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S11 ? if Zc complexe
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Hello
In the attached formula (ROS.png), as long as Zo is real, even if Zl is complex, everything is fine and S11 remains in a circle of radius 1. and the ROS is between 1 and infinity. If Zo is complex, this is no longer the case. While writing C# code looking for a (complex) adaptation by dichotomy, I was looking for a ROS of 1. It doesn't work. I had to use the modulus of the difference between my target and the current value. double optimiseur = Complex.Subtract(Zouth_, Application.Zcibleh).Magnitude; Is it legitimate to talk about S11 when Zo is complex? 73 -- F1AMM Fran?ois |
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On 6/26/23 2:22 AM, Fran?ois wrote:
Hello Sure it is: S11 is just a representation of the reflection coefficient on port 1 of the UUT. Generally people use just the log magnitude in dB, but it has a phase, too. Nothing prevents the Z0 being complex. |
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On 6/26/23 06:32, Jim Lux wrote:
On 6/26/23 2:22 AM, Fran?ois wrote:Definitely so. When working with telephone cable pairs at voice frequencies, Zo is always complex, generally with a phase angle approaching -45 degrees. Various graphical aids were published by the Bell System for solving transmission line problems with complex Zo, but today it's more common to simply solve the appropriate expressions using complex math in software.HelloSure it is: S11 is just a representation of the reflection coefficient on port 1 of the UUT.? Generally people use just the log magnitude in dB, but it has a phase, too. Various schemes were developed for "loading" cable pairs for voice transmission, the most common being the addition of inductors at intervals of a few thousand feet. The most common loading scheme in the Bell System added 88 mH every 6000 feet, but there were quite a few other loading schemes in use. Loading a pair made it's Zo close to, but not exactly, real across the frequency span of the load scheme, reduced the loss of the pair within that frequency span, and acted as a sharp cutoff low-pass filter, reducing significantly transmission at frequencies above the cutoff frequency. 73, Maynard W6PAP |
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I knew about the phone lines. My problem is that if we calculate the S11 with complex values, it leads, for example, to a ROS which can be negative. S11 is no longer in a circle of radius 1.
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Is this normal or am I mistaken? -- F1AMM Fran?ois -----Message d'origine-----De la part de Maynard Wright, P. E., W6PAP Envoy¨¦ : lundi 26 juin 2023 16:21 |
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My study (amateur) concerns the cascading of 2 adapters in 'L' one high pass and the other low pass to carry out two adaptations of impedance at two different frequencies.
To generalize the brought back impedance can be different for the two frequencies and complex. It works easily with a purely real impedance but when the impedance is complex, I tried to optimize the ROS by looking for its minimum value which I thought was ONE it does not work because the ROS varies from - infinity to + infinity. I had to change the search criteria to subtraction double optimizer = Complex.Subtract(Zouth_, Application.Zcibleh).Magnitude; And now it works. What is curious is that when the solution adapter is found, the ROS is indeed equal to ONE. A crazy story. -- F1AMM Fran?ois |
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On 6/26/23 7:26 AM, Fran?ois wrote:
I knew about the phone lines. My problem is that if we calculate the S11 with complex values, it leads, for example, to a ROS which can be negative. S11 is no longer in a circle of radius 1.It's possible, that with some active systems, you could get a reflection that is bigger than the incident wave (i.e. if the Zload were negative). But I think that for an entirely passive load, the reflected wave cannot be greater than the incident wave. |
It's possible, that with some active systems, you could get a reflectionWe have a little trouble understanding each other. It is not a question of measurement but of calculation. I have an adapter which for a given input impedance returns the impedance Zout = 1.96593624236642 -j 5601.37911926951)} The impedance I'm targeting is Ztarget = 50 +j 100 the results are recalculated in the Excel sheet attached S11 = 1.03617959022359 -j 0.0185190120640808 SWR = -56.028101163433028 negative |S11| = 1.036345067 -> |S11| > 1 -- F1AMM Fran?ois De la part de Jim Lux lundi 26 juin 2023 16:58 |
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On 6/26/23 11:12 AM, Fran?ois wrote:
It's possible, that with some active systems, you could get a reflectionWe have a little trouble understanding each other. It is not a question of measurement but of calculation. Indeed.. For instance if Z1 = 2 - 1000j and Z2 is 50 + 100j, then gamma = (Z1-Z2)/(Z1+Z2) gamma = 1.215 - 0.124j |gamma| > 1 (and it's not some sort of numerical precision thing) OK, we must have some interpretation problem. And it's not obviously "unphysical" - I can see a source with Z 50 + 100 j (A series inductor) and a load of 2-1000j (a series capacitor). Sure, it's not a conjugate match, but obviously power will be dissipated in the 50 ohm resistor. So this must be a case where "reflection coefficient" as defined by impedances isn't correct. There's no real transmission line here, so I'm not sure that the concept of a "traveling wave" (where gamma is reflected/incident) is valid. |
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R. A. Chipman, in "Schaum's Outline Series - Theory and Problems of Transmission Lines," McGraw-Hill, 1968, points out that when a lossy transmission line has Zo complex, minimum reflection from the load will occur when Zt equals Zo but maximum power transfer will occur when Zt is the complex conjugate of Zo. This seems nonituitive, but Chipman explains that the voltage and current relationships in the two or three eighth wavelengths adjacent to the load cause the loss of power to be reduced by exactly the amount necessary to provide the extra power required in the load (page 139).
