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High Pass filter

  1. Feb 5, 2015 #1
    1. The problem statement, all variables and given/known data
    ''A machineoutputs a 4kHz sine wave signal of amplitude 2 volts peak, plus a 100Hz interference component of amplitude 1 V peak''

    Given equation: Vout/Vin = 1/((1+(f0/f^2))^0.5 phase lead = inverse tan (f0/f)

    f0 = 1/(2*pi*R*C)


    I'm given a 4 kilo-ohm resistor and need to choose a capacitor to reduce the interference (100 Hz signal) to 0.05 V peak; I need to design a high pass RC circuit.



    2. Relevant equations

    given in the Q

    3. The attempt at a solution I'm not sure what f0 and f actually is. Equations I can find which are similar to this have 'critical frequencies' mentioned.
     
  2. jcsd
  3. Feb 5, 2015 #2

    gneill

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    Staff: Mentor

    In the equations given, f would be the frequency of interest that you are trying to suppress, while fo would be the "corner frequency" of the filter which is determined by the choice of R and C for the filter.
     
  4. Feb 5, 2015 #3
    Hi thanks for the reply. So would f would be 100 Hz? But I'm not sure how to cut the interference to 0.05 V only. Any ideas?
     
  5. Feb 5, 2015 #4

    gneill

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    Staff: Mentor

    Yes, 100 Hz is what you are trying to suppress from 1 V down to 0.05 V. Looks like a Vo/Vi ratio is close at hand ;)
     
  6. Feb 6, 2015 #5

    LvW

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    Vout/Vin = 1/[(1+(f0/f)^2)]^0.5

    Hello - at first, I have corrected the expression for the magnitude of a firdt-order highpass (see above).
    Because fo must be chosen so that at f=100 hz the damping factor is 1/0.05=20 we have the expression

    [(1+(f0/f)^2)]^0.5=20;
    (fo/f)^0.5=399~400.
    (fo/f)=20
    ;
    fo=20*f=2kHz.

    Hence, the highpass corner frequency is fo=2kHz.
    Therefore: T=RC=1/wo=1/(2Pi*fo)=1.6 ms.

    Now you have to find suitable values for R and C for realizing the time constant T=RC.

    Remark: Sorry - if I gave too many details in my answer. Only now I have realized that I am in the homework section.
     
    Last edited: Feb 6, 2015
  7. Feb 6, 2015 #6
    Thanks!!!!!!!!!!!
     
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