RC filter

Discussion in 'Homework Help' started by skusku, Aug 28, 2012.

  1. skusku

    Thread Starter Active Member

    Aug 9, 2009
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    Hi, this ain't homework though but i suspect something similar for a test.

    Last week i had a similar practical about an active filter...

    All i want to know is how to calculate the cut off freq of this 2stage filter with an Rload resistor.

    circuit - http://i49.tinypic.com/rbwsoz.jpg

    3db cut off - http://i49.tinypic.com/2em0qb4.jpg

    Above are the practical results.

    So the default equation goes Fc=1/(2∏RC)

    I understand that Rs and R1 is in series but my equation and practical doesn't match up.

    Fc=1/[(2∏(R1+Rs)//R2//R3)*(C1+C2)]
    = 212 Hz

    With orcad I get 122Hz. Can someone please help me with the theory cut off freq?

    Thank you
     
    Last edited: Aug 28, 2012
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I obtain a transfer function as follows

    G(s)=\frac{V_{out}}{V_2}=\frac{1}{a\(s^2+bs+c \)}

    where

    a={(R_s+R_1)R_2C_1C_2}

    b=\frac{1}{R_3C_2}+\frac{R_s+R_1+R_2}{(R_s+R_1)R_2C_1}+\frac{1}{R_2C_2}

    and

    c=\frac{R_s+R_1+R_2+R_3}{(R_s+R_1)R_2R_3C_1C_2}

    The -3dB point would be found for ω when

    \frac{c}{\sqr{{\( c-\omega^2\)}^2+b^2\omega^2}}=\frac{1}{\sqr2}

    For which I obtain ω=762.691 radians/sec or f=121.386Hz
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Can you tell me how you get this equation ?
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    The second RC filter on the right overloads the first RC filter on the left so they make a very poor filter. There should be a buffer amplifier between the RC filters.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    If

    G_{(s)}=\frac{1}{a \( s^2 +bs +c\)}

    the DC gain is simply

    G_{(0)}=\frac{1}{ac}

    At frequency ω [rads/sec]

    G_{\( j \omega \)}=\frac{1}{a\( -\omega^2+j \omega b+c\)}

    From which we obtain the magnitude

    |G_{\(\omega \)}|=\frac{1}{a sqrt{  {\( c-\omega^2 \) }^2 +\omega^2 b^2}}

    The -3dB point on the DC gain has a gain value

    |G_{-3dB}|=\frac{1}{\sqrt{2}} \times \frac{1}{ac}

    The value of ω which satisfies this condition is found by equating

    |G_{\(\omega \)}|=|G_{-3dB}|

    or

    \frac{1}{a sqrt{  {\( c-\omega^2 \) }^2 +\omega^2 b^2}}=\frac{1}{\sqrt{2}} \times \frac{1}{ac}

    which after multiplying through by ac gives

    \frac{c}{sqrt{  {\( c-\omega^2 \) }^2 +\omega^2 b^2}}=\frac{1}{\sqrt{2}}
     
  6. Jony130

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    Feb 17, 2009
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    OK I understand but I still have a question.
    Why don't you use a classical form of a transfer function?

    G_{(s)}=\frac{a}{\( s^2 +bs +c\)}
     
  7. t_n_k

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    I wasn't particularly concerned with standard forms. Variable a just happened to be the term I'd factored out in the denominator. No doubt I could have made one the inverse of the other to agree with convention.

    I must pose the obvious question - Would it have made any difference to the final relationship?
     
  8. Jony130

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    Feb 17, 2009
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    I think that the answer look like this

    \frac{a}{ sqrt{  {\( c-\omega^2 \) }^2 +\omega^2 b^2}}=\frac{1}{\sqrt{2}} \times \frac{a}{c}

    So if we multiplying through by c and divide by a the answer will look the same.

    So this equation holds for any second-order filter ?
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I guess it's definitely true of a low pass type without any pass-band ripple - such as the Butterworth filter.

    What about the Chebyshev LPF with a 3dB passband ripple say...? I'd have to think about it some more.
     
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