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Circuit analysis: find Network Function

  1. Sep 11, 2008 #1
    Given the following circuit: (see attached file)
    find the Network Function between I_L, flowing as drawn in picture, AND I_g.

    I guess all the calculation should lead up to an expression like this one:
    I_L(s)=F(s) I_g(s), where F(s) is a function in "s" which corresponds to the NETWORK FUNCTION we're looking for.
    This means we have to work in Laplace's domain.
    I know I have to "translate" the circuit elements into Laplace's domain so that all the resistors, capacitors and inductors contain Laplace's complex variable, "s".
    R --> R
    C --> 1/(sC)
    L --> sL
    "Y" network --> 1) I_1= 2V_1 -V_2 ; 2) I_2= -V_1 +2V_2

    I_1, I_2 are the currents flowing into the network while V_1, V_2 are the Voltage drops between the lateral points and the central point/junction.

    Ideal transformer (remember n=2 ?) ---> 1) I_3 = 2 * I_4 ; 2) V_3 = -1/2 * V_4

    I_3, V_3 are the current and voltage of the vertical inductor linked by the arrow in the picture, to the horizontal one in the upper part of the picture, I_4 and V_4.

    Also, I_1 = I_g and should be substituted in the 1st equation of the the "Y" network.

    Now, how do I proceed?
     

    Attached Files:

  2. jcsd
  3. Sep 11, 2008 #2

    CEL

    User Avatar

    Write Kirchoff's equations for the circuit (mesh or node) and solve the resulting system of equations for [tex]I_L[/tex]
     
  4. Sep 12, 2008 #3
    I'll go with node method. I'm attaching a new pic with the nodes I'm planning to consider. Before proceeding, could you pls check it out and tell me if it is a wise choice?
    By the way there was a mistake in my previous post regarding the ideal transformer's equation: not I_3 but V_3 is equal to 2 *V_4 like this:

    [tex]
    \left\{
    \begin{array}{ll}
    V_3(s) &= 2V_4(s)\\
    I_3(s) & =-\frac{1}{2}I_4(s)
    \end{array}
    \right.
    [/tex]

    PS: Is my idea of turning to Laplace's domain (i.e. working with Laplace's complex variable "s") correct?
     

    Attached Files:

    Last edited: Sep 12, 2008
  5. Sep 12, 2008 #4

    CEL

    User Avatar

    The nodes you have labeled [tex]E_3[/tex] and [tex]E_4[/tex] are in reality the same node. You don't need E_4.
    Since your excitation is a step function, you are right in working in the Laplace domain.
     
  6. Sep 12, 2008 #5
    ok, let me try out...

    [tex]
    \left[
    \begin{array}{ c c c }
    (\frac{1}{R} + sC) & 0 & -(\frac{1}{R} + sC) \\
    0 & 0 & 0\\
    -(\frac{1}{R} + sC) & 0 & (\frac{1}{R} + sC +\frac{1}{sL})
    \end{array} \right]

    \left[
    \begin{array}{ c }
    E_1 \\
    E_2 \\
    E_3
    \end{array} \right]
    =

    \left[
    \begin{array}{ c }
    -I_3-I_4-I_g \\
    I_1 \\
    I_2-I_L+I_4
    \end{array} \right]

    [/tex]

    Obviously there's a mistake somewhere since the second line is all "0" and then equals to I1. This cannot be. Please check the update drawing.
    Since I3=-½ I4, the currents' matrix becomes
    [tex]
    \left[
    \begin{array}{ c }
    -\frac{1}{2}I_4-I_g \\
    I_1 \\
    I_2-I_L+I_4
    \end{array} \right]

    [/tex]
     

    Attached Files:

  7. Sep 12, 2008 #6
    Where am I going wrong?
     
  8. Sep 12, 2008 #7

    CEL

    User Avatar

    Your equations are wrong. The only unknowns are the [tex]E_i[/tex]. In the independent vector you can only have linear combinations of [tex]I_g[/tex].
    Remember that in node 1 you have [tex]I_3+I_4+I_g=0[/tex] and since [tex]I_3=-\frac{I_4}{2}[/tex] you have [tex]I_3[/tex] and [tex]I_4[/tex] as functions of [tex]I_g[/tex].
    You have an equivalent relationship in node 3.
    I would like to see how you have developed your equations.
     
  9. Sep 12, 2008 #8

    CEL

    User Avatar

    Since your unknowns are the voltages, you should use [tex][Z=Y^{-1}[/tex], in order to have E2 and E3 as functions of the currents.
     
