Talk:Brune test

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More sources

This had been on my to do list for years, so well done for beating me to it. I never really got any further than collecting sources, so I'll just post them here before deleting from my list:

Melendez et al., "Limitations of feedback amplifiers analysis based on two port network modelling"
DH Horrocks, C Nightingale, "The compatibility of n-ports in parallel"
Wai-Kai Chen, Passive and Active Filters: Theory and Implementation
Wai-Kai Chen, Active Network Analysis.

Haven't checked to see how much is there that is not already in the article. SpinningSpark 19:10, 4 January 2017 (UTC)[reply]

Added diagrams

I added diagrams for parallel-parallel, series series and series-parallel tests, as I understand what is written. Not right? Let me know and I will adapt. Constant314 (talk) 20:53, 4 January 2017 (UTC)[reply]

This test is important for two-port networks that do not feature a through connection. -seems backward

Maybe I misunderstand. It seems to me that the test is likely to fail if there is a through connection such as a common ground.Constant314 (talk) 22:19, 4 January 2017 (UTC)[reply]

I'm baffled by that too. SpinningSpark 22:57, 4 January 2017 (UTC)[reply]

@Mathias magdowski: What did you mean by it? SpinningSpark 23:00, 4 January 2017 (UTC)[reply]

Constant314, I have now removed this statement from the article altogether. Your interpretation is definitely wrong (still not sure what Mathias originally meant). Through connections (ie zero resistance) from input pole to output pole is perfectly ok if the two-port being connected also has a through connection on the poles being connected. In general though, most passive two-ports without transformers will be disallowed, not just those with through connections. SpinningSpark 15:46, 5 January 2017 (UTC)[reply]

I'm fine with that. There are cases with through connections that would work. We should wait for a reliable reference, but I cannot imagine a case that fails that doesn't have a through connection. But that could just be my lack of imagination. Constant314 (talk) 17:58, 5 January 2017 (UTC)[reply]

Every two-port linear passive circuit can be reduced to a three-pole star topology by node reduction (I'll fix that redlink in a moment, I know we have somthing on this somewhere on Wikipedia) because one pole of each output port is not connected to anything in the test so we can throw that branch away for the purpose of analysis. The star node it was connected to can be thrown away as well because it is now just connecting two branches in series, so replace with a single branch. The two branches that are connected to the voltmeter draw no current (ideally) so can be replaced with short circuits (or anything) without changing the analysis. Essentially, that leaves just a voltage divider circuit, which will always output some voltage except when the branch the voltmeter is across is zero ohms, ie both poles are connected to through conductors and some pathological cases. One of those pathological cases is if the branch is in series ressonance, but very likely there are more that I haven't thought of. But the general result is that you are buggered if you actually make those connections. SpinningSpark 18:33, 5 January 2017 (UTC)[reply]

Not sure what your point is. If I misunderstand entirely, I apologize. If it is that some 2-ports with through connections will pass the Brune test, I agree with that. I also understand that a passive 2-port can be replaced with a star network (which I understand to be the same as a ”Tee” network in the 2-port case) without changing the 2-port equations. A three element start has three nodes, a 2-port has four nodes, so one node of the star has to be connected to two nodes of the 2-port. Obviously the two nodes that are connected must be on opposite ports of the 2-port. If we are free to choose which 2-port nodes share the common node then you can make a choice that will pass the Brune test for series-series or parallel-parallel (not sure about the hybrid case) and succeed if connected. I believe all that.

But, the Brune test and the success of connecting 2-ports in series-series, etc. depend on the actual implementation and the actual connection of the 2-ports. You are not free to replace a passive 2-port with its star equivalent for the purposes of the Brune test. The series-series connection of two passive 2-ports is not guaranteed to give the same behavior as connecting their star equivalent 2-ports in series-series. If it were, then you would not need the Brune test for passive 2-ports.

I added this example of two passive 2-ports (a pi network which be replaced with a star network). As drawn, the connection fails the Brune test. If 2-port #2 were flipped over so that the common connection was at the top, then the two 2-ports could be connected series-series. But that would be a different, non-equivalent connection. Constant314 (talk) 19:37, 5 January 2017 (UTC)[reply]

You are right that you have misunderstood. I have not connected two poles together. I've thrown one pole away - the one that is not connected to anything in the Brune test. I end up with a 3-pole that looks like this in nasty ASCII art,

    o------|
           |
           ZA   
           |  
    o---ZB----ZC----o

I can always reduce a 3-pole to this form no matter how complicated it started off. ZC doesn't do anything because the Brune test doesn't draw any current. That just leaves ZA and ZB acting as a potential divider. The same goes for the other two-port, and connecting the two input ports in series just just adds the two ZAs together and the two ZBs together, so still a potential divider, just with different values. In general, ZB is non-zero, hence in general the Brune test voltage will be non-zero, hence passive circuits, in general, fail the Brune test.

