Zeros, Poles and Bode Plots

Discussion in 'Homework Help' started by jegues, Dec 2, 2010.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    See figures attached for problem statement as well as my work.

    I managed to derive the transfer function without any troubles, however when it comes to part b) I'm not sure what route to take to solve it.

    Since,

    z_{1} = \frac{R}{1 + jRwc_{1}}

    and

    z_{2} = \frac{1}{jwc_{2}}

    My transfer function,

    H(w) = \frac{z_{1}}{z_{2} - z_{1}}

    turns into a big algebra mess. This algebra mess makes finding the values of C1 and C2 to create a pole at 200 rad.Hz difficult.

    I think if I could get a "clean" expression for the transfer function H(w) I would be able to solve the problem, but the algebra is so messy.

    Does anyone have any suggestions? Or perhaps tips to make the algebra a little bit easier?

    Thanks again!

    Edit: I decided to man up an try to tackle the algebra and here's what I came up see. (See 2nd and 3rd pages of my work attached)

    The 2nd page of my work is just finding z_{2} - z_{1}.

    The 3rd page shows me rewriting the transfer function and extracting the pole out of the denominator and solving for the difference between c_{1},c_{2}.

    Just a note when replying, WE DID NOT LEARN LAPLACE TRANSFORMS! When given a quadratic like in this example, we simply focused on the term with the highest power of w and solved it accordingly. We don't even consider if our the roots of our quadratic equation are real or complex.

    Sorry if that last part is confusing but that's how we learnt it, and it was very vaguely explained.
     
    Last edited: Dec 2, 2010
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You've probably overcomplicated matters.

    The first stage will have a gain G1=-Zf1/R where Zf1=R/(jωRC1+1)

    So G1=-1/(jωRC1+1)

    The second stage will have a gain G2=-R/Zi2 where Zi2=1/(jωC2)

    So G2=-jωRC2

    Call stage 1 output Vx

    Now noting that the R values are all the same

    Vx=G1*(Vo+Vi)=-{1/(jωRC1+1)}*(Vo+Vi)

    But Vo=G2*Vx=-jωRC2*Vx

    Hence Vx=-Vo/(jωRC2)

    Equate the two Vx expressions to eliminate Vx

    -Vo/(jωRC2)=-{1/(jωRC1+1)}*(Vo+Vi)

    giving

    Vo/(jωRC2)=Vo/(jωRC1+1)+Vi/(jωRC1+1)

    Collect the terms in Vo and you can find Vo/Vi

    I have |C1-C2|=1/(ω0*R)=1/20,000=50uF
     
    Last edited: Dec 3, 2010
  3. Georacer

    Moderator

    Nov 25, 2009
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    t_n_k is right. Just substitute z1 and z2 in the expression H=\frac{z1}{z2-z1} and try to bring it in the form H=\frac{Kj\omega}{j\omega+\omega_0},which is a High Pass filter by the way. \omega_0 is the circular frequency of the pole and it will contain the difference (C1-C2). Equate \frac{\omega_0}{2\pi}=200 and find the requested difference.
     
  4. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    I was told I couldn't do that by my TA because of the large resistor at the top wired to both op amps... He said because that resistor pulls some current across it, it won't be the simple inverting op amp, i.e.

    \frac{V_{*}}{V_{i}} = -\frac{z_{2}}{z_{1}}

    and that I must write KCL's at each node.

    Which one is correct I am confused now.

    Was my TA incorrect?

    Also, why are you multiplying both sides of G1 by (Vo+Vi)? That part I dont understand at all
     
    Last edited: Dec 3, 2010
  5. Georacer

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    Nov 25, 2009
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    I had another shot at it. Take a look.

    Proofreading is mandatory.
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Georacer, I think you got the sign of the transfer function wrong in post #5; you had it right in post #3, and jegues has it right.

    jegues, if you use the nodal method (which, of course, is just the systematic application of KCL) and let your calculator (or PC based math software) take care of the algebra, you can avoid most algebra mistakes.

    Number the nodes like this: V1 is the minus input of the first opamp, V2 is its output and V3 is the minus input of the second opamp.