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Chipman also shows that when Zo is complex, it is possible for the reflection coefficient to be greater than 1. He shows that the maximum value for the reflection coefficient is 1 + sqrt(2) (page 137) and states on page 138: "The conclusion is somewhat surprising, though inescapable, that a transmission line can be terminated with a reflection coefficient who magnitude is as great as 2.41 without there being any implication that the power level of the reflected wave is greater than that of the incident wave." In radio work, we often assume that a lossy line has a real characteristic impedance and this is a useful approximation in most cases. Chipman points out that the relationships among the primary constants make that relationship mathematically impossible except for a line (impractical physically) for which the loss due to R equals the loss due to G. 73, Maynard W6PAP On 6/26/23 07:58, Jim Lux wrote:
On 6/26/23 7:26 AM, Fran?ois wrote:I knew about the phone lines. My problem is that if we calculate the S11 with complex values, it leads, for example, to a ROS which can be negative. S11 is no longer in a circle of radius 1.It's possible, that with some active systems, you could get a reflection that is bigger than the incident wave (i.e. if the Zload were negative). ?But I think that for an entirely passive load, the reflected wave cannot be greater than the incident wave. |
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That's it, we have converged. In my application where I am looking for an equality between a value obtained by scanning and a target value, the distance between between the two points (in the complex plane) is a much better criterion for a good result than the ROS which only complicates by creating a non-monotonic function.
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When the result of the calculation is equal to the target, the ROS is indeed ONE and the distance zero. It is in the search for the optimal that the ROS disrupts everything. -- F1AMM Fran?ois -----Message d'origine-----De la part de Maynard Wright, P. E., W6PAP Envoy¨¦ : mardi 27 juin 2023 00:56 |
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Everitt and Anner, "Communication Engineering," Third Edition, McGraw-Hill, 1956, provide a general method (pages 330-331) for examining a mismatch between a source and load, such as a transmission line feeding an antenna. They factor the source and the load into series networks, one element of which in each case is Zo, the desired impedance.
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They then use "Compensation Theorem A" from page 107, which states": "Any impedance in a network (either linear or nonlinear) may be replaced by a generator of zero internal impedance, whose generated voltage at every instant is equal to the instantaneous potential difference produced across the replaced impedance by the current flowing through it." Using this theorem, they replace the added network segments by voltage generators so that the resultant equivalent network consists of a matched source and load with added voltage generators to represent the voltages generated by the mismatch. I applied this to the special case of the complex conjugate load in "More Octave for Complex Zo," QEX, September/October, 2019. In that case, no modification of the source voltage or impedance is needed and the complex conjugate load is replaced by a matched load (image impedance) in series with a reactance of opposite sign that is twice the value of the reactance of the complex conjugate load's reactance. That "extra" reactance may then be replaced by a complex voltage source of zero internal impedance that will be seen to be generating a voltage that corresponds to the value of the reflected signal. 73, Maynard W6PAP On 6/26/23 23:01, Fran?ois wrote:
That's it, we have converged. In my application where I am looking for an equality between a value obtained by scanning and a target value, the distance between between the two points (in the complex plane) is a much better criterion for a good result than the ROS which only complicates by creating a non-monotonic function. |
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Telecommunications engineers who work with transmission lines (telephone cable pairs) that exhibit complex characteristic impedances (Zo) generally use return loss and reflection loss to characterize those facilities, not SWR and reflection coefficient.
