Linear Algebra - Analysis of purely resistive DC circuit

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Discussion Overview

The discussion revolves around troubleshooting a purely resistive DC circuit simulation that fails to resolve correctly in MATLAB. Participants explore the application of linear algebra to analyze the circuit, addressing issues related to circuit consistency, grounding, and the behavior of current sources.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant reports an error in MATLAB indicating that the matrix is close to singular or badly scaled, leading to inaccurate results.
  • Another suggests debugging the simulation by disabling power sources to identify the issue with convergence.
  • Some participants propose using KCL (Kirchhoff's Current Law) equations to solve the circuit manually.
  • Concerns are raised about the circuit's consistency, particularly regarding current sources injecting current without a return path.
  • A participant mentions adding a ground at a specific node to allow current to leave the circuit, questioning whether this affects the incidence matrix.
  • Discussions arise about the implications of grounding and whether it alters the linear algebra equations used in MATLAB.
  • Some participants express confusion over the concept of a return path for current sources and how it relates to circuit functionality.
  • There is a suggestion that the grounded node should be reflected in the incidence matrix, raising questions about its structure.
  • Participants note that the determinant of the matrix is very close to zero, which may contribute to MATLAB's singularity warning.

Areas of Agreement / Disagreement

Participants generally agree that the circuit has issues related to grounding and current paths, but there is no consensus on the exact nature of the problem or the best approach to resolve it. Multiple competing views on grounding and circuit behavior remain unresolved.

Contextual Notes

Limitations include unresolved assumptions about circuit configuration, the dependence of results on specific grounding choices, and the implications of matrix singularity on the analysis.

YoshiMoshi
Messages
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Homework Statement



I'm trying to solve this circuit
circuit.PNG


when I try to simulate and run a dc sweep i get that it can't be solved. When I try to find the answer using linear algebra
circuit.PNG


Capture.PNG


I get this answer after I throw it into MATLAB with the warning

“[w]arning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 2.312965e-18”.

Homework Equations

The Attempt at a Solution



I don't understand what I'm doing wrong. Thanks for any help.
Capture.PNG


Capture.PNG

Capture.PNG


I'm stuck in the linear algebra part and not the electrical engineering aspect so i figured i'd post it here.

thanks.
 
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In the simulation, can you try to debug it by running separate simulations with one or two of the power sources disabled? Short the voltage source and open a current source to see if that simulation runs. Try each of the 3 simulations with only one power source enabled. Then try combinations of two power sources enabled. Offhand, I don't see why the circuit isn't resolving. You could also try forcing some initial conditions to see if that helps with convergence to the DC operating points...
 
BTW, are you able to just solve it by hand using the KCL equations?
 
i don't think that circuit is consistent with KCL. The current sources inject 5 A into the circuit, but there's no way for that current to leave the circuit.
 
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Likes   Reactions: berkeman
Vela - So I have to add in a ground at node 7 so that there's a place for the 5 amps to leave the circuit?

The simulation ran with the ground at node 7.

However won't the incidence matrix be the same? MATLAB would still give me that weird error. I think I'm doing something wrong with the linear algebra?

POSTING NEW PICTURE IN THIS POST OF CIRCUIT

Capture.PNG
 
upload_2016-4-15_20-11-13.png


upload_2016-4-15_20-12-10.png

upload_2016-4-15_20-13-19.png


upload_2016-4-15_20-16-27.png

upload_2016-4-15_20-16-50.png


I don't understand what I'm doing wrong or understand why MATLAB is giving me an error

>> N=[-1,1,0,0,0,0;0,-1,0,1,0,0;-1,0,1,0,0,0;0,0,1,-1,0,0;0,0,-1,0,1,0;0,0,0,0,-1,1];
>> R=[1,0,0,0,0,0;0,3,0,0,0,0;0,0,2,0,0,0;0,0,0,4,0,0;0,0,0,0,5,0;0,0,0,0,0,6];
>> E=[0;0;0;0;0;5];
>> I=[0;0;0;2;3;0];
>> A=[0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0];
>> inv([R,N;N',A])[I;E]
inv([R,N;N',A])[I;E]
↑
Error: Unbalanced or unexpected parenthesis or
bracket.

