Two port circuit

Discussion in 'Homework Help' started by naickej4, Jul 12, 2015.

  1. naickej4

    Thread Starter Member

    Jul 12, 2015
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    2
    Hello all,

    I hope all are well.

    Please can you help me with the below, I have a idea of doing it but I am so confused in determining the correct ANSWER. I am using the linear and nonlinear text book by Leon O. Chuao 1987, does anyone have the solution manual for this book?

    The question is for the two-port circuit determine the conductance matrix?

    thank you all.

    two port circuit.png
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    YOU need to show YOUR best attempt to solve YOUR homework problem. We can then see where you are going right and where you are going wrong and offer guidance accordingly.

    A good first step is to start with the equations that describe the conductance matrix parameters.
     
  3. naickej4

    Thread Starter Member

    Jul 12, 2015
    182
    2
    Hi Sir,

    thank you for the response. I did try something out if you can please find the attached solution. What confuses me is the dependent current source, how do we use it in the equations. I have googled to try and find similar examples but nothing helped.

    Do you know the text book Linear and Nonlinear Circuits by Leon O. Chauo 1987? my solution.png
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    The thing to focus on is the defining relationships:

    <br />
i_1 \; = \; G_{11}v_1 \; + \; G_{12} v_2<br />
i_2 \; = \; G_{21}v_1 \; + \; G_{22} v_2<br />

    If you FORCE v_2 to be zero, what do these relations reduce to?

    How do you FORCE v_2 to be zero?

    You now then simply have a circuit that needs to be analyzed that happens to have a dependent current source in it, which you should already be comfortable with.

    You then simply repeat this process by forcing v_1 to be zero.
     
  5. naickej4

    Thread Starter Member

    Jul 12, 2015
    182
    2
    Hi WBahn

    I did use the node 3 as the datum node.

    how will I incorporate the VCCS dependent current source in my equations? I know the constant value is for G is 2 and the current out will be equals to the GV which is 2V1 as per the circuit.

    But who do I get the VCCS current in my equation if it was in series I have an idea but since it is connect in parallel with the nodes I am confused.

    Please help me

    thank you
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    Please show your work? I cannot tell you what you are doing wrong unless I see what you are doing.
     
  7. naickej4

    Thread Starter Member

    Jul 12, 2015
    182
    2
    Good Evening WBahn

    Thank you for your response. I have done my homework and took a pic on my phone. I hope you able to open the image. The current in red is what I deduced and introduced into the circuit. Please help me to understand the VCCS which is in parallel to R3 and V2. I used the KCL to deduce the current will be I2 - 2V1 since the current source will be negative because of the direction of the current source is going the opposite direction of the node.
    my solution.jpg my solution.jpg
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Everything looks fine to me. The one thing that is out of kilter -- and the author set you up for this because of sloppiness on their part -- is that the coefficient on the current source is not '2', but rather '2 S' (or two siemens). If you multiply 2 by V1 you get a voltage, not a current. To get a current you need to either divide the voltage by a resistance or multiply it by a conductance. If you carry the units through properly, then in your G matrix the units will all work out.

    There are numerous checks you can do. For instance, if you open circuit V1, that forces I1 to be zero. The result is then

    <br />
i_1 \; = \; 0 \; = \; \( \frac{1}{R_1} \; + \; \frac{1}{R_2} \) V_1 \; - \; \( \frac{1}{R_2} \) v_2<br />
\;<br />
\( \frac{1}{R_1} \; + \; \frac{1}{R_2} \) V_1 \; = \; \( \frac{1}{R_2} \) v_2<br />
\;<br />
V_1 \; = \; \frac{\( \frac{1}{R_2} \)}{\( \frac{1}{R_1} \; + \; \frac{1}{R_2} \)} v_2<br />
\;<br />
V_1 \; = \; \frac{R_1}{R_1 \; + \; R_2 } v_2<br />

    which is just a voltage divider where v2 is applied across the series combination of R1 and R2 and the voltage v1 is taken across R1. A quick look at the schematic reveals that this is exactly the case that you have.

    You can (and should) do as many similar checks to confirm that you answer is correct.

    You can also get many of the coefficients (sometimes all of them) very easily by reducing the circuit to eliminate variables. For instance, we know that, by definition,

    <br />
i_1 \; = \; G_{11} v_1 \; + \; G_{12} v_2<br />

    If force v1 to be zero by shorting it, we have

    <br />
i_1 \; = G_{12} v_2<br />

    We know that the current in R1 will be zero and that the current flowing from right to left will be -i1. But this current is trivial to find because we have v2 applied to the right side of the resistor and ground applied to the left, so the current will be v_2/R2 (from right to left), making

    <br />
G_{12} \; = \; - \frac{1}{R_2}<br />

    Similarly, if we short v2 then all of the current from the dependent source will simply circulate through the short plus there will be no current in R_3 since Node 2 is force to be 0V. Hence the current i1 will be given by

    <br />
i_1 \; = \; G_{11} v_1 \; = \; \frac{v_1}{R_1 \; || \; R_2} \; = \; v_1 \( \frac{1}{R_1} \; + \; \frac{1}{R_2} \)<br />
\;<br />
G_{11} \; = \; \frac{1}{R_1} \; + \; \frac{1}{R_2}<br />
     
  9. naickej4

    Thread Starter Member

    Jul 12, 2015
    182
    2
    Hello Sir.

    THANK YOU so much it makes so much sense to me now. I really not sure why our lecturer is making us learn from this text book. It is even out of print.

    Thank you
    regards Joe
     
  10. naickej4

    Thread Starter Member

    Jul 12, 2015
    182
    2
    Hello Wbahn

    Can I convert the co-efficient of to 2 which you said was Simens (S) to 1/2 which will be in term of ohm? will this impact my solution?

    thank you once again.

    regards

    Joe
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    Yes, you can do it. No, it won't impact your solution. It will still be valid.

    <br />
2 \, S \; = \; \frac{1}{\frac{1}{2} \Omega}<br />
     
    Last edited: Jul 16, 2015
    naickej4 likes this.
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