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I was taught in class that 2 port RF networks have the following structure below, where V2+ is the reflected wave if load at end of port 2 does not match the characteristic impedance of lines at port 2:

schematic

simulate this circuit – Schematic created using CircuitLab

Let's take a simple circuit that we can make practically, it consists of 2 ports where input is being applied at port 1. The network consists of a single resistor whose value may not match with characteristic impedance of either lines resulting in reflection:

schematic

simulate this circuit

What I want to understand: If frequency of source is so low such that wavelength of input wave is much greater than L1 and L2 then the circuit doesn't need reflection consideration. What I mean is it would be like a 10 Hz source powering an appropriate load of wire length say 5 cm. This means voltage at load and source vary exactly as each other.

Statement 1: So, in above case, when we reduce it to simple network theory and we say that the current will travel thru Path 1 and return thru Path 2 (which we usually call ground path).

Statement 2: When we increase frequency such that wavelength is comparable to network, then also current should return thru Path 2. But according to first diagram, the reflected voltage/current returns thru path 2.

So, my confusion is does the reflected current wave, I1- (according to first diagram) returns on path 1 or path 2. If it returns on path 2, then does it add with the returning I1+?

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    \$\begingroup\$ In short, both. \$\endgroup\$ – Hearth Dec 11 '20 at 15:18
  • \$\begingroup\$ @Hearth both as reflected wave returns on path1 and path2? and when on path2 it adds up with the returning I1+? \$\endgroup\$ – Prasanjit Rath Dec 11 '20 at 15:47
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Statement 1: So, in above case, when we reduce it to simple network theory and we say that the current will travel thru Path 1 and return thru Path 2 (which we usually call ground path).

Correct, for low speed analysis we can consider current flowing in a loop: -

  • down path 1,
  • through the load R1 and
  • back along path 2.

We "say" (for the ease of the mathematics) that the current takes zero time to do this. Then it's easy to use basic circuit theories (like Kirchhoff's). But, strictly speaking we are not being absolutely correct.

Statement 2: When we increase frequency such that wavelength is comparable to network, then also current should return thru Path 2. But according to first diagram, the reflected voltage/current returns thru path 2.

No, this is not correct; we cannot talk in full circuit loops any more. What actually happens is that Path 1 and Path 2 form a transmission line (that's the name we use when analysing high speed data transmissions). Even before the signal has reached load R1, current is circulating back to the input source due to the input impedance of the transmission line called "the characteristic impedance" or \$Z_0\$.

If the load (R1) doesn't match the \$Z_0\$ impedance, we get a mismatch of voltage and current when they reach R1 and the "excess" (of both voltage and current) has to be sent back to the source in order to avoid ohm's law being violated. That excess is the reflection; it is both voltage and current.

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  • \$\begingroup\$ So, is it correct to say that the notion of current loop fails at very low wavelengths and each path will serve as a send and return path (for forward and backward wave)? \$\endgroup\$ – Prasanjit Rath Dec 11 '20 at 17:36
  • \$\begingroup\$ The notion of 1 singular current loop starts to fail when the cable length increases to about one-tenth of the wavelength. That's the rule of thumb. Both paths act together in terms of voltage and current as forward and backward waves. I don't like singling them out as "each path". \$\endgroup\$ – Andy aka Dec 11 '20 at 18:09
  • \$\begingroup\$ I thought about it a bit and when we are applying AC source then electrons will follow path 2 to load to path 1 (current opposite direction) in positive half cycle and path 1 to load to path 2 in negative half cycle. When wavelength is relatively large, then the electrons reach almost instantaneously at the load. Electron flow in opposite direction only starts when the current electrons have completed circling the loop. When, wavelength decreases however, electrons will start from path 2. Before they reach load/even complete their cycle, electrons from path 1 will also start. \$\endgroup\$ – Prasanjit Rath Dec 12 '20 at 10:31
  • \$\begingroup\$ So, if we see electrons are moving on path 2 towards load and on path 1 towards load, so they will have their respective reflections and we can say that each line acted as a send and return path, right? Is the logic correct or I am missing something? \$\endgroup\$ – Prasanjit Rath Dec 12 '20 at 10:32
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    \$\begingroup\$ @PrasanjitRath your final question ( If it returns on path 2, then does it add with the returning I1+?) is invalid because it is contingent on returning on path 2 i.e. if it returns on path 2 - and it doesn't return on path 2 because it returns on both paths hence, this is why I couldn't answer it and this is why I explained in my answer that reflections and forward waves occur on both paths. I have given you a lot of my time here. Your questions have been answered. \$\endgroup\$ – Andy aka Dec 12 '20 at 11:31

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