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Series Resistance per unit length of an infinite tranmission line?

  1. Sep 17, 2013 #1
    1. The problem statement, all variables and given/known data

    We are to assume an infinite transmission line with the following parameters:
    capacitance: 296 ρF/ft
    inductance: 8.0 × 10-7 H/ft

    We are told that a 100MHz signal is attenuated 31 dB per 100 feet and asked based on the information, assuming no leakage through the insulator, to find the series resistance per unit length.

    2. Relevant equations

    H(ω,l) = e-l[itex]\sqrt{(jωL+R)(jωC+G)}[/itex]

    20log(A) = -[attenuation in dB], where A is gain.

    3. The attempt at a solution

    I have worked on this problem multiple ways. I understand that the resistance will be frequency dependent, but for the life of me cannot figure out a way to set up the equation without getting a crazy resistance value. (e.g. -5050jΩ) I'm not even sure what value I am looking for in resistance as in, can it be imaginary?

    Also, is there some sort of way I could solve for the resistance by plotting this in MATLAB or some other software? I'm really struggling and frustrated with a problem that seems like it shouldn't be very difficult.
  2. jcsd
  3. Sep 18, 2013 #2
    Hi there! It seems to me that you do not have a solid background in Transmission Line (TL) Theory. I don't know exactly where to begin, since you made some imprecise statements and are probably walking the wrong path.
    Saying that "the resistance will be frequency dependent" is misleading: you are working at a fixed frequency. When deriving the telegrapher's equations from Maxwell's eq. (if you did it), do you remember the specifications that were made?

    The last question is really problematic: did you carefully consider it? I will be short: [itex]R[/itex] can not be imaginary! If you studied the Generalized Transmission Line, this result should be obvious, since [itex]R[/itex] is defined as a real part of an impedance. Even if you didn't, [itex]R[/itex] represents the (series) resistive component of the TL - it can't be complex, it can't be imaginary, it can only be real. For the relevant equations, you didn't write the characteristic impedance [itex]Z_0[/itex] (sometimes indicated with [itex]Z_c[/itex], probably as in your case). It is a very important equation:
    [tex]Z_0 = \sqrt{\frac{R+j \omega L}{G + j \omega C}}[/tex]
    The angular frequency ω can obtained simply by remembering that: [itex]\omega = 2\pi f[/itex], where f = 100 MHz. I suggest you to actually use this equation for this problem. Lastly, you know that the signal is attenuated 31 dB/100 feet (question: how much dB/ft is that?). It is probably more useful to know the attenuation in terms of its non-dB value: "??" 1/ft - or, using SI (just to be clear), in [itex]m^{-1}[/itex]. Can this information be used in conjunction with the equation of [itex]Z_0[/itex]? I think this clarifies things and should naturally lead to the solution.
  4. Sep 18, 2013 #3
    I have used that equation before, but I am having problems connecting it with the attenuation. When I fill in the equation I have:

    52 = [itex]\sqrt{(R+j(6.28×108)(8×107)/(j(6.28×108)(296×10-12)}[/itex]

    Having all of that filled in, I could actually solve for R. But in this case, R would most definitely be a complex number. And on top of that, I still haven't used the information that the signal is attenuated 0.31dB/ft.

    I'm not sure what you mean by "attenuation in terms of a non-dB value", could you elaborate on that as well?
  5. Sep 18, 2013 #4
    Wait a moment, why are you assuming that [itex]G[/itex] is equal to zero? In fact, if you assume [itex]G \neq 0[/itex], then you can easily prove that you don't necessary need [itex]R[/itex] to be a complex number for [itex]Z_0[/itex] to be real. Let [itex]R+j\omega L \triangleq \rho_1 e^{j \varphi_1}[/itex] and [itex]G + j\omega C \triangleq \rho_2 e^{j \varphi_2}[/itex]. Since both [itex]\rho_1 , \rho_2[/itex] are positive numbers, for [itex]Z_0[/itex] to be real, it is sufficient to let the phases "cancel each other out" (assuming that [itex]R,L,G,C[/itex] are all positive), in other words, the condition:
    [tex]\varphi_1 - \varphi_2 =0 \rightarrow \frac{L}{C} = \frac{R}{G}[/tex]
    is sufficient for [itex]Z_0[/itex] to be real. It is not a necessary condition, since another way for [itex]Z_0[/itex] to be real is simply [itex]R=G=0[/itex], which is the (very important) case of the lossless TL. But it is a sufficient one, i.e. you don't really need a complex [itex]R[/itex].

