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Where does it go? (mismatched power)
If you have a resonant antenna, supposedly all of the power is radiated.
(SWR 1:1) If you have a nonresonant antenna, some of the power is reflected back to the transmitter. (SWR 1:1) If you connect a tuner and it is "tuned", none of the power is reflected back to the transmitter. (SWR at transmitter 1:1, at antenna still 1:1) Obviously it has to go somewhere. Where? If you are designing a tuner, where would you design it to go? Thanks in advance Geoff. -- Geoffrey S. Mendelson, Jerusalem, Israel N3OWJ/4X1GM I do multitasking. If that bothers you, file a complaint and I will start ignoring it immediately. |
Where does it go? (mismatched power)
"Geoffrey S. Mendelson" wrote in
: Geoffrey, I have bad news for you, your post is based heavily on misconceptions, archetypal ham myths even. I wrote a short article entitled "A simple VSWR analysis without mirrors" at http://vk1od.net/blog/?p=1259 , which analyses a mismatched line example without resorting to dodgy concepts often (and almost exlusively) employed by hams. The article does deal with why there is a reflected wave, and the result of that reflected wave on line loss and transmitter loading. IMHO, you would do best by starting with a reputable text book, and studying the topic. Certainly, the S/N ratio in ham forums is low, 'facts' seem to be determined by vote, and your wrong concepts are more likely to be reinforced than seriously challenged. Owen |
Where does it go? (mismatched power)
Geoffrey S. Mendelson Inscribed thus:
If you have a resonant antenna, supposedly all of the power is radiated. Yes (SWR 1:1) With respect to what ! If you have a nonresonant antenna, some of the power is reflected back to the transmitter. (SWR 1:1) If you connect a tuner and it is "tuned", none of the power is reflected back to the transmitter. (SWR at transmitter 1:1, at antenna still 1:1) Obviously it has to go somewhere. Where? Its radiated, less losses. If you are designing a tuner, where would you design it to go? That would depend upon desired system characteristics. Thanks in advance Geoff. A antenna at resonance does not necessarily match the feeder impedance ! The whole idea of a tuner is to make one end match the other. Stating a VSWR of unity implies a 50 ohm measurement system, which an antenna at resonance rarely is. 73's -- Best Regards: Baron. |
Where does it go? (mismatched power)
Geoffrey S. Mendelson wrote:
If you have a resonant antenna, supposedly all of the power is radiated. (SWR 1:1) If you have a nonresonant antenna, some of the power is reflected back to the transmitter. (SWR 1:1) If you connect a tuner and it is "tuned", none of the power is reflected back to the transmitter. (SWR at transmitter 1:1, at antenna still 1:1) Obviously it has to go somewhere. Where? If you are designing a tuner, where would you design it to go? Thanks in advance Geoff. All of the power produced by a transmitter is either radiated or dissipated as heat, period. But you'll never understand where the "reflected power" goes until you understand where it comes from. Consider a 100 watt transmitter connected to a 50 ohm dummy load or antenna via a half wavelength of 50 ohm lossless transmission line. The transmitter sees an impedance of 50 ohms resistive. The transmitter delivers 100 watts to the transmission line. The transmission line delivers 100 watts to the load. The load dissipates 100 watts. The VSWR on the line is 1:1 The "forward power" in the transmission line is 100 watts. The "reverse power" in the transmission line is zero. Ok so far? Now replace the transmission line with one having 200 ohm impedance. The transmitter sees an impedance of 50 ohms resistive. The transmitter delivers 100 watts to the transmission line. The transmission line delivers 100 watts to the load. The load dissipates 100 watts. The VSWR on the line is 4:1. The "forward power" in the transmission line is 156.25 watts. The "reverse power" in the transmission line is 56.25 watts. By changing the line impedance, we somehow created an extra 56.25 watts of "forward power". Neither the transmitter nor the load was aware of this wonderful event, since the transmitter sees the same load impedance as before, and the load sees the same source impedance. The transmitter is delivering exactly the same amount of power as before, and the load is dissipating the same amount. (If the transmission line had loss, it would have increased very slightly, but probably not enough to measure, and certainly much, much less than 56 watts.) Likewise, neither the transmitter nor the load knows anything about the new "reverse power" which was created at the same time. So the answer to your question is that the "reverse power" goes to wherever the new excess "forward power" comes from. Once you figure out where that is, you'll have a better understanding of the topic than nearly everyone who has been arguing on this forum about it for years. As for a tuner, it has the magical property of having different amounts of "forward power" and "reverse power" at its input and output when the amount of actual power flowing is the same on both sides (neglecting tuner loss). So it can create or destroy "forward power" as well as "reverse power" by a simple twist of its knobs. Little does the tuner, transmitter, or load know that as soon as "reverse power" is created, interminable arguments will take place debating about where it goes. I now return you to your regularly scheduled programming. . . Roy Lewallen, W7EL |
Where does it go? (mismatched power)
On Wed, 9 Jun 2010 19:00:37 +0000 (UTC), "Geoffrey S. Mendelson"
