Bode Plots of first and Second Order Systems

First order circuits

• Voltage $V_C$ across C is treated as output. According to voltage divider rule, we have:
    $$H_C(j\omega)=\frac{V_C}{V_{in}}=\frac{Z_C}{Z_R+Z_C} =\frac{1/j\omega C}{R+1/j\omega C} =\frac{1}{j\omega RC+1}=\frac{1}{j\omega \tau+1}$$ (174)
  where $\tau=RC$.
• Voltage $V_R$ across R is treated as output:
    $$H_R(j\omega)=\frac{V_R}{V_{in}}=\frac{Z_R}{Z_R+Z_C} =\frac{R}{R+... ...mega C}=\frac{j\omega RC}{j\omega RC+1} =\frac{j\omega \tau}{j\omega \tau+1}$$ (175)

As $H_R(j\omega)$ can be written as:

  $$H_R(j\omega)= \frac{1}{j\omega \tau+1} j\omega \tau$$ (176)

The first term is just $H_C(j\omega)$. Now the log-magnitude is:

  $$Lm\;H_R(j\omega) =20\log_{10} \left\vert \frac{1}{j\omega \tau+1}... ...ft\vert j\omega \tau \right\vert =Lm\; H_C(j\omega) +20\log_{10} (\omega\tau)$$ (177)

The first term is the same as $H_C(j\omega)$ and the second plot is a straight line with slope of 20 dB/dec. at $\omega=\omega_c=1/\tau$, the first term is -3 dB and the second is 0 dB. The phase plot is:

  $$\angle H_R(j\omega)=\angle \left(\frac{1}{j\omega \tau+1}\right)+\angle j\omega \tau =\angle H_C(j\omega)+\frac{\pi}{2}$$ (178)

In the plots below, $\tau=0.01$, $\omega_C=100$ rad/sec.

Define $\omega_c=1/\tau=1/RC$ as the cut-off frequency, then when $\omega=\omega_c$, we have $\omega\tau=1$, and $\vert H_R(j\omega)\vert=\vert H_C(j\omega)\vert=1/\sqrt{2}$, i.e., $\omega_c$ is the half-power point, where $\vert H(j\omega)\vert$ is -3 dB.

Second order circuits

• Voltage $V_C$ across C is treated as output:
  

  $$H_C(j\omega) = \frac{V_C}{V_{in}}=\frac{Z_C}{Z_L+Z_R+Z_C} =\frac{1/j\omega C}{j\omega L+R+1/j\omega C} =\frac{1}{(j\omega)^2 LC+j\omega RC+1}$$

  $$= \frac{1/LC}{(j\omega)^2 +j\omega R/L+1/LC} =\frac{\omega_n^2}{(j\... ...^2_n} =\frac{1}{(1-\frac{\omega^2}{\omega_n^2})+j2\zeta\frac{\omega}{\omega_n}}$$ (179)

  where
    $$\omega_n=\frac{1}{\sqrt{LC}},\;\;\;\;\zeta=\frac{R}{2}\sqrt{\frac{C}{L}}$$ (180)
  The magnitude is
    $$\vert H_C(j\omega)\vert =\frac{1}{\sqrt{(1-\frac{\omega^2}{\omeg... ...2+4\zeta^2 \frac{\omega^2}{\omega_n^2}}} =\frac{1}{\sqrt{(1-u)^2+4\zeta^2 u}}$$ (181)
  where $u=(\omega/\omega_n)^2$. When $u=\omega/\omega_n=1$ or $\omega=\omega_n$, we have
    $$\vert H(j\omega_n) \vert=\frac{1}{2\zeta}=Q$$ (182)
• Voltage $V_R$ across R is treated as output:
  

  $$H_R(j\omega) = \frac{V_R}{V_{in}}=\frac{Z_R}{Z_L+Z_R+Z_C} =\frac{R}{j\omega L+R+1/j\omega C} =\frac{j\omega RC}{(j\omega)^2 LC+j\omega RC+1}$$

  $$= \frac{j\omega R/L}{(j\omega)^2 +j\omega R/L+1/LC} =\frac{2\zeta\o... ...2 +2\zeta\omega_n j\omega+\omega^2_n} =H_C(j\omega) \;2\zeta \omega_n\;j \omega$$ (183)

  Now we have:
    $$Lm\;H_R(j\omega)=Lm\; H_C(j\omega)+Lm\;(2\zeta\omega_n\;j\omega),\;\;\;\; \angle H_R(j\omega)=\angle H_C(j\omega)+\angle(2\zeta\omega_n\;j\omega)$$ (184)
  The log-magnitude of the second factor is a straight line with slope 20 dB/dec, and at $\omega=\omega_n$, its value is $20\log_{10} 2\zeta\omega_n^2$. The phase is $90^\circ$ for all $\omega$.

  The denominator can be written as $R+j(\omega L-1/\omega C)$, which is minimized when the imaginary part is zero, i.e, $j\omega L=1/j\omega C$. In other words, when $\omega=\omega_n=1/\sqrt{LC}$, $\vert H_R(j\omega)\vert$ reaches its peak value.

• Voltage $V_L$ across L is treated as output:
  

  $$H_L(j\omega) = \frac{V_L}{V_{in}}=\frac{Z_L}{Z_L+Z_R+Z_C} =\frac{j\omega L}{j\omega L+R+1/j\omega C} =\frac{(j\omega)^2 LC}{(j\omega)^2 LC+j\omega RC+1}$$

  $$= \frac{(j\omega)^2}{(j\omega)^2 +j\omega R/L+1/LC} =\frac{(j\omega)^2}{(j\omega)^2 +2\zeta\omega_n j\omega+\omega^2_n} =H_C(j\omega) \;(j\omega_n)^2$$ (185)

  Now we have:
    $$Lm\;H_L(j\omega)=Lm\; H_C(j\omega)+Lm\;(j\omega)^2,\;\;\;\; \angle H_L(j\omega)=\angle H_C(j\omega)+\angle (j\omega)^2$$ (186)
  The log-magnitude of the second factor is a straight line with slope 40 dB/dec, and at $\omega=\omega_n$, it's value is $20\log_{10} \omega_n^2$. The phase is $180^\circ$ for all $\omega$.

In the following plots, $\omega_n=100$ rad/sec and $\zeta=0.05$. At $\omega=\omega_n$, $20\log_{10} (2\zeta\omega_n^2)=20\log_{10} 1000=60$ dB, and $20\log_{10}(\omega_n^2)=20\log_{10} 10,000=80$ dB.

Example, a Band-pass filter:

  $$H(j\omega)=-\frac{Z_2(j\omega)}{Z_1(j\omega)} =-\frac{R_2\vert\ve... ..._1)/j\omega C_1} =-\frac{j\omega \tau_3}{(1+j\omega \tau_1)(1+j\omega \tau_2)}$$ (187)

where $\tau_1=R_1C_1$, $\tau_2=R_2C_2$, $\tau_3=R_2C_1$.