# convert transfer function into circuit

#### ahmedOnsy

Joined May 14, 2020
4
now i have this equation
g(s)=4/(s^2+3*s+4)
and need to Design the equivalent electric circuit (using op-amps) describing open loop transfer function of this process G(s).
assuming ideal op-amp

#### Papabravo

Joined Feb 24, 2006
19,574
now i have this equation
g(s)=4/(s^2+3*s+4)
and need to Design the equivalent electric circuit (using op-amps) describing open loop transfer function of this process G(s).
assuming ideal op-amp
You have already asked this question in Homework Help. Multiple posts are considered bad form, and you still need to show us your work because we don't do homework for you. You need to show us some effort towards a solution.

#### MisterBill2

Joined Jan 23, 2018
13,746
I will not do homework for you, either. But I will offer an explanation of how to do this process: First, factor that polynomial term into a product of first order terms, then create the circuit with the response of each of those terms, snf then put the circuits in series. If you are not able to do that then you should have paid better attention to the lecture during class.

#### Papabravo

Joined Feb 24, 2006
19,574
I will not do homework for you, either. But I will offer an explanation of how to do this process: First, factor that polynomial term into a product of first order terms, then create the circuit with the response of each of those terms, snf then put the circuits in series. If you are not able to do that then you should have paid better attention to the lecture during class.
I'm not sure that is a viable strategy. 1st order realizations have a pole on the real axis which this transfer function does not.

#### MisterBill2

Joined Jan 23, 2018
13,746
I'm not sure that is a viable strategy. 1st order realizations have a pole on the real axis which this transfer function does not.
OK, then second order terms, with a resistor and either an inductor or a capacitor. We used that method back in my feedback systems class in 1972. I may have forgotten some of the details since then.

#### ahmedOnsy

Joined May 14, 2020
4
I will not do homework for you, either. But I will offer an explanation of how to do this process: First, factor that polynomial term into a product of first order terms, then create the circuit with the response of each of those terms, snf then put the circuits in series. If you are not able to do that then you should have paid better attention to the lecture during class.
now because of covid-19 , the all lesson being self study and i can't understand this part.

#### Papabravo

Joined Feb 24, 2006
19,574
OK, then second order terms, with a resistor and either an inductor or a capacitor. We used that method back in my feedback systems class in 1972. I may have forgotten some of the details since then.
Quadratic equations with real coefficients can have roots that are
1. Complex conjugate pairs
2. Real and distinct
3. Real and identical
In case 1, I don't think there is a distinct realization of one member of a complex conjugate pair. In the other thread Mr. Al sketched out the method for a Tow Thomas realization. That is actually a pretty slick piece of work. It is the way we used analog computers with the 50 lb patchboards (ca. 1968-69)

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#### MrAl

Joined Jun 17, 2014
9,633
now i have this equation
g(s)=4/(s^2+3*s+4)
and need to Design the equivalent electric circuit (using op-amps) describing open loop transfer function of this process G(s).
assuming ideal op-amp
Hello AGAIN

So you are saying that you have absolutely no idea whatsoever how to do this?
What have you been taught so far...anything related to this at least?

#### Papabravo

Joined Feb 24, 2006
19,574
Hello AGAIN

So you are saying that you have absolutely no idea whatsoever how to do this?
What have you been taught so far...anything related to this at least?
I'll be making book on the non-return of the TS. We didn't conform to his world view.
Since getting face time with an instructor could be problematical, maybe we could be bit bit more Socratic, and answer questions with one or more. Or as Yoda might say: "Question relentlessly we should".

#### MisterBill2

Joined Jan 23, 2018
13,746
I'll be making book on the non-return of the TS. We didn't conform to his world view.
Since getting face time with an instructor could be problematical, maybe we could be bit bit more Socratic, and answer questions with one or more. Or as Yoda might say: "Question relentlessly we should".
I defined a process, and without anybody factoring the expressson to see if it would work for that one, the attack hit hard and heavy. BUT not one person tried the evaluation to see if it would work.

#### Papabravo

Joined Feb 24, 2006
19,574
I defined a process, and without anybody factoring the expressson to see if it would work for that one, the attack hit hard and heavy. BUT not one person tried the evaluation to see if it would work.
Not quite true. Factoring the denominator is a good thing, and I did that. The factors are a complex conjugate pair. Knowing the pole locations as α ± jβ, there are any number of 2nd order circuits than can be implemented. This particular pair of poles is not on the unit circle, or the Chebyshev Ellipse, and it does not meet the criteria for a 2nd order Bessel, which still leaves many other biquad realizations. The only part you missed was not being able to implement complex poles as independent circuits. No biggie!

#### MisterBill2

Joined Jan 23, 2018
13,746
Not quite true. Factoring the denominator is a good thing, and I did that. The factors are a complex conjugate pair. Knowing the pole locations as α ± jβ, there are any number of 2nd order circuits than can be implemented. This particular pair of poles is not on the unit circle, or the Chebyshev Ellipse, and it does not meet the criteria for a 2nd order Bessel, which still leaves many other biquad realizations. The only part you missed was not being able to implement complex poles as independent circuits. No biggie!
Given that it appears to have been a homework assignment my presumption was that the problem would have a solution. I only had one instructor at one college who would give non-solvable homework. That did teach us to examine the data of problems to verify that an answer could exist. Of course it may also be that the function is copied incorrectly.

