Bandwidth

The bandwidth $\triangle \omega$ of a system is defined as the difference between the two -3dB frequencies $\omega_1$ and $\omega_2$ on each side of the peak frequency, i.e.,

$$\triangle \omega \stackrel{\triangle}{=}\omega_2-\omega_1$$

where $\omega_1$ and $\omega_2$ satisfy

$$20 log_{10} \frac{\left\vert H(\omega_{1,2}) \right\vert}{\left\vert H(\omega_p) \right\vert}=-3$$

or

$$\frac{\left\vert H(\omega_{1,2}) \right\vert}{\left\vert H(\omega_p) \right\vert}=10^{-3/20}=0.707$$

or

$$\frac{\left\vert H(\omega_{1,2}) \right\vert^2}{\left\vert H(\omega_p) \right\vert^2}=0.5$$

We see that the output power at the -3dB frequency is half of that at the peak frequency. As shown before,

$$\left\vert H(j\omega) \right\vert^2=\frac{1}{G(j\omega)}=\frac{1}{(u-1)^2+4\zeta^2u}$$

and

$$\left\vert H(\omega_p) \right\vert^2=\frac{1}{4\zeta^2(1-\zeta^2)}$$

so we have

$$\frac{\left\vert H(\omega_{1,2}) \right\vert^2}{\left\vert ... ...ht\vert^2}= \frac{4\zeta^2(1-\zeta^2)}{(u-1)^2+4\zeta^2u}=0.5$$

This can be rewritten as

$$u^2+(4\zeta^2-2)u+1-8\zeta^2(1-\zeta^2)=0$$

which can be solved for $u$ to get

$$\left\{ \begin{array}{l} u_1=(1-2\zeta^2) - 2\zeta\sqrt{1-\z... ...t{1-\zeta^2}=u_p+ 2\zeta\sqrt{1-\zeta^2} \end{array} \right.$$

and

$$\left\{ \begin{array}{l} \omega_1=\sqrt{u_1} \omega_n \\ \omega_2=\sqrt{u_2} \omega_n \end{array} \right.$$

and

$$\triangle \omega=\omega_2-\omega_1=(\sqrt{u_2}-\sqrt{u_1})\omega_n$$

To simplify this result, consider

$$u_2-u_1=(\sqrt{u_2}+\sqrt{u_1})(\sqrt{u_2}-\sqrt{u_1})=4\zeta\sqrt{1-\zeta^2}$$

We further assume the peak frequency to be in the middle of the two -3dB frequencies (not exactly the case, only an approximation)

$$\sqrt{u_p}\approx \frac{\sqrt{u_2}+\sqrt{u_1}}{2}$$

then we have (assuming $\zeta \ll 1$):

$$\sqrt{u_2}-\sqrt{u_1}=\frac{u_2-u_1}{\sqrt{u_2}+\sqrt{u_1}} ... ...rac{2\zeta\sqrt{1-\zeta^2}}{\sqrt{1-2\zeta^2}} \approx 2 \zeta$$

Now we see that

$$\triangle \omega=\omega_2-\omega_1=(\sqrt{u_2}-\sqrt{u_1})\omega_n \approx 2\zeta \omega_n$$

In summary, we rewrite the peak frequency and the bandwidth of a this second order system as

$$\left\{ \begin{array}{ll} \mbox{Peak Frequency: } & \omega_... ...\triangle \omega \approx 2\zeta \omega_n \end{array} \right.$$

and the transfer function can be rewritten as

$$H(s)=\frac{\omega_n^2}{s^2+\triangle \omega s + \omega_n^2}$$