Sallen Key Filter Nodal Analysis

Discussion in 'Homework Help' started by jegues, Jan 9, 2012.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    I'm confused as to how they defined the vector of unknowns for this problem.

    I would have added another unknown in the form of an auxiliary variable describing the current coming from the input voltage source Vi.

    From here I could then write KCL statements for all 3 nodes.

    If you didn't include it, I don't see how you could write KCL statement for V1, specifically the current flowing across R1. (Assuming you are only using the unknowns V1, V2, Vo)

    Am I missing something?
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    How about showing us your efforts to solve the problem so far?

    hgmjr
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    It should be fairly obvious if you simply write

    S=\left[ \begin{array}{c} \vec{V_i} \\ 0 \\ 0 \end{array} \right]
     
  4. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    Here's my attempt so far, hopefully you guys can tell me whether the equations I've written are correct or not.

    I am then asked to solve for the transfer function,

    \frac{\vec{V_{o}(\omega)}}{\vec{V_{i}(/omega)}}

    Is this suppose to be simplified from the my results found in the nodal analysis?
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I agree with your first equation (1) at V1 node.

    It may not matter, but I would have expressed things differently for V2 vector.

    One case for V2 at the -ve terminal would be

    \vec{V_2}=\left ( \frac{R_3}{R_3+R_4}\right)\vec{V_o}

    hence

    0=-\vec{V_2}+\left (\frac{R_3}{R_3+R_4}\right ) \vec{V_o} \ ... \ equation(2)


    The other relates V1 & V2 at the +ve terminal & would be

    \frac{\vec{V_1}-\vec{V_2}}{R_2}=j\omega C_1 \vec{V_2}

    or after re-arranging

    \vec{V_1}=(1+j\omega R_2 C_1)\vec{V_2}

    and then

    0=-\vec{V_1}+(1+j\omega R_2 C_1)\vec{V_2}  \ ... \ equation(3)
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    To find Vo(ω)/Vi(ω) one would 'simply' invert the A matrix and express the output vector array V in terms of the input array S.
     
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  7. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    I don't understand the portion I had put in bold, can you clarify further what you mean by this?
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Since we have

    [A][V]=[S]

    \left[ \mathbf{A} \right] \left [ \begin {array}{c} \vec{V_1} \\ \vec{V_2} \\ \vec{V_o}\end{array} \right ] = \left [ \begin {array}{c} \vec{V_i} \\ 0 \\ 0\end{array} \right ]
    then

     \left [ \begin {array}{c} \vec{V_1} \\ \vec{V_2} \\ \vec{V_o}\end{array} \right ] = {\left[ \mathbf{A} \right]}^{-1} \left [ \begin {array}{c} \vec{V_i} \\ 0 \\ 0\end{array} \right ]

    Expanding the RHS matrix equation and noting the 3rd value in the V array will give Vo for Vi at freq ω.
    It's easier to invert A using real values with a matrix handling application like Matlab but I guess you can do it by hand in symbolic form.

    If I have to find the transfer function in symbolic form I'd prefer to do so by straight algebraic manipulation of the symbolic nodal equations - rather than resorting to the vector form and then having to find the inverse A matrix in symbolic form.
     
  9. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Your A matrix does not give the correct result.

    Note that Vo and V2 are simply related by the gain of the opamp (as set by R3 and R4), so the third equation should just express this relationship.

    I don't think your second equation is quite correct, either.
     
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