wrote: If you are designing a tuner, where would you design it to go? Hi Geoff, You would put it closer to the antenna. If you connect a tuner and it is "tuned", none of the power is reflected back to the transmitter. Obviously it has to go somewhere. From the perspective of the transmission line in between, when it hits the discontinuity of the tuner, it is reflected back to the antenna. The antenna drains away some of that power (just as it did in the original pass). The process repeats (the antenna is still partially reflective) and is further sustained with new cycles of energy. There are finer details that shift these dynamics. 73's Richard Clark, KB7QHC |
Where does it go? (mismatched power)
Geoffrey S. Mendelson wrote:
If you have a resonant antenna, supposedly all of the power is radiated. (SWR 1:1) If you have a nonresonant antenna, some of the power is reflected back to the transmitter. (SWR 1:1) If you connect a tuner and it is "tuned", none of the power is reflected back to the transmitter. (SWR at transmitter 1:1, at antenna still 1:1) Obviously it has to go somewhere. Differentiate between an antenna that happens to present the wrong non-reactive resistance... the tuner is just a transformer (whether done with linked magnetic fields, or by a narrow band matching network of Ls and Cs) and an antenna that presents a reactive feedpoint impedance. In this case: energy circulates between the tuning network and the antenna. Say your antenna presents a Z that is inductive and you put a parallel capacitor across the feed to exactly cancel the "inductance". What's really happening is that you have the equivalent of an LC tank where the energy moves back and forth between L and C every cycle. In the classic LC, the energy moves between the magnetic field of the L and the electric field of the C. In the antenna case, it's somewhat more complex: the antenna stores energy in the near field in both electric and magnetic fields. That answers the question... The problem is that nothing is lossless, so as it moves, there's a loss. If it's a resistive loss, it goes as I^2, so doubling the current results in 4 times the loss. And, if you have an antenna with high stored energy (about which more, later), this square of the current means bad news. Many antennas don't have an explicit separate matching network, but do the cancelling by doing it within the antenna structu say by changing element lengths, etc... so now that "circulating current" is circulating between different parts of the antenna. In fact, the ration between that stored energy and the amount flowing "through" (i.e. radiated away) is related to the directivity of the antenna: high directivity antennas have high stored energy (large magnetic and electric fields): the ratio of stored to radiated energy is "antenna Q" (analogous to the stored energy in a LC circuit leading to resonant rise). So, high directivity = high stored energy = high circulating energy = high I2R losses. It circulates between tuner and antenna. Where? If you are designing a tuner, where would you design it to go? Thanks in advance Geoff. |
Where does it go? (mismatched power)
On Jun 9, 2:44*pm, Roy Lewallen wrote:
So the answer to your question is that the "reverse power" goes to wherever the new excess "forward power" comes from. Once you figure out where that is, you'll have a better understanding of the topic than nearly everyone who has been arguing on this forum about it for years. Let's be careful to point out that "forward power" is a measurement *at a fixed point* on a transmission line of the rate of flow of the forward energy which is traveling at the speed of light in the medium. Same concept for "reflected power" - reflected energy flowing in the reverse direction at the speed of light in the medium. EM waves cannot stand still. It's easy to figure out where the excess forward energy comes from especially in an ideal lossless system which I will assume for the rest of this posting. Consider the following thought experiment. An ideal lossless system with a one-second long Z0=300 ohm feedline. Why one-second long? Because the watts (joules/second) and the joules will be the same magnitude, i.e. If it's a matched line with 100 watts of forward power, it is easy to show that a one-second feedline has 100 joules of energy in it that has been sourced but not yet delivered to the load - like a delay line. Here's an example of a Z0-matched system with an SWR of 6:1 on the Z0=300 ohm feedline. *Steady-state* conditions a forward power = 204 watts and reflected power = 104 watts. Power delivered to the load = 100 watts. There is a steady-state Z0-match at '+'. Zero reflected energy reaches the source, i.e. the SWR on the 50 ohm line is 1:1. 100w source---50 ohm line---+---1-sec-long 300 ohm line---50 ohm load During the transient key-down state, the source supplies a number of joules to the transmission line that do not reach the load. At the beginning of steady-state that value is 204 joules plus 104 joules equals 308 joules. When steady-state is reached there are 308 joules of energy "stored" in the one second long transmission line that have not been delivered to the load. Only after the transmission line is charged with steady- state energy does the load start to accept the same amount of power as is being sourced. There is no magic here - just a simple accounting for all the energy. Why otherwise intelligent, educated RF engineers cannot perform that simple second-grade math is a subject for another thread. During steady-state, there are 308 joules of energy "stored" in the transmission line that are performing the Z0-match function. There are 204 joules in the forward wave and 104 joules in the reflected wave. Since RF cannot actually be stored as in a battery, this energy is moving at the speed of EM waves - the speed of light in the medium. 