Please find Cutoff frequency

Discussion in 'General Electronics Chat' started by dileepchacko, Sep 1, 2009.

  1. dileepchacko

    Thread Starter Active Member

    May 13, 2008
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    Dear All

    Please calculate the cutoff frequency of this circuit. I think the R2 will not depend on the cutoff frequency, but it depends on the other parameter such as pass band gain and quality factor.
     
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  2. t06afre

    AAC Fanatic!

    May 11, 2009
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    Can you tell more about why you struggle. You have drawn your figure in some sort of ecad system. Do this tool have a simulator you can use for simulation?
     
  3. dileepchacko

    Thread Starter Active Member

    May 13, 2008
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    Yes I have drown this circuit in LT SPICE. I have the option for simulation, I can get the cutoff frequency by simulating the circuit, but I need the mathematical steps to solve this. I know the Simple LPF cutoff frequency ( 1/(2*pi*R*C)). How R2 is depends on this circuit characteristics.
     
  4. t06afre

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    May 11, 2009
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    You have to find the transfer function. This is equal to Vout/Vin. In this case I think nodal analysis is the essayist way to solve this problem. Are you familiar with this form of circuit analysis?
     
  5. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    A clue: The value of R2 will drastically affect the cutoff frequency, it is as important as R1.
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    Are you planning on using the classic 3db level? This is typically ½ the input signal voltage. This question is more important than it looks, because you have a voltage divider whose voltage out is entirely dependent on frequency, and you need to set a point where this magic level (3db) happens.
     
  7. Electronworks

    New Member

    Sep 1, 2009
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    Combined impedance of C1 and R2 is product/sum, thus:

    (R2/jwC)/(R2+1/jwC) = R2/(1+jwCR2).

    Then use resistive divider equation to find transfer function:

    [R2/(1+jwCR2)] / ([R2/(1+jwCR2)] + R1) = R2/(R2+R1(1+jwCR2))

    cut off frequency is when denominator real = denominator imaginary

    ie when (R2 + R1) = jwCR1R2 (expanding out the denominator)

    If R1 = R2 = 1k and C = 1uF

    2k = 2pi x f x 1uF x 1k x 1k (w = 2pi x f)

    318 Hz

    As a rough check, your 2 1k resistors are in parallel with a voltage source, so the cut off freq is the same as a single RC filter with a 500 Ohm resistor and a 1uF cap - 318 Hz according to my Casio..

    is this what Linear tech say...?

    Interested to know..

    Bill Naylor

    <snip>
    Electronic Kits for Education and Fun
     
    Last edited by a moderator: Sep 1, 2009
  8. t06afre

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    May 11, 2009
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    Is not the 3 Db level equal to 1/square root(2) or approximately 0.707
     
  9. Audioguru

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    Dec 20, 2007
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    -3dB is half the output power or 0.707 times the voltage.
     
  10. Wendy

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    Mar 24, 2008
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    Sheesh. I stepped in that one.
     
  11. Tronster

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    Dec 29, 2008
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    An LT Spice IV simulation of this circuit agrees precisely with your estimate.
     
  12. dileepchacko

    Thread Starter Active Member

    May 13, 2008
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    Yes I need 3dB bandwidth. The 3dB bandwidth cutoff frequency is my requirement.
     
  13. dileepchacko

    Thread Starter Active Member

    May 13, 2008
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    If the circuit is in cutoff frequency how we can write (R1+R2) = 2Pi*R1R2*f ?

    Can you please explain me , Bill Naylor
     
  14. dileepchacko

    Thread Starter Active Member

    May 13, 2008
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    The LT-SPICE simulation tool is showing the Cutoff frequency is = 314.534Hz. this value is based on the 3dB reference point. I have attached the file for the reference. It shows different value than 318hz.
     
  15. Jony130

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    Feb 17, 2009
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    The cut-off frequency is equal
    Vo/Vin=1/√2=0.7071067812=-3.010299957dB
    And for this circuit is equal
    Fc=1/ ( 2*PI*(R1||R2*C) ) = 318.3098862Hz
    And simulation show:
    Fc=318.313Hz
     
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  16. Ron H

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    Apr 14, 2005
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    As Jony130 pointed out, the cutoff frequency occurs at -3.010299957dB. Another, easier to remember point, is the phase shift. It will be exactly -45.0000 degrees.
     
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