Suppose we have a two-port network. Then, we connect voltage sources V1 and V2 to the left and right end. For the transmission parameters, t11=V1/V2, then could one use the voltages given by the voltages sources and not care about the network? That doesn't make sense to me, and one should use the actual network to determine t11 (and all the other parameters), however I am not sure. Could someone shed light on this confusion?


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    \$\begingroup\$ Consider this; take a totally different two port network and connect those two voltage sources as you mentioned. Does the ratio of V1 and V2 now become anything else other than what they were with your original two port network and, if you agree, does that ratio of V1 and V2 tell you anything about either of those two port networks? \$\endgroup\$
    – Andy aka
    Jan 14, 2021 at 17:10
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    \$\begingroup\$ It does not. So, taking the ratio of V1/V2 is not correct, right? and I should actually use the network itself? \$\endgroup\$
    – Schach21
    Jan 14, 2021 at 17:28

1 Answer 1


No, when you're measuring network parameters, you have to terminate the network correctly for the kind of parameter you're measuring.

Your \$T_{11}\$ parameter measures the voltage response at port 1 when the network is stimulated by a voltage at port 2. If you're measuring a voltage response, you need to terminate the response port with something that allows the voltage to vary freely. In the case of T-parameters, where current on port 1 is also a response parameter rather than a stimulus, you also need to allow the current to vary. That means terminating port 1 with a fixed impedance of some kind rather than with a source.

Similarly, when you measure Z parameters and you want to know \$Z_{21}\$ you're applying a current at port 1 as stimulus and getting voltage at port 2 as the response. Therefore you would (as the most convenient option, since you don't need to measure port 2 current and voltage simultaneously) leave port 2 open (aka terminate with a 0-A current source).


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