    If you use Z1 and Z2 as in post #1, you will have:

    \left[ \begin{array}{4}\frac{2}{R}+ \frac{1}{Z1}&\frac{-1}{Z1} & 0 &\frac{-1}{R}\\1&0&0&0\\\0&\frac{-1}{Z2}&\frac{1}{R}+\frac{1}{Z2}&\frac{-1}{R}<br />
\\0&0&1&0\end{array}\right]*\left[ \begin{array}{4}V1\\V2\\V3\\Vo\end{array}\right]=\left[ \begin{array}{4}\frac{Vi}{R}\\ 0\\0\\0\end{array}\right]

    The result is:

    \left[ \begin{array}{4}V1\\V2\\V3\\Vo\end{array}\right]=\left[\begin{array}{4}0\\-\frac{Z2*Z1}{R(Z2-Z1)}Vi\\0\\\frac{Z1}{Z2-Z1}Vi\end{array}\right]

    You can do the whole thing all at once without using Z1 and Z2 to represent subcircuits:

    \left[ \begin{array}{4}\frac{3}{R}+ j\omega C1&-(\frac{1}{R}+ j\omega C1) & 0 &\frac{-1}{R}\\1&0&0&0\\\ 0&-j\omega C2&\frac{1}{R}+ j\omega C2&\frac{-1}{R}<br />
\\0&0&1&0\end{array}\right]*\left[ \begin{array}{4}V1\\V2\\V3\\Vo\end{array}\right]=\left[ \begin{array}{4}\frac{Vi}{R}\\ 0\\0\\0\end{array}\right]

    The result is:

    \left[ \begin{array}{4}V1\\V2\\V3\\Vo\end{array}\right]=\left[\begin{array}{4}0\\\frac{-1}{1+j\omega R(C1-C2)}Vi\\0\\\frac{j\omega RC2}{1+j\omega R(C1-C2)}Vi\end{array}\right]

    From there it's fairly easy to answer part b).
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Jegues,

    I have exactly the same expression for Vo/Vi as Electrician's - using the approach I suggested.

    I believe your TA is incorrect - so long as the op-amps are ideal it is a valid approach.
     
  8. t_n_k

    AAC Fanatic!

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    My solution is attached if you want to use it as a discussion point with your TA.
     
  9. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    Thanks everyone for all your help. I was able to redo my KCL equations, rework my algebra and obtain the same result.

    Now it asks me to plot the magnitude Bode approximation of the transfer function.

    I was able to plot the graph just fine, my only problem being how to solve for the dB when the graph levels off. (See figure attached)

    I picked a frequency where I know for sure the graph as leveled off, say w = 300.

    So, 20log|H(300)| = x\text{ dB}

    I found,

    |H(w)| = \frac{wc_{2}R}{\sqrt{1+R^{2}(c_{1}-c_{2})^{2}w^{2}}}

    So, |H(300)| = 0.83

    Then, x = -1.59dB

    Is this correct?
     
  10. t_n_k

    AAC Fanatic!

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    If you think about it the graph levels out when

    \omega^2 R^2 (C_1-C_2)^2 >> 1

    At which point the gain approaches

    H(\omega)=\frac{C_2}{(C_1-C_2)}

    So you'd need to know the value of C2 as well as the difference (C1-C2) to find the actual gain & hence the value of X[dB] you seek.
     
  11. Georacer

    Moderator

    Nov 25, 2009
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    I noticed I had a different answer than the previous posts but I ran through the calculations and couldn't find anything.

    Guess I was too tired or distracted to locate the error...
     
  12. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    So then I probably wouldn't have to indicate it if that's the case. Just the general form of the plot would need to be shown.

    I have that part correct, right?
     
  13. t_n_k

    AAC Fanatic!

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    I guess so. Maybe exercise some initiative in the matter. Believe in what you are producing. It's like being Neo in The Matrix. Teachers presumably want us to think independently, synthesize ideas etc.... Perhaps they expect too much (or too little).

    Sorry about the rant.

    You could make an appropriate comment as to why you haven't written a specific value. Or perhaps state the general form.

    X[dB]=20log([C2/(C1-C2)])
     
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