# Operational Amplifier from a third order Transfer function.

#### Muzwakhe

Joined May 16, 2023
4
Using the transfer function, simulate the system using operational amplifiers. Please help, I don't have any idea where to start.

#### crutschow

Joined Mar 14, 2008
32,848
Did they not discuss this in your class?

#### Papabravo

Joined Feb 24, 2006
20,378
1. Do you know what poles and zeros are?
2. Do you know how to find them?

#### MrAl

Joined Jun 17, 2014
10,588
Using the transfer function, simulate the system using operational amplifiers. Please help, I don't have any idea where to start.
View attachment 294374
Hi,

The simplest way to handle this is to turn this into an equation where the form is in the form of all integrators. You'll find 3 integrators will be used. You can then implement using op amps set up as integrators. You can then simplify that if you like but may not be necessary because that will in fact form the system just like that.
Does that help or do you want more help?

#### Muzwakhe

Joined May 16, 2023
4
Did they not discuss this in your class?
Hi Zapper
In the class the only examples that was discussed were first order transfer function, that is why I don't know how to work with third order function.

#### Muzwakhe

Joined May 16, 2023
4
Hi,

The simplest way to handle this is to turn this into an equation where the form is in the form of all integrators. You'll find 3 integrators will be used. You can then implement using op amps set up as integrators. You can then simplify that if you like but may not be necessary because that will in fact form the system just like that.
Does that help, or do you want more help?
Thanks a lot

#### Muzwakhe

Joined May 16, 2023
4
1. Do you know what poles and zeros are?
2. Do you know how to find them?
Yes I know how to find Poles and Zeroes, the only thing I don't know are the actually steps I need to follow in order to solve the problem.

#### LvW

Joined Jun 13, 2013
1,667
Yes I know how to find Poles and Zeroes, the only thing I don't know are the actually steps I need to follow in order to solve the problem.
Step 1: It is an easy task to find the poles of the function (zeros of the denominator). All three poles are simple integer numbers (one pole must be guessed - start with the most simple assumption: A negative real pole). Knowing one pole, you can find the remaining two poles.
Step 2: Knowing the poles, you can rewrite the transfer function as a product of a first order and a second order function (or, as an alternative, as a product of three 1st-order functions if all poles are real).
Step 3: Now you can split the whole expression into new functions which are added (numerators "s" as well as "4")
Step4: Now we have simple first and second order functions which can be realized (using well-established filter blocks) and combined according to step 3.

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#### MrAl

Joined Jun 17, 2014
10,588
Thanks a lot
You're welcome.

To get into a form with integrators, you can proceed like this example. Note that if you also have to provide feedback then you must use the feedback formula first that involves both G and H not just G, but assuming for now that is not needed i'll show a simple second order system which i'll make up just for illustration.
This is so easy you dont have to know anything much just follow a step by step procedure.

Starting with:
y/x=(s+a)/(s^2+b*s+c)

(and remember if you need to add feedback because you have "G" not "y/x" then you need to do that first).

Multiply by both denominators and get:
(s^2+b*s+c)*y=(s+a)*x

and then expand that to get:
s^2*y+b*s*y+c*y=s*x+a*x

then divide by the highest power of 's', which here is '2' so we get:
(b*y)/s+(c*y)/s^2+y=x/s+(a*x)/s^2

Now subtract everything on the left that is divided by any 's', and that puts the first two terms on the left to the right side:
y=-(b*y)/s-(c*y)/s^2+x/s+(a*x)/s^2

Now factor to group those terms with 'y' and those with 'x' and get:
y=-y*(b/s+c/s^2)+x*(1/s+a/s^2)

Now place two integrators such that one feeds the other, and note that 'y' is the output and 'x' is the input, and have a summing junction in between the two.
The second integrator is the single 's' integrator, and the first integrator is the squared 's' integrator, because if a signal goes through both integrators it ends up being integrated twice.
Now looking at the coefficients for 'x' we have:
1/s and a/s^2
This means that 'x' is applied to the second integrator with a gain of '1', and applied to the first integrator with a gain of 'a'.
Looking at them for 'y' we have:
-b/s and -c/s^2
This means that 'y' is applied to the second integrator with a gain of -b and applied to the first integrator with a gain of -c. These are negative feedback paths.

That's about it except when you have more than one signal going into an integrator you use a summer or subtractor as needed.

The final system would look like the bottom drawing in the attachment. Note any joining lines are summing junctions.
You can then implement this as op amp integrators with possible inverting stages also.

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#### LvW

Joined Jun 13, 2013
1,667
Starting with:
y/x=(s+a)/(s^2+b*s+c)
Am I wrong - or is the function to be realized (post#1) of third order?
However, when ther 3rd-order function is split into a series connection of a 1st as well as 2nd-order expression, the proposed approach seems to be a good method for realizing the 2nd-order part.

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#### MrAl

Joined Jun 17, 2014
10,588
Am I wrong - or is the function to be realized (post#1) of third order?
However, when ther 3rd-order function is split into a series connection of a 1st as well as 2nd-order expression, the proposed approach seems to be a good method for realizing the 2nd-order part.
Hi there,

Yes, the original was third order but i didnt want to hand out the entire solution so i did an illustration of a second order system.
A 3rd order system would require three integrators and since there would be more coefficients there would be more feedback paths.
The procedure is the same for any order though, just need more integrators for higher order.

