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Low Pass Filter

  1. Jun 5, 2013 #1
    Exams are coming up and i've been doing some revision from past papers. Having issues with working out part c of the attached problem. I've already calculated part a and b, possibly correct. Basically I simplified resistances and capacitance. And to have a constant frequency I let Z equal the inverse of the simplified resistance to cancel ω the frequency component. I then used this information to determine a new resistance and capacitance that had no ω component.

    Any hints or tips would be greatly appreciated.

    Thanks
     

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  2. jcsd
  3. Jun 5, 2013 #2

    rude man

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    If you did parts a and b correctly (they're really the same question) then you got R1C1 = R2C2 where 1/Z = G1 + sC1, G1 = 1/R1, and R2 = 1/G2 = 10K and C2 = 1 nF.

    Now R1 → 2R1 so what is your new transfer function? And same question when R1 → R1/2. Do you know how to draw Bode plots?

    Part 1 c iii does not specify the circuit so can't be answered.
     
  4. Jun 6, 2013 #3
    Not sure exactly what you meant by your answer to the first two parts.. However the way i calculated it was by simplifying the resistor and capacitor in parallel. This gave me 1x10^4 + 1x10^6/jw. Then I assumed that to remove jw, since Z is now in series with this equivalent resistance, it would have to equal -1x10^6/jw.

    For part b. I assumed a resistor of 1Ω in parallel with a capacitor of 0.1x10^-6. Simplifying this to obtain 1x10^6/jw + 1. I assume this answer would be negative as that's how it would cancel if it was added in series. This is the new realised Z which then adds to the result of the previous question which gave a result of 9999Ω. Which has no ω component.

    Is this the correct approach?

    Also, I'm not sure how to draw a bode plot. I've done them in practical classes with a simulation program called SPICE, however the book doesn't explain how to draw them very well.. Would you mind giving me an overview of the process?
     
    Last edited: Jun 6, 2013
  5. Jun 6, 2013 #4

    rude man

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    What do you mean by 'simplifying'? Do you mean determining the combined impedance of the 10K resistor and 1 nF capacitor in parallel? If so, you did that wrong. Y = G + jwC so Z = 1/Y where G = 1/R. The real part of the complex impedance of the 10K and 1 nF is frequency-dependent.

    BTW 1 nF = 10^(-9)F, not 10^(-6).

    So you don't have the correct network for frequency-independent gain yet.

    ******************************************************************
    In this situation I know from experience that it's much easier to deal in admittance Y, conductance G and susceptance B than impedance Z, resistance R and reactance X. The relations are simple: Y = 1/Z, B = 1/X and G = 1/R. Example: for a capacitor, B = wC and Y = jB.

    You'll recall that for parallel-connected components, susceptances are added: so for your R and C, Y = G + jB = G + jwC, G = 1/R.

    Given the above, what is Vout/Vin of your diagram if we replace Z with Y = 1/Z?

    This is much too long a subject for me to discuss here. Look at

    http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/Freq/Freq5.html or

    www.dartmouth.edu/~sullivan/22files/Bode_plots.pdf
     
  6. Jun 6, 2013 #5
    Ok thanks for all your help!

    Just to get me back on track, Z is meant to equal Z0 which is: 1/(1x10^(-9)jw + 1/10000)?

    I just used the relationships you were describing to find: Y = 1 x 10^(-9)jw + 1/10000

    Then to find Impedance: Z = 1/Y, therefore Z = 1/(1x10^(-9)jw + 1/10000)

    Is this correct so far?
     
  7. Jun 6, 2013 #6

    rude man

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    Fine so far! But forget Z of your network. You'll be dealing with Y's instead.

    Also, don't put in numbers yet. Call the 1 nF = C2 and the 10K = R2 = 1/G2.

    Now, what is Vout/Vin in terms of this Y2 = G2 + jwC2 and the Y equivalent of the Z as marked on your diagram? Defie Y1 = 1/Z.
     
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