# cascading two single pole low pass filters

Discussion in 'Homework Help' started by PG1995, Jun 2, 2012.

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1. ### WinterstoneBanned

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PG1995,

The equation as derived by you includes 3 resistors and 2 capacitors (including the element I call Y5 which in your drawing is Y5=sC1).
That means: All 5 elements are present.
After element allocation for lowpass behaviour (proper selection of R and C) you will notice that there is an expression (Y1+Y5), which simply means: Y1=1/R1 in parallel to Y5=1/R5 .
Thus: Rp=R1||R5.
Thus, to realize a particular lowpass transfer function the element Rp must have a certain value which can be realized by R1 only (why use two R in parallel when one R can do the job?).
Please note the word "CAN". Of course, if you like you can design also with a resistor R5.
That´s what I mean with "redundant".
Perhaps you have dropped R5 from the beginning (because somebody or Wikipedia told you this).
But how could you know? When you like to understand a circuit and the process of proper dimensioning you need the equation including all elements. Only then you can decide if one part can be dropped or not.
I hope I was able to explain it clearly.

REMARK: By the way, the fact that - after proper grouping - you see that Y5 appears only tgether with Y1 (in a sum) is one of the advantages of using admittances Y rather than impedances Z.
In case of Z it would me not easy to see that R1||R5.

Last edited: Jun 10, 2012
2. ### PG1995Active Member

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Thank you very much, Winterstone.

I understand it somehow now. But one point really puzzling me. Please have a look on the attachment and kindly help me. Thanks.

Kind regards
PG

3. ### WinterstoneBanned

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The answer is based on the fact that all these active filters rely on RC feedback.
Let me start from the beginning - how could somebody (like Mr. Sallen and Key) know (or hope) that the generic circuit is able to produce 2nd order filter functions? The key is the requirement that a general 2nd order filter must be able to generate a complex pole pair.
They startet with a simple 2nd order RC circuit (with two real poles), which was used as a frequency dependent feedback path of an amplifier, because feedback can change real poles into complex poles (movement to the imaginary axis).
Therefore, the feedback path dominates the principle behaviour.
Now to our circuit: Looking into the circuit from the opamp output (and grounding the normal input) both elements R1 and R5 act as a parallel combination.
In short: The feedback factor (more correct: The feedback transfer function) contains R5 only in parallel to R1.
Thus, R5 can be set to infinity if R1 is selected properly.

Perhaps it is good to repeat the steps again:
* Derive H(s) based on admittances Y for the particular generic circuit topology,
* Select two capacitive admittances with the following aims:
The numerator must contain s^0 (low pass), s^1 (band pass), or s^2 (high pass).
For this allocation the remaining 3 elements are chosen so that the denumerator is a 2nd order polynominal (including s^0, s^1 and s^2).
* Then, the several expressions are grouped (using brackets) - and then you will see that a 2nd order polynominal is possible even in case of Y5=0.
* Only now (after proper simplification) you insert for Y=1/sC and Y=1/R.
* Finally, a normal standard form should be generated by division.
Here you have two choices: Either you produce a "1" (s^0 without a factor) or you produce a (s^2) without any factor.
* Such a standard form allows a direct identification of the expressions for the pole Q as well as the pole frequency.

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5. ### PG1995Active Member

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Thank you very much, Winterstone. I'm much obliged for all the help.

With best wishes
PG

6. ### WinterstoneBanned

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PG1995, you are welcome.
As you may have noticed, in the past I had the opportunity to gain some experience with active filters.

7. ### PG1995Active Member

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Hi Winterstone

I totally agree with almost everything which you have said above. Actually my book doesn't provide any information on the topic we have discussed in previous posts. Perhaps, the reason for this is that the book isn't about active filters. The book is electronic devices by Floyd. I had also consulted another book electronic principles by Malvino and it too didn't provide any help.

You are right in saying that nothing can beat good textbook because this way we learn things in sequence and the learning approach is linear. So, I searched the internet only because I couldn't find any information in the book. But the reason for mentioning "I have searched the net for more one hour" in the attachment was for those persons who instead of helping you are more ready to ask counter questions of these kinds, "Have you searched the net?", "Have you googled it?", "Why don't you use Wikipedia?", "The net is full of information, why don't you search?", etc. But I have to mention one thing here that Electro-Tech-Online is full of helpful members like you and MrAl who help other without any air of arrogance. Many thanks.

With best wishes
PG

8. ### MrAlWell-Known MemberMost Helpful Member

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Hi PG,

I just have to wonder, did you try to do this the way the original filter was done in that paper? That should work too.

In reality, the most modern method would be to use a matrix. The various elements enter into the matrix in a set fashion and then matrix algebra is used to solve the system. This is the most modern method and of course using a computer to do the math.

9. ### PG1995Active Member

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Hi MrAl

The method I used was more or less similar to the one in that page. I'm sure there would be many methods to solve that problem as the matrix method you have mentioned but the method used by me made more sense to me because it used simple ideas such as nodal equations etc. But I'm sure someday I would be able to master other methods too with yours help!