>> inv([R,N;N',A])*[I;E]
Warning: Matrix is close to singular or badly
scaled. Results may be inaccurate. RCOND =
2.312965e-18.ans =

1.0e+16 *

0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
-4.5036
-4.5036
-4.5036
-4.5036
-4.5036
-4.5036

My simulator says that

upload_2016-4-15_20-31-52.png

upload_2016-4-15_20-34-34.png


I don't see what I'm doing wrong in MATLAB =(
 
YoshiMoshi said:
Vela - So I have to add in a ground at node 7 so that there's a place for the 5 amps to leave the circuit?
Did you have a reason to choose node 7 in particular? It seems kind of arbitrary.
 
No reason. I'm just use to putting the ground at the negative lead of the voltage source. It shouldn't matter right where we put the ground? I was able to simulate the circuit and got a solution.

I think MATLAB gets the error when it performs the following calculation
>>([R,N;N',A])^(-1)

Indeed just putting this into MATLAB returns the same error (except with different numbers of course). I don't see what's wrong with the calculation. I'm trying to get it to take the inverse of a 12x12 matrix split up in block form so it shouldn't have a problem.

I even try this
>>F=([R,N;N',A])
it does indeed return a matrix that has 12 collumns and 12 rows. I try this calculation
>>inv(F)
and I get the same error
"Matrix is close to singular or badly scaled. REsults may be inaccurate..."

I don't see what's wrong. =(
 
Oh... so MATLAB thinks the matrix is singular? I just tried det(F) and it returned
1.1102*10^(-15) so even though it's not singular MATLAB thinks it is because the determinant is almost zero?
 
  • #10
YoshiMoshi said:
Oh... so MATLAB thinks the matrix is singular? I just tried det(F) and it returned
1.1102*10^(-15) so even though it's not singular MATLAB thinks it is because the determinant is almost zero?

Your incidence matrix is singular. Perhaps when you ground a node, the grounded node should vanish from the incidence matrix?
 
  • #11
@vela had it right in post #4. There are only two ground connections in the circuit and both are affixed to current sources. Ground symbols in this case are really just labels used in place of wiring the common connections together. If you replace the symbols with wires:
upload_2016-4-16_11-47-11.png


You can see that there are two problems:

1. 5 A are injected into the circuit with no return path (might make an excellent Van DeGraff generator replacement!).
2. Two current sources with different current values are in series. Flash! Bang! Smoke! :smile:
 
  • #12
Hi Gnelli,

Thanks for your response. Yeah I think your right but I find it strange that the circuit would run in the simulator when it is indeed two current sources in parallel. But is it really in parallel if they are separated by resistors? I have redrawn the circuit and have tried to use MATLAB again by modifying the circuit to remove one of the current sources and to have the only ground in the circuit be the one at the other end of the current source. MATLAB still doesn't seem to like it =(. I don't see what I'm doing wrong. Thanks for any help.
upload_2016-4-16_21-57-48.png

Equations on modified circuit
upload_2016-4-16_21-58-43.png

upload_2016-4-16_21-59-5.png

upload_2016-4-16_21-59-18.png

upload_2016-4-16_21-59-34.png

upload_2016-4-16_21-59-54.png

putting my equations into MATLAB provides with the following answer as well as a warning that the matrix is singular
upload_2016-4-16_22-1-25.png

