Can someone explain to me how to use the transmission line (TL) formula (from telegrapher's equations) to calculate the DC resistance of the TL?

Here's my work so far:

V(x,s) = A * exp(-gamma*x) + B * exp(gamma*x)

where A and B = complex amplitudes of forward going and backward going waves, respectively

gamma = sqrt((R+sL)(G+s*C))


  • Telegrapher's model (L,C,R,G constants)
  • A DC source (Vs) is attached at one end of the transmission line, and the output of transmission line is a shorted.
  • G = 0 (no shunt resistance)

I expect the DC resistance of the transmission line to be equal to R*(length of transmission line).

I should be able to calculate the constants A and B from the boundary conditions (voltage = Vs at one end, 0V at other).

However, when I plug s = 0 (DC) and G = 0 into formula for gamma, I get gamma = 0. When using gamma = 0 in transmission line equation, I get V(x,s) = A + B

This implies that V(x,s) is not a function of x at all, which contradicts the boundary conditions.

Please help me figure out where my reasoning went wrong. Do TL equations not apply at DC, and if so, why not?

Thank you.

  • \$\begingroup\$ what is a DC wave? \$\endgroup\$
    – user16324
    Commented Mar 7, 2020 at 19:20
  • \$\begingroup\$ A DC Step is infinite waves with just the DC resistance steady state as the result with s=0 \$\endgroup\$ Commented Mar 7, 2020 at 19:39
  • \$\begingroup\$ @tony-stewart-sunnyskyguy-ee75 Why do I have a DC step? I'm interested in the steady state behavior (not when DC source is first turned on) and why the TL equations don't predict seem to predict it. I'm expecting a voltage of Vs at one end, with the voltage continuously decreasing down to 0V at the end. This can be represented as a line, V(x) = C *x + D (a straight line), with C and D constants. There's no discontinuity in the steady state response (at least not that I can tell). \$\endgroup\$
    – jds
    Commented Mar 7, 2020 at 20:58
  • \$\begingroup\$ @Brian, are you saying that I shouldn't be able to predict the steady state DC resistance of a TL from the TL equation I gave? \$\endgroup\$
    – jds
    Commented Mar 7, 2020 at 21:01
  • \$\begingroup\$ @jds there is an error in your assumptions or formula. WHen you measure voltage you a[pply a step function and read steady state current or V/I=R and s=0 \$\endgroup\$ Commented Mar 7, 2020 at 21:55

1 Answer 1


I figured out the answer to my question. Yes, you can use the Telegrapher's equations to compute the DC resistance when a transmission line is terminated with a short and when G (shunt conductance) = 0. The key to using the equations is to keep G as a term but assume it to is very small at the end so that you can use the asymptotic behavior of the functions that is in. I'll explain below:

Using the boundary conditions for the short terminated wave equation, the constants A and B for the voltage TL equation can be determined. These are A = Vs/(2*sinh(gamma*Lwg)) and B = -A.

Input impedance is simply V(x,s) / I(x,s),


I(x,s) = A/Zc * exp(-gamma*x) - B/Zc * exp(gamma*x)

x = 0 at shorted end and x = -Lwg at source end.

Zc = sqrt((R + sL)/(G + sC)) (see Telegrapher's equations)

gamma = sqrt((R + sL)(G + sC)) (see Telegrapher's equations)

Plugging A and B into V and I and calculating V/I gives:

V(x,s)/I(x,s) = -tanh(gamma*x) * Zc

At DC, Zc = sqrt(R/G)

At DC, gamma = sqrt(RG)

With x = -Lwg, we get

V/I = tanh(sqrt(RG)*Lwg) * sqrt(R/G)

if G is very small, tanh(sqrt(RG)*Lwg) will be approximatley sqrt(RG)*Lwg

And thus V/I = sqrt(RG)*Lwg * sqrt(R/G) = R*Lwg

This is the result I was expecting.


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