104 joules/second is continuously being reflected at the load. 104 joules/second is continuously being redistributed back toward the load at the Z0-match. At key-up, steady-state ends and the key-up transient state occurs. During this transient state, all of the 308 joules in the forward wave and reflected wave during steady-state will be dissipated. During the key-down transient state, 308 joules are stored in the one second long transmission line. During the key-up transient state, the 308 joules are dissipated. For those who feel overwhelmed, it is just simple second-grade math In a Z0-matched system, the entire analysis can be performed using an energy analysis instead of a voltage analysis. The common way to analyze the above system is to assume that 70.7 volts is present on the 50 ohm side of the Z0-match with 1.414 amps flowing into the Z0- match. But with only an energy analysis, we can reverse-engineer the voltage and current conditions at any point in a Z0-matched system. Fact of Physics: There are exactly enough joules of energy in the mismatched transmission line to support the forward wave power and the reflected wave power. It's not magic nor rocket science. Anyone willing to count the joules that have not reached the load will realize that the energy in the forward and reflected waves consists of real-world joules that cannot be swept under the rug just because some alleged RF guru is ignorant of the laws of physics. Here's a simple math problem for all of the readers having trouble with this concept. On 10 MHz, we have a 50 ohm antenna being fed with one wavelength of lossless 300 ohm line from a 100w source. During steady-state, how many microjoules of energy are "stored" in the transmission line that have not yet reached the load? How are those microjoules split between the forward wave and reflected wave? -- 73, Cecil, w5dxp.com |
Where does it go? (mismatched power)
On Jun 9, 5:38*pm, Jim Lux wrote:
It circulates between tuner and antenna. Just a nit: A certain magnitude of energy circulates between tuner and antenna. Experiments with TV signal ghosting prove that it is not the identical energy, just the same magnitude of energy. -- 73, Cecil, w5dxp.com |
Where does it go? (mismatched power)
On Jun 9, 2:44*pm, Roy Lewallen wrote:
Consider a 100 watt transmitter connected to a 50 ohm dummy load or antenna via a half wavelength of 50 ohm lossless transmission line. ... (clipped) ... Ok so far? Now replace the transmission line with one having 200 ohm impedance. The transmitter sees an impedance of 50 ohms resistive. The transmitter delivers 100 watts to the transmission line. The transmission line delivers 100 watts to the load. The load dissipates 100 watts. The VSWR on the line is 4:1. The "forward power" in the transmission line is 156.25 watts. The "reverse power" in the transmission line is 56.25 watts. Could you or someone please post the same analysis when the electrical length of the 200 ohm transmission line is 1/4-wave rather than 1/2- wave, including the physical locations/components where r-f output power is dissipated? RF |
Where does it go? (mismatched power)
On Jun 9, 10:38*pm, Jim Lux wrote:
In fact, the ration between that stored energy and the amount flowing "through" (i.e. radiated away) is related to the directivity of the antenna: high directivity antennas have high stored energy (large magnetic and electric fields): *the ratio of stored to radiated energy is "antenna Q" (analogous to the stored energy in a LC circuit leading to resonant rise). So, high directivity = high stored energy = high circulating energy = high I2R losses. this is a relationship i haven't heard of before... and would be very wary of stating so simply. it may be true for a specific type of antenna, MAYBE Yagi's, MAYBE rhombics or or close coupled wire arrays, but some of the most directive antennas are parabolic dishes which i would expect to have very low Q and extremely low losses. you could also have an antenna with very high Q, very high i^2r losses, but very low directivity, so i would be careful about drawing a direct link between the two. |
Where does it go? (mismatched power)
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Where does it go? (mismatched power)
On Jun 11, 6:16*am, Owen Duffy wrote:
... The transmitter output power is probably different ... Thank you, Owen. Do your comments apply to a transmitter designed/adjusted for, and expecting a 50 + j 0 ohm load? IOW, if the net output power of such a transmitter (which equals that dissipated in the load) probably is different with such a mismatch, do you expect the reason for that to be related to "reflected power?" RF |
Where does it go? (mismatched power)
On Jun 11, 11:32*am, Richard Fry wrote:
On Jun 11, 6:16*am, Owen Duffy wrote: *... The transmitter output power is probably different ... Thank you, Owen. Do your comments apply to a transmitter designed/adjusted for, and expecting a 50 + j 0 ohm load? IOW, if the net output power of such a transmitter (which equals that dissipated in the load) probably is different with such a mismatch, do you expect the reason for that to be related to "reflected power?" RF NO! NO! NO! the difference is not due to 'reflected power'... any difference is due to the impedance change seen by the transmitter at its terminals. A VERY important method of analyzing sinusoidal steady state response of a system like a transmission line with a load on the end takes the load impedance and transforms it using the transmission line equations (with or without loss) back to the source terminals and then replacing it with an equivalent impedance. as long as the load is constant and linear this is a well known and easily proved substitution. you can then solve for conditions seen by the source ignoring the length of the line and load and using only a lumped impedance, therefore any reflected wave is irrelevant when analyzing the response of the source. |