    Let [itex]x_{dB} =10 \log_{10} |x|^2[/itex] (I must specify: squared or not squared, depending on the actual dB definition used). Then, if you have [itex]x_{dB}[/itex], the "non-dB value" I was referring to is simply [itex]x[/itex]. Hope this clarifies my previous explanation. Anyway, show me what you did with the attenuation: for example, how are you using the fact that the TL is infinite?
  6. Sep 18, 2013 #5

    rude man

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    Because the problem stated to ignore leakage thru the insulator. That makes G = 0.

    Zc = 52 ohms suggests that Zc is complex = R0+ jX0. So |Zc| = 52 ohms.

    Probably a Z0 = 50 + j0 ohm line (typical coax) but with G then forced to zero.

    Then attenuation α = α(R, ω, L, C) only, where R, L and C are per unit length.
  7. Sep 18, 2013 #6
    Correct. The 52 ohms are also in the LC region which I forgot to mention before. So I have the equation:

    H(ω,l) = e^-l*sqrt((jωL+R)(jωC+G))

    After plugging the numbers in, my professor suggested that I take the magnitude of the equation and plot it in MATLAB since the actual calculations would be very difficult. He said to have the x-axis as resistance, and the y-axis as the magnitude, and where I found 0.31 dB/ft would be my answer.

    The problem now is when I plot (abs(h), R), I get a simple linear line with the resistance hovering around 1.002 ohms. I feel as though this can also not be the correct resistance per length of the line.
  8. Sep 19, 2013 #7
    I did thought of that possibility, but I was not sure of it. Let me explain: rifleman_d should have specified the precise definition of [itex]Z_c[/itex]. If (big "if") you define: [itex]Z_c \triangleq |Z_0|[/itex] (which is a bit misleading), then everything is fine (but see below). If, however, you simply state "Zc=52Ω", then I do assume you are saying: "the characteristic impedance is 52Ω", which is very different from saying: "the modulus of the characteristic impedance is 52Ω". Precision (both of language and notation) is critical here.

    When he typed the formula, he did write: 52 = (...), thus (mis)leading me to think: ok, it is in fact: ZC=Z0 (a simply different notation). As it appears, it is not. However, if so, then he did make a mistake, since he confused [itex]Z_0[/itex] (a complex number) with [itex]Z_c[/itex] (a real, non-negative number). Thus, he didn't correctly use the formula I wrote in my first post.

    In any case, some points must be made clear:
    1. The characteristic impedance, denoted with [itex]Z_0[/itex], is generally a complex number: [itex]Z_0 \triangleq R_0 + jX_0[/itex]. In some special but very important cases (e.g. lossless TL), it reduces to a real number.
    2. The primary line constants, i.e. [itex]R, G, L, C[/itex], are always real (and also non-negative) numbers.
    3. Given positive R,L,C, and G=0, it does not follow that R is complex. In fact, what follows is that Z0 is strictly complex, i.e. [itex]X_0 \neq 0[/itex].
    4. If R,G,L,C are all strictly positive, it does not follow that Z0 is complex. In fact, for Z0 to be real, it is sufficient to satisfy the condition: [itex]\frac{L}{C} = \frac{R}{G}[/itex].

    This should be basic theory. This is also the reason why, when writing a problem, notation must be clarified! Since I would like to avoid another "notational surprise", I will wait a post from rifleman_d in order to validate everything that was said.

    @rude man: This is an edit: you can forget everything I wrote.
    Last edited: Sep 19, 2013
  9. Sep 19, 2013 #8
    I apologize for the confusion but, the homework is actually written that way. It simply states that "[itex]Z_c[/itex] = 52Ω in the LC Region" and that is it. This is my first course that uses transmission lines, and the professor seems to be trying to push half a semester's worth of material into 4 lectures. I was pretty sure he said that [itex]Z_c[/itex] and [itex]Z_0[/itex] were interchangeable during one of the lectures, but I might have misheard him.

    Either way, I ended up calculating the gain (A), using the 20log(A) = -(attenuation) equation. This calculation came out to be about 0.0285.

    Now, since the magnitude of the H(ω, l) equation should be equal to the gain, I plot the magnitude of the H(ω,l) versus resistance in MATLAB and look for where the value is around 0.028. This appears to be the resistance value of about 3.74Ω which seems like a much more valid value.

