Electronic – 3-dB Frequency of second order transfer function

bandwidthcutoff frequencyfilterfrequency responsetransfer function

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:

Is there any approximate approach for obtaining the same?

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

$$|G_1(j\omega)|^2=\frac{|G(0)|^2}{2}=\frac{1}{2b^2}\tag{1}$$

With

$$G_1(j\omega)=\frac{1}{-\omega^2+ja\omega+b}\tag{2}$$

you get

$$|G_1(j\omega)|^2=\frac{1}{(b-\omega^2)^2+a^2\omega^2}\tag{3}$$

Plugging (3) into (1) gives

$$(b-\omega^2)^2+a^2\omega^2=2b^2\tag{4}$$

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

$$x^2+(a^2-2b)x-b^2=0\tag{5}$$

with the positive solution

$$x_0=b-\frac{a^2}{2}+\sqrt{\left(b-\frac{a^2}{2}\right)^2+b^2}\tag{6}$$

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

$$\omega_c=\sqrt{x_0}=\sqrt{b-\frac{a^2}{2}+\sqrt{\left(b-\frac{a^2}{2}\right)^2+b^2}}\tag{7}$$