Where does it go? (mismatched power)
On Jun 11, 9:13*am, K1TTT wrote:
NO! NO! NO! *the difference is not due to 'reflected power'... any difference is due to the impedance change seen by the transmitter at its terminals. If the "forward" power that can be generated by a transmitter is dependent on the degree of match of the output load to the load specified for that transmitter, would that not mean that there would be little/no risk of damage to the transmitter if it was operated into a load with ~infinite SWR relative to the load it expects, such as a short or open directly at its r-f output connector? RF |
Where does it go? (mismatched power)
On Jun 11, 10:45*am, K1TTT wrote:
On Jun 9, 10:38*pm, Jim Lux wrote: In fact, the ration between that stored energy and the amount flowing "through" (i.e. radiated away) is related to the directivity of the antenna: high directivity antennas have high stored energy (large magnetic and electric fields): *the ratio of stored to radiated energy is "antenna Q" (analogous to the stored energy in a LC circuit leading to resonant rise). So, high directivity = high stored energy = high circulating energy = high I2R losses. this is a relationship i haven't heard of before... and would be very wary of stating so simply. *it may be true for a specific type of antenna, MAYBE Yagi's, MAYBE rhombics or or close coupled wire arrays, but some of the most directive antennas are parabolic dishes which i would expect to have very low Q and extremely low losses. *you could also have an antenna with very high Q, very high i^2r losses, but very low directivity, so i would be careful about drawing a direct link between the two. and yes, this does work for complex loads and multiple stubs and connections to the line. this is a reasonable description of the derivation of these techniques, study especially the Thevenin equivalent impedance representation on page 2-13 and how it is applied: http://ee.sharif.edu/~comcir/readings/tran%20line.pdf |
Where does it go? (mismatched power)
On Jun 11, 9:13*am, K1TTT wrote:
NO! NO! NO! *the difference is not due to 'reflected power'... any difference is due to the impedance change seen by the transmitter at its terminals. Please let's be careful to give all the details. The *virtual* impedance, the ratio of Vtotal/Itotal, seen by the transmitter at its terminals is: Z = Vtotal/Itotal = (Vfor + Vref)/(Ifor + Iref) where all math is phasor (vector) math. Vref and Iref are components of the reflected wave. So the mismatch is certainly related to the magnitude and phase of the reflected wave. If the Z0 of the transmission line is the impedance for which the transmitter was designed, we can go as far as to say that the reflected wave causes the mismatch, the virtual impedance that deviates from the Z0 of the line. Before the reflected wave arrives back at the transmitter, the transmitter sees the Z0 of the transmission line. The mismatch develops when the reflected wave arrives. -- 73, Cecil, w5dxp.com |
Where does it go? (mismatched power)
On Jun 11, 3:03*pm, Cecil Moore wrote:
On Jun 11, 9:13*am, K1TTT wrote: NO! NO! NO! *the difference is not due to 'reflected power'... any difference is due to the impedance change seen by the transmitter at its terminals. Please let's be careful to give all the details. The *virtual* impedance, the ratio of Vtotal/Itotal, seen by the transmitter at its terminals is: Z = Vtotal/Itotal = (Vfor + Vref)/(Ifor + Iref) where all math is phasor (vector) math. Vref and Iref are components of the reflected wave. So the mismatch is certainly related to the magnitude and phase of the reflected wave. If the Z0 of the transmission line is the impedance for which the transmitter was designed, we can go as far as to say that the reflected wave causes the mismatch, the virtual impedance that deviates from the Z0 of the line. Before the reflected wave arrives back at the transmitter, the transmitter sees the Z0 of the transmission line. The mismatch develops when the reflected wave arrives. -- 73, Cecil, w5dxp.com i wouldn't call it a 'virtual' impedance, it is a very real impedance. it is the steady state impedance seen by the transmitter at its output terminals. once the transient response rings down the waves in the line beyond the terminals are irrelevant and the constant impedance can be used to calculate the voltage, current, and power in the source. in the steady state (which except in special cases is entirely adequate for amateur use) the mechanism of generating that impedance is irrelevant, waves or not it can be represented by a constant value and the source won't know the difference. |
Where does it go? (mismatched power)
On Jun 11, 10:36*am, K1TTT wrote:
