If the RC must be 1, how can i fulfill my requirements which the s^0 term which is 198025=RC?

Why do you think that "198025=RC" ?
It`s wrong! The correct answer is given by the general transfer function - that is the reason I have asked you to find it.

 

Hello MrAl,
only now I have seen your comment to my post.
I like to mention that I did not "introduce another method" (and, hence, I did not make "things more difficult").
The questioner (Aarontan0219) has shown a working circuit in his post'17 (well-known structure with two integrators) and with my post'18 I gave him a formula for the pole frequency wp which involves the parts values for this circuit. That`s all!!
With this formula - together with the specified transfer function - he would be able to immediately calculate the corresponding parts values.....

Hello,

Oh yes very good. I also had seen that he accepted learning two different ways to approach this kind of problem so all is well in AAC Land
I recently noticed he may be missing some of the basics. After he comes up to speed we should probably do that method too once he learns how to do the most basic structures.

 

If the RC must be 1, how can i fulfill my requirements which the s^0 term which is 198025=RC?

Hello again,

Compare this next diagram with the one in post #34 and see if you can figure out the value of Rx that makes the two system implementations the same. Look at the bottom circuit in each drawing.

 

Hello again MrAl
everything OK....and I fully agree to your last sentece. I`ve got the impression that he even does not know the meaning of the fundamental pole parameters wp and Qp

 

It can be hard to solve this. Have you already found the solution? I am interested in the result. If I were in your place I think I'll hire coursework service to get help from their experts if they can solve such a problem. Maybe you can try to get some help form them. It is not a spam. Maybe they will provide you with some support materials to get some help.

 

It can be hard to solve this. ...

No - in contrary. If someone is familiar with filter basics (general form for a 2nd-order lowpass transfer function) itis a rather quick and simple calculation....

 

It can be hard to solve this. Have you already found the solution? I am interested in the result. If I were in your place I think I'll hire coursework service to get help from their experts if they can solve such a problem. Maybe you can try to get some help form them. It is not a spam. Maybe they will provide you with some support materials to get some help.

Hi,

That's interesting. You joined at 8:27am ET today 11/25/2019 and posted just ONE message at 8:29am ET and it's a link to a paid help service.

 

Hi there,

If you have no other constrains on how to implement this then the simplest way is purely algebraic where we cross multiply and then separate into individual integrators. You can then use op amp integrators for the implementation.

This is so easy to do i almost hate to mention it :--)
But if you need more help that's ok it might be your first time doing this.

Here's a quick example....

System:
Vout/Vin=(s+1)/(s^2+2*s+3)

Cross multiply and get:
(s^2+2*s+3)*Vout=(s+1)*Vin

Expand and get:
s^2*Vout+2*s*Vout+3*Vout=s*Vin+Vin

Divide by highest power of 's' and get:
(s^2*Vout+2*s*Vout+3*Vout)/s^2=(s*Vin+Vin)/s^2

Expand that and get:
(2*Vout)/s+(3*Vout)/s^2+Vout=Vin/s+Vin/s^2

Solve that for the lone "Vout" term and get:
Vout=-2*Vout/s-3*Vout/s^2+Vin/s+Vin/s^2

That is the final math form which is sort of canonical in that we have Vout alone on the left and various terms with both Vin and Vout on the right and divisions involving powers of 's' as well as constant gains (-2, -3, and 1).
The interpretation of this final form is that anything on the right side with Vin in it is in the forward path and anything on the right side with Vout in it is in the feedback path, and any power of 's' represents an integration where the number of cascaded integrators is equal to that power of 's'.
So Vin gets integrated twice and so does Vout and the terms with Vout are negative so there is also an amplification by minus 2 or minus 3. The circuit therefore has two integrators in it and possibly some amplifiers that implement the gains -2 and -3. There are various ways to implement the gains (such as combining them) although that's the most straightforward.

Now try this on your system.

The attachment is a comparison of the step response of the example system, first a simulation after implementation as above and then as the time function found from the Inverse Laplace Transform.

Sir Some what better than others but please tell me more clear.

 

Sir Some what better than others but please tell me more clear.

I love to see that there are still people interested in this stuff.

You start with the transfer function and use algebra to get it into a form where you can see transformed integrations present. Every time you see a 1/s that means an integrator.

So say we have:
T(s)=1/(s^2+s+1)

That means we have this:
Vout/Vin=1/(s^2+s+1)

Now start by multiplying out the denominators and get:

(s^2+s+1)*Vout=Vin

then expand that to get:
s^2*Vout+s*Vout+Vout=Vin

then divide by the highest power of 's' and get:
(s^2*Vout+s*Vout+Vout)/s^2=Vin/s^2

then finally expand that to get:
Vout/s+Vout/s^2+Vout=Vin/s^2

and get the lone term of Vout on the left and everything else on the right and get:
Vout=-Vout/s-Vout/s^2+Vin/s^2

and here we can see that Vin is integrated twice 1/s^2=(1/s)*(1/s) so that means we start by drawing two integrators in tandem with Vin on the left and Vout on the right. That is the start of the implementation.
Next we have to include the effects of Vout. Vout on on the right represents feedback. Since Vout on the right is divided by both 's' and by 's^2' we first have to get Vout integrated two times and have that added with the effect from Vin (because that signal gets integraded twice and we also need minus Vout to be integrated twice) then later we can include the effect from Vout/s. That means we have summers (or subtractors in this case) that sum both the signal that is already present with the feedback signal. For example, since we have Vin on the far left and that gets integrated twice and one of the effects we have to include is the minus Vout/s^2 term, we have to sum Vin with minus Vout to get the full input to the first integrator in the tandem connection. The term with minus Vout/s means Vout gets integrated one time,so we have to feed Vout into the input of the second integrator also, and to do that we need another summer that sums the output of the first integrator and the new signal minus Vout.
After all that, we can notice that the actual output Vout satisfies the transfer function if we use network analysis.