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cascading two single pole low pass filters

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

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  1. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Wasnt the text referring to the two second order filters? I'll re read it and see.

    A FEW MINUTES LATER:
    Yes the text refers to two *second* order filters in cascade, not two *first* order filters. So perhaps you'll want to re read that text yourself. They refer to the diagram just below the text which shows two second order filters.

    In other words if we were to rewrite that section of text it could read something like this:
    "If we used resistor and capacitor values to make the configurations in fig 15.17 both Butterworth filters, then..."
     
    Last edited: Jun 4, 2012
  2. Winterstone

    Winterstone Banned

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    Wasnt the text referring to the two second order filters? I'll re read it and see.

    MrAl, I was referring to Q1 of post#15. For my opinion, the question was related to the first line in the table which is a 1st order filter. However, it is not to important - perhaps a minor misunderstanding.
     
  3. PG1995

    PG1995 Active Member

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    Thanks a lot for helping with the queries in this posting, Winterstone, MrAl.


    Would someone please help with the queries included in the attachment? I would really appreciate your help. Thank you.

    Regards
    PG
     
  4. dave

    Dave New Member

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  5. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Q1:
    The VCVS topology is different than SK. That filter is using a different topology therefore it doesnt have to have the impedances in the same places. This i think ruins the flow of the paper. One minute they are talking about SK then the next VCVS, without a proper transitory introduction.

    Q2:
    There are a number of ways to analyze an op amp circuit. You might be able to follow their example for the SK circuit. You can also replace the op amp with a voltage controlled voltage source and let the gain be undetermined as A, then later let the gain A go to infinity finding the limit, and you end up with the ideal transfer function. You might use normal circuit analysis techniques for this or you can handle it as a multiple source circuit which makes it a little easier to understand i think. If you like i can show you this method.
    Another thing that helps is to think of the op amp and Ra and Rb as being an amplifier apart from the rest of the circuit, which has a gain of 1+Rb/Ra. This means you can solve for the non inverting input alone, then multiply that by 1+Rb/Ra and you'll get Vout. You'll end up with Vout on two sides of the equation so then you solve for Vout explicitly and there's your transfer function. Not too hard to do in theory but there is a bit of algebraic manipulation involved which you have to do carefully. I'd be happy to show how this is done in detail.
     
    Last edited: Jun 8, 2012
  6. Winterstone

    Winterstone Banned

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    PG1995,
    I agree wth you. The circuit called "generic" filter in Fig.1 is a simplified version of the real "generic topology".
    That means:
    1.) There must be an additional element Z5 from the node Vx to ground,
    2.) The opamp gain is not in any case G=1, it may be larger (positive and finite).

    The derivation of the general transfer function is not to complicated.
    We need just two node equations (see your Fig.1 with Y=1/Z):

    (a) (Vin-Vx)*Y1=(Vx- V-)*Y2+(Vx-Vout)*Y3+Vx*Y5
    (b) (Vx- V-)*Y2=Y4*V- with Vout=G*V-

    From this you can derive

    H(s)=Vout/Vin=N(s)/D(s)

    with N(s)=G*Y1*Y2 and

    D(s)=Y1Y4+Y3Y4+Y2Y4+Y4Y5+Y1Y2+Y2Y5+Y2Y3*(1-G).

    Proper allocation of all Y=1/Z (resistors, capacitors) leads to lowpass, highpass, bandpass.

    Note that for lowpass and highpass the element Y5 is redundant to Y1 and can be dropped.
    Hope this is of little help to you.

    Remark 1: During dimensioning you have several degree of freedom:
    (1) You can choose equal component values - and calculate the necessary gain G=1+Ra/Rb to realize the desired response (Q value),
    (2) You can fix the gain (preferred G=1 or G=2) and calculate the corresponding element values.

    Remark 2: Start dimensining with N(s)=G*Y1*Y2 because you know that the numerator is independent on frequency for a lowpass (Y1=1/R1 and Y2=1/R2); in case of a bandpass one of both must be Y=sC.
    The other elements - in both cases: low and bandpass - have to be selected with the aim to produce a second order denominator D(s) containing expressions with s^0 , s^1 and s^2 .
    __________
    As expected - each alternative has its advantages/disadvantages. That`s quite normal in electronics: Each circuit alternative and each dimensioning is a trade-off between some conflicting requirements.
     
    Last edited: Jun 9, 2012
  7. Winterstone

    Winterstone Banned

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    Comment to MrAl:

    As far as I know the relevant literature, there is no difference between the so called "VCVS topology" and the "S+K topology" because Mr. Sallen and Mr. Key have proposed this filter structure based on finite gain stages (VCVS). That means: The original S+K topology consists of 5 passive elements and can be realized as such. However, in some cases (low and high pass) the element Z5 is redundant and CAN be dropped - however, it has also advantages NOT to drop it (element spread).
    More than that, the gain of the VCVS may be also negative. But note, in this case the element allocation differs from the pos. gain case.
     
