Hello.
I am given a transfer function and I am asked to form a real circuit with OAs that respects this function.
\[ G(s)= (s-2)/(s^2+3s+2) \]
I know it may not be that hard, but I didn't get this topic that well and I hope to get some help and advices to solve this question. It's a really important assignment and time is quite short for me.
Thanks a lot in advance.

 

Hi Reno,
Welcome to AAC.
As this is a homework assignment we need to see your best attempt at answering, we then can help you.
E

 

Reno21 - have you any idea?
Do you have some experience with filter functions?

 

Do you understand what that transfer function does?

 

Reno21 - have you any idea?
Do you have some experience with filter functions?

Not that much. I know I can simplify the function a bit and get G(s)= (s-2)*(1/(s+1))*(1/(s+2)). Something like G(s)=(G1(s)-G2(s))*G3(s)*G4(s).
I know that G3 and G1 are a basic integrator and differentiator, but I don't know how to work with G4 and G2.

 

Do you understand what that transfer function does?

Can you be a bit more specific?

 

Can you be a bit more specific?

What the transfer function would do with a signal if you implemented it with circuits.

 

What the transfer function would do with a signal if you implemented it with circuits.

It would fluctuate a bit and then stabilise around a value after some time. I've done it in matlab for a step signal and I obtained a slope signal.

 

It would fluctuate a bit and then stabilise around a value after some time. I've done it in matlab for a step signal and I obtained a slope signal.

When we speak about the term "transfer function " we refer to a frequency-dependent function in the frequency domain.
This can be senn in the function you have shown in your first post (s is the complex frequency variable).
When you use a step signal as a test signal you are operating in the time domain.

 

When we speak about the term "transfer function " we refer to a frequency-dependent function in the frequency domain.
This can be senn in the function you have shown in your first post (s is the complex frequency variable).
When you use a step signal as a test signal you are operating in the time domain.

I see, but signals aside (that is not that important for the assignment), I am just asked to make the skematic, and eventually to provide some explanations to why I chose those configurations.

 

You might want to start by finding the roots to the denominator polynomial. So you should have two poles in the left half-plane. The numerator describes a right half-plane zero. Then you need circuits for a second order low pass and a circuit for the zero which should give you a first order lead.

 

Hello.
I am given a transfer function and I am asked to form a real circuit with OAs that respects this function.
\[ G(s)= (s-2)/(s^2+3s+2) \]
I know it may not be that hard, but I didn't get this topic that well and I hope to get some help and advices to solve this question. It's a really important assignment and time is quite short for me.
Thanks a lot in advance.

Hi,

Did you say you had to use integrators and differentiators?

I am not sure what you did in the past, but if that's true then you need to get the terms in the form of:
y=An*x/s^n+Bn*y/s^n

where 'n' can be either 1 or 2 or both. For 1st order you just have one integrator, for 2nd order you have two integrators, etc.
In that form, any term with 'x' in it is in the feedforward path, and any term with 'y' in it is the feedback path.

This does turn out to be the simplest method too because there is very little you have to think about, it's just a matter of algebraic manipulation into the required form, then draw up the signal paths.

Here's a fictitious first order example (note the inclusion of 'Vin'):
T(s)=Vin*1/(s+1)
For simplicity first change this to:
y=x*1/(s+1) which is also just y=x/(s+1).
Now multiply by the denominator on the right side and get:
y*(s+1)=x
Now expand:
y*s+y=x
now divide by 's' because that's the highest power of 's':
y+y/s=x/s
Now isolate that lone 'y' on the left by subtracting:
y=x/s-y/s

Here we are done with the algebra and we see we have x/s which is 'x' being integrated once, and -y/s the feedback being integrated once. We need one integrator, and since 'x' is the forward path that means 'x' goes to the input of the integrator, and since -y/s is the feedback, that means the output 'y' has sign flipped and also goes into the input. Thus we end up with:
(x-y)
being integrated once over time.
Now since x=Vin and y=Vout then we have:
(Vin-Vout) being integrated once and the output is Vout.

Does that make sense to you?