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How can i obtain the 3db frequency of the each transfer function ?

My attempt:

My guess is both transfer functions have the same 3db frequency.But equating the magnitude to 0.707 gives a fourth order equation which can be reduced to second order to obtain square of 3db frequency.

However, i am not able to simplify it.The exact expression however is as follows:

enter image description here

Is there any approximate approach for obtaining the same?

  • \$\begingroup\$ For G1(s), write the numerator as 'b' \$\endgroup\$
    – Chu
    Dec 26 '15 at 17:56
  • \$\begingroup\$ just need to be careful as dc gain is 1/b right? i tuk that into account while calculating . \$\endgroup\$ Dec 26 '15 at 17:57
  • \$\begingroup\$ It's 4th order, but reduces to 2nd order if you let \$\omega^2=\Omega\$. Then factorises easily. \$\endgroup\$
    – Chu
    Dec 26 '15 at 18:01
  • \$\begingroup\$ does factor,but not into a simple expression , there will be quadratic terms under the root thou \$\endgroup\$ Dec 26 '15 at 18:03
  • 2
    \$\begingroup\$ Why do you think it's a HPF? \$\endgroup\$
    – Chu
    Dec 26 '15 at 18:17

I show you how to obtain the 3dB cut-off frequency for the low pass filter \$G_1(s)\$. You can calculate the cut-off frequencies of the band pass filter \$G_2(s)\$ in a similar way, as long as you know that its maximum magnitude is attained at \$\omega=\sqrt{b}\$, as pointed out in a comment by robert bristow-johnson. The latter fact can be derived by setting the derivative of \$|G_2(j\omega)|^2\$ to zero. (Note that \$b>0\$ is always satisfied for \$G_1(s)\$ and \$G_2(s)\$ to be transfer functions of causal and stable filters.)

To compute the 3dB cut-off frequency of \$G_1(s)\$ you have to solve




you get


Plugging (3) into (1) gives


With the substitution \$x=\omega^2\$, you get the following quadratic equation


with the positive solution


From (6), the value of the 3dB cut-off frequency is



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