i wouldn't call it a 'virtual' impedance, it is a very real impedance. Yes, it meets the (B) non-dissipative definition of impedance from "The IEEE Dictionary". It doesn't meet the (A) dissipative definition and it is not an impedor as it only exists as a V/I ratio. That's why I call it a virtual impedance. There is no resistor, there is no inductor, and there is no capacitor. There is only a V/I ratio caused by something else. it is the steady state impedance seen by the transmitter at its output terminals. * And it is a V/I ratio caused by the superposition of the forward wave and reflected wave, i.e. it would NOT be the same impedance without the reflected wave. That fact of physics is undeniable. When we have an actual impedor, i.e. a resistor plus an inductor or a capacitor, the voltage/current ratio is caused by the impedor. When we have a virtual impedance, the cause/effect procedure is reversed and the impedance is caused by the voltage/current ratio which may (or may not) contain both forward and reflected values. For the two definitions of impedance, (A) and (B), given in "The IEEE Dictionary", cause and effect are reversed. Let's take an example. Assume a 50 ohm load resistor fed with 1/2WL of 300 ohm lossless line. If we assume a 100 watt (50 ohm) source, the forward power will be 204 watts and the reflected power will be 104 watts. 100w Source----1/2WL 300 ohm feedline----50 ohm load In the feedline, a forward power of 204 watts is a forward voltage of 247.3 volts and a forward current of 0.825 amps. In the feedline, a reflected power of 104 watts is a reflected voltage of 176.6 volts and a reflected current of 0.589 amps. The reflected voltage is 180 degrees out of phase with the forward voltage at the transmitter, so the superposed voltage at the transmitter is 70.7 volts. The reflected current is in phase with the forward current at the transmitter so the superposed current is 1.414 amps. Since everything is either in phase or 180 degrees out of phase, phasor addition is not needed. So I repeat, the impedance seen by the transmitter is: Z = (Vfor - Vref)/(Ifor + Iref) Z = (247.3 - 176.6)/(0.825 + 0.589) Z = 70.7/1.414 = 50 ohms, NON-DISSIPATIVE! It is clear that the superposition of the forward wave with the reflected wave is the CAUSE of the 50 ohm impedance. That particular impedance cannot exist without the effects of the reflected wave. Using a math model for answers to problems for so long that one comes to believe that the model dictates reality is a major contributor to the myths and old wives' tales that exist in amateur radio today. It's time to get back to the basics even if it causes a few old rusty brains to be put in gear for the first time in decades. -- 73, Cecil, w5dxp.com |
Where does it go? (mismatched power)
Richard Fry wrote:
On Jun 11, 6:16 am, Owen Duffy wrote: ... The transmitter output power is probably different ... Thank you, Owen. Do your comments apply to a transmitter designed/adjusted for, and expecting a 50 + j 0 ohm load? IOW, if the net output power of such a transmitter (which equals that dissipated in the load) probably is different with such a mismatch, do you expect the reason for that to be related to "reflected power?" RF We've had this discussion many times before, but with no apparent change in your perceptions of the nature of "reflected power". But why would you expect the "reflected power" to have a different effect when the transmission line is a quarter wavelength than when it's a half wavelength? Can you write an equation giving the supposed dissipation caused by "reflected power" as a function of transmission line length? Roy Lewallen, W7EL |
Where does it go? (mismatched power)
On Jun 11, 4:18*pm, Cecil Moore wrote:
On Jun 11, 10:36*am, K1TTT wrote: i wouldn't call it a 'virtual' impedance, it is a very real impedance. Yes, it meets the (B) non-dissipative definition of impedance from "The IEEE Dictionary". It doesn't meet the (A) dissipative definition and it is not an impedor as it only exists as a V/I ratio. That's why I call it a virtual impedance. There is no resistor, there is no and that is part of the problem in here... 'you call it'. what is the definition of 'virtual impedance' in the ieee dictionary? |
Where does it go? (mismatched power)
On Jun 11, 12:09*pm, Roy Lewallen wrote:
But why would you expect the "reflected power" to have a different effect when the transmission line is a quarter wavelength than when it's a half wavelength? If reflected power is fictitious, and the number wavelengths of transmission line of any random impedance compared to the load connected to it makes no difference in the load seen by the transmitter, the output power produced by the transmitter, and the power dissipated in the far-end termination, then what is the reason you chose a 1/2 wavelength of transmission line in your quoted post? RF |
Where does it go? (mismatched power)
On Jun 11, 12:29*pm, Richard Fry wrote:
If reflected power is fictitious, etc Followup: Those denying the existence of reflected signals within an antenna system may wish to view the measurement of such signals, at the link below. http://i62.photobucket.com/albums/h8...easurement.gif RF |
Where does it go? (mismatched power)
On Jun 11, 12:09*pm, Roy Lewallen wrote:
But why would you expect the "reflected power" to have a different effect when the transmission line is a quarter wavelength than when it's a half wavelength? Good Grief! Because the interference has the opposite result between the two examples? -- 73, Cecil, w5dxp.com |
Where does it go? (mismatched power)
On Jun 11, 12:28*pm, K1TTT wrote:
and that is part of the problem in here... 'you call it'. *what is the definition of 'virtual impedance' in the ieee dictionary? "The IEEE Dictionary" has no such definition. Those are my words adopted from "Reflections", by Walter Maxwell, and designed to differentiate between the (A) and (B) definitions of "impedance" in "The IEEE Dictionary". Calling the V/I (B) definition of impedance, "virtual", is much more descriptive than calling it "the (B) definition". Walt is arguing that the impedance of an RF source is "non- dissipative". Ratios of V/I are non-dissipative if they exist devoid of an impedor. Walt adopted the word "virtual" from the optics "virtual image". It is an image that is not really there in reality. A virtual impedance would therefore be the image of an impedor that is not really there. I'm not hung up on the word "virtual". What adjective would you use to differentiate between a dissipative impedor and a V/I non-dissipative impedance? I am not trying to be difficult - just trying to communicate. I'm willing to adopt any convention that you suggest for the duration of this discussion. -- 73, Cecil, w5dxp.com |
