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Inverse Engineer a RLC circuit from a transferfunction
- Thread starter Fredje
- Start date
Since you cannot achieve DC gain with a passive circuit, an actual gain stage is required. A gain of 100 on the output will duplicate the transfer function. The rest of the derivation is substantially correct out to frequencies that are above the operating range of normally available opamps; and where inductors and capacitors stop behaving as lumped components.
Hi,
I am not sure what you are showing there and if it complies with your original problem specification. I assume you have to have those variables independent but i could be wrong.
Anyway, if this helps, since the convolutions is so simple with the buffer i went ahead and calculated the transfer function in two separate stages:
Hs1: the first stage: (s*C1*R1*R2+R2)/(s*C1*R1*R2+R2+R1)
Hs2: the last stage: 1/(s*C2*R3+s^2*C2*L+1)
and the convolution is then just Hs1*Hs2.
Now since Hs2 denominator is in a convenient form for this problem, if you divide top and bottom by C2 then we would see that the variable 'C' in the problem spec would be simply 1/C2, and so the variable C is easily controlled by C2 alone. However, that means 1/C2 is also in the numerator so you would have to find out if you can devolve the numerator of Hs1 into a unique set of variables as well as the denominator to fit the problem spec.
I think you might be able to but it will mean that 1/C2 will be included in at least one of the other variables. I am not sure if that is acceptable, but to get anything where each variable has to be independent of any other may not be possible with any circuit.
I am showing these functions because they are easy to talk about because the three required variables in the denominator of Hs2 are easy to determine by simple inspection.
See what you think about this.
BTW in a numerical simulation like Papa did, it is a good idea to do that with some random values to see if the response looks right. You can then try some other values too if you like and that should work too with reasonable real world values of course.
Referring back to post #17, if you rearrange the terms of the transfer function to be B/A and C/A you can determine the natural frequency and Q of the 2nd order system. You have:
\( \cfrac{C}{A}\;=\; {\omega_{0}}^2 \)
and
\( \cfrac{B}{A}\;=\;\cfrac{\omega_0}{Q} \)
Now we can set the corner and the Q of the filter for the desired characteristics. If we want a Butterworth responses at 1200 Hz., we know how to do that.
\( \cfrac{C}{A}\;=\; {\omega_{0}}^2 \)
and
\( \cfrac{B}{A}\;=\;\cfrac{\omega_0}{Q} \)
Now we can set the corner and the Q of the filter for the desired characteristics. If we want a Butterworth responses at 1200 Hz., we know how to do that.
Hi,
Where are you getting C3 from?
If we start out with the two sections separately we have:
Hs1=(s*C1*R1*R2+R2)/(s*C1*R1*R2+R2+R1)
Hs2=1/(s*C2*R3+s^2*C2*L+1)
Now note that the second part has to match 1/(A*s^2+B*s+C), and since they are so much of the same form we dont even have to divide anything because the coefficients A,B,C are already present in that Hs2 part, so it is obvious now that
A=C2*L
B=C2*R3
C=1
And note that C has to be the constant equal to 1. This means that C can not be arbitrary at least not in that circuit.
Now for the part Hs1, we have to convert that into the form:
(s+a)/(s+b)
in order to make this work.
Now note that
(s*C1*R1*R2+R2)/(s*C1*R1*R2+R2+R1)
would end up in the correct form if we just divided the numerator and denominator by C1*R1*R2.
Ok, so now take it from there see if you can get the first part into the correct form.
The only remaining question now is can we find a circuit such that C can also be arbitrary. With the present circuit it looks like if we try to make C an arbitrary parameter it messes up the first part Hs1 so it can not be factored into the correct form.
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