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The Rest of the Story
On Mar 20, 1:07*pm, Roger Sparks wrote:
Keith Dysart wrote: On Mar 16, 10:21 am, Cecil Moore wrote: snip Would you please explain how energy is conserved in the following example at the zero-crossing point for Vs? * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl * * * * *+----/\/\/-----+----------------------+ * * * * *| * *50 ohm * * * * * * * * * * * * * | * * * * *| * * * * * * * * * * * * * * * * * * | * * * * Vs * * * * * * * * 45 degrees * * * * *| Shorted * * *100v RMS * * * * * * *50 ohm line * * * * | Stub * * * * *| * * * * * * * * * * * * * * * * * * | * * * * *| * * * * * * * * * * * * * * * * * * | * * * * *+--------------+----------------------+ * * * * gnd At the zero-crossing of Vs, Ps(t) = 0, i.e. the source is supplying zero watts at that time but Prs(t) = 100w. Where is the 100 watts coming from? For the first 90 degrees of time, the circuit can be represented as * * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl * * * * * *+----/\/\/-----+----------------------+ * * * * * *| * *50 ohm * * * * * * * * * * * * * | * * * * * *| * * * * * * * * * * * * * * * * * * / * * * * * Vs * * * * * * * * * * * * * * * * * * \ 50 ohm resistor * * * *100v RMS * * * * * * * * * * * * * * * * */ * * * * * *| * * * * * * * * * * * * * * * * * * \ * * * * * *| * * * * * * * * * * * * * * * * * * | * * * * * *+--------------+----------------------+ * * * * * gnd After 90 degrees of time has passed, the circuit can be represented as * * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl * * * * * *+----/\/\/-----+----------------------+ * * * * * *| * *50 ohm * * * * * * * * * * * * * | * * * * * *| * * * * * * * * * * * * * * * * * * | * * * * * Vs * * * * * * * * * * * * * * * * * *--- * * * *100v RMS * * * * * * * * * * * * * * * * --- 50 ohm inductive * * * * * *| * * * * * * * * * * * * * * * * * * | * * * * * *| * * * * * * * * * * * * * * * * * * | * * * * * *+--------------+----------------------+ * * * * * gnd The sudden switch in circuit design at time 90 degrees is not unique to start up, but is true for any adjustment made to Vs and any returning wave from the shorted stub,. *As a result, a true stable circuit will never be found unless some voltage adjustment is allowed for the 90 timing shift caused by the shorted stub. *Keith (in his analysis of the circuit) recognizes that Vs drives into a reactive circuit. If we want to understand how constructive and destructive interference act to cause a 50 ohm resistor to evolve into a 50 ohm capacitor, then we need to examine how traveling waves might do this. *It would be nice to have a formula or wave sequence that fully addressed this evolution. *From circuit theory, we have the inductive reactance of a short-circuited line less than 1/4 wavelength long is * * *XL = Zo * tan (length degrees) * * * * = 50 * tan(45) * * * * = 50 ohms *From traveling wave theory, we would have the applied wave from the source arriving 90 degrees late to the stub side of resistor Rs. *This ignores the fact that current must already be passing through resistor Rs because voltage has been applied to Rs from the Vs side 90 degrees earlier. *Whoa! *Things are not adding up correctly this way! We need to treat the wave going down the 50 ohm line as a single wave front. *The wave reverses at the short-circuit, reversing both direction of travel and sign of voltage. *When the wave front reaches the input end, 90 degrees after entering (for this case), the voltage/current ratio is identical to the starting ratio (the line was 45 degrees long, tan(45) = 1), and the returning voltage directly adds to the voltage applied from Vs. *As a result, the current flowing through Rs will increase, and Vs will see a changed (decreased) impedance. After 90 degrees of signal application, we should be able to express the voltage across Rs as * * *Vrs(t) = Vs(t) - Vg(t) + Vref(t) Just to ensure clarity, you are using a different convention for terms and signs than I do. For me, Vg(t) is the actual voltage at point Vg. In terms of forward and reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at point g is sum of the forward and reverse voltage at point g. In my terms, this leads to Vrs(t) = Vs(t) - Vg(t) = Vs(t) - Vf.g(t) - Vr.g(t) Your Vg seems to be same as my Vf.g and you seem to be using a different sign for Vref than I do for Vr. You must be carrying these differences through the equations correctly because the final results seem to agree. Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2 Substitute so Vrs(t) = Vs(wt + 90) - Vs(wt + 90)/2 + Vs(wt)/2 * * *= Vs(wt + 90)/2 + Vs(wt)/2 Allow Vs to be represented by a sine wave, we have A small aside. It is conventional to use cos for sinusoidal waves because it maps more readily into phasor notation. * * 2Vrs(t) = Vs*sin(wt + 90) + Vs*sin(wt) * * * * * * = 2Vs*(sin(wt + 45)(cos(45)) * * Vrs(t) = Vs(sin(wt + 45)(cos(45)) Vs is defined as 100v RMS, which equals 100 * 1.414 = 141.4v Peak. *The maximum voltage would occur when the sin term was 90 degrees and equals 1, which would occur at wt = 45 degrees. * We would have * * *Vrs(45) = 141.42 * sin(90)(cos(45) * * * * * * *= 141.42 * 1 * 0.7071 * * * * * * *= 100v Now consider the current. *After the same 90 degrees of signal application, we should be able to express the current through Rs as * * *Irs(t) = Is(t) + Iref(t) Is(t) = Is(wt + 90), Iref(t) = Is(wt) The reflected current has been shifted by 90 degrees due to the reflection so we must rewrite Iref(t) to read * * *Iref(t) = Is(wt + 90) Substitute, * * * Irs(t) = Is(wt + 90) + Is(wt) Allow Is to be represented by a sine wave, we have * * * Irs(t) = Is*sin(wt + 90) + Is*sin(wt) * * * * * * *= 2*Is(sin(wt + 45)(cos(45)) How do we find Is? *Is is the initial current found by dividing the applied voltage at peak (141.42v) by the initial resistance (100 ohms). * * *Is = 141.42/100 = 1.4142a The maximum current would occur when the sin term was 90 degrees and equals 1, which would occur at wt = 45 degrees. * We would have * * * Irs(t) = 2*Is(sin(wt + 45)(cos(45)) * * * * * * *= 2 * 1.4142 * 1 * 0.7071 * * * * * * *= 2a These results agree with the results from Keith and from circuit theory. We have a theory and at least the peaks found from the theory agree with the results from others. * How about Cecil's initial question which is * At the zero-crossing of Vs, Ps(t) = 0, i.e. the source * is supplying zero watts at that time but Prs(t) = 100w. * Where is the 100 watts coming from? We will use the equation * *Vrs(t) = Vs(t) - Vg(t) + Vref(t) Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2 The challenge looks at the time when Vs(t) = Vs*sin(wt + 90) = 0, which occurs when wt = -90. *When wt = -90, all the power to the source resistor Rs is from the reflected portion described as Vref(t) = Vs(wt)/2. *The voltage across Rs would be * *Vrs(-90) = Vref(-90) * * * * * * = Vs*sin(-90)/2 * * * * * * = 141.4/2 * * * * * * = 70.7v The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming from the reflection. In summary, power to the resistor Rs comes via two paths, one longer than the other by 90 degrees (in this example). * This is not quite complete. The power to Rs does come from two places but they are the voltage source and the line. Ps(t) = Prs(t) + Pg(t) where Pg(t) is the power at point g flowing out of the generator into the line and Ps(t) is the power flowing from the source to the source resistor and the line. The power into the line can be seaparated into a forward component and a reflected component: Pg(t) = Pf(t) + Pr(t) (Sometimes this is written as P = Pf - Pr, but that is just a question of how the sign is used to represent the direction of energy flow.) Substituting we have Ps(t) = Prs(t) + Pf.g(t) + Pr.g(t) or Prs = Ps(t) - Pf.g(t) - Pr.g(t) Whenever Ps(t) equal 0, then Pf.g(t) also equals 0 so at those times, this simplifies to Prs = - Pr.g(t) which is the same conclusion you came to. But it is wise not to forget the Pf.g(t) term because at times when Ps(t) is not equal to zero, the Pf.g(t) term will be needed to balance the energy