# Intuitive explanation of shorted stub behaviour

Can someone please give me an intuitive explanation of shorted stub behaviour?

NOTE: This is not a duplicate of Intuitive explanation of open stub behaviour which has been narrowed to focus on open stubs.

I have the length-based mathematical derivation and explanation from my textbook, but I am looking for an intuitive one.

For example, for an open stub (circuit 2), I think of it as an open ended transmission line through which the wave travels and gets reflected due to the open end, hence travelling double the distance. The reflected wave (from the open stub) will come back, get evenly distributed to both sides (term 3 and term 4), hence adding to the total reflection and transmission. Hence, S33 should be greater than S11 (which it is as shown below) at every frequency.

But:

How does shunt stub work? (I am not able to explain its behavior as to me it is a transmission line taking the signal straight to Ground)

• Hi @DaveTweed, that question was put on hold stating the reason that it lacked focus and had multiple questions. That has already been edited and is only "focused" on open stubs while this "focuses" only on shorted stubs. Commented Feb 1 at 13:17
• Fair enough, but that certainly wasn't obvious. We'll try to get them reopened. Commented Feb 1 at 16:33

You should read into quarter wave transformers and reflection coefficients. But it may be more intuitive to draw the circuit like first principles.

RF doesn't work like DC. Imagine the transmission lines as coaxial cables. A short to GND is a 0 ohm resistor connecting inner/outer conductors. The power has to travel back to the source before it's really "gone".

First things first:

• Open circuits reflect back in phase - the reflection stays on the center conductor.
• Short circuits reflect back 180 degrees out of phase. The signal travels from the short onto your outer conductor and then back to the source. This is equivalent to a negative signal traveling backwards on the center conductor.
• Quarter wave transmission lines turn opens into shorts and shorts into opens (explained below).

Circuit 2 (Open stub):

• The open at TL6 causes a reflection to come back in phase.
• TL6 is 90 degrees, therefore the reflection sees a 180 degree phase shift before getting back to Node A.
• The reflection add destructively at Node A which then looks like a virtual short. TL6 transformed an open into a short.
• The virtual short is then transformed into an open circuit by TL5 (same logic as above but explained more later).
• Your Smith Chart shows an open circuit at 1 GHz for S33.

Circuit 3 (Short stub):

• The short at TL9 causes signal to go into the return path. Or, said differently, a signal reflects back 180 degrees out of phase.
• TL9 is 90 degrees, therefore the reflection sees another 180 degree phase shift before getting back to Node A.
• The reflection adds constructively at Node A which looks like a virtual open. TL9 transformed a short into an open.
• An open at Node A is similar to if TL9 didn't exist and power is transferred to the load.
• Your Smith Chart shows perfect 50 ohms at 1 GHz for S55.

This repeats at integer multiples of 1 GHz.

Note: I've ignored impedance for simplicity. This is just for an intuitive explanation - no math :).

• Many thanks for the great explanation @Jason. That cleared many of my doubts. Commented Feb 2 at 22:21
• One more question though. At node A, the cicuit splits into 2 subcircuits. So, the wave should also split into 2 ? Now, one half of the wave will be completely transferred, but the other half, the reflected one, which will constructively or destructively interfere, will have half the amplitude ? Assuming that, it can't create a complete short/open ? Commented Feb 2 at 23:33
• @CuriousCosmopolitan In a bounce diagram, the voltage would split, yes. However, in steady state (for 1 GHz) the quarter-waves transform the opens/shorts. Commented Feb 3 at 23:40