Below is a two-port network representation:

enter image description here

And what I understand so far according to the definition of a two-port network, there must be two independent variables and two dependent variables as in the above figure. And a port's input and output current should be same to call it a port.

But how about the following circuit from this text?:

enter image description here

Is the above circuit a "passive two-port network" as the text claims? Or is it a one-port passive network?

There is only one independent source in that network above which is Vin. Vout is dependent on Vin; so it is not independent for me at least.

So is this type of passive networks are two-port or one port network?

Is there a practical example for a two-port passive network ?

  • \$\begingroup\$ Why do you think a source couldn't be connected to the output terminals? \$\endgroup\$ – The Photon May 14 '17 at 1:07
  • \$\begingroup\$ I'm imagining I hook up the Vout to a scope to observe the waveform. I mean Vout is just the output, it is not connected to an independent source like power supply or current source. Vout means output response, Vin means we use an independent source. Is that correct? What is the second independent variable there besides Vin? \$\endgroup\$ – HelpMee May 14 '17 at 1:11
  • \$\begingroup\$ I mean if I use it how I described, would it be one-port network? And if one applied independent source to Vout why to call it "out" it would be another input isnt it? Out stands for response for an input. \$\endgroup\$ – HelpMee May 14 '17 at 1:15
  • \$\begingroup\$ The scope loads the circuit. It will affect the voltage and current at the output port. Therefore you need to consider it a two-port network. \$\endgroup\$ – The Photon May 14 '17 at 1:21
  • 1
    \$\begingroup\$ You're overthinking things. The variables are independent as concerns the behavior of the 2-port. They might not be independent once you add models for the generator and load that are connected and consider the whole system. \$\endgroup\$ – The Photon May 14 '17 at 1:30
  1. It is possible to connect a source to the "output" port and measure the response at the input port. For example, you'd do this to measure the reverse isolation characteristic of the network if you were considering using it as a filter.

  2. If you connect something other than the source to the output port, it will load the circuit, and the voltage and current at the output port will depend on this load. Analyzing the network as a two-port allows you to understand this effect.

  3. Even if the thing you connect to the output port behaves like an ideal open circuit (for example, an idealized oscilloscope probe), that is only one special case of all the possible things you could connect to the output port. Modeling the network as a two-port allows you to predict the behavior with any linear load connected to the output.

  4. When you model something as a one-port, the internal nodes of the circuit are considered not accessible. Therefore, with a one-port model, you'd have no information about what voltage and current appear at the output port.

One way to think of this, whatever load you connect to the output port "reflects" some signal back to the 2-port. This reflection behavior is explicit if you use an S-parameter representation of the networks in your model, but it is still there regardless (because the different network representations just represent the same behavior in different ways). The "reflected" signal from the load interacts the same with the 2-port as if it were generated by an independent source.

  • \$\begingroup\$ When analyzing two-port networks for example they first short V1(independent input voltage source) and obtain parameters; then they short V2 and obtain rest of variables as you know better than me. What I dont get is that: If V2 is just an output voltage and not connected to an independent source; can we still obtain parameters(notice now there is one independent variable)? What does it mean in that case(when V2 is nothing but the output voltage not a source) shorting V2? Im sorry to dig it but I really cannot find a clear explanation to this yet in texts. \$\endgroup\$ – HelpMee May 14 '17 at 1:52
  • \$\begingroup\$ Shorting V2 means connect a 0-V ideal voltage source between the output terminals. \$\endgroup\$ – The Photon May 14 '17 at 1:54
  • \$\begingroup\$ So even there is one independent variable V1. And V2 is nothing but an output voltage and not an independent source we can still treat that circuit as two-port and obtain parameters? \$\endgroup\$ – HelpMee May 14 '17 at 1:56
  • \$\begingroup\$ BTW, you'd only use this procedure (shorting the ports) if you're looking for a Y-parameter representation, that is when you're choosing \$V_1\$ and \$V_2\$ as the independent variables. If you want a different representation, you'd use other "experiments" to determine the parameters. \$\endgroup\$ – The Photon May 14 '17 at 1:56
  • \$\begingroup\$ Because in that case when we short V1 there is no source in the circuit to relate, since V2 is not a source. \$\endgroup\$ – HelpMee May 14 '17 at 1:57

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