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Graphing Transfer Function

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

  1. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    On the graphs make sure to find the pass band gains and the -3db points as that is what this is all about.
     
    Last edited: Sep 21, 2010
  2. EN0

    EN0 Member

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    I apologize for not coming on sooner! I fixed my error in the other post, so it should look ok now.

    I’ve found some math software’s, which I’m looking into. I’ll let you know when the graphs are produced.
     
  3. EN0

    EN0 Member

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    Hey MrAl,

    I am very sorry this took so long to generate, but I had a lot of school work and then I needed to learn the syntax of Freemat. Anyway, here it is:

    -3dB Point of Ordinary LPF: 15915.49 Hz
    -3dB Point of LPF With Load: 15915.49 Hz

    Given:

    R1 = 1kΩ
    C = 10nF
    R2 = 1kΩ
    Vin = 5V

    Is that correct?
     

    Attached Files:

    Last edited: Sep 28, 2010
  4. dave

    Dave New Member

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  5. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    Something doesnt look right in the graphs. How does the amplitude get as high as 5v in the top trace for example?
     
  6. EN0

    EN0 Member

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    Well the top graph is without a load, so it should start from 5V, shouldn't it? The lower graph is with the load so it begins at 2.5V. I should have told you beforehand which was which, sorry.
     
  7. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,

    So you are using an input voltage of 5v then? Usually you would use 1v to graph it. Of course it helps to graph in db gain too rather than voltage so you can easily spot the -3db point.
     
  8. EN0

    EN0 Member

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    After all that, I need to change the graphs again? ;)

    Alright, I'll fix it up and put it in terms of dB. One volt for the input voltage is a good idea, I'll implement that as well.

    Till Then,

    Austin
     
    Last edited: Oct 1, 2010
  9. EN0

    EN0 Member

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    Hey MrAl,

    Apparently there is a mistake in our mathematics? Hayato provided the graph that is attached and his equations are different.
     

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  10. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello,

    I'll double check and get back here.
     
  11. EN0

    EN0 Member

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    Hi MrAl,

    I believe our first transfer function is correct, but the second one isn't. I went over the math myself and will show you my results.

    Conventional Low-Pass Filter

    Final Equation

    [Error: Syntax\ error : /usr/bin/latex --interaction=nonstopmode 182e04b2a5fea3adb4a88a85e92dc47d.tex && /usr/bin/dvipng -q -D 300 -T tight -gamma 2.0 -bg Transparent -o 5e0bfaa73c137e992b3eb5d28e591195-2.png 182e04b2a5fea3adb4a88a85e92dc47d.dvi]

    Low-Pass Filter With Load

























    Final Equation



    I believe that's correct?

    [​IMG]
     
    Last edited: Oct 23, 2010
  12. Hayato

    Hayato Member

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    Plot the graph again, if they are similar to the ones I've plotted, they are ok.
     
  13. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,


    Ok since you only have a problem with the circuit with the load resistor we'll concentrate on that one.

    With R1 being the input resistor and R2 being the load resistor and C the only capacitor, the transfer function is:
    Vout=(Vin*R2)/(s*C*R1*R2+R1+R2)

    and the amplitude (after subst s=j*w) is:
    VoutAmpl=(Vin*R2)/sqrt(w^2*C^2*R1^2*R2^2+R2^2+2*R1*R2+R1^2)

    Now this last result has been verified in a circuit simulator, which by the way you should be doing too. You can compare your result to that VoutAmpl and if you dont get that or an equivalent expression then it can not be correct so you'll have to see what went wrong.
    You might note that both R1 and R2 appear in the denominator too. I also think that using Latex just complicates matters rather than making them simpler. You should probably avoid that until you get the final correct solution.
     
    Last edited: Oct 24, 2010

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