I have several transmission lines, each with different impedances. These lines are attached to an ADC receiver with a known impedance of 400ohm.

I would like to have an impedance matching network that matches the impedance for an arbitrary transmission line to my 400ohm receiver.

Is this possible? If not, does this mean I need a new impedance matcher for each transmission line of different impedance? I'm new to this so I just want a general idea of what I'm thinking about is possible. If not what do people typically do?

  • \$\begingroup\$ What signal frequencies and what cable lengths are you talking about? \$\endgroup\$
    – Andy aka
    Commented Aug 17, 2017 at 11:38

1 Answer 1


It depends what you are trying to achieve by matching.

If you want maximum power transfer, some particular attenuation, lab-grade measurements, or to terminate an LC filter, then you have to use the right impedance for each line.

If you only want to transmit signals with flat attenuation, that is no ringing on the line, then only one end of the transmission line needs to be terminated in the line's impedance. But that means each arbitrary line still needs to be matched in its impedance on at least one end.

If the transmission lines are shorter than \$\frac{\lambda}{20}\$ or so at the maximum frequency, then you can pretty much ignore the transmission line properties, and treat them as lumped C or L, depending on whether the impedance is lower or higher than your 400 ohm receiver.

If you want the line to be 'tame', that is no excess ringing for transitions, no big suck-outs or peaks (for reasonable values of 'big'), then precise matching is not necessary, especially if it is approximately matched at both ends. Use a simulator to see just how much mismatch you have to apply to get behaviour you don't like. As the range of available line impedances is quite strongly limited by geometry (it tends to go as the log of a dimension ratio) you can probably define a fairly small range of likely impedances that will ever be connected, and approximately terminate them in the geometric mean of that range.


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