Gain in dB

Discussion in 'Homework Help' started by Hello, Mar 30, 2009.

  1. Hello

    Thread Starter Active Member

    Dec 18, 2008
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    Calculate (i) the gain in dB at 10kHz and (ii) the cut-off frequency for
    R = 100W and C = 1μF if the input voltage = 10V.


    Any help is appreciated.
     
    Last edited: Mar 30, 2009
  2. ABoul

    Member

    Mar 30, 2009
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    can you get the gain itself? because the gain in dB is simply 20log |G|, where |G| is the modulus of the gain, and that's log to the base 10.
     
  3. Hello

    Thread Starter Active Member

    Dec 18, 2008
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    But how would i work out the output voltage?
     
  4. ABoul

    Member

    Mar 30, 2009
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    what you have there is a potential divider. let X denote the impedence of the capacitor, so that X = 1/jwC. the output voltage would therefore be V_out = V_in[X / (X + R)]. from there, you can divide by V_in to get V_out/V_in, plug in your numbers, find the modulus and 20log it.
     
  5. scythe

    Active Member

    Mar 23, 2009
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    you're asking for help on your exam?
     
  6. ABoul

    Member

    Mar 30, 2009
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    past papers are a great way to prepare for your REAL exams. in fact, it's the only way i prepare for them.
     
  7. Hello

    Thread Starter Active Member

    Dec 18, 2008
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    You're right! Thanks for the help.
     
  8. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    If so, I think he already failed about 10 months ago (exam dated May/08). :p
     
  9. ABoul

    Member

    Mar 30, 2009
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    what goes inside the log is |Vo/Vi|, NOT Vo.
     
  10. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    @ 10kHz ≈ -16dB (I believe you had a typo)
    -3dB / 45° cutoff ≈ 1.6kHz
    Filter is 2nd order low pass filter, attenuation -6dB/octave after cutoff point.

    Always check other calculations to see if they "make sense", i.e. 10kHz is about 4x 1.6kHz, so either -dB per octave is way too low for attenuation, or the 10kHz calculation is incorrect. :)
     
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