Where does it go? (mismatched power)
On Jun 11, 12:56*pm, Richard Fry wrote:
Followup: *Those denying the existence of reflected signals within an antenna system may wish to view the measurement of such signals, at the link below. Or at this link. Scroll down to "Using Dielectric Beamsplitters to find the "missing energy" in destructive interference". http://www.teachspin.com/instruments...eriments.shtml I guarantee that every optical physicist who is reading this thread is laughing at the ignorance of the alleged RF gurus. -- 73, Cecil, w5dxp.com |
Where does it go? (mismatched power)
On Jun 11, 12:09*pm, Roy Lewallen wrote:
Can you write an equation giving the supposed dissipation caused by "reflected power" as a function of transmission line length? I'll just quote the following from Chapter 24-11 of REFERENCE DATA FOR RADIO ENGINEERS, 1975 Edition: QUOTE A load of 0.4-j2000 ohms is fed through a length of RG-218/U cable at a frequency of 2.0 MHz. What are the input impedance and efficiency for a 24-foot length and a 124-foot length? (Omitting the number crunching shown...) Tabulating the results, Input Length Impedance Efficiency Loss (feet) (ohms) ( % ) ( dB ) 24 0.106-j95 1.1 19.6 124 1.8+j55 0.03 35 The considerably greater loss for 124 feet compared with 24 feet is because the transmission passes through a current maximum, where the loss per unit length is much higher than at the current minimum. END QUOTE As the input Z for each of these lengths is not the same as their 0.4- j2000 ohm termination, this shows the dependence of the performance of such systems on the electrical lengths of the transmission line in use. I realize we have strayed into the real world here, because RG-218/U transmission line is not lossless even when perfectly matched at both ends. RF |
Where does it go? (mismatched power)
Richard Fry wrote in news:510d2db0-df70-4433-a216-
: On Jun 11, 12:09*pm, Roy Lewallen wrote: Can you write an equation giving the supposed dissipation caused by "reflected power" as a function of transmission line length? I'll just quote the following from Chapter 24-11 of REFERENCE DATA FOR RADIO ENGINEERS, 1975 Edition: QUOTE .... The considerably greater loss for 124 feet compared with 24 feet is because the transmission passes through a current maximum, where the loss per unit length is much higher than at the current minimum. END QUOTE The author acknowledges that loss under standing waves is not uniform, that in this case, it is higher in the region of a current maximum. That is simply explained the I^R effects on the conductors (which account for most of the loss in most practical coax cables at that frequency). If you treated the forward and reflected waves as travelling independently, and each attenuated by the matched line loss characteristic of the line, you would get a different (and incorrect) result. It happens that we sometimes make that approximation, and under many practical scenarious, it gives a result with acceptable error, it is nevertheless less accurate because the model is poorer. As the input Z for each of these lengths is not the same as their 0.4- j2000 ohm termination, this shows the dependence of the performance of such systems on the electrical lengths of the transmission line in use. Yes, it does, and not simply on the VSWR as often inferred. But that does not change the transmitter's behaviour in its output power being sensitive the the impedance at its load terminals, however that impedance is derived, and your response has not answered Roy's question, well at least in my mind. I realize we have strayed into the real world here, because RG-218/U transmission line is not lossless even when perfectly matched at both ends. Indeed, under some circumstances, the line can have less loss when not perfectly matched. Yes, throw those treasured graphs of ExtraLoss vs VSWR away, they depend on assumptions not usually stated. Cases similar to the example you quoted can be solved using TLLC at http://www.vk1od.net/calc/tl/tllc.php . An interesting case is to explore the loss in 1m of RG58C/U at 3.6MHz with 50+j0, 500+j0, and 5+j0 loads. The latter two are VSWR=10, but have quite different loss. The first two cases demonstrate that VSWR does not necessarily cause extra loss (compared to the matched line case). Owen |
Where does it go? (mismatched power)
On Jun 11, 7:19*pm, Cecil Moore wrote:
On Jun 11, 12:28*pm, K1TTT wrote: and that is part of the problem in here... 'you call it'. *what is the definition of 'virtual impedance' in the ieee dictionary? "The IEEE Dictionary" has no such definition. Those are my words adopted from "Reflections", by Walter Maxwell, and designed to differentiate between the (A) and (B) definitions of "impedance" in "The IEEE Dictionary". Calling the V/I (B) definition of impedance, "virtual", is much more descriptive than calling it "the (B) definition". Walt is arguing that the impedance of an RF source is "non- dissipative". Ratios of V/I are non-dissipative if they exist devoid of an impedor. Walt adopted the word "virtual" from the optics "virtual image". It is an image that is not really there in reality. A virtual impedance would therefore be the image of an impedor that is not really there. what is an 'impedor' in this context? that is a relatively rarely used term in circuit and wave analysis, but is generically defined as anything that has an impedance. that doesn't seem to fit your definition though if you can qualify an impedance as non-dissipative if they don't have one. I'm not hung up on the word "virtual". What adjective would you use to differentiate between a dissipative impedor and a V/I non-dissipative impedance? I am not trying to be difficult - just trying to communicate. I'm willing to adopt any convention that you suggest for the duration of this discussion. the ieee dictionary qualifiers of dissipative and non-dissipative seem adequate to me. no need to make up any other terms or qualifiers. |
