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#1
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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. |
#2
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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 |
#3
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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. |
#4
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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 |
#5
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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 |
#6
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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. |
#7
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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 |
#8
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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 |
#9
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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 |
#10
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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. |
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