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Unsure Which Method To Solve Linear System Was Used

  1. Sep 16, 2012 #1
    Hi, I have a problem from a textbook on control systems along with the solution, but I'm not certain how the answer was derived. The only thing I'm confused by is the method by which the system of equations was solved, which appears to be something similar to, but not quite, Cramer's rule. I've attached the problem and part of the solution.


    I began by converting the circuit elements into impedances and then using nodal analysis at each node, the bottom node as ground, to obtain the system of equations, as in the matrix listed. The transfer function is equivalent to Vo(s)/Vin(s).

    Now, I can tell that the method isn't Cramer's rule, because, for the equation Ax=b, the b is the zero vector, meaning any determinant replacing a column with it would be equal to zero. For reasons unknown to me, the 1st row and 3rd column were eliminated to create a 2x2 matrix (presumably for which the determinant was calculated) in order to solve for the 3rd variable, and the 1st row and 1st column were removed to similarly solve for the 1st variable. I'm unfamiliar with any reason why these rows and columns were removed, or the procedure as a whole, for that matter.


    Thank you for your assistance.
     

    Attached Files:

  2. jcsd
  3. Sep 23, 2012 #2

    The Electrician

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    In the solution, an expression Vj=Δij/Δ*I1 appears.

    The ratio Δij/Δ is an element of the inverse of the matrix. Would you know how to derive the transfer function from the inverse of the nodal matrix?
     
  4. Sep 23, 2012 #3
    Since then, someone informed me that this method is using an "admittance matrix" to solve for the transfer function.
    I'm still not completely sure why the numbers of the cofactor are what they are, but I'm told it pertains to using a test current course, indicating that this may not be purely mathematical.
     
  5. Sep 23, 2012 #4

    The Electrician

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    It is purely mathematical, and it is a method using the admittance matrix. Do you want any further explanation, or is this a moot point for you now?
     
  6. Sep 23, 2012 #5
    I'm not sure how figure out the ij in ∆ij.
     
  7. Sep 23, 2012 #6

    The Electrician

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    The image from your post is apparently using the "i" suffix to represent the input node, and the "j" suffix is for the output node.
     
  8. Sep 23, 2012 #7
    That doesn't explain it to me, though.

    Why is V1 ∆11 and V3 ∆13 ?

    I don't see how this is just input and output.
     
  9. Sep 24, 2012 #8

    The Electrician

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    Assume you've got the admittance matrix (Y matrix) for the circuit, with 3 nodes. Node 1 on the left, node 2 in the middle and node 3 on the right. You'll have a 3x3 admittance matrix, which can be denoted in general terms as:

    [ y11 y12 y13 ]
    [ y21 y22 y23 ]
    [ y31 y32 y33 ]

    Next invert this matrix, obtaining an impedance matrix (a Z matrix):

    [z11 z12 z13 ]
    [z21 z22 z23 ]
    [z31 z32 z33 ]

    The elements of the Z matrix, which is the inverse of the Y matrix, can be found by various methods. The solution from your text found each element as a term of the form Δij/Δ, but any suitable method can be used.

    The elements of the Z matrix have these properties:

    1. The elements on the main diagonal are the driving point impedances of each node. If a current I is injected at a node k, the voltage appearing at that node will be I*zkk.

    2. The elements off the main diagonal are the transfer impedances. These impedances tell us the voltage appearing at a node resulting from a current injected at another node. For example, if we inject a current I at node 1, the voltage appearing at node 3 will be I*z31.

    The transfer impedance z31 gives us immediately the ratio of the voltage appearing at node 3 when a current I is injected at node 1; V3 = I*z31. But, we want the ratio of the voltage at node 3 to the voltage at node 1 (not the current at node 1).

    If we inject a current at node 1, the voltage appearing at that node will be I*z11, and the voltage appearing at node 3 will be I*z31. The ratio of the output voltage to the input voltage, V3/V1, will be (I*z31)/(I*z11) = z31/z11.

    This is what the text solution is doing.
     
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