Where does it go? (mismatched power)
On Jun 11, 4:29*pm, K1TTT wrote:
what is an 'impedor' in this context? Hopefully, the same IEEE Dictionary definition as any other context: "impedor - a device, the purpose of which is to introduce impedance into an electric circuit." Note that it has a material existence. the ieee dictionary qualifiers of dissipative and non-dissipative seem adequate to me. *no need to make up any other terms or qualifiers. OK, I will change "virtual impedance" to "non-dissipative impedance" although if the resistance is zero, that still doesn't solve the semantic problem. The word "virtual" as used by Walter Maxwell over the past half-century conveys the meaning as well as any other words, IMO. The fact remains that a dissipative impedor is something that exists in the material world and can cause an outcome. A non-dissipative impedance is a *result* of a superposed V/I ratio, not a cause of anything. Roy once challenged me to detect the difference between a 50 ohm antenna and a 50 ohm dummy load. I said, "Simple, use a field strength meter." -- 73, Cecil, w5dxp.com |
Where does it go? (mismatched power)
On 11 jun, 16:27, Cecil Moore wrote:
On Jun 11, 12:56*pm, Richard Fry wrote: Followup: *Those denying the existence of reflected signals within an antenna system may wish to view the measurement of such signals, at the link below. Or at this link. Scroll down to "Using Dielectric Beamsplitters to find the "missing energy" in destructive interference". http://www.teachspin.com/instruments...eriments.shtml I guarantee that every optical physicist who is reading this thread is laughing at the ignorance of the alleged RF gurus. -- 73, Cecil, w5dxp.com Hello, here we are...! :) At the risk of naïve and conciliatory I still thinking that to some extent this is a problem of "same cat seen from different points of view". What if the question is formulated in terms apart, for example = generator responses to load differences, and by what mechanism a transmission line transforms impedances to presents to the generator those different loads? In that formulation I think there are room to simple, basic, and understandable electrical laws to account for generator behaviour and TL travelling waves interference phenomenom to account for Z transformings. From my perspective your main differences are reducible 73 Miguel Ghezzi - LU6ETJ |
Where does it go? (mismatched power)
On Jun 11, 10:03*pm, lu6etj wrote:
On 11 jun, 16:27, Cecil Moore wrote: On Jun 11, 12:56*pm, Richard Fry wrote: Followup: *Those denying the existence of reflected signals within an antenna system may wish to view the measurement of such signals, at the link below. Or at this link. Scroll down to "Using Dielectric Beamsplitters to find the "missing energy" in destructive interference". http://www.teachspin.com/instruments...eriments.shtml I guarantee that every optical physicist who is reading this thread is laughing at the ignorance of the alleged RF gurus. -- 73, Cecil, w5dxp.com Hello, here we are...! *:) At the risk of naïve and conciliatory I still thinking that to some extent this is a problem of "same cat seen from different points of view". What if the question is formulated in terms apart, for example = generator responses to load differences, and by what mechanism a transmission line transforms impedances to presents to the generator those different loads? In that formulation I think there are room to simple, basic, and understandable electrical laws to account for generator behaviour and TL travelling waves interference phenomenom to account for Z transformings. From my perspective your main differences are reducible 73 Miguel Ghezzi - LU6ETJ of course, but that is no fun! |
Where does it go? (mismatched power)
On Fri, 11 Jun 2010 14:29:17 -0700 (PDT), K1TTT
wrote: you can qualify an impedance as non-dissipative It's called reactance. There are certainly some contortions that have evolved from an argument that the source lacks the ability to dissipate. Impedance = (R ± jX) Ohm This is well known by all and yet it seems unsatisfactory and impossible to measure in a Tube even when R is exhibited both by measurement and by heat - something that by all normal accounts is evidence of dissipation. That same heat seems to be unaccountable because it's non-linear? If you use this heat on an ice cube, do you get harmonics? Even or odd? If it doesn't dissipate it must be because it has NOhms. If we embark further on this mysterious Load Conjugation with a loss-less resistor, what would we see for the Load Objurgation formula? rraaZ = (Rr ± iR ± jX) NOhm We must now have a Nimpedance measured along a second (hither too unreported) imaginary axis. I can imagine the dawn of the new Photon Ninterferences that will emerge from this. KEWEL ! 73's Richard Clark, KB7QHC |
Where does it go? (mismatched power)
On Fri, 11 Jun 2010 20:31:42 GMT, Owen Duffy wrote:
