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#31
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Derivation of Reflection Coefficient vs SWR
Suzy wrote:
"JOHN PASSANEAU" wrote in message ... There is no such thing as DC! How's that you say? You have to turn it on sometime and someday you may turn it off. There for, DC is just very low frequency AC. John Passaneau W3JXP That's false reasoning OM! Alternating current is not the same as discontinuous current. In the example you provide, a DC supply is either off or on; it does not reverse polarity! You do not make AC by switching on and off DC, even at 50 Hz (or your 60 Hz) But through the techniques of linear circuit analysis, we can split the pulsed DC into two components, a pure steady DC component and a symmetrical AC component. We can do two separate analyses with the two (AC and DC) excitations, and sum the results. The answer will be exactly the same as if we had done the calculations directly. One can then reasonably claim that the switched DC is the sum of an AC waveform and a pure DC component. Roy Lewallen, W7EL |
#32
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Derivation of Reflection Coefficient vs SWR
Arrgh! I did it again!
Roy Lewallen wrote: Ah is a unit of charge, not energy. The battery, making the simplifying (and invalid) assumption of constant voltage during discharge, contains 2.16 Mj of energy. That's 2.16 MJ, not Mj. It'll sink in eventually . . .I hope. Roy Lewallen, W7EL |
#33
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Derivation of Reflection Coefficient vs SWR
Roy Lewallen wrote:
"Ah is a unit of charge, not energy." The most common 12-volt battery is the lead-acid storage battery used in automobiles. It should not be allowed to become completely discharged nor to remain less than fully charged for a long time. The battery`s capacity is rated in amperes x hours. The discharge rate is assumed to be 8 hours. Slower discharges can be supported for more ampere-hours and faster discharges likely won`t meet the rated ampere-hour product. A 50 Ah battery would be expected to deliver about 6.25 A for a period of 8 hours provided its electrolyte does not rise to more than 110 degrees F. 2-volt cells can be discharged down to 1.75 volts per cell or a voltage of 10.5 volts for the 6 cells of a 12-volt battery. Best regards, Richard Harrison, KB5WZI |
#34
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Derivation of Reflection Coefficient vs SWR
Roy Lewallen wrote:
"So the rate at which a transmission line transfers energy depends on its length?" No, it depends on the power fed into the line. Storage in the line depends on its length, plus the incident and reflected energies per unit length of the line. Velocity is fixed by construction of the line so a slow velocity factor allows more energy storage as the source energy output is constant and independent of the line`s velocity factor. Best regards, Richard Harrison, KB5WZI |
#35
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Derivation of Reflection Coefficient vs SWR
On Jan 27, 9:45*am, Cecil Moore wrote:
Keith Dysart wrote: Not at all. The equations don't just stop working at 0 frequency. As a matter of fact, EM waves cannot exist without photons. There is zero wave activity at DC. Therefore, the forward power and reflected power is zero at DC steady-state. And yet all the circuit theory derivations of reflection coefficient, power, voltage and current distributions work just fine for DC (or, if you prefer, low rate pulses). Digital designers use exactly that for solving real world problems. They do not refuse to solve problems when the conditions approach DC. The equations all work. You really should try to stop thinking about photons for just a short while. Yep, you guys would like to sweep the technical knowledge from the field of optical physics and quantum electro- dynamics under the rug. One wonders why. An intriguing accusation. Transmission lines can be understood well, and real world problems solved without reference to photons and EM waves. Just use the well known circuit theory based equations. And they extend all the way to DC. For those who like EM waves and photons... Why do you want to limit yourselves? Why won't you use the circuit theory bases equations to solve problems for which they work? Just because EM waves and photons do not? Think of it as a long pulse. That should satisfy your need to have 'waves'. No, only photons will satisfy the definition of EM waves. There are no photons. There are no waves. There are no forward and reflected powers. There are no changing E-fields or H-fields. There is not even any movement of electrons associated with the source voltage. May be true. But why do you want to use that as an excuse not to solve solveable problems? ...Keith |
#36
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Derivation of Reflection Coefficient vs SWR
On Jan 27, 10:26*am, Cecil Moore wrote:
Keith Dysart wrote: When the generator is matched to the line so that the reflected wave does not encounter an impedance discontinuity when it arrives back at the generator (and therefore is not reflected), ... On the contrary, it is redistributed back toward the load in the process of destructive interference and becomes constructive interference associated with the forward wave. Whether you call that a reflection or not, the fact that the forward power equals the source power plus the reflected power tells us that reflected power being dissipated in the source would violate the conservation of energy principle. Unfortunately, this is quite wrong. And I continue to be surprised that you argue that there is a reflection where there is not an impedance discontinuity. Some parts of the rest of your post are correct by coincidence, but since the underlying premise of reflections where there is no discontinuity is incorrect, I have snipped it. But this debate has been had before. You do not want to understand how the output impedance of a generator affects a returning signal. I have offerred references and you have refused to look. I have offerred spice simulations, and you have refused to look. When the discussion moves to simpler generators so that the behaviour can be studied, you will declare them uninteresting because they do not represent "real ham transmitters". You will make jokes about 10 cent resistors, not realizing that is how real test equipment prevents re-reflection. (How well would a TDR work, if any substantial amount of the return was reflected?) When you decide that you do not want to argue that reflections occur where there is no impedance discontinuity, and are willing to study output immpedance, the learning can begin. ...Keith |