MATLAB commands

>> N=[-1,1,0,0,0,0;0,-1,0,1,0,0;-1,0,1,0,0,0;0,0,1,-1,0,0;0,0,-1,0,1,0;0,0,0,0,-1,1];
>> R=[1,0,0,0,0,0;0,2,0,0,0,0;0,0,3,0,0,0;0,0,0,4,0,0;0,0,0,0,5,0;0,0,0,0,0,6];
>> A=[0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0];
>> E=[0;0;0;0;0;5];
>> I=[0;0;0;0;3;0];
>> inv([R,N;N',A])*[I;E]
Warning: Matrix is close to singular or badly
scaled. Results may be inaccurate. RCOND =
2.312965e-18. ans =

1.0e+16 *

0
0
0
0
0.0000
0.0000
4.5036
4.5036
4.5036
4.5036
4.5036
4.5036

and when I try to take the determinant of the matrix [R,N;N',A] I get

-1.1102e-15

which may be to close to zero and very close to singular causing the error that I'm getting in MATLAB.

When I check the values in my simulator it tells me this time that it's unable to get the solution but I'm not sure what's wrong with my circuit. Thanks for any help!
 
  • #13
Where is the return path for the current source?
upload_2016-4-16_21-57-48-png.99226.png
 
  • #14
I'm not exactly sure what you are talking about by return path to the current source or even what one would like. Can you help? I'm not sure what you are describing.
 
  • #15
I1, if it is a current source then their are electrons flowing out of one of its terminals. What route do they take to get back in the other terminal?
 
  • #16
So your saying that I need to add a ground to the circuit? is node 6 ok to place the ground? This would allow the circuit to be simulated I think?

Also I think my equations would still be the same and MATLAB would still give me an error? I'm not sure it would effect the linear algebra equations.

Thanks for all of your help!
 
  • #17
Put a ground anywhere you wish. As it stands, the lower end of I1 is goimg nowhere, as far as the ckt. is concerned it is floating.
 
  • #18
Ok sounds good, but wouldn't my linear equations all be the same and MATLAB still provide me with an error?

I think I'll place it at node 6 since I'm used to having a ground at the negative end of the battery terminal mostly.

I just don't see though does this effect the equations at all?
 
  • #19
Well, if you try using KCL you'll find that you can't solve for the lower end of I1.
 
  • #20
You mean the current coming out of the battery?
 
  • #21
I mean the current from the current source, that thing labeled I1.
 
  • #22
Oh so your talking about the lower part of the current source? I thought that there was no current coming from the ground to the current source and that the current is generated from the current source? At node 4 I have current coming into the node from R2 current going into the node from I1 and current leaving the node into R4?

I was able to simulate the circuit after placing a ground at node 6. However I was unable to place the pinpointer on the wire that connects the current source to the ground. I thought that there was 0 current at this point in the wire and that there was the 2A coming out of the current source?

Sorry I guess I never thought about that part of the wire.

Although I do remember learning something about avoiding applying KCL at ground
 
  • #23
YoshiMoshi said:
I thought that there was no current coming from the ground to the current source and that the current is generated from the current source?
YoshiMoshi said:
I thought that there was 0 current at this point in the wire and that there was the 2A coming out of the current source?
That seems to be the basis of the problem.
You may recall that for current to flow there must be a closed path from the source, thru a load, back into the source. That's why devices have at least two terminals, one for electrons to leave from and one for electrons to return to. The present circuit does not have a way for electrons to get back to the current source lower terminal. It is as if you connected a light bulb to only one terminal of a battery, the bulb won't light because there is no return path.
 
  • #24
Ok I think it makes since now. So if I update the circuit with a ground at node 6, than the ground acts a return bath to the current source? Like so

upload_2016-4-16_23-50-16.png

This should be ok correct? With the ground at node 6 acting as a return to the current source.

The only problem is that I don't think this changes my equations and MATLAB will still give me an error saying that the matrix is singular and still through the error. I'm not sure how to get around that. What can I do?

Thanks for all of your help!
 
  • #25
YoshiMoshi said:
...I don't think this changes my equations...
So run it in the simulator. If it runs or gives a different error message then the equations changed.
Let us know the results.