Kind regards
PG

10. ### MrAlWell-Known MemberMost Helpful Member

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Hi PG,

The idea with circuits like this (but really of any kind) is to find a method that you like that allows you to analyze *ANY* circuit that you might come across, or at least those which you are most interested in. If you are interested in filter circuits, then you should look for a general method that allows you to analyze any filter circuit you can find, or any that are of the type you wish to know about. Once you find a way to do any of these circuits you'll never have another question to ask because you'll be able to do it yourself for any circuit

11. ### WinterstoneBanned

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Hello MrAl,

I appreciate your recommendation - and I understand the aim behind it.
However, I doubt if it is possible to find (quote) "a general method that allows you to analyze any filter circuit you can find"

Perhaps your recommendation is good for active filters based on the cascade approach (cascade of S&K or MFB second-order stages) - but I think for designing/analyzing alternative filter topologies like active ladder (active L, FDNR), leap-frog, FLF and state variable (coupled integrators) other mathematical methods are appropriate (or even necessary).

12. ### MrAlWell-Known MemberMost Helpful Member

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Hi Winter,

What i was really talking about was a method that uses a matrix to solve the circuit. If you look at the matrix entries you'll see that many components enter into the matrix in an orderly fashion. This allows a general method of analysis more or less. We're talking in the frequency domain here too, not the time domain, but there are interesting things there too.

I do understand your point i think about the coupled integrators for example, where a simpler method could be used, but i guess what i am talking about here is to have the end goal be one of computer automation where the computer does all the work. There's a decent book out there on the subject but i'd have to look up the name/authors now as i cant remember from back in the 80's when even then it was known about these ideas.

13. ### WinterstoneBanned

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I like to point to the following:
Perhaps both of us speak about two different things - that means: perhaps there is a misunderstanding?
I think, it is quite different if we speak about
(1) Analysis of a known circuit or a specific topology (in general terms Y or Z); or
(2) Synthesis of a certain filter function after choice for a particular topology.

Anyway, if you could identify the title of the book you have mentioned, I am interested.
Thank you.
W.

14. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Analysis of a known circuit in terms of the Y matrix.

I'll have to look around for the title and/or authors. It's been so long since i read that book. It's pretty big, about 2 inches thick and size of a phone book. Talks about computer generated circuit analysis. One simple topic i remember was Gaussian Elimination. Drawing a blank on just about everything else in that book (chuckle)
Another good book i read was on non linear circuit/system analysis. That was an expensive book too. I have it somewhere.

Last edited: Jun 12, 2012
15. ### PG1995Active Member

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Hi again

Thank you for the help.

Regards
PG

Last edited: Jun 18, 2012
16. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

They are simply referring to the feedback circuit, which is a pi filter. That's where the Q comes in.

17. ### PG1995Active Member

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Thank you, MrAl.

So, Q is in fact 'quality factor' and loading is simply connecting of the output of a filter to something.

Regards
PG

18. ### WinterstoneBanned

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Hi PG,

the circuit as shown in your attachement requires some comments because it does not work - if the triangle in Fig. 16-17 represents an opamp.
This circuit resembles an adaptation of the original BJT-Colpitt circuit to opamp techniques - forgetting that the BJT mainly acts as a current source with a finite source resistance.
Therefore, you must add a resistor (some kohms) beween opamp output and the node (in).
This is obvious because otherwise the first capacitor acts as an opamp load only (assuming zero output resistance) without influence on the fedback characteristic.
Secondly, the opamp must be wired as an inverting amplifier because the feedback circuit (by the way: it is not a Pi filter, but a bridged-T filter, see correction 1) produces a phase shift of 180 deg at the oscillation frequency.

To the role of Q: The filter Q is directly related to the damping of a filter network - and the amount of damping directy influences the steepness of the phase response. A higher Q value causes a phase cross-over (180 deg) with a rather high slope. All resistors which load the filter cause more damping and, thus, decrease of the Q slope (correction: phase slope). At the same time the 180-deg frequency is slightly shifted. This is in contrast to the classical tank circuit - but realize: The shown bridged-T filter exhibits not the classical bandpass character.

Regards
W.

1) Sorry, I am in error: For this configuration with negative gain the filter is simply a 3rd-order low pass with one real and two complex poles.

Last edited: Jun 19, 2012
19. ### PG1995Active Member

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Thank you, W. I really appreciate your help.

Best wishes
PG

20. ### PG1995Active Member

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Hi

Regards
PG

Last edited: Jun 22, 2012
21. ### WinterstoneBanned

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Hi PG1995,

Q1: I see no discrepancy if compared with wikipedia. The transfer function expression in the middle of the last line (not the yellow one) of your scanned paper is identical to the wiki function, is it not?

Q2: The rule that both opamp input terminals have (nearly) identical voltages applies for negative feedback only! That is the basic result of negative feedback (or if negative feedback dominates over pos. feedback). If there is positive feedback instead (or if pos. feedback dominates over some negative feedback) the circuit does not operate in its linear region and this rule does NOT apply.

Q3: Your question touches the "secret" behind the complexe variable s. Sometimes it is used as s=(sigma+jw) and sometimes simply as s=jw.