#### Papabravo

Joined Feb 24, 2006
19,574
Given that it appears to have been a homework assignment my presumption was that the problem would have a solution. I only had one instructor at one college who would give non-solvable homework. That did teach us to examine the data of problems to verify that an answer could exist. Of course it may also be that the function is copied incorrectly.
The transfer function is in a standard form with ω₀ = 2, and Q= 2/3. So the implementation follows from those parameters, or you can get them from the pole locations. The existence of a solution was never even a question.

#### MrAl

Joined Jun 17, 2014
9,633
Given that it appears to have been a homework assignment my presumption was that the problem would have a solution. I only had one instructor at one college who would give non-solvable homework. That did teach us to examine the data of problems to verify that an answer could exist. Of course it may also be that the function is copied incorrectly.
Hi,

Sorry but i found it to be unfactorable too except in the form of complex poles.
I dont think there is a way to resolve this such that we can obtain two first order realizations. The reason for this is because complex poles always come from 2nd degree expressions which come only from circuits that can not be directly cascaded.
We can look at this though and see if we can find some augmentation that would allow us to proceed in that manner just with a slight modification. I have a feeling though that we might end up with the state space solution once again.

#### Papabravo

Joined Feb 24, 2006
19,574
Hi,

Sorry but i found it to be unfactorable too except in the form of complex poles.
I dont think there is a way to resolve this such that we can obtain two first order realizations. The reason for this is because complex poles always come from 2nd degree expressions which come only from circuits that can not be directly cascaded.
We can look at this though and see if we can find some augmentation that would allow us to proceed in that manner just with a slight modification. I have a feeling though that we might end up with the state space solution once again.
There are well known and common implementations for 2nd order transfer functions with a complex conjugate pair of poles including Tow Thomas, Biquad, Sallen-Key and others. The implementations follow directly from the complex poles, or from the parameters ω₀. and Q, which are simple functions of the coefficients.

#### MrAl

Joined Jun 17, 2014
9,633
There are well known and common implementations for 2nd order transfer functions with a complex conjugate pair of poles including Tow Thomas, Biquad, Sallen-Key and others. The implementations follow directly from the complex poles, or from the parameters ω₀. and Q, which are simple functions of the coefficients.
Yes i realize that, but he was looking for a piecewise circuit solution where we find part of the total solution and then another part, then combine them with simple interconnections. This could come from factoring in the completely real case, but i dont think there is a solution for the complex case. Of course there is a solution for the entire problem taken at once without trying to factor the transfer function, but the implementation goal was different for this approach.

#### Papabravo

Joined Feb 24, 2006
19,574
Yes i realize that, but he was looking for a piecewise circuit solution where we find part of the total solution and then another part, then combine them with simple interconnections. This could come from factoring in the completely real case, but i dont think there is a solution for the complex case. Of course there is a solution for the entire problem taken at once without trying to factor the transfer function, but the implementation goal was different for this approach.
Yes there is. It is called the Tow Thomas filter and it works by rewriting the transfer function to show the use of two integrators, exactly as you would do it if you were programming an analog computer to solve the problem. Just like we did 50 years ago.

#### MrAl

Joined Jun 17, 2014
9,633
Yes there is. It is called the Tow Thomas filter and it works by rewriting the transfer function to show the use of two integrators, exactly as you would do it if you were programming an analog computer to solve the problem. Just like we did 50 years ago.

The more general method is the state space, which TT is just one case of, and if you notice i mentioned that and that was the original method i posted in my first post in this thread.

But that's not the same as factoring, creating two completely separate circuits, then connecting them in cascade and winding up with the transfer function you were after all along.
I thought you were already familiar with this as you seemed to understand the difference in your previous posts.
To show the difference, simply start with a denominator that *is* factorable into real factors, create two first order filters, then simply tie them together in cascade without any other connections including no feedback. Compare that to TT and see the difference. However we both know that does not work with a factoring that turns out to be complex.

Did i make this clear yet? If not i guess i could show a circuit or two.

#### Papabravo

Joined Feb 24, 2006
19,574
The more general method is the state space, which TT is just one case of, and if you notice i mentioned that and that was the original method i posted in my first post in this thread.

But that's not the same as factoring, creating two completely separate circuits, then connecting them in cascade and winding up with the transfer function you were after all along.
I thought you were already familiar with this as you seemed to understand the difference in your previous posts.
To show the difference, simply start with a denominator that *is* factorable into real factors, create two first order filters, then simply tie them together in cascade without any other connections including no feedback. Compare that to TT and see the difference. However we both know that does not work with a factoring that turns out to be complex.

Did i make this clear yet? If not i guess i could show a circuit or two.
No we are clear. I'm not that familiar wit the state space method. I guess I read your post too fast.
What I thought of was reducing what should be a 2nd order ODE to a system of 1st order equations. I never really grokked that method.

#### MisterBill2

Joined Jan 23, 2018
13,746
The transfer function is in a standard form with ω₀ = 2, and Q= 2/3. So the implementation follows from those parameters, or you can get them from the pole locations. The existence of a solution was never even a question.
OK, there is a solution, probably I should have qualified that as a simple, or easy, or convenient solution. Typically homework problems get more complicated as one advances down the listing.