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Insertion Loss / VSWR of RF Lines

  1. Nov 17, 2011 #1
    Hello all,

    I'm not sure if any of you have ever used a device that measures RF transmission line loss and presents it as a graph of Insertion Loss (dB) vs. Frequency (GHz), but that is exactly what I'm trying to calculate. I've been looking through equations for days, and I just don't understand how these tools work. The equation I see everywhere for Insertion Loss is IL = 10 * log_10(P_t/P_r) where P_t is the transmitted power before insertion, and P_r is power received after insertion. But if power isn't a function of frequency (only impedance, current, and voltage), then how are they related/calculated? I'm a software engineer, so my knowledge of this kind of this is limited, but I'd really appreciate if someone could help me understand!

    Thanks again!
     
  2. jcsd
  3. Nov 17, 2011 #2

    Born2bwire

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    Power is a function of frequency. The input and output impedance of your device is frequency dependent. As such, the reflection of the input power and output power will vary with frequency thus giving you the frequency dependence of the insertion loss. All you network analyzer is doing here is sending in an electromagnetic wave at a known frequency out of it's output port into the device and then it measures the wave that is transmitted out of the device into the input port. The ratio between the transmitted and received power is used to calculate the insertion loss. The analyzer does this over a bandwidth of frequencies to give you the relevant plot.
     
  4. Nov 18, 2011 #3
    Are there some general equations that relate power and frequency? I haven't been able to find any useful equations. Thanks again for your help.
     
  5. Nov 18, 2011 #4

    Born2bwire

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    I don't know what you're asking for. The insertion loss depends upon the device under test.
     
  6. Nov 20, 2011 #5

    davenn

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    Yes that may be so for say an amplifier etc, but in the case of a transmission line as in the OP's query. The transmission line impedance is fixed regardless of the freq its still say 50Ohms at 30MHz or 3GHz. So for a fixed power in and a fixed impedance. The variable is the frequency. As the freq increases so does the insertion loss.

    hence we dont need to work out the insertion loss of our favourite coax cables etc, as the manufacturer has already given us those figures over a wide bandwidth

    PRELIC....
    here is a link to insertion loss measurement methods that may help you :)

    http://downloadfile.anritsu.com/Ref...ation-Notes/Application-Note/11410-00276.pdf"

    cheers
    Dave
     
    Last edited by a moderator: Apr 26, 2017
  7. Nov 21, 2011 #6

    Born2bwire

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    A transmission line's impedance is frequency dependent. The characteristic impedance is given as

    [tex] Z_0 = \sqrt{ \frac{R+j\omega L}{G+j\omega C} } [/tex]
     
    Last edited by a moderator: Apr 26, 2017
  8. Nov 21, 2011 #7
    Network analyzer is the tool to measure this. Insertion loss almost always is frequency dependent. Even a straight tx line is frequency dependent as series resistance, and parallel conductance all come into play.

    I don't quite follow what you want as insetion loss is device dependent. The way we do it is we go through a process to de-embard the input coax line and the output coax line using dummy load to get the s-parameters of the line and back calculate to the input and output of the DUT so you can just look at from the input to the output. This is build into the network analyzer and you just follow the step where they tell you to short the end, open the end and 50 ohm termination. Don't quote the exact step as I have not done it for 7 years. But it is something like this.

    I use Microwave Office simulation tool and it has the de-embard feature to get rid of the connecting lines.

    Those to me is the difficult part, after that, you just pretty much measure the s-parameters and calculate the insertion loss. De-embarding the lines would be the first step.

    If you talk about measuring and find the insertion loss of the DUT, you use 2 port s-parameter. You just follow the 4 steps of of getting the 4 parameters. It is way too long to go into the steps for the devices that's what a RF book is for. You should get an RF book like Micro-Wave Engineering by David Pozar or other ones that talk about network analyzer and measuring s-parameters. This is all dancing on the Smith Chart. Smith Chart is the key of RF design, where you get the s-parameters, input output impedance, forward gain, reflections VSWR............ I spent years dancing on the Smith Chart.........by inputing a circuit, draw the chart, change values of different components and see how the trace move to adapt the mind to start thinking in Smith Chart, to "see" the trace movement.
     
    Last edited: Nov 21, 2011
  9. Nov 22, 2011 #8

    davenn

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    so you are saying that all the 50 Ohm coax etc that has ever been produced only works on 1 frequency ? and what is that magic frequency? and that we really need millions different types of transmission lines for every freq that is ever going to be used ??

    please explain why my single piece of 50 Ohm coax can support 1MHz to 20GHz ? I dont see any freq dependence :)

    Dave
     
  10. Nov 22, 2011 #9
    No, there is a relation between L and C, and the impedance comes out to be constant. It is the characteristic impedance.

