Lets say we have two transmission lines in parallel as shown in the figure below, what will be the total impedance of their combination? That rectangle is the way I am representing TL and Z is the characteristic impedance and theta is the electrical length.

enter image description here

I was thinking whether I can use the same formula as for the case of resistors. So, the characteristic impedance of two parallel transmission lines will be as shown below and electrical length is the same, theta: $$ Z_{total} = \frac{Z_1*Z_2}{Z_1+Z_2} $$ Is this correct?

  • \$\begingroup\$ It's not clear, from your diagram how the transmission lines are connected at the sending end and receiving end. There is more than one way of doing this. \$\endgroup\$
    – Chu
    Jul 14, 2019 at 0:07
  • \$\begingroup\$ Is the length of both transmission lines the same? \$\endgroup\$ Jul 14, 2019 at 0:11
  • \$\begingroup\$ Yes, they have similar electrical length. \$\endgroup\$
    – BGA
    Jul 14, 2019 at 5:37

2 Answers 2


Your schematic would be difficult to realize, because normally the two ends of a transmission line are far enough apart that you can't connect a lumped source across them. The whole point of a transmission line is that it is "long" relative to the wavelength of the signals it carries.

If you connect two transmission lines in parallel (and terminate the far ends with matched loads) like this:


simulate this circuit – Schematic created using CircuitLab

then you could use the formula you proposed to obtain the equivalent input impedance.

  • \$\begingroup\$ I agree with The Photon that the configuration shown by the OP really doesn't make any sense. What The Photon showed is more likely to show up in the real world, and his analysis is correct. \$\endgroup\$
    – SteveSh
    Dec 16, 2019 at 14:08
  • 1
    \$\begingroup\$ For more information try to find a book by one of the high speed design experts such as Dr. Howard Johnson or Dr. Eric Bogatin. By Howard Johnson, "High-Speed Digital Design: A Handbook of Black Magic (1993), ISBN 978-0133957242", and "High-Speed Signal Propagation: Advanced Black Magic (2003), ISBN 978-0130844088.". I have both in my library. \$\endgroup\$
    – SteveSh
    Dec 16, 2019 at 14:13

Suppose you have photon's arrangement. Since the loads match the line impedance, there is no reflected power on either line, and the voltage on each of the loads will be the same as the voltage at the splitting point (assuming equal propagation times and dispersions, but not equal impedances) for the two lines. And if the voltages are the same, the two points can be connected without any consequence at all. Therefore, the two loads can be replaced with a single load per your formula, and, since there is no reflected power in either of the two lines, the equivalent line impedance must also be according to your formula for this truth to continue to hold.

Another configuration of interest is where the Photon's two loads are connected, but not matched to the lines. What will be the equivalent Zo of the paralleled lines? To see this, just imagine what you have when the two lines are infinitely long: an impedance that is Z1 || Z2, not dependent on the loads, which matches your formula.

Such arrangements have been used by radio hams since the advent of coaxial cable for impedance matching when cable of the desired impedance (25 or 37.5 Ohms) did not exist.


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