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Linear Algebra - Analysis of purely resistive DC circuit

  • Engineering
  • Thread starter YoshiMoshi
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
berkeman
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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...
 
  • #3
berkeman
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BTW, are you able to just solve it by hand using the KCL equations?
 
  • #4
vela
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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.
 
  • #5
127
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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
 
  • #6
127
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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 =(
 
  • #7
vela
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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.
 
  • #8
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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. =(
 
  • #9
127
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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
The Electrician
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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
gneill
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@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
127
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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
Tom.G
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Where is the return path for the current source?
upload_2016-4-16_21-57-48-png.99226.png
 
  • #14
127
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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
Tom.G
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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
127
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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
Tom.G
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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
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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
Tom.G
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Well, if you try using KCL you'll find that you can't solve for the lower end of I1.
 
  • #20
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You mean the current coming out of the battery?
 
  • #21
Tom.G
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I mean the current from the current source, that thing labeled I1.
 
  • #22
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Oh so your talking about the lower part of the current source? I thought that there was no current comming 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 comming 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
Tom.G
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I thought that there was no current comming from the ground to the current source and that the current is generated from the current source?
I thought that there was 0 current at this point in the wire and that there was the 2A comming 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
127
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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
Tom.G
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...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.})
 

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