flows. The short path is the series path of two resistors composed of the source resistor Rs and the input to the 50 ohm transmission line measured by Vg. *The long path is the series path of one resistor Rs and one capacitor composed of the shorted transmission line. *Both paths are available at all times. Power flows through both paths to Rs at all times, but because of the time differential in arrival timing, at some point Rs will receive power only from Vs, and at another point, receive power only from Vref. And at still other times, the power in the source resistor will depend on all of Ps(t), Pf.g(t) and Pr.g(t), or, more simply, it will depend on Ps(t) and Pg(t). ...Keith |
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On Mar 20, 2:07*pm, Cecil Moore wrote:
Roger Sparks wrote: Vrs(t) = Vs(t) - Vg(t) + Vref(t) Good job, Roger. Let's simplify the equation through substitution. Let V1(t) = [Vs(t) - Vg(t)] Let V2(t) = Vref(t) Vrs(t) = V1(t) + V2(t) The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming from the reflection. Yes, because when Vs(t)=0, there is zero instantaneous interference. In summary, power to the resistor Rs comes via two paths, one longer than the other by 90 degrees (in this example). *The short path is the series path of two resistors composed of the source resistor Rs and the input to the 50 ohm transmission line measured by Vg. *The long path is the series path of one resistor Rs and one capacitor composed of the shorted transmission line. *Both paths are available at all times. Power flows through both paths to Rs at all times, but because of the time differential in arrival timing, at some point Rs will receive power only from Vs, and at another point, receive power only from Vref. What Keith is missing is that: Vrs(t)^2 = [V1(t) + V2(t)]^2 NOT= V1(t)^2 + V2(t)^2 In his math, Keith is asserting that since Prs(t) NOT= P1(t) + P2(t), then the reflected power is not being dissipated in the source resistor. But every sophomore EE knows NOT to try to superpose powers like that. Roger, I'll bet you know not to try to superpose powers? Since Keith doesn't listen to me, would you pass that technical fact on to him? When Keith uses the correct equation: Prs(t) = P1(t) + P2(t) +/- 2*SQRT[P1(t)*P2(t)] he will see that the equation balances and therefore 100% of the reflected energy is dissipated in the source resistor since the interference term averages out to zero over each cycle. It would be excellent if you would correct my exposition using the correct formulae. We could then see if your proposal actually provides accurate results. ...Keith PS. It won't. |
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On Fri, 21 Mar 2008 05:10:01 -0700 (PDT)
Keith Dysart wrote: On Mar 20, 1:07*pm, Roger Sparks wrote: Keith Dysart wrote: On Mar 16, 10:21 am, Cecil Moore wrote: snip Would you please explain how energy is conserved in the following example at the zero-crossing point for Vs? * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl * * * * *+----/\/\/-----+----------------------+ * * * * *| * *50 ohm * * * * * * * * * * * * * | * * * * *| * * * * * * * * * * * * * * * * * * | * * * * Vs * * * * * * * * 45 degrees * * * * *| Shorted * * *100v RMS * * * * * * *50 ohm line * * * * | Stub * * * * *| * * * * * * * * * * * * * * * * * * | * * * * *| * * * * * * * * * * * * * * * * * * | * * * * *+--------------+----------------------+ * * * * gnd At the zero-crossing of Vs, Ps(t) = 0, i.e. the source is supplying zero watts at that time but Prs(t) = 100w. Where is the 100 watts coming from? For the first 90 degrees of time, the circuit can be represented as * * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl * * * * * *+----/\/\/-----+----------------------+ * * * * * *| * *50 ohm * * * * * * * * * * * * * | * * * * * *| * * * * * * * * * * * * * * * * * * / * * * * * Vs * * * * * * * * * * * * * * * * * * \ 50 ohm resistor * * * *100v RMS * * * * * * * * * * * * * * * * */ * * * * * *| * * * * * * * * * * * * * * * * * * \ * * * * * *| * * * * * * * * * * * * * * * * * * | * * * * * *+--------------+----------------------+ * * * * * gnd After 90 degrees of time has passed, the circuit can be represented as * * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl * * * * * *+----/\/\/-----+----------------------+ * * * * * *| * *50 ohm * * * * * * * * * * * * * | * * * * * *| * * * * * * * * * * * * * * * * * * | * * * * * Vs * * * * * * * * * * * * * * * * * *--- * * * *100v RMS * * * * * * * * * * * * * * * * --- 50 ohm inductive * * * * * *| * * * * * * * * * * * * * * * * * * | * * * * * *| * * * * * * * * * * * * * * * * * * | * * * * * *+--------------+----------------------+ * * * * * gnd The sudden switch in circuit design at time 90 degrees is not unique to start up, but is true for any adjustment made to Vs and any returning wave from the shorted stub,. *As a result, a true stable circuit will never be found unless some voltage adjustment is allowed for the 90 timing shift caused by the shorted stub. *Keith (in his analysis of the circuit) recognizes that Vs drives into a reactive circuit. If we want to understand how constructive and destructive interference act to cause a 50 ohm resistor to evolve into a 50 ohm capacitor, then we need to examine how traveling waves might do this. *It would be nice to have a formula or wave sequence that fully addressed this evolution. *From circuit theory, we have the inductive reactance of a short-circuited line less than 1/4 wavelength long is * * *XL = Zo * tan (length degrees) * * * * = 50 * tan(45) * * * * = 50 ohms *From traveling wave theory, we would have the applied wave from the source arriving 90 degrees late to the stub side of resistor Rs. *This ignores the fact that current must already be passing through resistor Rs because voltage has been applied to Rs from the Vs side 90 degrees earlier. *Whoa! *Things are not adding up correctly this way! We need to treat the wave going down the 50 ohm line as a single wave front. *The wave reverses at the short-circuit, reversing both direction of travel and sign of voltage. *When the wave front reaches the input end, 90 degrees after entering (for this case), the voltage/current ratio is identical to the starting ratio (the line was 45 degrees long, tan(45) = 1), and the returning voltage directly adds to the voltage applied from Vs. *As a result, the current flowing through Rs will increase, and Vs will see a changed (decreased) impedance. After 90 degrees of signal application, we should be able to express the voltage across Rs as * * *Vrs(t) = Vs(t) - Vg(t) + Vref(t) Just to ensure clarity, you are using a different convention for terms and signs than I do. For me, Vg(t) is the actual voltage at point Vg. In terms of forward and reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at point g is sum of the forward and reverse voltage at point g. In my terms, this leads to Vrs(t) = Vs(t) - Vg(t) = Vs(t) - Vf.g(t) - Vr.g(t) Your Vg seems to be same as my Vf.g and you seem to be using a different sign for Vref than I do for Vr. You must be carrying these differences through the equations correctly because the final results seem to agree. Thanks for looking at my posting very carefully, because I think you are drawing the same conclusions as I, at least so far. I think you are correct in saying that For me, Vg(t) is the actual voltage at point Vg. In terms of forward and reverse: Vg(t) = Vf.g(t) = Vr.g(t), meaning the actual voltage at point g is sum of the forward and reverse voltage at point g. but doesn't this describe the standing wave? At point g, the forward wave has passed through resistor Rs, loosing part of the original wave energy, and is inroute beginning the long path to the resistor Rs. The reflected wave has arrived back to one side of Rs and the voltage is already in series with the source voltage acting on Rs. I think we need to subtract the two terms so that we can separate the apples and oranges. Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2 Substitute so Vrs(t) = Vs(wt + 90) - Vs(wt + 90)/2 + Vs(wt)/2 * * *= Vs(wt + 90)/2 + Vs(wt)/2 Allow Vs to be represented by a sine wave, we have A small aside. It is conventional to use cos for sinusoidal waves because it maps more readily into phasor notation. This is an important observation. cos(0) is 1 so to me it seems like we are saying that we begin at time zero with peak voltage or peak current. I wanted to begin with zero current at time zero so I used the sine. Am I missing something important here? * * 2Vrs(t) = Vs*sin(wt + 90) + Vs*sin(wt) * * * * * * = 2Vs*(sin(wt + 45)(cos(45)) * * Vrs(t) = Vs(sin(wt + 45)(cos(45)) Vs is defined as 100v RMS, which equals 100 * 1.414 = 141.4v Peak. *The maximum voltage would occur when the sin term was 90 degrees and equals 1, which would occur at wt = 45 degrees. * We would have * * *Vrs(45) = 141.42 * sin(90)(cos(45) * * * * * * *= 141.42 * 1 * 0.7071 * * * * * * *= 100v Now consider the current. *After the same 90 degrees of signal application, we should be able to express the current through Rs as * * *Irs(t) = Is(t) + Iref(t) Is(t) = Is(wt + 90), Iref(t) = Is(wt) The reflected current has been shifted by 90 degrees due to the reflection so we must rewrite Iref(t) to read * * *Iref(t) = Is(wt + 90) Substitute, * * * Irs(t) = Is(wt + 90) + Is(wt) Allow Is to be represented by a sine wave, we have * * * Irs(t) = Is*sin(wt + 90) + Is*sin(wt) * * * * * * *= 2*Is(sin(wt + 45)(cos(45)) How do we find Is? *Is is the initial current found by dividing the applied voltage at peak (141.42v) by the initial resistance (100 ohms). * * *Is = 141.42/100 = 1.4142a The maximum current would occur when the sin term was 90 degrees and equals 1, which would occur at wt = 45 degrees. * We would have * * * Irs(t) = 2*Is(sin(wt + 45)(cos(45)) * * * * * * *= 2 * 1.4142 * 1 * 0.7071 * * * * * * *= 2a These results agree with the results from Keith and from circuit theory. We have a theory and at least the peaks found from the theory agree with the results from others. * How about Cecil's initial question which is * At the zero-crossing of Vs, Ps(t) = 0, i.e. the source * is supplying zero watts at that time but Prs(t) = 100w. * Where is the 100 watts coming from? We will use the equation * *Vrs(t) = Vs(t) - Vg(t) + Vref(t) Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2 The challenge looks at the time when Vs(t) = Vs*sin(wt + 90) = 0, which occurs when wt = -90. *When wt = -90, all the power to the source resistor Rs is from the reflected portion described as Vref(t) = Vs(wt)/2. *The voltage across Rs would be * *Vrs(-90) = Vref(-90) * * * * * * = Vs*sin(-90)/2 * * * * * * = 141.4/2 * * * * * * = 70.7v The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming from the reflection. In summary, power to the resistor Rs comes via two paths, one longer than the other by 90 degrees (in this example). * This is not quite complete. The power to Rs does come from two places but they are the voltage source and the line. Ps(t) = Prs(t) + Pg(t) where Pg(t) is the power at point g flowing out of the generator into the line and Ps(t) is the power flowing from the source to the source resistor and the line. The power into the line can be seaparated into a forward component and a reflected component: Pg(t) = Pf(t) + Pr(t) (Sometimes this is written as P = Pf - Pr, but that is just a question of how the sign is used to represent the direction of energy flow.) Substituting we have Ps(t) = Prs(t) + Pf.g(t) + Pr.g(t) or Prs = Ps(t) - Pf.g(t) - Pr.g(t) Whenever Ps(t) equal 0, then Pf.g(t) also equals 0 so at those times, this simplifies to Prs = - Pr.g(t) which is the same conclusion you came to. No, I think my Pr.g(t) was positive, not negative. The difference is important because of the timing of the energy flows. Here is where it is important to observe that the forward wave entering the transmission line will have no further effect on resistor Rs until it has traveled 90 degrees of time. On the other hand, the reflected wave Pr.g(t) is only an instant of time away from entering the resistor Rs, but has not actually entered it because you are measuring it as being present at Vg, not as