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The reason might best be made clear by an example from voice frequency transmission where the angle of Zo is close to -45 degrees. An ideal case might be a cable pair that presents Zo = 600 - j600 ohms at 1000 Hz. To transmit maximum power to the load, we should make the load (Zl) the conjugate of the line, 600 + j600 ohms. Then the reflection coefficient = rho = (Zl - Zo) / (Zl + Zo) = 0 - j1 ohms SWR = (1 + |rho|) / (1 - |rho|) = infinity If you use the replacement theorem I mentioned previously to replace a portion of the load's impedance with an equivalent voltage generator, you will find that it generates a voltage that is in quadrature with the current and satisifies the calculation of the reflection coefficient. An examination of the voltage waveform as a function of position along the line away from the load will show that the voltages are periodically out of phase and add vectorially to zero as is required by the denominator of the SWR expression. So both the reflection coefficient and the SWR are calculated correctly for such an extreme case, but they are not meaningful as measures of performance as are return loss and reflection loss. If you deal with Zo complex, but with a smaller imaginary component, just be aware that for Zo other than real, the reflection coefficient and SWR will not be exact measures of what is happening in the line. Chipman points out that, from an exact standpoint, SWR has meaning only for a losless line. When the line is lossy, adjacent peaks and nulls in voltage will not be measured (or calculated) from the same signal voltages and their ratios will not be a mathematically accurate measure of the performance of the line. We can ignore this in most practical cases. He points out that we often assume that Zo = sqrt(L / C) and, for the same line, assume some loss, which means that either R or G or both cannot be zero. The small errors caused by these assumptions can usually be ignored. 73, Maynard W6PAP On 6/28/23 08:37, Maynard Wright, P. E., W6PAP wrote:
Everitt and Anner, "Communication Engineering," Third Edition, McGraw-Hill, 1956, provide a general method (pages 330-331) for examining a mismatch between a source and load, such as a transmission line feeding an antenna.? They factor the source and the load into series networks, one element of which in each case is Zo, the desired impedance. |
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Thank you very much for your explanations.
1/ We found in old spectrum analyzers for RADAR (1960) delay lines made of a coaxial cable with losses. Our teacher had told us that it allowed to have a lower speed of propagation. I guess we can say that it is a lossy coaxial therefore presenting a complex characteristic impedance (true or false?). If we use a nanaoVNA, what will be the consequences? 2/ More recently I have seen the use of radiating coaxial cable to retransmit radios in the tunnels. This is a cable where the outer conductor is split. This type of cable also has a complex characteristic impedance. Also, what would be the consequence with a nanoVNA? -- F1AMM Fran?ois De la part de Maynard Wright, P. E., W6PAP Envoy¨¦ : vendredi 30 juin 2023 20:23 |
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For HF and higher frequencies, L >> R and C >> G are generally good assumptions. We can then simplify the general expressions for complex characteristic impedance and complex propagation constant to:
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Zo = sqrt(L / C) ohms gamma = j * omega * sqrt(L * C) radians per unit length Because the power dissipating elements R and G are assumed to be zero, the real part of gamma, the attention constant, is zero. Note that if we increase L, we increase the number of radians of wavelength that occupy each unit of length. We can therefore delay propagation by increasing L and that has been done with coax by using a spiral center conductor. Voice frequency telephone cables pairs are often inductively loaded by placing load coils at intervals. The most popular scheme in the Bell System was 88 mH every 6000 feet, but there are other schemes. The velocity of propagation is pretty low for such loaded cables, although the reasons for loading are to reduce the attenuation and make it relatively constant below a cutoff frequency. I don't know anything in particular about the special cables for tunnels. I would assume that the "leaked" energy would show up as an increase in attenuation, but maybe someone here knows more about those lines. 73, Maynard W6PAP On 6/30/23 20:47, Fran?ois wrote:
Thank you very much for your explanations. |
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