You have also conflated "node" and "pole". For node I mean nodes that are internal to the network, as well as the port terminals. For pole I mean only the external connections of the network. SpinningSpark 22:50, 5 January 2017 (UTC)[reply]

I agree with your conclusion, "passive circuits, in general, fail the Brune test", because, I think, in general, they have a continuous path through wires or components from the left side to the right side, unless they are isolated by a transformer in the middle, except at high frequency where the transformer's inter-winding capacitance becomes a through path. I still think that my statement was correct that the Brune Test only fails when there is a connection that goes through from left to right. But, no reference, so it stays out. I've guess we've beat this to death.Constant314 (talk) 04:36, 6 January 2017 (UTC)[reply]

Spinningspark, I have meant the following: If two-port networks have a through connection (e.g. each one has the same potential at the lower terminal at the input and output) they can be combined without any problems in series, parallel or any hybrid connection, as long as the through connection will not short-circuit any other circuit element of any of the other combined two-port networks. Therefore, it might be necessary to vertically flip some of the networks (e.g. if two T-shaped two-port network are connected in series, the lower one must be flipped vertically to allows for the combination). This vertical flipping will not alter the parameters of the two-port. Therefore, for two-port networks with a through connection, a combination is always possible and valid, no Brune's test is necessary. See the following example:

T-shaped two-port network:

    o---Z--*---Z----o
           |
           Z   
           |  
    o------*--------o

Vertically flipped T-shaped two-port network:

    o------*--------o
           |
           Z  
           |  
    o---Z--*---Z----o

These two can be connected in series without problem, no Brune's test is necessary.

Different things will happen, if the two-ports to be combined don't feature such a through connection. Then the port condition of each two-port will usually be broken after the combination and a Brune's test is always necessary prior to it.

Example of a two-port network without any through connection:

    o---Z--*---Z----o
           |
           Z   
           |  
    o---Z--*---Z----o

@Spinningspark: Is it now understandable, what I meant? Mathias magdowski 11:47, 7 January 2017 (UTC)[reply]

Mathias, I have moved your post down so it doesn't break the chronological order (even though I asked you the question earlier). That is very true, but if "series connection" is specified without qualification we would normally assume that the circuit was not to be flipped. It would be a very odd thing to do because when this kind of unbalanced topology is being used it is normally because it is being connected to unbalanced circuitry that will take one pole of the port down to ground. The output could not be grounded and you would be back to inserting isolating transformers. SpinningSpark 19:42, 7 January 2017 (UTC)[reply]

I see now that we meant different things by "through connection". Mathias means a zero ohm impedance between a terminal on the left and a terminal on the right, whereas I meant an impedance between any terminal on the right and any terminal on the left that was less than infinite. I would regard the H pad as having a through connection. Perhaps we could use the words "direct connection" or "common connection" for the zero ohm connection. Anyhow, it is a special case in which the success of the Brune Test can be determined by inspection. I have no objection to such a statement as long as the terms are defined clearly enough that someone like me won't misinterpret the meaning. I added an example of failure under the series-series connection. I'll change that heading from "example of failure" to just "examples" and add a picture of the same ports with port #2 flipped so that it passes the Brune test. If I get ambitious I may add an example with a transformer. Also, I am ready to create any example that anyone wants. Just let me know. Constant314 (talk) 23:10, 7 January 2017 (UTC)[reply]

How about a real life example such as an emitter follower amplifier? That's a real example of two-ports being connected in something other than cascade. The common emitter resistor constitutes a feedback network two-port connected in series-parallel arrangement with the amplifier. Here's a source so it can be cited [1]. Here's where it's generalised as a pair of two-ports in the same source [2]. SpinningSpark 01:40, 8 January 2017 (UTC)[reply]

Are you thinking an NPN emitter follower as the upper port and a PNP as the lower port with a series series connection? Or maybe two NPN EFs in parallel-parallel? Constant314 (talk) 06:36, 8 January 2017 (UTC)[reply]

No, something much simpler than that. Did you not follow the link to the circuit diagram? SpinningSpark 17:33, 8 January 2017 (UTC)[reply]

I did follow the link and it took me to a blank page. But I just did it again and now there circuit and text, etc. I'll see what I can do with it. Constant314 (talk) 02:48, 9 January 2017 (UTC)[reply]