voltage Vrs. But it is wise not to forget the Pf.g(t) term because at times when Ps(t) is not equal to zero, the Pf.g(t) term will be needed to balance the energy flows. Pf.g(t) is the remaining power after the wave generated by Ps(t) has passed through Rs and left behind Prs(t). Another Prs(t + x) will be generated by the returning reflected wave. The short path is the series path of two resistors composed of the source resistor Rs and the input to the 50 ohm transmission line measured by Vg. *The long path is the series path of one resistor Rs and one capacitor composed of the shorted transmission line. *Both paths are available at all times. Power flows through both paths to Rs at all times, but because of the time differential in arrival timing, at some point Rs will receive power only from Vs, and at another point, receive power only from Vref. And at still other times, the power in the source resistor will depend on all of Ps(t), Pf.g(t) and Pr.g(t), or, more simply, it will depend on Ps(t) and Pg(t). ...Keith We differ on how to utilize the term Pg(t). I think it describes the standing wave, not the voltage across Rs. -- 73, Roger, W7WKB |
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Roger Sparks wrote:
So it looks to me like Keith is right in his method, at least in this case. Roger, do you understand why EE201 professors admonish their students not to try to superpose powers? -- 73, Cecil http://www.w5dxp.com |
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Keith Dysart wrote:
But you need to clearly state your limitations and stop flip flopping. What you are calling "flip flopping" is me correcting my errors. Once I correct an error, I don't flip-flop back. My error was in assuming that the power-density (irradiance) equation only works on average powers. K7ITM convinced me that it works on instantaneous powers also. I am surprised, this being 2008, that I could actually be offering a new way to study the question, but if you insist, I accept the accolade. I'm sure you are not the first, just the first to think there is anything valid to be learned by considering instantaneous power to be important. Everyone except you discarded that notion a long time ago. Analysis has shown that when examined with fine granularity, that for the circuit of Fig 1-1, the energy in the reflected wave is not always dissipated in the source resistor. Yes, yes, yes, now you are starting to get it. When interference is present, the energy in the reflected wave is NOT dissipated in the source resistor. Those facts will be covered in Part 2 & 3 of my web article. Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A) Which of the two needs the 'cos' term? Ps(t) = Prs(t) + Pg(t) or Pg(t) = Pf.g(t) + Pr.g(t) In fact neither do. For instantaneous values of voltage, the phase angle is either 0 or 180 degrees so the cosine term is either +1 or -1. The math is perfectly consistent. That you don't recognize the sign of the instantaneous interference term as the cosine of 0 or 180 is amazing. Non-the-less do feel free to offer corrected expression that include the 'cos(A)' term. I did and you ignored it. There is no negative sign in the power equation yet you come up with negative signs. That you don't recognize your negative sign as cos(180) is unbelievable. The math holds as it is. But I invite you to offer an alternative analysis that includes cos(A) terms. We can see how it holds up. It it unfortunate that you don't comprehend that the cos(0) is +1 and the cos(180) is -1. The sign of the instantaneous interference term is the cosine term. The math certainly does hold as it is - you are just ignorant of the "is" part. -- 73, Cecil http://www.w5dxp.com |
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Keith Dysart wrote:
In my terms, this leads to Vrs(t) = Vs(t) - Vg(t) = Vs(t) - Vf.g(t) - Vr.g(t) How about expanding those equations for us? Vs(t) = 141.4*cos(wt) ???? Vg(t) = ____*cos(wt+/-____) ???? Vf.g(t) = ____*cos(wt+/-____) ???? Vr.g(t) = ____*cos(wt+/-____) ???? If you ever did this before, I missed it. Given the correct voltage equations, I can prove what I am saying about destructive and constructive interference averaging out to zero over one cycle is a fact. -- 73, Cecil http://www.w5dxp.com |