#37
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Derivation of Reflection Coefficient vs SWR
Keith Dysart wrote:
Transmission lines can be understood well, and real world problems solved without reference to photons and EM waves. If that were true, we would not be having this argument. The real world problem is - where does the reflected EM energy go? Since you have not "solved that real world problem", your methods are suspect. OTOH, optical physicists solved that same problem long before any of us were born. Why won't you use the circuit theory bases equations to solve problems for which they work? The main reason not to use your methods is that you use them to arrive at wrong concepts. EM waves cannot exist without energy. If there exists no EM wave energy, there are no EM waves. If EM waves exist, they are necessarily associated with energy and momentum, both of which must be conserved. The amount of that energy flowing past a measurement point/plane in a unit-time is the power (density) associated with the reflected wave. Even the energy and momentum of a single photon can be calculated. Any length of transmission line with reflections contains exactly the amount of energy necessary to support the forward and reflected waves. That amount of energy exists in the transmission line and is not delivered to the load during steady-state. Because your model doesn't tell you where the reflected energy goes, you assume there is zero energy in reflected waves. Again, I challenge you to use your fingers to replace the reflected power circulator load in a system with a KW source driving an open-circuit. That shock therapy will, no doubt, change your mind about the non-existence of reflected power. -- 73, Cecil http://www.w5dxp.com |
#38
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Derivation of Reflection Coefficient vs SWR
Keith Dysart wrote:
Unfortunately, this is quite wrong. And I continue to be surprised that you argue that there is a reflection where there is not an impedance discontinuity. Since an absence of reflections violates the conservation of energy principle, there is something wrong with your assertion and your earlier example was proved to contain a contradiction. Psource = Pfor - Pref = Pload Pfor = Psource + Pref = Pload + Pref Those equations are true only if reflected energy does not flow back into the source. I suspect that, contrary to your assertions, the actual real-world source presents an infinite impedance to reflected waves. When you decide that you do not want to argue that reflections occur where there is no impedance discontinuity, ... You simply fail to recognize the impedance discontinuity. Please perform the following experiment to prove there is no impedance discontinuity and no distortion and get back to us. zero ohm TV RCVR--+--TV source--+--Z0 stub----------open | Z0 source impedance resistor | GND The source output goes into the stub. Reflections occur at the open end of the stub and flow back through the zero impedance source to be dissipated in the Z0 source resistor and displayed without distortion on the TV RCVR. If what you say is true, it should be duck soup for you to prove. -- 73, Cecil http://www.w5dxp.com |
#39
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Derivation of Reflection Coefficient vs SWR
Keith Dysert wrote:
"(How well would a TDR worh, if any substantial amount of return was reflected?)" A properly terminated line does not make reflections. I would imagine multiple reflections on an oscilloscope produce the same smear as as they do on TV. TDR is suggested for monitoring bridge integrity and performance. Where was it when we needed it? TDRs are likely cheaper than replacing all the questionable bridges. TDR is used to determine the characteristics of electrical lines by observing reflected waveforms.Tektronnix is a leader with its "I Connect" software. Roy can likely describe it. Agilent suggests TDR is the most general approach to evaluating the time domain response of any electromagnetic system is to solve Maxwell`s equations in the time domain. The foregoing is courtesy of Wikipedia. The incoherent structure of the last sentence was theirs too. Best regards, Richard Harrison, KB5WZI |
#40
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Derivation of Reflection Coefficient vs SWR
Roger Sparks wrote:
"It is not too hard to use the concept of traveling waves to derive the familiar reflection coefficient to SWR relationship." Yes. Terman has done it for us in his 1955 opus. He just uses the letter S to represent SWR. On page 86: "The voltage and current of the incident wave at the load must satisfy Eq. (4-8) = Eprime / Iprime = Zo." And at the top of page 86: "The reflected wave.----is identical with the incident wave except that it is traveling toward the generator. Eq. (4-11) = Edouble prime / Idouble prime = -Zo." "The load voltage is the sum of the voltages of the incident and reflected waves at the load,.... The load current is the sum of the currents of the incident and reflected waves at the load,.... The vector ratio EL / IL must equal the load impedance ZL." "The vector ratio E2 / E1 of the voltage of the reflected wave to the voltage of the incident wave at the load is termed the reflection coefficient of the load. Reflection coefficient = rho = E2/E1 = (ZL/Zo)-1 / (ZL/Zo)+1." And on page 97: "Standing-wave ratio = S = Emax/Emin, Eq.(4-20) And: S= [E1] + [E2] / [E1] - [E2], Eq. (4-21)" The standing-wave ratio S is one means of expressing the magnitude of the reflection coefficient; the exact relation between the two is S = 1+[rho] / 1- [rho] or [rho] = S-1 / S +1 This relationship is illustrated graphically in Fig. 4-9. Best regards, Richard Harrrison, KB5WZI |
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