Calculating the half power frequency

Discussion in 'Circuit Help' started by kdoug, Mar 1, 2011.

  1. kdoug

    kdoug

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    I would like to calculate the half power frequency of the circuit attached.

    I have combined the impedances of R2 and C1 with the 1/Z = 1/Z_1 + 1/Z_2 ... forumla. To get R2 / (R2*j*w*C + 1).

    Then use the voltage divider on this combination and R1 to find that,

    Vin / Vout = 1 / ( R1jwC + (R1/R2 +1) )

    I believe the gain now is |Vin / Vout| = 1 / sqrt( (R1wC)^2 + (R1/R2 + 1)^2 )

    At this point, I'm unsure how to proceed. The Internet has told me that the half power frequency occurs when | Vin / Vout | = 1/sqrt(2). If I set what's contained the the radical above equal to 2 and solve for w, I can find the frequency f. I turns out to 141.5 Hz which doesn't agree with a computer generated Bode plot.

    Can anyone suggest what I'm doing wrong?

    Sorry for the poor Maths formatting.

    Thanks for your time,
     

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    kdoug, Mar 1, 2011
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  2. kdoug

    Laplace

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    The gain of a network is Vo/Vi. The gain of this network is a function of frequency. Since this is a low-pass filter, the highest gain will be found at f=0 so the gain at the cutoff frequency will be half the power of the gain at zero frequency.

    An alternative method is to realize that the cutoff frequency of a simple low-pass filter is Fc=1/(2piRC) where C=C1 and R=R1| R2. My calculator says Fc=175 kHz. What does the computer generated Bode plot say?
     
    Laplace, Mar 2, 2011
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  3. kdoug

    kdoug

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    Thank you for the response, Laplace.

    I've attached the generated Bode plot. I've highlight the cursor representing your answer in green, and what I would have though the right answer is in red (70.7% of 5V, makes for a frequency of approx 325 kHz), though, what I think to be the right answer, appears to have occured after the main "break down" in the curve.

    It's likely I've misunderstood something again, or I'm not using my tools properly. I used PSpice to generate this Bode plot from the circuit I attached previously.

    Thanks for your time & help.
     

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    kdoug, Mar 2, 2011
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  4. kdoug

    Laplace

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    Here are my calculations doing it the long way.
     

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    Laplace, Mar 3, 2011
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