The Bode plot of second order systems

We first consider the transfer function of a 2nd order system (e.g., RCL circuit with voltage across the capacitor C as the output):

$$H(s)=k_0 \frac{\omega_n^2}{s^2+2\zeta \omega_n s + \omega_n^2}$$

where $k_0$ is an arbitrary gain factor. If we let $s=j\omega$, we get the frequency transfer function

$$H(j\omega)=\frac{\omega_n^2}{-\omega^2+2\zeta \omega_n j \om... ...-(\frac{\omega}{\omega_n})^2+j2\zeta(\frac{\omega}{\omega_n})}$$

Specially, when $\omega=\omega_n$, we have

$$H(\omega_n)=\frac{1}{j2\zeta} =\frac{-j}{2\zeta}$$

with magnitude

$$\left\vert H(\omega_n) \right\vert = \frac{1}{2\zeta} \stackrel{\triangle}{=} Q$$

The Bode plot of this transfer function can be found as

$$20 log_{10} \left\vert H(j\omega) \right\vert =-10 log_{10} [... ...ac{\omega^2}{\omega_n^2})^2+(2\zeta\frac{\omega}{\omega_n}^2)]$$

and

$$\angle H(j\omega) = -tan^{-1} \frac{2\zeta \omega / \omega_n}{1-(\omega/\omega_n)^2}$$

To obtain the asymptotic approximation of this function, consider the following three cases:

• When $\omega \ll \omega_n$ i.e., $\omega/\omega_n \rightarrow 0$, we have
  
  $$20 log_{10} \left\vert H(j\omega) \right\vert \rightarrow -10 log_{10} \left\{ 1 \right\} =0$$
  
  
  and
  
  $$\angle H(j\omega) \rightarrow -tan^{-1}(0) = 0$$
  
  

• When $\omega \gg \omega_n$, we have
  
  $$20 log_{10} \left\vert H(j\omega) \right\vert \rightarrow -... ...a_n^2}]^2 \right\} = -40 log_{10} ( \frac{\omega}{\omega_n} )$$
  
  
  and
  
  $$\angle H(j\omega) \rightarrow -tan^{-1} (-0) = - \pi$$
  
  

• When $\omega=\omega_n$, we have
  
  $$20 log_{10} \left\vert H(j\omega) \right\vert = -20 log_{10} 2 \zeta = 20 log_{10} Q$$
  
  
  and
  
  $$\angle H(j\omega) = -tan^{-1}(+\infty)=-\frac{\pi}{2}$$
  
  
  We see that the smaller $\zeta$, the larger $Q \stackrel{\triangle}{=} 1/2\zeta$, and the sharper and narrower the peak.

Next consider a second type of 2nd order system (e.g., RCL circuit with voltage across the resistor R as the output):

$$H(j\omega)=\frac{j2\zeta\omega_n}{-\omega^2+2\zeta \omega_n j... ...-(\frac{\omega}{\omega_n})^2+j2\zeta(\frac{\omega}{\omega_n})}$$

As the denominator of the frequency response function is already considered above, all we need to do now is to find the log-magnitude of the numerator:

$$20 log_{10} \left\vert j2\zeta \omega/\omega_n \right\vert=20 log_{10} ( 2\zeta \omega/\omega_n )$$

The is a straight line with a slope of 20 dB/decade. In particular, when $\omega=\omega_n$, it is $20 log_{10} (2\zata)$, which cancels the denominator $-20 log_{10} (2\zata)$, in other words, at $\omega=\omega_n$, $H(j\omega)=1$ and $20 \log_{10} \left\vert H(j\omega)\right\vert=0$. The phase of the numerator is

$$\angle (j2\zeta \omega/\omega_n)=\pi/2$$

Finally consider a second type of 2nd order system (e.g., RCL circuit with voltage across the inductor L as the output):

$$H(j\omega)=\frac{(j\omega)^2}{-\omega^2+2\zeta \omega_n j \om... ...-(\frac{\omega}{\omega_n})^2+j2\zeta(\frac{\omega}{\omega_n})}$$

the log-magnitude of the numerator:

$$20 log_{10} \left\vert (-(\omega/\omega_n)^2 \right\vert=40 log_{10} ( \omega/\omega_n )$$

The is a straight line with a slope of 40 dB/decade. In particular, when $\omega=\omega_n$, it is $20 log_{10} (1)=0$. The phase of the numerator is

$$\angle (-(\omega/\omega_n)^2)=\pi$$