  8. Winterstone

    Winterstone Banned

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    Hi PG1995,
    sorry, but I cannot resist to comment again on your inquiry.
    In your attachement you ask for help because you have spent more than 1 hour searching in the net - without any success.
    I think this is again a good example to demonstrate that one should not only rely on "bits and pieces", which certainly can be found in the Internet. However, such a mosaic - derived from several information sources - never can replace a good textbook. I am sure, that every book dealing with active filters can give you the answer to your problem. According to my experience there are many students believing that "everything is available in the net". And they waste their time with searching over hours. And as a consequence - the technical libraries are empty!
    I have severe doubts that this is the right and systematic way of learning.
     
  9. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi Winterstone,

    Sorry but that doesnt make sense. One has impedances from the unknown node 'x' as follows:
    1. From input to node x
    2. From output to node x
    3. From non inverting input to node x
    4. From ground to node x

    while the other only has the first three.

    A 'topology' is just a method of connection. If you connect two resistor in parallel that's one topology, if you connect them in series then that is another topology. Since the second circuit has an impedance from node x to ground and the first circuit doesnt, they have to be taken as different topologies. It's true that it is not 'that' much different, but it is in fact different so it can not be called the same topology. It's really quite simple. Also, one may be a degenerate case of the other but it's still different. This is especially true because of the way the first circuit was introduced as a 'general' network for that topology. If it is the general network then nothing can be left out. If something is left out, then it's not the general network. If that node x to ground impedance was originally included in the 'general' network, then that would be different.
    For example, i dont tell you that i have a set of parallel resistors as a general network:
    Rp=R1*R2/(R1+R2)
    and then show you two resistors in series and say it is just part of that general network. For one thing we would need a different way to calculate the total resistance with this new network.
    On the other hand, if i show you a group of resistors that includes both parallel AND series resistors and claim that to be the general network, then i can leave out some resistors and still claim that it was derived from the general network.
    What the paper implies is that somehow that new network with more impedances somehow came from the more simple network with less impedances. That's not the way a general network is to be described.
    To put it another way, it is much harder to derive the equations for a network with ADDED components than it is to do with components that are later removed. The general network should included any and all possible components and if the end user wants to simplify that by removing a component that's fine.

    So the bottom line is that they introduce a network with N components, claim it to be the general network, then later show a network with N+1 components and imply that it came from the N component model.
    We can do it the other way around, we can start with an N+1 model and then regress back to the N model, but we'd have to call the N+1 model the general network not the N model.

    As a really comical example, we know that *most* circuits contain at least one single resistor, possibly more than that. So i have a resistor R1 drawn on my schematic in front of me. Now i claim that is the general network for (almost) every possible electrical circuit that could possibly be built in the history of mankind. Yes, a single resistor :)
    Doesnt make sense does it?
    So if we were allowed to *add* to the general network then it would not be general enough to be describing anything that made anything else more simple to understand. If so, we could claim that a blank page was the most general network.
     
    Last edited: Jun 9, 2012
  10. Winterstone

    Winterstone Banned

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    Hi MrAl,

    thank you for your reply - and I think, I can agree to everything you wrote in post #28.
    I am afraid, you did misunderstand something in my contribution #26 or - most probably - I did not express myself clear enough.
    Therefore, I repeat the first sentence from my posting #25:

    The circuit called "generic" filter in Fig.1 is a simplified version of the real "generic topology".

    I completely agree with you that (quote) the general network should included any and all possible components and if the end user wants to simplify that by removing a component that's fine.

    No doubt about it.

    However, the main pupose of my comment #26 to your contribution #24 was to point to the fact that - according to my knowledge - the term "VCVS topology" in some books and articles is used instead of "S+K" (Sallen-Key) topology. That means: Both are interchangeable.
    I think, this happens because, indeed, the circuits as proposed by Sallen and Key (in1955, based on transistor amplifiers) are the only filter structures which are based on finite gain amplifiers (VCVS). All other known filter circuits use other principles (MFB/infinte gain, integrators, GIC, FDNR, active ladder). But I also agree with you that it is not a good idea to use both terms within one chapter resp. one paragraph.

    Additional remark: In the context as discussed above I like to mention that the very well known MFB (multi-feedback) band pass topology contains 5 passive elements. However, if one accepts some specific constraints, one grounded resistor can be dropped. But it is still based on the same topology, isn´t it?

    With regards
    W.
     
  11. PG1995

    PG1995 Active Member

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    Thank you very much, MrAl, Winterstone. I genuinely appreciate your help.

    I have tried to solve the circuit the way Winterstone did but my solution doesn't look right. Please have a look on the attachment and kindly help me. Thanks a lot.

    Regards
    PG
     
  12. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi there Winterstone,

    Thanks for the reply there. That clears things up a lot so i can understand you much better now.

    There is one small point however, and that is that my notes were with regard to that *very* paper that PG has linked to, not some other book on the subject. If another book wants to picture a circuit with 100 resistors and 100 capacitors and put them in a given configuration and state that is a general network for some good reason then that is just fine with me. I was simply saying that the way that one paper put it i think was misleading. This lead to a problem for PG so i think that was self evident.
    I understand your point about the other circuit being general, and that's ok with me :) and i appreciate you pointing that out too.

