Two Low pass filter cascaded together

Discussion in 'General Electronics Chat' started by reachme09, Jun 22, 2010.

  1. reachme09

    Thread Starter New Member

    Jun 22, 2010
    6
    0
    Hi,

    Can anyone tell me how to calculate the cuttoff frequency and circuit delay in cascaded low pass RC filter.

    1. Cuttoff frequency calculation
    2. Equation for KCL in Laplace form for calcualting circuit delay

    I have contacted many people, but i have not got the desired answer.
    Please anyone help me in solving my problem.

    I am attaching the schematic for your reference..
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I get 's' domain transfer function .....

    \frac{V_o}{V_i}=\frac{1}{R_1R_2C_1C_2\[s^2+\[\frac{1}{R_1C_1}+\frac{1}{R_2C_2}+\frac{1}{R_2C_1}\]s+\frac{1}{R_1R_2C_1C_2}\]}

    Where R1, R2, C1 & C2 are from left to right on your schematic

    which for your values gives cut-off at f0=1865.4 Hz [-9.134dB] based on ....

    \omega_o^2=\frac{1}{R_1R_2C_1C_2}

    The -3dB point is at about 753 Hz.

    The rise time [10% to 90% FV] to a pulse input is about 478.7 uS - done by simulation.
     
    Last edited: Jun 22, 2010
  3. reachme09

    Thread Starter New Member

    Jun 22, 2010
    6
    0
    Could you please explain me
    How [​IMG] and The -3dB point is at about 753 Hz.

    I am able to understand how the below transfer function came


    [​IMG]
    but not the first and second frequency
     
  4. reachme09

    Thread Starter New Member

    Jun 22, 2010
    6
    0
    Could you please explain me
    How [​IMG] and The -3dB point is at about 753 Hz.

    I am able to understand how the below transfer function came
    Could you please explain me
    How [​IMG] and The -3dB point is at about 753 Hz.

    I am able to understand how the below transfer function came


    [​IMG]
    but not the first and second frequency

    [​IMG]
    but not the first and second frequency
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    From the general 2nd order transfer function denominator term of the form

    s^2+2\zeta\omega_o s+\omega_o^2

    the phase shift is 90° at

    \omega=\omega_o

    where the substitution s=jω is made in the second order denominator term.

    The assumption used is that the cut-off is defined at that value of ω where the transfer function phase shift value is 90°.

    The 753 Hz -3dB is obtained by solving for the required ω value at the -3dB value for the transfer function magnitude.

    It depends on what one defines as the cut-off frequency. The normal convention is to take the -3dB frequency. The issue is reconciling the mathematical interpretation with respect to the transfer function at the frequency ωo.
     
  6. ue418

    New Member

    Jun 22, 2010
    17
    1
    I tried this a few years ago, but eventually gave up and had to go to active filters. I actually built some 4th-order active lowpass filters (i.e. cascaded) using formulas from Horowitz' book. But eventually I found that even this didn't give sufficient "sharpness" of cut-off frequency. So I went to an off-the-shelf 8th-order Bessel filter. It's also programmable, allowing me to change the cutoff frequency digitally from 1 to 256 Hz...so it's pretty expensive. But they also make less glamorous 8th-order filters, etc. This was my experience, in a very condensed nutshell. Unless your application is able to handle an extremely "soft" knee or cut-off region, just cascading two passive filters together will not give good results.
     
    reachme09 likes this.
Loading...