shown the superposition principle does not apply to power as it is a non-linear relationship! 4. in steady state the net power flowing past any point in the line is zero... that is, there is as much power flowing one direction as the other.. ok, i'm done with my lecture... you guys can now ignore me if you want and go back to your regularly scheduled misconceptions and circular arguments. 4. Are you really prepared to throw away P = VI? In my analysis, P(x,t) = V(x,t) * I(x,t) is the equation that means the power at any point and time can by obtained by measuring the actual voltage and current on the line at the point and time of interest. Are you sure you want to throw away this ability? Are you sure you want to claim that instantaneous power can NOT be obtained by multiplying the instaneous measured voltage by the instanteous measured current? Throwing this away will invalidate much. ...Keith |
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