This approach can be found in a U.S. Navy handbook. It's also quite intuitive.
The differential form views the integrators as differentiators by viewing the signal flow in reverse.

#### LvW

Joined Jun 13, 2013
1,667
This approach can be found in a U.S. Navy handbook. It's also quite intuitive.
The differential form views the integrators as differentiators by viewing the signal flow in reverse.
Yes, this is really a good and very effective method to derive a block diagram from a given transfer function - based on (positive) integrator units.

#### MrAl

Joined Jun 17, 2014
10,588
Yes, this is really a good and very effective method to derive a block diagram from a given transfer function - based on (positive) integrator units.
Yeah, that's the only drawback, we need noninverting integrator units. That's a shame because it's so much easier to form an op amp summing junction on the inverting terminal. That may leave us with the requirement that we also need two regular op amp inverting amplifiers.

#### LvW

Joined Jun 13, 2013
1,667
Yeah, that's the only drawback, we need noninverting integrator units. That's a shame because it's so much easier to form an op amp summing junction on the inverting terminal. That may leave us with the requirement that we also need two regular op amp inverting amplifiers.
I just have tried to replace both positive integrators with negative units (-1/s), which are basic units (Miller integrator) in circuit theory.
The consequences are:
* The x-signal must be multiplied with "-1" (instead of "+1) before it is fed into the 2nd integrator;
* The feedback factor "b" for the 2nd integrator must now be positive (instead of "-b")

#### MrAl

Joined Jun 17, 2014
10,588
I just have tried to replace both positive integrators with negative units (-1/s), which are basic units (Miller integrator) in circuit theory.
The consequences are:
* The x-signal must be multiplied with "-1" (instead of "+1) before it is fed into the 2nd integrator;
* The feedback factor "b" for the 2nd integrator must now be positive (instead of "-b")
Hi,

Yes, there will be a number of sign flips required.
I think there is a way to do it with positive integrators using op amps but i'd have to remember how it is done. For the second order system that would mean two op amps. I'll see if i can remember how to do this or you could try it yourself if you like. It introduces some gains that may have to be compensated for too i think.

#### LvW

Joined Jun 13, 2013
1,667
I think there is a way to do it with positive integrators using op amps but i'd have to remember how it is done.
In principle, there are three topologies for a non-inverting integrator:
* Use of a second opamp (inverting) - either in series with the inverting Miller integrator or within its feedback loop.
* NIC-integrator (Negative Impedance Converter): Risky design, close to instability
* BTC integrator (Balanced Time Constants): R-C lowpass at the non-inv. input together with R-C negative feedback (like Miller integrator).

#### MrAl

Joined Jun 17, 2014
10,588
In principle, there are three topologies for a non-inverting integrator:
* Use of a second opamp (inverting) - either in series with the inverting Miller integrator or within its feedback loop.
* NIC-integrator (Negative Impedance Converter): Risky design, close to instability
* BTC integrator (Balanced Time Constants): R-C lowpass at the non-inv. input together with R-C negative feedback (like Miller integrator).
Hi,

Yes it's not that easy to do this without using inverting stages, and no matter what we do we still need to invert the output to get the negative constants, or at least one. So the implementation may as well stay with integrator stages combined with inverting stages.
If you feel like trying another way post it here so we can see it too.

I just saw that i recently reached 10000 posts here

#### MrAl

Joined Jun 17, 2014
10,588
In principle, there are three topologies for a non-inverting integrator:
* Use of a second opamp (inverting) - either in series with the inverting Miller integrator or within its feedback loop.
* NIC-integrator (Negative Impedance Converter): Risky design, close to instability
* BTC integrator (Balanced Time Constants): R-C lowpass at the non-inv. input together with R-C negative feedback (like Miller integrator).
Hello again,

Yes i remembered now that you mentioned what you are calling the BTC integrator.
It is a little tricky figuring out the feedforward resistors and the feedback resistors though because of the need for summing junctions.
In the diagram, without the feedforward gain from R9 resistor R3 would have been 1k, but because of the combined impedance R3 had to be reduced to 1/2 it's original value. That allows the junction of R9 and R2 to act as an actual summing junction. That makes the circuit work for the forward paths but i did not get to the feedback paths yet. I imagine they can go to the inverting inputs of the two op amps, with another change in resistances. For example, if we end up with y/s then probably R10 would be 1k and R3 would have to go back to 1k, but i did not try that yet. For R11 and R7, they both probably have to be 2k each.
Note this is only if all the feedback gains are 1 and so are the forward gains. If the gains are other than that, the resistor values would have to be adjusted again, and i think that would work out but i did not get that far yet it started to take a little too much time.

The simplest way is to just use inverting sections and integrators, but i wanted to see if this would be possible too, and i think it is but i haven't verified it yet.
Another way would be to just analyze that circuit and then calculate all the resistor values as i think that would be the right topology.
The drawback is that we need twice as many resistors and capacitors just for the basic integration mechanism, and they probably have to be matched to some degree because in the transfer functions part of the denominators must be able to cancel the numerators so we end up with a pure integrator 1/(sRC).

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