Saturday, 19 May 2018

operational amplifier - How to realize this transfer function with OP-AMPs?


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}\$


enter image description here


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}\$



Answer



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): -



enter image description here


The transfer function for it is: -


enter image description here


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: -


enter image description here


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: -


enter image description here


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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