# Sallen LP derivation from Time-domain

Discussion in 'Homework Help' started by Tera-Scale, May 8, 2012.

1. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
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Hi,
I am trying to derive the transfer function of the attached LP filter. I already was successful starting directly from the s-domain. Now I want to start in the time domain - simplifiy as much as I can and convert to s-domain.

Attached is document of what I managed to do till now. I would appreciate some hints of what I might be doing wrong. thanks

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2. ### steveb Senior Member

Jul 3, 2008
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I didnt' look at your file because I don't open doc files on the internet.

However, the straightforward way to solve linear systems in the time domain is to put the system equations in state-space form, and then use the state transition matrix solution.

http://en.wikipedia.org/wiki/State-transition_matrix

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3. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
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• ###### Sallen Low Pass.pdf
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4. ### steveb Senior Member

Jul 3, 2008
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OK, I can read it now.

It is not so easy to follow your work because currents are not clearly defined and the directions are not obvious. However, it seems the equations are likely to be correct.

My above recomendation is to use the state space representation of the sytem. Are you familiar with this? If so, try to use that approach. If not, I recommend you learn it because it is the proper way to handle analysis in the time domain.

To get you started, note that there are 2 energy storage devices. This means you will have 2 first order differential equations for the 2 state-space equations and your system is of order two.

You can directly write these two state space equations in terms of variables. Then, it is a simple matter to express variables in terms of state variables (the capacitor voltages) and the input variables (X).

After you have the state equations, you can write the output equation and then get the A, B, C and D matrices for the state space system. Then the state transition matrix approach will give you the time domain solution.

Last edited: May 8, 2012
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5. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
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I appreciate your suggestion and I intend to learn this method in the futur. Unfortunately i do not have enough time now. The thing is that in the question it does not specify this method but it asks to use kirchhoff's Laws. I will attach a more detailed schematic with gain and currents labelled.

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6. ### steveb Senior Member

Jul 3, 2008
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You need to specify the direction of the current also. According to your KCL, they are either all going into the node, or all going out of the node.

7. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
5
Current direction specified aswell. With these currents I already managed to get a good answer when using the s-domain from the very beginning. Now I need it in time-domain from the very beginning.

• ###### Sallen LP.png
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Last edited: May 8, 2012
8. ### steveb Senior Member

Jul 3, 2008
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OK, first of all, I have a better idea of what you are trying to do. So, my above suggestion is not necessary if you are just trying to get the transfer function from the differential equation. I thought you wanted to get the solution in the time domain, but that is not the case.

So, for the current directions you specified, aren't your KCL equations wrong?

I get I3=I4 and I1=I2+I3

9. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
5
In fact I do want it in the time domain in the first steps as in the pdf attached earlier.

They should be correct since when derived using s-domain I got the correct transfer function which is written at the end of the pdf.

Now my challenge is to start from the time-domain...

10. ### steveb Senior Member

Jul 3, 2008
2,431
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I'm still confused on what your goal is. You say you want "it" in the time domain. What is the "it" you want?

I don't see consistency between your current equations and the way you defined them in your drawing. Applying KCL to your drawing yields the two equations I show above, which differ from the two you use.

11. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
5
You are right. These are the new directions: attached.

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12. ### steveb Senior Member

Jul 3, 2008
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OK, that looks consistent. I can now follow the top of page one, however at the middle of page one you say, "therefore ... " and give an equation. It is not clear how you arrive at this equation.

13. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
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Look at this pdf.

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14. ### steveb Senior Member

Jul 3, 2008
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It looks like you are equating zero with zero to match up the two equations. There are two issues with this.

First, it looks like you have a typo in what you wrote, the second term does not look correct.

Second, it is dangerous to do a zero equals zero step. It is prone to errors if units don't match, and you are likely to be throwing information away.

I'm not sure if the second issue will be a problem unless we trace it through. However, you first need to correct the error, and then see if you run into trouble. After you correct the error, you might want to try doing a variable substitution, rather than a zero=zero relation.

EDIT: On tracing it through, it looks like your method is OK. Just fix the typo and follow it through. You should be all set.

Last edited: May 8, 2012
15. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
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You are right there is an error in the second term. I corrected it and worked all the way again. There was not much improvement as you can see.

What variable do you suggest substituting?

Edit: check this pdf

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16. ### steveb Senior Member

Jul 3, 2008
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I don't see the correction. Did you put up the wrong pdf?

I made the correction and then worked it through. It seemed to work out fine.

Double check what you did.

I'll post another way in a few minutes. I just need to scan it to PDF.

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17. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
5
Corrected again, term in last step still missing.

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18. ### steveb Senior Member

Jul 3, 2008
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I think you misunderstood my correction.

The first red equation you have shows a term with C*d(Y-Z)/dt. In the middle of the page, after the word "therefore", you put in C*d(W-Z)/dt.

If you change the W to a Y, then it should work out.

I've attached the way I did it, which is a little more concise than your method, but it looks to me that your way is fine.

EDIT: I made an update to the attachment

• ###### PTDC.pdf
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Last edited: May 8, 2012
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19. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
5
thanks a lot steve. I really appreciate your help I will go through and correct my method while following your method tomorrow as it is midnight here. thanks again.

Would the state space method still be feasible for these applications? It looks interesting!

20. ### steveb Senior Member

Jul 3, 2008
2,431
469
Note that I updated the PDF to correct a mistake I made.

The state-space method is always applicable since it is more general and more powerful. Eventually, you are likely to need it for advanced work. If your goal is only to derive a transfer function for a single input, single output system, then the method you used is perfectly adequate.