(I'll take bets on this.:wink: {Well, probably not on this site.})
 
  • #26
Ya that's the problem though. I got it to run in the simulator however my MATLAB equations stay the same and I still get the singularity matrix problem. How can I get MATLAB to give me the correct answer?
View attachment 99227
View attachment 99228
View attachment 99229
View attachment 99230
View attachment 99231
putting my equations into MATLAB provides with the following answer as well as a warning that the matrix is singular
View attachment 99232
MATLAB commands

>> N=[-1,1,0,0,0,0;0,-1,0,1,0,0;-1,0,1,0,0,0;0,0,1,-1,0,0;0,0,-1,0,1,0;0,0,0,0,-1,1];
>> R=[1,0,0,0,0,0;0,2,0,0,0,0;0,0,3,0,0,0;0,0,0,4,0,0;0,0,0,0,5,0;0,0,0,0,0,6];
>> A=[0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0];
>> E=[0;0;0;0;0;5];
>> I=[0;0;0;0;3;0];
>> inv([R,N;N',A])*[I;E]
Warning: Matrix is close to singular or badly
scaled. Results may be inaccurate. RCOND =
2.312965e-18.ans =

1.0e+16 *

0
0
0
0
0.0000
0.0000
4.5036
4.5036
4.5036
4.5036
4.5036
4.5036

and when I try to take the determinant of the matrix [R,N;N',A] I get

-1.1102e-15

I have to restart my simulator I'll edit this post with what it gives me. I just don't understand what's wrong with MATLAB telling me that it's singular and throwing me that error?

EDIT:

My simulator says
V1 = 5 V
V2 = 6.244 V
V3 = 5.710 V
V4 = 8.733 V
V5 = 3.115 V
V6 = 0.000 V
I_R1 = 1.244 A
I_R2 = 1.244 A
I_R3 = 236.6 mA
I_R4 = 755.7 mA
I_R5 = 519.1 mA
I_R6 = 519.1 mA

But I can't seem to figure out why MATLAB is giving me an error and not allow me to get the correct answer =( what do I do.
 
Last edited:
  • #27
I'm confused. You change a circuit and the describing equations did not change? How can that be? The equations describe the circuit topology.

Lets see if @The Electrician can come back and handle the MATLAB stuff.
 
  • #28
Yeah exactly. That's why I was confused about. I originally posted this in the calculus and beyond section because my problem was with the linear algebra part and not necessarily the electrical engineering part. Yeah I don't think it changes.

The incidence matrix N remains the same. Right?
The resistance matrix definitely stays the same.
I and E stay the same.

I guess for the incidence matrix, once I add in the ground at node 6, the current on edge 6 (R6) is still leaving node 5 and entering node 6 right? So yeah I don't think it changes.

upload_2016-4-16_21-58-43-png.99227.png

upload_2016-4-16_21-59-5-png.99228.png

upload_2016-4-16_21-59-18-png.99229.png

upload_2016-4-16_21-59-34-png.99230.png

upload_2016-4-16_21-59-54-png.99231.png

I'm strugglign with the linear algebra part =(
 
  • #29
Unfortunately it's been too many decades since I've done any matrix algebra so I can't be of much help.
I did notice your node numbering changed between posts #5 & #6 though. If you haven't yet, how about writing the MATLAB stuff from scratch again from your current schematic? We all get blind to our own errors after about the third pass. :oldfrown:
 
  • #30
Oh dear your right! Thanks for catching that. Let me see for my current schematic
upload_2016-4-16_23-50-16-png.99234.png

it appears that the incidence matrix is
upload_2016-4-17_1-25-47.png

I think my incidence matrix was right for my current schematic in this post! Was wrong for post in 5 Let me retry in MATLAB
http://blob:https%3A//mail.google.com/59fc45b5-dd1e-46bd-a6c6-e12a02f50259
equation 1.PNG

equation 2.PNG

equation 3.PNG

equation 4.PNG

equation 5.PNG

equation 6.PNG

=( I'm stuck
 

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