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Graphing Transfer Function

Discussion in 'Mathematics and Physics' started by EN0, Aug 18, 2010.

  1. EN0

    EN0 Member

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    Hi MrAl,

    I acknowledge the fact that the R and C values should be different for HP than for LP, with respect to various frequencies. What other properties do you think I should analyze?

    Clearly, I'm the student here, in which I have no idea as to what path I should take in order to learn filter design. I will always remain indulgent, since you know more than I do about how to learn this material. I don't mind at all if you tell me what you think I should learn, and what order to learn it. In fact, it would be all the more gratifying for me because I would then know what I need to do. In other words, you don't need to ask my opinion about anything relevant to filters, because I don't know much. After all, I'm only a high school student who needs some orders now and again. ;)

    That being said, I'd be grateful if you would let me know what characteristics I should learn about for each filter? Perhaps I should get a filter design book that I can follow along with, and one that has practice problems. Not sure what is out there, but hopefully you know of something good? Even if it is an old filter design textbook, that's alright with me; since the fundamentals of math and filter design have already been discovered, the methods shouldn't change.

    My conventional plan was to learn how to apply mathematics to each filter I encounter, in addition to examining certain properties they exhibit (impedance, frequency characteristics, etc). I'd like to be able to narrow down the mathematics for each filter to a single equation, like I did with the HP and LP filters. That enables me to simply plug in the R, C, and Vin values and produce the results. However, my plans certainly aren't the most coherent means of achieving my filter design goal. Therefore, what do you think I should do next? Not meaning what filter I should learn about next, since we already talked about that, but rather the specific properties the filter I am learning about undergoes? I believe I still should learn more about the HP filter, is that correct?

    So to begin with, maybe you and I could contrive a plan that I can follow along with? I always feel better when a plan is made because it gives me a sense of direction.

    I am extremely thankful for your assistance regarding this matter, it means a lot to me!

    Thanks,

    Austin
     
  2. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    Well maybe we should jump into nodal analysis. Im not sure if you had this yet or not, so i guess i have to ask you that. Have you done any nodal analysis yet? That's probably a good place to start.
    I guess i should also ask you, what is your ultimate goal for these filters, and do you want to stick with passive or do you want to move into active too?
     
  3. EN0

    EN0 Member

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    Yeah, I've done some nodal analysis, which is based on currents. To prove my understanding I'll give you an example: If I have a resistor with a voltage Va going to one terminal and a voltage Vb going to the other, the current through that resistor is (Va - Vb)/R.

    My ultimate goal is to master filter design. That's somewhat of a equivocal answer, but ideally if I'm designing an RF circuit (FM transmitter, for instance) and I'm faced with a noise problem, I'd like to know what filter I need in order to mitigate the noise. Also, I'd like to produce the mathematics to note the affects my filter will have, given a voltage input and various frequencies. In terms of education, I would like to familiarize myself with all types of filters.

    I thought it would be best for me to start out with passive filters, and then move on to active filters. That decision is up to you, but I'm sure that you would want me to learn about both passive and active though?

    What is the order in which college students learn about filters? Maybe it would be a good idea for me to follow their plan?
     
  4. dave

    Dave New Member

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  5. EN0

    EN0 Member

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    Maybe it would be better if I analyzed RLC circuits with nodal analysis before I do filters? I believe that's called "Network Analysis"?

    The problem is that I can't do calculus yet, so I'd have to stay with complex numbers.
     
    Last edited: Sep 8, 2010
  6. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,

    Oh ok, so i guess you know what nodal analysis is so lets do another circuit then. Passive filters are the starter filters yes, and this next circuit problem will address the issue of output loading. Many times the load on a filter will degrade its performance. What you could do next is look at another low pass filter, but this time load the output (from output to ground) with a resistor R2. You'll have to analyze a low pass filter again, but this time it will have another resistor R2 in parallel to the capacitor. You should find the transfer function again.
    If that turns out to be too easy, perhaps you can go back to the BP filter and come up with an expression for the transfer function given two different resistors and two different capacitors. Note this will be a little more difficult than simply convolving the two filter functions, so lets see how you do with that one and we'll go from there.

    When it comes to design work, there are a number of approaches.

    Also, maybe we should stick with RC filters for now and move into inductor filters later.
     
  7. EN0

    EN0 Member

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    Sounds great MrAl, I'll get right to it and let you know how things are going.

    Stay tuned! ;)
     
  8. EN0

    EN0 Member

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    Hopefully this is correct:























    Thanks,

    Austin
     

    Attached Files:

    Last edited: Sep 10, 2010
  9. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    It looks like you're on the right track again, but something seems inconsistent when you go to simplify. For example, the first three lines look ok, but then on the fouth line there seems to come in a new variable "R" and the variable "Vin" seems to get left out.
    You certainly started out right, so i think all you have to do now is go back over it starting from line three and redo the simplification.