  10. Sep 13, 2008 #9
    [tex]
    \left[
    \begin{array}{ c c c }
    (\frac{1}{R} + sC) & 0 & -(\frac{1}{R} + sC) \\
    0 & 0 & 0\\
    -(\frac{1}{R} + sC) & 0 & (\frac{1}{R} + sC +\frac{1}{sL})
    \end{array} \right]

    \left[
    \begin{array}{ c }
    E_1 \\
    E_2 \\
    E_3
    \end{array} \right]
    =

    \left[
    \begin{array}{ c }
    -\frac{1}{2}I_4-I_g \\
    I_1 \\
    I_2-I_L+I_4
    \end{array} \right]
    [/tex]



    following your suggestion about node 1:

    [tex]

    \left \{
    \begin{array}{rl}
    I_3 & = -\frac{1}{2}I_4 \\
    I_3 + I_4 +I_g & = 0
    \end{array}
    \right.

    \left \{
    \begin{array}{cc}
    I_3 & = -\frac{1}{2}I_4 \\
    -\frac{1}{2}I_4 + I_4 +I_g & = 0
    \end{array}
    \right.


    \left \{
    \begin{array}{cc}
    I_3 & = -\frac{1}{2}I_4 \\
    \frac{1}{2}I_4 +I_g & = 0
    \end{array}
    \right.

    \left \{
    \begin{array}{cc}
    I_3 & = -\frac{1}{2}I_4 \\
    I_4 & = -2I_g
    \end{array}
    \right.
    [/tex]

    take the 2nd equation and substitute in the currents matrix above:

    [tex]
    \left[
    \begin{array}{ c c c }
    (\frac{1}{R} + sC) & 0 & -(\frac{1}{R} + sC) \\
    0 & 0 & 0\\
    -(\frac{1}{R} + sC) & 0 & (\frac{1}{R} + sC +\frac{1}{sL})
    \end{array} \right]

    \left[
    \begin{array}{ c }
    E_1 \\
    E_2 \\
    E_3
    \end{array} \right]
    =

    \left[
    \begin{array}{ c }
    -\frac{1}{2}(-2I_g)-I_g \\
    I_1 \\
    I_2-I_L+(-2I_g)
    \end{array} \right]
    =
    \left[
    \begin{array}{ c }
    0 \\
    I_1 \\
    I_2-I_L-2I_g
    \end{array} \right]
    [/tex]

    Before going on, is the third place of the currents' matrix correct?

    [tex]
    I_{i,j}=I_{3,1}=I_2-I_L-2I_g
    [/tex]

    Which are the currents entering/going out of node 3? Is it as I wrote above?

    Now is the turn of I2. From the Y matrix we get:

    [tex]
    \left \{
    \begin{array}{cc}
    I_g & = 2V_1 - V_2 \\
    I_2 & = -V_1 +2V_2
    \end{array}
    \right.

    \left \{
    \begin{array}{cc}
    I_g & = 2E_2-E_3 \\
    I_2 & = -E_2 +2E_3
    \end{array}
    \right.

    \left \{
    \begin{array}{clc}
    E_2 & = \frac{I_g}{2} +\frac{E_3}{2} &\\
    I_2 & = -\frac{I_g+E_3}{2}+2E_3 &= -\frac{1}{2}I_g+\frac{3}{2}E_3
    \end{array}
    \right.
    [/tex]

    I then take I2 and substitute it in the current's matix.
    By the way, isn't it E3= V of the inductance L?
     
  11. Sep 13, 2008 #10
    I'm not sure whether there are any initial conditions on the capacitor and the inductance.
    After all Ig=1 for t ≤ 0. This certainly leads to some initial condition:

    L has [tex]i(0^{-})\neq 0[/tex] and

    C has [tex]v(0^{-})\neq 0[/tex]
     
  12. Sep 13, 2008 #11

    CEL

    User Avatar

    In your second equation you have [tex]I_1=0E_1+0E_2+0E_3[/tex]. Where is I1 in your circuit?
    E3 is the voltage across the inductor.
     
  13. Sep 13, 2008 #12

    CEL

    User Avatar

    You are right. The fact that the current is nonzero for t<0 imposes initial conditions to the circuit.
     
  14. Sep 13, 2008 #13
    As I said in a previous post...


    What to do now? How to find their values? Help, I'm stuck!
     
  15. Sep 13, 2008 #14

    CEL

    User Avatar

    There is no I1 in your circuit. No wonder it equates to zero.
    Your three unknowns are E1, E2 and E3. You must write them as functions of the currents. As I mentioned previously, change the quadrupole matrix from Y to Z, in order to write E2 and E3 as functions of Ig and I2. Use the equations a node 1 to write I3 and I4 as functions of Ig. Use the branch equation of the inductor to write IL as a function of E3. Since I2=I4-IL, you can write I2 as a function of Ig and E3.
     