    PG:
    Another way to do this is to separate the sources and use superposition. This leads to a pretty simple to understand technique.
    If you look at the op amp and Ra and Rb you'll see it is just a gain stage with gain G=1+Rb/Ra. This helps reduce the work. We can redraw the circuit so that the op amp and Ra and Rb form a simple gain block without showing the resistors and calling the gain simply G.

    Now that we have that in place, here's what we do next...we'll call the non inverting input to the op amp (which is just the input now) by the variable v2, and the voltage at the node of the junction of all the components to the left of that v1.

    First, short the output to ground. Now analyze for v2 using only the input voltage as source. Note that by shorting the output to ground that puts the feedback element in parallel with the capacitor going to ground. This forms one impedance which simplifies things a little.

    Second, short the input to ground, using Vout as the source. Now again analyze for v2. Note this time the input resistor is in parallel with the cap to ground.

    To check your work, note that the first analysis for v2 and the second analysis both have the same denominator. Also, one analysis contains Vin and the other contains Vout.

    Now take the two separate analysis for v2 and add them together. This equation now has both Vin and Vout in it.
    Now multiply that by the gain we talked about before, G, then equate that to Vout. You end up with an equation in this form:
    Vout=G*f(Vin,Vout,R,C)
    where f is the function you've just found and it has parameters of Vin and Vout and all the R's and C's.
    Now solve explicitly for Vout and you end up with:
    Vout=G*F(Vin,R,C)
    so on the right all you have now is a function that has parameters only Vin and not Vout anymore, and all the R's and C's. That's the transfer function.

    The result doing it this way comes out exactly as the solution shown in the Wikipedia article.
     
    Last edited: Jun 9, 2012
  13. Winterstone

    Winterstone Banned

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    PG1995,

    I couldn`t check your calculation - but two comments from my side:
    (1) According to my experiences it is more easy to start with admittances Y rather than with impedances Z.
    Insert the corresponding impedances (Y=1/Z) in the final H(s) expression (as given in my response #2).
    (2) Do NOT insert the gain formula G=1+Ra/Rb into the H(s) expression. Thus, you make the calculation more involved. Instead, the opamp gain should be expressed with the term G. Then finally, you have the choice to select the proper value (G=1, 2, or something else).

    Another important aspect:
    Reduce H(s) to the so called "normal standard form" - which means: The term with s^0 in the denominator must simply be "1".
     
    Last edited: Jun 9, 2012
  14. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Just a quick note, the gain is 1+Rb/Ra right? Oh ok, he's got the Ra as feedback, so nevermind.
    In the original diagram however Rb is the feedback resistor, not Ra, so there it would be 1+Rb/Ra.
     
    Last edited: Jun 9, 2012
  15. Winterstone

    Winterstone Banned

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    OK, I suppose this shouldn`t be cause any confusion.

    Quote: I was simply saying that the way that one paper put it i think was misleading. This lead to a problem for PG so i think that was self evident.

    100% agreement :)
     
  16. Winterstone

    Winterstone Banned

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    OK, I suppose this shouldn`t cause any confusion.

    Quote: I was simply saying that the way that one paper put it i think was misleading. This lead to a problem for PG so i think that was self evident.

    100% agreement :)
     
  17. PG1995

    PG1995 Active Member

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    Many, many thanks, MrAl. I think I should also try to solve the problem using the superposition method you have just described once the method I'm already using produces correct solution.

    Thanks a lot for the suggestions, W. But could you or MrAl please check my three equations, (1), (2), and (3), errors? If they are correct, then I will replace 1/Z with Y and would simply use G. Thanks.

    Best wishes
    PG
     
    Last edited: Jun 9, 2012
  18. Winterstone

    Winterstone Banned

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    PG1995,
    yes, I think all three equations are correct and after suitable and proper manipulation you should arrive at the H(s) as given in one of my former postings.
     
  19. PG1995

    PG1995 Active Member

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    Thank you for letting me know that the equations were correct.

    Now there exists only one fat problem with the solution I have ended up with. Kindly have a look on the attachment and help me to trace where the things went wrong. Thanks a lot.

    Regards
    PG
     
  20. Winterstone

    Winterstone Banned

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    Hi G1995,

    At first, the general form for H(s) expressed in Y terms is correct.
    But I have detected one severe error: You have written Y=s/C.
    Correct: Z=1/sC and Y=sC .

    At the end, there is a simple method to detect errors: The dimensions for the several terms must be identical.
    Example: If the denominator contains s^2 (without any factor) all s^1 terms must have a factor like 1/RC and the s^0 term has a factor (1/RC)^2.
    Regards and good luck.
     
    Last edited: Jun 10, 2012
  21. PG1995

    PG1995 Active Member

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    Thank you, Winterstone.

    I have corrected the mistake and now everything fits together well.

    In post #25 Winterstone said this:

    I don't understand the part in red. When lowpass and highpass filters are derived, what Y5 is set equal to? In highpass and lowpass, Y5 doesn't exist, so how do we get rid of it? Please let me know. Thank you.

    Best wishes
    PG
     
    Last edited: Jun 10, 2012
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