    If you look at lossless approximation where R and G =0,

    [tex] Z=\sqrt{\frac L C}[/tex]

    Where U is constant velocity for a given line.

    They are related by the speed of propagation...

    [tex]U=\frac 1 {\sqrt{LC}}[/tex]
     
  11. Nov 22, 2011 #10

    davenn

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    hi yungman

    Yes I understand Zo characteristic impedance, I know my 50 Ohm cable is useable on any freq. but I fail to understand why born2bwired says its freq dependant when it obviously isnt

    Dave
     
  12. Nov 22, 2011 #11

    davenn

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    maybe there's something else I just havent realised all these years lol

    D
     
  13. Nov 22, 2011 #12

    f95toli

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    You are talking about different things. The characteristic impedance of a standard 50 ohm line is (more or less) independent of frequency as long as you don't start to excite higher order modes in the transmission line, but this is by design ; i.e. the manufacturer have picked the 4 parameters of the cable in such a way that the impedance is more or less constant over a fairly wide BW. But this is a property of specific lines, not a property of transmission lines in general.
    Moreover, the losses are of course always frequency dependent.
     
  14. Nov 22, 2011 #13

    Born2bwire

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    It is dependent. If you assume a lossless transmission line then the characteristic impedance is constant. However, once you assume that it is lossy, which has to be if we are to talk about insertion loss, then it is frequency dependent. And yes, they do measure the characteristic impedance at a specific frequency. This frequency, or the bandwidth over which the characteristic impedance is valid, is given to you in the cable specifications. For example, I just randomly pulled off the spec sheet for Belkin's Cat-5 patch cable. Note that for the characteristic impedance they define a variance and a bandwidth over which this is valid.

    http://www.belkin.com/cables/pdf/CAT5eUTPPatchA3L791.pdf

    More importantly, one needs to remember that a transmission line is a one-dimensional approximation of a three-dimensional waveguide. The TL model does not take into account the nitty-gritty details of wave propagation in a waveguide nor does it appreciate the difference in behavior over a bandwidth between different waveguide structures. But as f95toli points out, manufacturers are aware of this and so they design their cables or waveguides to work over a desired bandwidth so that these dependencies are reduced. To take an extreme example, if you were given a rectangular waveguide, blindless devotion to the TL model would lead one to believe that it could work down to DC when that is most obviously not the case. Transmission line model is just an approximation which is why one needs to perform measurements to get the true behavior of the waveguide or get a more accurate approximation over the bandwidth of interest.

    So in my mind there are going to be two factors in insertion loss. There is the inherent loss in the line which will increase with frequency since that increases the electrical length of the line. The second is the variation in the impedances that cause small amounts of reflections.
     
  15. Nov 22, 2011 #14
    The conductance and resistance are frequency dependent, but unless you are talking about very high frequency, it is not important. resistance is from skin effect that it get thinner and thinner as freq goes up. Conductance is from loss tangent of the dielectric. Usually resistance is not a gating factor, you choose the dielectric ( money!!!) for the frequency you are working with.

    In a straight sense, characteristic impedance is frequency dependent and the equation. But I have calculation program that take into all these and they don't vary that much. The lossless approx still pretty much hold.
     
    Last edited: Nov 22, 2011
  16. Nov 22, 2011 #15
    I forgot to mention. All the de-embardment are just to find the s-parameter of the input feed line and the output line. So the network analyzer take the total measurement as a cascade of three 2 port s-parameters:

    1) the input feed line s-parameter.
    2) DUT s-parameter.
    3) output line s-parameter that hook the DUT back to the network analyzer.

    By finding the s-parameter of the two lines, you essentially can eliminate them by calculation so you can measure as if you all directly connect to the DUT and measure the s-parameter. It is all done by the instrument. Good luck on hand calculating the whole thing.
     
    Last edited: Nov 22, 2011
  17. Nov 23, 2011 #16
    To be honest, most of this thread has gone way over my head. I do appreciate all of your comments, and I have been researching and reading everything I can find on the subject, but my goal is to somehow produce the output of a network analyzer (IL/VSWR vs. Frequency) manually by using characteristics of the setup (cables/connectors/etc). What I got from this thread was that this isn't really possible because IL/VSWR measurements are specific to the actual DUT, and not just specific to identical devices? So if I have 2 independent RF setups that consist of identical components, and then I measure IL/VSWR with a network analyzer, the measurements may not be the same? Is this assumption true?
     
    Last edited: Nov 23, 2011
  18. Nov 23, 2011 #17
    I don't follow what you said in red.
    1) the result is DUT dependent.
    2) If you have the same DUT, you should get identical result using different RF setup...IF you properly de-embard the connections. This imply if you use the same analyzer but change to a new set of connecting lines. OR even change the analyzer all together. You just have to go through the pain of calibration, de-embarding.

