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Math Help - Low pass filter

  1. #1
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    Low pass filter

    With R = 1k ohm, C = 3.1827x10^-8, Find 6db cutoff freq.

    How do i get the eqn?
    Attached Thumbnails Attached Thumbnails Low pass filter-low-pass-filter.bmp  
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  2. #2
    Senior Member Peritus's Avatar
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    The simple answer to your question is that you have to write the equations of the circuit according to Kirchhoff's circuit laws, and then use the relation of the change of voltage across the capacitor and the current flowing through it:


    I(t) = C\frac{{dV(t)}}<br />
{{dt}}
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  3. #3
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    If i am using voltage divider rule?
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  4. #4
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    R = R
    C = 1/jwC
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  5. #5
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    ok, but do you have to find the cutoff frequency of Vout or Vout1?
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  6. #6
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    Vout
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  7. #7
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    Z_{out1}  = \left( {R + \frac{1}<br />
{{jwc}}} \right)||\frac{1}<br />
{{jwc}} = \frac{{1 + Rjwc}}<br />
{{2jwc - Rw^2 c^2 }}

    now we use voltage divider:

    <br />
V_{out1}  = V_{in} \frac{{Z_{out1} }}<br />
{{Z_{out1}  + R}}<br />

    now we'll use the voltage divider again:

    <br />
V_{out}  = V_{out1} \frac{{\frac{1}<br />
{{jwc}}}}<br />
{{R + \frac{1}<br />
{{jwc}}}} = V_{out1} \frac{1}<br />
{{1 + Rjwc}} = <br />

    <br />
 = V_{in} \frac{{Z_{out1} }}<br />
{{Z_{out1}  + R}}\frac{1}<br />
{{1 + Rjwc}}<br />

    <br />
H_{out} (jw) = \frac{{Z_{out1} }}<br />
{{Z_{out1}  + R}}\frac{1}<br />
{{1 + Rjwc}}<br />
    Last edited by Peritus; February 27th 2008 at 05:56 AM.
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  8. #8
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    hi,

    your Vout is the voltage across Zout

    but my Vout is the voltage across the 2nd Cap(RHS)

    are both the same?
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  9. #9
    Senior Member Peritus's Avatar
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    Quote Originally Posted by MilK View Post
    hi,

    your Vout is the voltage across Zout

    but my Vout is the voltage across the 2nd Cap(RHS)

    are both the same?
    A little blunder on my part, but I've fixed it, check my post again.
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  10. #10
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    I see, thks

    If i am told to find the 6db cutoff freq,

    how do i translate this statement into your eqn?

    i got this eqn:

    Vout[ (-1/jw^3c^3) + R^3 + (4R^2/jwc) - (4R/w^2c^2) ] = Vin[ (-R/w^2c^2) - (1/jw^3c^3)]

    how do i solve for f?
    Last edited by MilK; February 27th 2008 at 06:49 AM.
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  11. #11
    Senior Member Peritus's Avatar
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    you can easily verify that the passband gain is 0db thus we are interested in finding the frequency in which the gain of the transfer function is -6db:

    <br />
20\log _{10} \left| {H_{out} \left( {jw} \right)} \right| =  - 6

    solve for w...
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  12. #12
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    I see...

    But how can i simplifiy my eqn:

    Vout[ (-1/jw^3c^3) + R^3 + (4R^2/jwc) - (4R/w^2c^2) ] = Vin[ (-R/w^2c^2) - (1/jw^3c^3)]

    to get the magnitude? struck at here.
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  13. #13
    Senior Member Peritus's Avatar
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    Quote Originally Posted by Peritus View Post
    Z_{out1}  = \left( {R + \frac{1}<br />
{{jwc}}} \right)||\frac{1}<br />
{{jwc}} = \frac{{1 + Rjwc}}<br />
{{2jwc - Rw^2 c^2 }}

    now we use voltage divider:

    <br />
V_{out1}  = V_{in} \frac{{Z_{out1} }}<br />
{{Z_{out1}  + R}}<br />

    now we'll use the voltage divider again:

    <br />
V_{out}  = V_{out1} \frac{{\frac{1}<br />
{{jwc}}}}<br />
{{R + \frac{1}<br />
{{jwc}}}} = V_{out1} \frac{1}<br />
{{1 + Rjwc}} = <br />

    <br />
 = V_{in} \frac{{Z_{out1} }}<br />
{{Z_{out1}  + R}}\frac{1}<br />
{{1 + Rjwc}}<br />

    <br />
H_{out} (jw) = \frac{{Z_{out1} }}<br />
{{Z_{out1}  + R}}\frac{1}<br />
{{1 + Rjwc}}<br />
    after substituting Zout1 we get:

    <br />
\left| {H_{out} \left( {jw} \right)} \right| = \frac{1}<br />
{{\left[ {1 - \left( {Rwc} \right)^2 } \right]^2  + \left( {3wc} \right)^2 }}<br /> <br />
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  14. #14
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    Hi,

    I got struck after simplify to this eqn:

    (-2.03(10^-9))w^2 - (1.026(10^-18))w^4 -1 =0
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  15. #15
    Senior Member Peritus's Avatar
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    just substitute: x = w^2 and solve the quadratic equation...
    Last edited by Peritus; February 27th 2008 at 09:52 PM.
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