  16. Sep 13, 2008 #15
    Ok. Z is the inverse matrix of Y:

    [tex]

    Y= \left[ \begin{array}{cr}
    2 & -1 \\
    -1 & 2
    \end{array} \right]

    \rightarrow

    Z=\left[ \begin{array}{cr}
    \frac{2}{3} & \frac{1}{3} \\
    \frac{1}{3} & \frac{2}{3}
    \end{array} \right]
    [/tex]

    So this would be:
    [tex]
    [Z]=[E][/tex]


    [tex]
    \left[ \begin{array}{cr}
    \frac{2}{3} & \frac{1}{3} \\
    \frac{1}{3} & \frac{2}{3}
    \end{array} \right]

    \left[ \begin{array}{r}
    I_g \\
    I_2
    \end{array} \right]

    =
    \left[ \begin{array}{c}
    E_2\\
    E_3
    \end{array} \right]

    \rightarrow

    \left \{ \begin{array}{lr}
    \frac{2}{3}I_g+\frac{1}{3}I_2 &=E_2 \\
    \frac{1}{3}I_g+\frac{2}{3}I_2 &=E_3
    \end{array} \right.

    [/tex]

    This is the branch equation of the inductor:

    [tex]
    I_L=\frac{V_L(s)}{sL}+ \frac{i(0^-)}{s}= \frac{E_3}{2s}+\frac{i(0^-)}{s}
    [/tex]

    However I don't know the value of [tex]\frac{i(0^-)}{s}[/tex]
    Is it right so far?
    I don't quiet understand why you say I2=I4-IL
    How's the node 1 currents' direction? I mean which is going in(+) and which is coming out (-)? Could you write it for me please?
     
  17. Sep 13, 2008 #16

    CEL

    User Avatar



    To obtain IL(0-) substitute the capacitor by an open circuit and the inductor and the coils of the transformer by short circuits and make Ig = 1A.
    In node 3 you have I4 in and I2 and IL out, so I4 = I2 + IL.
    By your drawing, in node 1 you have Ig, I3 and I4 are all going out.
    You must understand that the directions are arbitrary. I am using the ones you chose. If the directions you chose are the right ones, you will have positive values. If not, the values will be negative.
     
  18. Sep 14, 2008 #17
    Please see the pic I've attached. Did I get your suggestion right?
    I thought that since the current is not flowing in the upper part of the circuit because of the open circuit represented by the capacitor, I could just eliminate it with a big red "X". This means that for t>0 the initial condition of the capacitor is i(0-)=0. Right? Or is it v(0-)=0?

    I4 -I2 - IL=0 (now I got it, thanks for explaining)
    0 -I2 = IL because of what I said above, I4=0 so finally IL=-I2 and that is like saying

    [tex]i_L(0^-)=I_L=-I_2[/tex]

    The point is, how do I get IL in numbers? Write the set of equations I wrote previously?

    [tex]
    \left \{ \begin{array}{lr}
    \frac{2}{3}I_g+\frac{1}{3}I_2 &=E_2 \\
    \frac{1}{3}I_g+\frac{2}{3}I_2 &=E_3
    \end{array} \right.
    [/tex]

    I_g = 1 for t<=0, but I don't know E_2 and E_3 and I need them to find I2 . How do I get them?

    What to do next? Do I have to find E1, E2, E3? E3=V_L=?
    I'm getting confused and frustred w/ this problem. Help me please :(
    I don't have much time.
     

    Attached Files:

  19. Sep 14, 2008 #18

    CEL

    User Avatar

    Initial conditions for capacitors are always voltages and for inductors, currents. The fact that the current in the capacitor is zero tells nothing about the voltage. Since there is no current, there is no voltage drop across the resistor, so Vc(0-) = E1(0-) - E3(0-).
    Since the nodes 1 and 3 are connected to ground by short circuits, E1 = E3 = 0.
    You have 2 equations relating Ig, I2, E2 and E3. Since you know Ig and E3, you have two equations with two unknowns.
     
  20. Sep 14, 2008 #19
    [tex]

    \left \{ \begin{array}{ll}
    \frac{2}{3}I_g+\frac{1}{3}I_2 &=E_2 \\
    \frac{1}{3}I_g+\frac{2}{3}I_2 &=E_3
    \end{array} \right.

    \left \{ \begin{array}{ll}
    \frac{2}{3}+\frac{1}{3}I_2 &=E_2 \\
    \frac{1}{3}+\frac{2}{3}I_2 &=0
    \end{array} \right.

    \left \{ \begin{array}{ll}
    E_2 &= \frac{2}{3}-\frac{1}{6}=\frac{5}{6} \\
    I_2 &=-\frac{1}{2}
    \end{array} \right.

    [/tex]

    Is that right? Now it should be

    [tex]
    i_L(0^-)= -I_2= \frac{1}{2}
    [/tex]

    Later on, for t>0 the branch current for the inductor L is

    [tex]
    I_L=\frac{E_3}{2s}+\frac{i_L(0^-)}{s}=\frac{E_3}{2s}+\frac{1}{2s}
    [/tex]

    Now what should I do? I wonder if all this trouble is really necessary, I mean do I really need to know E1, E2, E3 in order to know the NETWORK FUNCTION? Isn't there any other way to get the network function?
     
  21. Sep 14, 2008 #20

    CEL

    User Avatar

    You must write E3 as a function of Ig for t > 0 and replace it in the equation for IL. Then you calculate the network function.
     
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