Low Pass filter

Discussion in 'Homework Help' started by MilK, Mar 1, 2008.

  1. MilK

    Thread Starter Member

    Mar 1, 2008
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    Hi,

    Eqn for low pass RC filter

    (Vout/Vin) = (1/jwc) / ((1/jwc)+R)

    R = 1khz
    C=3.1847 x 10^-8
    f= 50khz
    Vin= 2v

    Find Vout express in Db

    My working:

    i find the magnitude of both side which gives

    vout = 2/ sqrt(1+R^2 w^2 c^2)
    sub in all values i get vout = 4.997 x 10^-5

    change to db i get : 20log(4.997x10^-5) = -86.02db which is wrong:confused:

    answer shld be around 5db

    what when wrong?

    Thks.

    to admin: Sorry for the "Urgent help needed" topic......pls remove it
     
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    I don't think the answer is around 5 dB.

    I would think Vo = (Vin * Xc) / Z
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    This is wrong. How did you get this? Show your steps.


    This is also wrong. Where did you get it?

     
  4. MilK

    Thread Starter Member

    Mar 1, 2008
    25
    0
    I've solve it:) Thanks
     
  5. rwmoekoe

    Active Member

    Mar 1, 2007
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    db = 20 log( vo/vi )
    = 20 log( zc / (zc+r) )
    = 20 log( (1/wc) / (1/wc+r) )
    = 20 log( (1 / (2 pi 50k 3.18e-8) / (1 / (2 pi 50k 3.18e-8) + 1k) )
    = 20 log( 100 / (100+1k) )
    = -20 db
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    This calculation actually comes out to -20.83dB, which is wrong.

    You forgot that the capacitor's reactance is 90 degrees out of phase with the resistance. You have to find the vector magnitude of the denominator.
    Gain = 20 log( (1/wc) / sqrt((R^2) + (1/wc)^2)
    Gain = -20.05dB

    The OP wanted Vout when Vin=2V.
    Vout=-14.03dB.
     
  7. JoeJester

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    Apr 26, 2005
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    Ron,

    I got 198.9 mV as Vo, which is approximately -20.05 dB with respect to 2 Vi.

    I didn't understand your Vo being -14.03 dB
     
  8. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    The OP wrote:
    You got Vout=198.9 mV.
    20 log (.1989) = -14.03dB.
    You are correct in showing gain as -20.05dB, which will of course be true for any input voltage. I realize that dB is a unitless gain measurement (a ratio), but the OP said
    I suppose to be correct, we should say
    Vout=-14.03dBV, i.e., relative to 1 volt, which identifies it as an absolute measurement rather than a ratio.
     
  9. JoeJester

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    Apr 26, 2005
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    That is true Ron. Had you stated dBV, there wouldn't be an issue. At least you didn't use dBf as the reference ... :D
     
  10. Ron H

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    Apr 14, 2005
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    Database Format? Divorced Black Female? Decibels (frequency)? I guess I'm not up-to-date on acronyms.:(
     
  11. JoeJester

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    Apr 26, 2005
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    femtowatt.

    I saw it on an HP signal generator back in the early 80s. Not one of the more popular references. :D
     
  12. Ron H

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    Apr 14, 2005
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    That's funny! Definitely engineering humor, though. My wife doesn't think much of any of my jokes or "funny" stories. I guess I won't try that one on her.:rolleyes:
     
  13. JoeJester

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    Apr 26, 2005
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    If I didn't have to look it up, I would have never remembered it. It was on the old HP-8640 signal generator series.

    And to think 1 uV (50 ohms) is 130 dBf. The stuff people remember is amazing sometimes. :D
     
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