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This is a homework question.

I know that: Given a transfer function of \$H(s)\$ below, we can realize it with an OP-AMP as follows.

\$H(s)=-\dfrac{2}{s+2}=-\dfrac{\dfrac{1}{2}}{\dfrac{s}{4}+\dfrac{1}{2}}=-\dfrac{Z_f}{Zi}=-\dfrac{\dfrac{R_f}{R_f*s*C_f+1}}{Rin}\$

where \$R_{in}=R_f=\dfrac{1}{2}\Omega\$ and \$C_f=1\text{F}\$

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However, now that I have to realize a transfer function with complex numbers, I am puzzled on how to do so. Could you lead me to the correct direction on realizing the following transfer function using OP-AMP(s)?

\$H(s)=\dfrac{1}{s + 0.383 + j*0.924}\$

Above equation is a part of:

\$H(s)=\dfrac{1}{s^2 + 0.765*s + 1}=\dfrac{1}{s + 0.383 + j*0.924}*\dfrac{1}{s + 0.383 - j*0.924}\$

Note: In the big picture, I have to realize a HPF of 4th order using cascaded(serial) decomposition method. Normalized transfer function of the filter is given as:

\$H(s)=\dfrac{s^4}{s^4 + 2.613*s^3 + 3.414*s^2 + 2.613*s + 1}\$

This can be written as:

\$H(s)=\dfrac{s^2}{s^2 + 0.765*s + 1}*\dfrac{s^2}{s^2 + 1.848*s + 1}\$

\$=\frac{s}{s + 0.383 + j*0.924}*\frac{s}{s + 0.383 - j*0.924}*\frac{s}{s + 0.924 + i*0.383}*\frac{s}{s + 0.924 - i*0.383}\$

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Here is the vital part of your question: -

enter image description here

This tells me that you'll need two cascaded 2nd order high-pass-filters. Cascading the filters is the same as the multiply in the middle of the bottom equation.

Here's a sallen-key high-pass filter (remember you'll need two cascaded): -

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The transfer function for it is: -

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Now you need to convert your individual 2nd order equations into a form that suits the sallen-key formulas. From experience (and with a little help from google and wiki) your formulas are of the form: -

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And this means that for the left-hand part of your equation, 0.765 = Wo/Q AND 1 = (Wo)^2

By my reckoning, this means Q = 1/0.765 and Wo = 1. Equate these values to the sallen-key formulas for Wo and Q to get the resistor and capacitor values for the left hand stage. Then repeat for the right hand stage of your formula. This isn't as easy as it sounds and a little trial and error will be needed. Assume both capacitors are the same value and that R1 is half of R2 - try and get values that match Q and Wo - if Q is too low make R1 a bit smaller and repeat/iterate.

Alternatively, use a website where you can enter F (Wo/2Pi) and Q. Here is one that looks suitable. It gave the following result for the first part of your transfer function: -

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Note that there is a little tiny discrepency in the numbers due to the suggested website using standard resistor and capacitor sizes. Maybe you can find one that doesn't default to using standard values.

Then it is just a simple matter of cascading the output from the left-hand sallen-key filter into the input of the right hand sallen-key filter and you have your answer.

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  • \$\begingroup\$ Thank you for your answer. I have to use a Butterworth type filter, sorry I have not mentioned this. Your suggested method can be easily applied to Butterworth topology, I am aware of that. But I have to move to the circuit from the equation, as I showed an example in my question. I am stuck on realizing transfer functions with complex numbers using opamps. \$\endgroup\$ – abdullah kahraman May 24 '13 at 21:22
  • \$\begingroup\$ I am sorry I have just realized that sallen-key topology can use Butterworth transfer function. \$\endgroup\$ – abdullah kahraman May 24 '13 at 21:28
  • \$\begingroup\$ I know that this is my third comment, please forgive my amateurism. Your answer addresses my problem spot on. I will write an answer on how I will accomplish this, tomorrow. \$\endgroup\$ – abdullah kahraman May 24 '13 at 21:43
  • \$\begingroup\$ @abdullahkahraman no probs about the comments. That's what they are there for. One point - A butterworth function will only fit if the original transfer function has the correct values of Q in both stages and this totally depends on the coefficients - the original transfer function may be butterworth by design but, if it isn't then the answer will not be a butterworth function - it is what it is and butterworth is just a particular set of Qs in each section that gives the butterworth maximally flat frequency response (much over-egged in my book). \$\endgroup\$ – Andy aka May 24 '13 at 22:27
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Instead of looking at the problem as \$H(s)=\dfrac{1}{s + 0.383 + j*0.924}\$, take it as:

\$H(s)=\dfrac{s^2}{s^2 + 0.765*s + 1}\$ . Realizing this is easy, and was already done by Sallen and Key. Consider the following circuit and derivation of its transfer function:

enter image description here

By Kirchhoff's current law (KCL) applied at the \$Vx\$ node,

\$\dfrac{Vin-Vx}{Z_1}=\dfrac{Vx-Vout}{Z_3}+\dfrac{Vx-V^+}{Z_2}\$ (Eq.1),where \$V^+=V^-=Vout\$ because of the negative feedback.

Also, by applying KCL on OP-AMPs non-inverting input,

\$\dfrac{Vx-Vout}{Z_2}=\dfrac{Vout}{Z_4}\$ ,then: \$Vx=Vout*(\dfrac{Z_2}{Z_4}+1)\$

If we re-write all these in (Eq.1), then, we get:

\$\dfrac{Vin-Vout*(\dfrac{Z_2}{Z_4}+1)}{Z_1}=\dfrac{Vout*(\dfrac{Z_2}{Z_4}+1)-Vout}{Z_3}+\dfrac{Vout*(\dfrac{Z_2}{Z_4}+1)-Vout}{Z_2}\$

If we rearrange this equation, we will end up with:

\$\dfrac{Vout}{Vin}=\dfrac{Z_3*Z_4}{Z_1*Z_2+Z_3*(Z_1+Z_2)+Z_3*Z_4}\$

If \$Z_1\$ and \$Z_2\$ are selected as capacitors and others as resistors, we will have:

\$\dfrac{Vout}{Vin}=\dfrac{R_3*R_4}{\dfrac{1}{s^2*C_1*C_2}+\dfrac{R_3*(C_1+C_2)}{s*C_1*C_2}+R_3*R_4}\$

Rearranging this will give us a more meaningful equation:

\$\dfrac{Vout}{Vin}=\dfrac{s^2*R_3*R_4*C_1*C_2}{s^2*R_3*R_4*C_1*C_2+s*R_3*(C_1+C_2)+1}\$

Dividing numerator and denominator with \$R_3*R_4*C_1*C_2\$:

\$\dfrac{Vout}{Vin}=\dfrac{s^2}{s^2+s* \dfrac{R_3*(C_1+C_2)}{R_3*R_4*C_1*C_2}+\dfrac{1}{R_3*R_4*C_1*C_2}}\$

Simplifying this to:

\$\dfrac{Vout}{Vin}=\dfrac{s^2}{s^2+s*\dfrac{C_1+C_2}{R_4*C_1*C_2}+\dfrac{1}{R_3*R_4*C_1*C_2}}=\dfrac{s^2}{s^2 + 0.765*s + 1}\$

If we select \$C_1=C_2=1\text{F}\$, then \$R_4=2.6144\Omega\$ and \$R_3=0.3825\Omega\$.

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