    Can either of your confirm? Thanks in advance!
    Last edited: Sep 19, 2013
  10. Sep 19, 2013 #9
    Don't worry, it is not the worst of what I saw. You can hardly imagine interpreting an equation where [itex]n[/itex] indicates a real number and [itex]\varepsilon[/itex] was an integer (without anyone telling you, obviously). See edit.

    EDIT: "LC Region" - is that, perhaps, the Constant-Loss Region? I'm afraid I did miss one of your posts and this very important piece of information (which changes everything). I'm also afraid that I can't help you anymore, since I do not have MATLAB to check it. Anyway, it seems you (at least in principle) solved it.
    Last edited: Sep 19, 2013
  11. Sep 19, 2013 #10

    rude man

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    EDIT:Major screw-up on my part! But, this does not really change anything except R is 1/10 of what I previously posted, and agrees closely with the OP's value in post # 8.

    Assume for the moment low attenuation approximation, i.e. R/ωL << 1:

    Then α = (R/2)/(L/C)1/2 so

    R = 2α(L/C)(1/2) = 2*0.31*0.1151*(8e-7/296e-12)(1/2)
    = 2*0.31*0.1151*52 = 3.71 ohms/ft

    Notice that R0 = Re{Z0} = (L/C)(1/2) = 52 ohms! Assuming low attenuation, Z0 is indeed real and = 52 ohms.

    Check on low-attenuation assumption: R/ωL = 3.71/[6.28*100e6*8e-7] = 0.007 so we go along with it.

    (Otherwise we'd have to solve a polynomial :rolleyes:)

    Maybe you can check the math? Thanks.
    Last edited: Sep 19, 2013
  12. Sep 19, 2013 #11
    Okay, so [itex]\sqrt{(jωL+R)(jωC+G)}[/itex] is our Propagation Coefficient. Our transfer function can be calculated as H(ω,l) = e-l*sqrt{(jωL+R)(jωC+G)} where l is the length.

    Gain was only defined as it's relationship to attenuation, which was given. The relationship is 20log(A) = -(attenuation).

    Now, we can split the transfer function into it's real an imaginary parts. The real part of the function will be equal to our attenuation and the imaginary part will be equal to our delay. Since we assume no leakage, we will not have to worry about the imaginary part of the equation.

    The attenuation given was 31dB per 100 feet. So to calculate our gain we have: 20log(A) = -31. Which solving gives us A = 0.0285.

    So now that we have our gain, we should be able to set this value equal to the magnitude of our transfer function. So:

    0.0285 = |e-l*sqrt{(jωL+R)(jωC+G)}|

    So what I did instead was make a function f, that holds |e-l*sqrt{(jωL+R)(jωC+G)}| and graph that versus resistance. This gave me a decaying curve that I traced down 0.028 on the Y-axis and found the R value on the X-axis.

    The professor suggested MATLAB as calculating the magnitude by hand would be very difficult.
  13. Sep 19, 2013 #12

    rude man

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    You have to convert dB into nepers. The relationship is 20log10(A) dB = ln(A) Np So 1 dB = 0.1151 Np and for us, 0.31 dB/ft = 0.31*0.1151 Np/ft = 0.036 Np/ft = α.

    And your A was for 100 ft but you need attenuation per unit length, i.e. ft.
  14. Sep 19, 2013 #13
    Ehm... are you sure? 296=37*8 (exactly). Thus:
    [tex]\sqrt{ \frac{L}{C} } = \sqrt{ \frac{ 8*10^{-7} }{ 296*10^{-12} } } = \sqrt{ \frac{10^5}{37} } \approx 51,9875[/tex]
    I'm aware that low attenuation approximation is pretty good (when it holds), but it can't be that good! :smile:

    P.S. Strictly speaking, in fact, [itex]Z_0[/itex] is complex (only approximately real). Anyway, if you use that value in your computations, you should get a slightly better estimate of R, but I think that (if you read my past edits) it is now beyond the point, since this is a closed case.
  15. Sep 19, 2013 #14

    rude man

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    51.9875 is 52 considering the level of precision given for L and C.

    Of course there is a slight imaginary component to Z0! There is for every cable! In order for Z0 to be 100% real, an exact relationship between R and G would have to obtain. This of course is impossible since both are parasitics.

    This may be a closed case for you, but is it for the OP?

    Yoo hoo, rifleman?
  16. Sep 19, 2013 #15
    Haha, yeah. It's a closed case for me, I'm pretty sure. Thank you both very much.
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