there is a proposition that a transmitter "designed/adjusted for, and expecting a 50 + j 0 ohm load" can be well represented by a Thevenin equivalent circuit and naturally has Zeq=50+j0. However, that proposition is easily proven wrong by valid experiments in the real world Using the same transmitter originally "designed/adjusted for, and expecting a 50 + j 0 ohm load"? I don't suppose this anecdote has any data behind it, does it? The easy proof has yet to wrong Walt's proposition - or did he state something to the contrary my reading that he represented a Thevenin equivalent circuit in the first two steps of his recent report? 73's Richard Clark, KB7QHC |
Where does it go? (mismatched power)
Richard Clark wrote in
: On Fri, 11 Jun 2010 20:31:42 GMT, Owen Duffy wrote: .... I don't suppose this anecdote has any data behind it, does it? I have performed many tests on many radios. One documented example is at http://vk1od.net/blog/?p=1045 . The concept is dealt with in more depth at http://vk1od.net/blog/?p=1028 .. Thing is that those supporting the Zeq=50 proposition, declare that since the test showed that Zeq (or Zs in the article) is not approximately 50 +j0, then the radio isn't "designed/adjusted for, and expecting a 50 + j 0 ohm load". Thing is that you cannot use the measured VSWR to reasonably predict the reduction in power output (Mismatch Loss) on many HF ham radio transceivers. I invite you and others to perform the same test. You will realise that one, or even 100 supporting tests do not prove the proposition, but one valid test to the contrary is damaging. Owen |
Where does it go? (mismatched power)
On Jun 11, 11:27*pm, Richard Clark wrote:
On Fri, 11 Jun 2010 14:29:17 -0700 (PDT), K1TTT wrote: you can qualify an impedance as non-dissipative It's called reactance. not always. there is a non-dissipative resistance. a lossless transmission line has a pure real impedance, but no dissipation. |
Where does it go? (mismatched power)
On Jun 12, 12:19*am, Owen Duffy wrote:
Richard Clark wrote : On Fri, 11 Jun 2010 20:31:42 GMT, Owen Duffy wrote: ... I don't suppose this anecdote has any data behind it, does it? I have performed many tests on many radios. One documented example is athttp://vk1od.net/blog/?p=1045. The concept is dealt with in more depth athttp://vk1od.net/blog/?p=1028 . Thing is that those supporting the Zeq=50 proposition, declare that since the test showed that Zeq (or Zs in the article) is not approximately 50 +j0, then the radio isn't "designed/adjusted for, and expecting a 50 + j 0 ohm load". Thing is that you cannot use the measured VSWR to reasonably predict the reduction in power output (Mismatch Loss) on many HF ham radio transceivers. I invite you and others to perform the same test. You will realise that one, or even 100 supporting tests do not prove the proposition, but one valid test to the contrary is damaging. Owen define "mismatch loss". and why would you expect that vswr is a good predictor of power output? other than radios that use vswr sensing circuits to reduce power output to protect their output transistors? and why would anyone expect the output of an rf amplifier to be linear except in a narrow range of selected loads and tuning conditions? thats just plain crazy! |
Where does it go? (mismatched power)
K1TTT wrote in
: define "mismatch loss". Mismatch loss is the reduction in power available in a mismatched load compared to a matched load. Mismatch Loss can be calculated in linear systems with Zs purely real as 1-rho^2. I wrote some notes at http://vk1od.net/blog/?p=659 which discusses the underlying assumptions. and why would you expect that vswr is a good predictor of power output? other than radios that use vswr sensing circuits to reduce power output to protect their output transistors? The concept of Mismatch Loss is a sound one, but application to a typical ham transmitter is flawed because the underlying assumptions are violated .. and why would anyone expect the output of an rf amplifier to be linear except in a narrow range of selected loads and tuning conditions? thats just plain crazy! Yes. Owen |
Where does it go? (mismatched power)
On Jun 11, 5:03*pm, lu6etj wrote:
From my perspective your main differences are reducible The basic argument revolves around what math shortcuts can be used to solve a particular problem vs what is actually happening in reality according to the accepted laws of physics. I agree one doesn't necessarily need to understand the laws of physics to solve a problem but one should probably know enough physics to recognize when those laws of physics are being violated by one's argument. -- 73, Cecil, w5dxp.com |
Where does it go? (mismatched power)
On 11 jun, 23:26, Cecil Moore wrote:
On Jun 11, 5:03*pm, lu6etj wrote: From my perspective your main differences are reducible The basic argument revolves around what math shortcuts can be used to solve a particular problem vs what is actually happening in reality according to the accepted laws of physics. I agree one doesn't necessarily need to understand the laws of physics to solve a problem but one should probably know enough physics to recognize when those laws of physics are being violated by one's argument. -- 73, Cecil, w5dxp.com ............ of course, but that is no fun! I agree :D :D ...... As a courtesy to me, a foreigner tourist ham, would you mind stop for a brief moment your more general differences and tell me if you agree on the behavior of a Thevenin generator with a series resistance of 50 ohms in relation to changes in impedance of a lossless TL predicted by the Telegrapher's equations solutions in terms of the power dissipated on the load resistance and series resistence of Thevenin source? I am pretty serious about this: until today I could not know if you agree in that!! :) |
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