    I afraid if you don't know what we are talking about, it would be very hard if not impossible for you to write the simulation program. To do this, the minimum you need to know are, network parameters, 2-port matrix (linear algebra), s-parameters and how it work. That would be the absolute bare minimum to even attempt to write the program. In reality, you pretty much need to study 2/3the standard RF text book if not the complete book.
     
    Last edited: Nov 23, 2011
  19. Nov 23, 2011 #18
    So regardless of my technical knowledge for a minute, if I had a technical document for an RF system that listed the components and specifications like voltage, current, IL limits, etc...could I come up with the output of the network analyzer without having a network analyzer or the actual system?
     
  20. Nov 23, 2011 #19
    If you just want to simulate the VSWR and insertion loss of a tx line, not the network analyzer, that is a whole lot easier. Try to simulate the network analyzer is hard. Now you are talking about just the line. If you have the loss tangent of the tx line, you can find the conductance, if you have the dimension of the line and the material of the conductor, you can get the resistance. From that, you can simulate use phasor or other equation. The equation you need to look at is "lossy transmission". It will show you how to find the attenuation constant [itex] \alpha\;\hbox{ and phase constant }\;\beta[/itex]. An Engineering Electromagnetics text book will have this in detail. The book I recommend is "Field and Wave Electromagnetics" by David K Cheng. If you can limit yourself to parallel plate tx line, coax line, parallel wire tx line like those flat cable used in the old tv antennas. You can actually get the formulas of the attenuation constant or even the whole tx line equation.

    Look around first, if I have time, I'll dig up some notes, this is just not a good time as turkey day is tomorrow!!!!:rofl:. And at that, it is still not a straight forward equation...... more like a collection of equations to find the attenuation constant, phasor calculation, complex numbers. But you just make the whole thing easier now. You can get away without all the s-parameters.

    If you can limit to one type of tx line, reply back and let people think about it. Also if the impedance of the source and load is the same as the characteristic impedance of the tx line, it will simplify further. But if you are asking about VSWR, that implies you want to have different impedance.
     
    Last edited: Nov 24, 2011
  21. Nov 24, 2011 #20
    Turkey in the oven and I have a little time!!!

    Let's consider a standard tx line model with R' series with L' and then with the shunt capacitance C' parallel with conductance G'. where all are values per unit length. Using solution of differential equation of the voltage and current phasor, we agreed on

    [tex]Z_0=\sqrt {\frac {R'+j\omega L'}{G'+j\omega C'}}\;\hbox { and we define } \delta = \alpha + j\beta\;=\; \sqrt{(R'+j\omega L')(G'+j\omega C')}\;\Rightarrow Z_0=\frac {R'+j\omega L'} \delta[/tex]

    [tex]\delta = \alpha + j\beta\;=\; \sqrt{(R'+j\omega L')(G'+j\omega C')}\;=\; j\omega\sqrt{L'C'}\sqrt{1+\frac{R'}{j\omega L'}}\sqrt{1+\frac {G'}{j\omega C'}}≈j\omega\sqrt{L'C'}\left(1+\frac{R'}{2j\omega L'}\right)\left(1+\frac{G'}{2j\omega C'}\right)\;\hbox { using binomial approx for low loss.}[/tex]

    Expand this out, you'll get:

    [tex] \delta ≈ j\omega\sqrt{L'C'}\left[1+\frac 1 {2j\omega}\left(\frac {R'}{L'}+\frac{G'}{C'}\right)\right]\;\Rightarrow \alpha =\frac 1 2 \left(R'\sqrt{\frac {C'}{L'}}+G'\sqrt{\frac {L'}{C'}}\right) \;\hbox { and }\;\beta=\omega\sqrt{L'C'}[/tex]

    [tex]\hbox { velocity =}v_p=\frac {\omega}{\beta}\;\hbox{ therefore it is independent to R' and G'.}[/tex]

    [tex]Z_0=R_0+jX_0 =\frac {R'+j\omega L'} {\delta}≈\sqrt{\frac {L'}{C'}}\left[1+\frac 1 {2j\omega} \left( \frac {R'}{L'}+\frac {G'}{C'}\right) -\frac {R'G'}{4\omega^2L'C'}\right][/tex]

    [tex]\hbox {For low loss, R'G'<< }4\omega^2L'C' \; \Rightarrow \; R_0≈ \frac{L'}{C'} \hbox { and } \;X_0=-\sqrt{\frac{L'}{C'}}\frac 1 {2\omega}\left( \frac {R'}{L'}+\frac {G'}{C'}\right) ≈0 [/tex]

    Therefore the characteristic change very little for low loss tx line. We usually don't use high loss tx lines!!! Excuse me if there is any typo, turkey is calling me, so.......................

    For the op, this might be the place to start.
     
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