    Also, always keep in mind that for any equation with a real part and imag part it will be in the form:
    RealPart+ImagPart*j

    which simply means that the real part is a group of one set of terms and the imag part is another set all of which were multiplied by 'j'.
    For equations that have a denominator, we can always get it into the form:
    (RealPart+ImagPart*j)/D

    which of course can be broken up into:
    RealPart/D+ImagPart*j/D

    and if we think of RealPart/D as being RealPartA and ImagPart/D as being ImagPartA, we can rewrite the original equation as:
    RealPartA+ImagPartA*j

    which means of course that it is back in the simpler complex form:
    realpart+imagpart*j

    It's just that we would have a denominator also for each part instead of just a numerator when we write it out. This means we can note that when the denominator is real it does not affect whether or not the numerator is real or imag...if it was real it stays real, and if it was imag it stays imag. That's one benefit of making the denominator real when it happens to be complex to start with.

    One more quick little note:
    If we multiply A+B+C*j times 'j' we have to get A*j+B*j+C*j^2 (and the j^2 can be simplified more) but we dont only multiply the one term C*j times j to get A+B+C*j^2, which would be incorrect. To get the denominator to be real we have to multiply by the conjugate of the denominator.
     
    Last edited: Sep 10, 2010
  10. EN0

    EN0 Member

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    Yeah, I forgot to do
    instead of
    .

    I'll try and simplify it better, and post it.

    Thanks,

    Austin
     
  11. EN0

    EN0 Member

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    Alright, hopefully this is correct now:





















    I didn't get
    on both sides of each term in the numerator, which on the other transfer functions I have. Nevertheless, my results seem logical. Although, I'm still not convinced it is correct.

    Thanks,

    Austin
     
    Last edited: Sep 10, 2010
  12. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again Eno.


    You're getting closer now. I think all you have to do is be a little more careful when you do your algebraic simplification.
    For example, the first line:
    Vo/RL=(Vin-Vo)/ZR-Vo*j*w*C

    looks good, but then in the second line you left out the "Vo" in the far right hand term and left out the "RL" completely. It might be better for you if you try to simplify one thing at a time instead of two at once. For example, you could first simplify the right side:
    Vo/RL=Vin/ZR-Vo/ZR-Vo*j*w*C

    and then next simplify the left side:
    Vo=RL*Vin/ZR-RL*Vo/ZR-RL*Vo*j*w*C

    or even instead first collect terms with "Vo" in them:
    Vo/RL+Vo/ZR+Vo*j*w*C=Vin/ZR

    and then go from there.

    Sound good to you? Give it another shot and this time take care to keep all the terms intact. If you have any doubts, you can at the very least plug random values into your equation before and after, and make sure you get the same exact result. For example, with:
    Vo/RL=Vin/ZR-Vo/ZR-Vo*j*w*C

    plug in 3 for RL, 4 for Vin, 5 for ZR, 6 for w, and 7 for C, and then calculate what Vo is:
    Vo=6/(315*j+4)

    Now even though those substitute values are not true values, we have something to check out our final answer with later. If the two results do not match perfectly, we know we made a mistake. If they do match, we probably didnt make a mistake.
    In the above example and set of substitutions, we should get Vo=6/(315*j+4) before AND after simplification.
     
    Last edited: Sep 11, 2010
  13. EN0

    EN0 Member

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    I think this might be more like it:













    If that looks good, I'll go ahead and finish doing the real and imaginary parts.
     
  14. Hayato

    Hayato Member

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    Just some suggestions, and a little mistake correction you don't have that '1' over the denominator:
    You don't need to make your denominator pure real, for now (because the phase and magnitude plots are separated), so just keep this one:


    Use a math gimmick, by calling
    and






    Magnitude:






    Phase:
    Another math gimmick:














    If I didn't make any algebra mistake, that is it.
     
    Last edited: Sep 13, 2010
  15. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,

    If you take your second line and multiply both sides by RL you dont get your third line. It looks like RL is missing in some of the terms in the third line.

    For example:
    A=B/C+D+E

    if we multiply both sides by C we get:
    A*C=B+C*D+C*E

    We got rid of the denominator part C but we also have to distribute the C to the other two terms as well.

    Take it from the second line and im sure you'll get there this time :)
     
  16. Hayato

    Hayato Member

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    You know,
    Sometimes you just need to sketch your transfer function, with 3 base-points, take a look:

    For w=0 (DC):
    Zc = infinite.
    So your Vo = Vin*ZL / (ZR + ZL)

    For w=infinite:
    Zc = 0.
    So your Vo = 0.

    For w=fc (-3 dB point with no load):
    Zc = Zcfc.
    So your Vo =Vin*(Zcfc//ZL) / (ZR + Zcfc//ZL).
     
    Last edited: Sep 12, 2010
  17. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello Hayato,

    Oi tudo bem.

    I was going to get into Bode later after we did a few general analyses.
     
  18. Hayato

    Hayato Member

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    Tudo bem, e com vocĂȘ?

    Sorry, l didn't mean to interrupt you.
     
  19. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    Muito bem, obrigado.

    I guess i didnt want to introduce 100 different ways of doing it just yet hoping to not confuse :)

    Actually it is nice of you to join in.
     
    Last edited: Sep 13, 2010
  20. EN0

    EN0 Member

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    Hey MrAl,

    Ok, I think this may be it:





















    I don't think I can simplify it anymore? I could, however, if I didn't have the
    term.

    Edit: I tried graphing the filter, and it produces one that starts at 25V, while
    .

    Thanks,

    Austin
     
    Last edited: Sep 14, 2010
  21. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    You're getting closer now. Just a simple error in line 5, the denominator is not quite correct, but it's close.
     

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