This post will ask how to distribute loss in a transmission line so that the line has a known total loss, while dissipating the least amount of power. We'll refer to "gain" of a transmission line, but we're thinking of the case where the line is lossy, so the gain is always less than one.

This post is in a sense a warmup for a somewhat more relevant and complex question that I will post after this one is resolved.

Discrete case

Consider a short section of transmission with a gain $G$, meaning that if a signal goes into that section with squared amplitude $A^2$, then it comes out with squared amplitude $G \, A^2$. If this gain is really coming from losses in the line, then $G<1$.

If we cascade many section of transmission line with gains $\{G_1, G_2\ldots \}$, then the total gain is $$\prod_{i=1}^n G_i = \exp \left(\sum_{i=1}^n \ln G_i \right) \, .$$

Each section of line dissipates power $P_i$ where $$P_i = P_\text{in} - P_\text{out} = A^2 - G A^2 = A^2 ( 1 - G ) \, .$$

Continuous case

Now suppose we have a continuous transmission line of length $L$ where the gain per length at each point $x$ along the line is $g(x)$. Extending the formula for the discrete case given above, it's clear that the total gain of the line is (remember that $g(x)<1$)$^{[a]}$ $$G = \exp \left( \int_0^L dx \, \ln g(x) \right) \, . \tag{$\star$}$$

The power dissipated in a bit of line of length $\varepsilon$ at position $x$ is \begin{align} P(x) =& A(x)^2 \left[ 1 - \exp \left( \int_x^{x+\varepsilon} dx \, \ln g(x) \right) \right] \\ \approx & A(x)^2 \left[ 1 - \left( 1 + \varepsilon \ln g(x) \right) \right] \\ =& -A(x)^2 \varepsilon \ln g(x) \, . \end{align}

The total power dissipation is of course $$P \equiv \int_0^L dx \, P(x) = - \int_0^L dx A(x)^2 \ln g(x) \, . $$

The problem

Given a fixed value of $G$, calculate $g(x)$ that minimizes $P$.

This is a constrained optimization problem and I think some kind of variational calculus is needed. Before we get to that, however, we should write the thing we're minimizing, $P$, in a better way by replacing $A(x)$ with an expression involving $g(x)$. In particular, for an input amplitude $A_\text{in}$, the amplitude at a particular point $x$ along the line is $$A(x)^2 = A_\text{in}^2 \exp \left( \int_0^x dx \, \ln g(x) \right) \, .$$ Therefore, \begin{align} P =& - \int_0^L dx \, A(x)^2 \ln g(x) \\ =& -A_\text{in}^2 \int_0^L dx \, \ln g(x) \exp \left( \int_0^x dx' \ln g(x') \right) \end{align}

How do we minimize $P$ subject to the constraint $(\star)$?

It's pretty obvious that if the total gain is fixed, then the power dissipation is also fixed because they're the same thing. In other words, the form of $g(x)$ should not matter. Therefore, I suppose a rewording of this question could be "how do we prove using variational calculus that the form of $g(x)$ doesn't matter?".

$[a]$: It's weird to have $\ln g(x)$ because $g$ has dimensions of length$^{-1}$. I suppose we can imagine multiplying $g$ by some length unit and dividing $dx$ by that same unit.

  • $\begingroup$ It used to be I squared times R = power $\endgroup$ – blacksmith37 Oct 10 '17 at 21:03
  • $\begingroup$ @blacksmith37 I don't understand the relevance of your comment. $\endgroup$ – DanielSank Oct 10 '17 at 22:35
  • $\begingroup$ Have you seen math.harvard.edu/archive/115_fall_06/lagrange_multiplier.pdf ? $\endgroup$ – Ryan Unger Oct 10 '17 at 22:45
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    $\begingroup$ Comment to the post (v1): Define $\Phi(x):=\int_0^x dx' \ln g(x')$ for notational convenience. So $\Phi(0)=0$. Then the constraint $(\star)$ reads $e^{\Phi(L)}=G$. OP's last expression $-P/A_\text{in}^2 = e^{\Phi(L)}-e^{\Phi(0)}=G-1$ is then completely fixed and independent of allowed functions $g(x)$. $\endgroup$ – Qmechanic Oct 11 '17 at 9:28
  • $\begingroup$ From the OP: "This post will ask how to distribute loss in a transmission line so that the line has a known total loss, while dissipating the least amount of power.". Question: "loss" of what? Loss of voltage, loss of power transmitted, or something else? $\endgroup$ – David White Oct 12 '17 at 22:07

Define $$\Phi(x) \equiv \int_0^x dx\, \ln g(x) \, .$$ Note that $\Phi(0)=0$ and $\Phi(L)=\ln G$. Then \begin{align} -\frac{P}{A_\text{in}^2} &= \int_0^L dx \, \Phi(x)' \exp ( \Phi(x) ) \\ &= \int_0^L dx \, \frac{d \exp \Phi }{dx} \\ &= \exp (\Phi(L)) - \exp (\Phi(0)) \\ &= G - 1 \, . \end{align} Therefore, $P$ doesn't depend on $g(x)$, so the distribution of gain doesn't matter.

This answer was motived by a comment from Qmechanic.

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The solution is to take $\ln g(x)$ to be equal to a delta function concentrated at the point where $A(x)^2$ is minimal.

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  • $\begingroup$ As explicitly explained in the post, $A(x)$ depends on $g(x)$. $\endgroup$ – DanielSank Oct 11 '17 at 17:07

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