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Keith Dysart wrote:
It would be excellent if you would correct my exposition using the correct formulae. We could then see if your proposal actually provides accurate results. Roger didn't understand your terms and subscripts and neither do I. I doubt that anyone understands your formulas well enough to discuss them. What is the equation for the forward voltage component dropped across Rs? What is the equation for the reflected voltage component dropped across Rs? Given valid equations for those two voltages, I can prove everything I have been saying. -- 73, Cecil http://www.w5dxp.com |
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On Fri, 21 Mar 2008 11:22:22 -0500
Cecil Moore wrote: Roger Sparks wrote: So it looks to me like Keith is right in his method, at least in this case. Roger, do you understand why EE201 professors admonish their students not to try to superpose powers? -- 73, Cecil http://www.w5dxp.com No, I really don't. I do understand that you had better not try to find the power flowing from superposed voltages because they may be the sum of apples and oranges. That is, the superposed voltages may be flowing in opposite directions so they will each be carrying power but the power will be available/reachable only for the direction toward which the wave is traveling. An SWR meter comes to mind here. -- 73, Roger, W7WKB |
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Roger Sparks wrote:
Cecil Moore wrote: Roger, do you understand why EE201 professors admonish their students not to try to superpose powers? No, I really don't. It is because (V1 + V2)^2 usually doesn't equal V1^2+V2^2 because of interference. Keith's addition of powers without taking interference into account is exactly the mistake that the EE201 professors were talking about. One cannot validly just willy-nilly add powers. It is an ignorant/sophomoric thing to do. If we add two one watt coherent waves, do we get a two watt wave? Only in a very special case. For the great majority of cases, we do *NOT* get a two watt wave. In fact, the resultant wave can be anywhere between zero watts and four watts. The concepts behind Keith's calculations are invalid. If you are also trying to willy-nilly add powers associated with coherent waves, your calculations are also invalid. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
On Fri, 21 Mar 2008 19:43:12 GMT
Cecil Moore wrote: Roger Sparks wrote: Cecil Moore wrote: Roger, do you understand why EE201 professors admonish their students not to try to superpose powers? No, I really don't. It is because (V1 + V2)^2 usually doesn't equal V1^2+V2^2 because of interference. Keith's addition of powers without taking interference into account is exactly the mistake that the EE201 professors were talking about. One cannot validly just willy-nilly add powers. It is an ignorant/sophomoric thing to do. If we add two one watt coherent waves, do we get a two watt wave? Only in a very special case. For the great majority of cases, we do *NOT* get a two watt wave. In fact, the resultant wave can be anywhere between zero watts and four watts. The concepts behind Keith's calculations are invalid. If you are also trying to willy-nilly add powers associated with coherent waves, your calculations are also invalid. -- 73, Cecil http://www.w5dxp.com OK, yes, I agree. It is OK to add powers when you are adding the power used by light bulbs. It is not OK to willy nilly multiply the voltage or current by the number of bulbs to learn the power used. You must carefully consider the circuit that connects the bulbs before selecting the proper method of calculating power, especially the possibility that the bulbs may be connected to phased power as in 3 phase or in traveling waves. -- 73, Roger, W7WKB |
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