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    low-pass RC filter design with attenuation known

    Hello,

    I want to ask this question stated in my textbook:

    Can an low-pass RC filter be used to provide 1dB attenuation at 21 kHz and at least 22 dB attenuation at 820 kHz?

    It is written in the solution that the 3 dB frequency is 41,27 kHz. I don't know where this comes from?

    Can you please help m? Thank you.

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    Re: low-pass RC filter design with attenuation known

    Hi,

    What informations about filters do you have?
    Do you know how they are defined?
    Can you draw a frequency response chart? And insert the given informations as points and lines.

    Klaus
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    Re: low-pass RC filter design with attenuation known

    The attenuation is easy enough to calc: Xc/SQRT((Xc)^2+(R)^2), 20log Answer for deci-Bell, where Xc = 1/( 2.pi.F.C)

    at 820kHz the L of the cap and the wiring must be taken into consideration - as well as any L in the R.



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    Re: low-pass RC filter design with attenuation known

    The combination of capacitor and resistor create a time constant (R x C). This is associated with a phase change. The phase difference results in reducing the amplitude. A given frequency is attenuated according to a formula.
    To see the relationships it is helpful to draw the parallelogram angles. That might be the means to determine what time constant is suitable, and what R & C values will solve the exercise.



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    Re: low-pass RC filter design with attenuation known

    The ratio of R/C is dependent on the value of any load R... heavy loading = large C, and vice versa ....


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    Re: low-pass RC filter design with attenuation known

    I applied this equation for the 1dB attenuation:
    20xlog(Xc/sqrt(R^2+Xc^2)=-1 dB

    At the end I got RC=3.79x10^-6 which is for 21kHz

    Then I computed f3dB=1/(2*pi*RC)= 41993 Hz
    which is quite similar to the textbook solution: 41270 Hz

    Then I computed the attenuation for 820kHz and got 25.81 dB which is again similar to the textbook solution 25.97 dB.

    Does that make sense? Because I did many other computations maybe with better accuracy but I lost them as there were many of meaningful and meaningless computations :)



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    Re: low-pass RC filter design with attenuation known

    Your home work is really your concern, not to clog up message boards like this one ...


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    Re: low-pass RC filter design with attenuation known

    Quote Originally Posted by Easy peasy View Post
    Your home work is really your concern, not to clog up message boards like this one ...
    I respect your opinion, however i was meaning to check out the accuracy with your help. A criticism is also not a part of the scientific discussion.



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    Re: low-pass RC filter design with attenuation known

    Hi,

    I did not calculate the values...

    But regarding the accuracy:
    With low frequencies you are in an about horizontal line. This means a small error in attenuation causes a big error in frequency.
    With higher frequencues you are in an steap area of the curve, then an error in attenuation causes a much smaller error in frequency.

    Thus I assume, if you start your calculations from the higher frequencies you get more exact results.

    I'd do such test calculations with Excel, because it's just repeating the same formula with different input values.
    And you easily can calculate 50 points (or so) and draw a chart to visualize the results.

    Klaus
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    Re: low-pass RC filter design with attenuation known

    Graph of ordinary RC low-pass filter (first order). The rolloff curve occupies a comparatively narrow range of the frequency spectrum. At low frequencies there is slight attenuation. At high frequencies there is extreme attenuation.

    Click image for larger version. 

Name:	RC low pass filter 1 to 1 MHz (falstad).png 
Views:	2 
Size:	13.2 KB 
ID:	147170

    By convention, 3 dB attenuation designates the cutoff frequency. Your task is to calculate RC values which produce the desired frequencies for dB attenuation. A suitable R value to start with is 1k. This assumes the exercise can be solved for a first-order filter. If you cannot then a second-order filter might be needed.


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    Re: low-pass RC filter design with attenuation known

    The numbers in calculated in post #6 are correct, but not very accurate. Pocket calculator gives me RC = 3.8565e-6.



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    Re: low-pass RC filter design with attenuation known

    Quote Originally Posted by KlausST View Post
    Hi,

    I did not calculate the values...

    But regarding the accuracy:
    With low frequencies you are in an about horizontal line. This means a small error in attenuation causes a big error in frequency.
    With higher frequencues you are in an steap area of the curve, then an error in attenuation causes a much smaller error in frequency.

    Thus I assume, if you start your calculations from the higher frequencies you get more exact results.

    I'd do such test calculations with Excel, because it's just repeating the same formula with different input values.
    And you easily can calculate 50 points (or so) and draw a chart to visualize the results.

    Klaus
    I have tried the 820 kHz and got RC=2.42e-6 so it's better than the previous value gotten by 21kHz.

    Anyway I don't have my own PC. I use public computer. Can you please suggest an online application that can help me with this? I think, however, that my calculations are not too bad as they are done manually.



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