First and Second Order Low/High/Band-Pass filters

• Low-pass filter:
  

  $$H(j\omega) = -\frac{Z_2}{Z_1}=-\frac{R_2\;\vert\vert\;(1/j\omega C)}{R_1} =-\... ...frac{R_2/j\omega C}{R_2+1/j\omega C} =-\frac{R_2}{R_1}\frac{1}{j\omega R_2C+1}$$

  $$= -H(0)\frac{1}{1+j\omega \tau}=-H(0)\frac{1}{1+j\omega/\omega_c}$$

  
  
  where $H(0)=R_2/R_1$ is the DC gain when $\omega=0$, $\tau=R_2C$, $\omega_c=1/\tau=1/R_2C$ is cut-off or corner frequency, at which $\vert H(j\omega_c)\vert=H(0)/\sqrt{2}$. Intuitively, when frequency is high, $Z_2(j\omega)$ is small and the negative feedback becomes strong, and the output is low.
  
  $$\vert H(j\omega)\vert=\frac{H(0)}{\sqrt{1+(\omega/\omega_c)... ...{2} & \omega=\omega_c\\ 0 & \omega=\infty\end{array}\right.$$
  
  

  At the cut-off frequency $\omega=\omega_c$, we have
  

  $$\vert H(j\omega_c) \vert=\frac{H(0)}{\vert 1+j \vert}=\frac{H(0) }{\sqrt{2}}$$
  
  
  and
  
  $$Lm H(j\omega_c)=20\log_{10} \left(\frac{H(0)}{\sqrt{2}} \ri... ... =20\log_{10} H(0) -20\log_{10} \sqrt{2}\approx Lm H(0)-3.01$$
  
  
  i.e., the log-magnitude of $H(j\omega_c)$ is approximately $3\;dB$ lower than that of $H(0)$.

  Also, when $\omega\gg \omega_c$, we have
  

  $$Lm H(j\omega)=20\log_{10}\left(\frac{1}{\sqrt{1+(\omega/\om... ...ght) \approx 20\log_{10}\left(\frac{\omega_c}{\omega}\right)$$
  
  
  When frequency is a decade (10 times) higher
  
  $$Lm H(j 10\omega)\approx 20\log_{10}\left(\frac{\omega_c}{1... ...ac{\omega_c}{\omega}\right)-20\log_{10} 10 =Lm H(j\omega)-20$$
  
  
  i.e., the log-magnitude of $H(j 10\omega)$ is approximately 20 dB lower than that of $H(j\omega)$.

  For example, when $\tau=10^{-3}$, $\omega_c=1/\tau=10^3$, the Bode plots are shown below:

  

• High-pass filter:
  

  $$H(j\omega) = -\frac{Z_2(j\omega)}{Z_1(j\omega)} =-\frac{R_2}{R_1+1/j\omega C} =-\frac{R_2}{R_1}\;\frac{1}{1+1/j\omega R_1C}$$

  $$= -H(\infty)\;\frac{1}{1+1/j\omega\tau} =-H(\infty)\frac{1}{1-j\omega_c/\omega}$$

  
  
  where $H(\infty)=R_2/R_1$ is the gain when $\omega\rightarrow \infty$, $\tau=R_1C$, and $\omega_c=1/\tau=1/R_1C$ is the cut-off or corner frequency, at which $\vert H(j\omega_c)\vert=H(\infty)/\sqrt{2}$. Intuitively, when frequency is low $Z_1(j\omega)$ is large and the signal is difficult to pass, therefore the output is low.
  
  $$\vert H(j\omega)\vert=\frac{H(\infty)}{\sqrt{1+(\omega_c/\o... ...mega=\omega_c\\ H(\infty) & \omega=\infty\end{array}\right.$$
  
  
  For example, when $\tau=10^{-6}$, $\omega_c=1/\tau=10^6$, the Bode plots are shown below:

  

  If we let $R_1=R_2$, i.e., $H(0)=1$, and ignore the negative sign ($180^\circ$ phase shift), the low-pass and high-pass filters can be represented by their transfer functions with $s=j\omega$:
  

  $$H_{lp}(j\omega)=\frac{\omega_c}{j\omega+\omega_c}, =-H(\in... ...;\;\;\;\;\;\;H_{hp}(j\omega)=\frac{j\omega}{j\omega+\omega_c}$$
  
  

• Second Order Band-pass Filters:

  

  We let
  

  $$Z_1(\omega)=R_1+(1/j\omega C_1),\;\;\;\;\;\;\; Z_2(\omega)=R_2\vert\vert(1/j\omega C_2)$$
  
  
  and get the FRF of this inverting amplifier as
  

  $$H_{BP}(j\omega) = -\frac{Z_2(\omega)}{Z_1(\omega)} =-\frac{R_2\vert\vert(1/j\omega... .../j\omega C_1} =-\frac{R_2/j\omega C_2}{(R_1+1/j\omega C_1)(R_2+1/j\omega C_2)}$$

  $$= -\frac{j\omega R_2C_1}{(j\omega R_1C_1+1)(j\omega C_2R_2+1)} =-\frac{j\omega \tau_3}{(1+j\omega \tau_1)(1+j\omega \tau_2)}$$

  $$= -\left(\frac{1/\tau_1}{j\omega+1/\tau_1}\right)\; \left(\frac{1/\tau_}{j\omega+1/\tau_2}\right)\; \left(\frac{j\omega}{1/\tau_3} \right)$$

  $$= -\left(\frac{\omega_{c_1}}{j\omega+\omega_{c_1}}\right)\; \left(... ...c_2}}{j\omega+\omega_{c_2}}\right)\; \left(\frac{j\omega}{\omega_{c_3}}\right)$$

  
  
  where $\omega_{c_1}=1/\tau_1=1/R_1C_1$, $\omega_{c_2}=1/\tau_2=1/R_2C_2$, and $\omega_{c_3}=1/\tau_3=1/R_2C_1$.

  We assume both $\omega_{c_1}$ and $\omega_{c_2}$ are higher than $\omega_{c_3}$, i.e., $\omega_{c_1},\;\omega_{c_2}>\omega_{c_3}$, then we have a band-pass filter, as can be seen in the Bode plot.

  For example, when $\omega_{c_1}=10^6$, $\omega_{c_2}=10^8$, and $\omega_{c_3}=10^3$, the Bode plots are shown below:

  

• Second Order Band-stop Filters:

  If we swap $Z_1(\omega)$ and $Z_2(\omega)$ in the op-ammp circuit of the band-pass filter, we get:
  

  $$H_{BS}(j\omega) = -\frac{Z_1(\omega)}{Z_2(\omega)} =-\frac{R_2+1/j\omega C_2}{R_1\... ...1/j\omega C_1} =-\frac{R_2+1/j\omega C_2}{R_1/j\omega C_1/(R_1+1/j\omega C_1)}$$

  $$= \frac{(1+j\omega C_1R_1)(1+j\omega C_2R_2)}{j\omega R_1C_2} =\frac{(1+j\omega\tau_1)(1+j\omega\tau_1)}{j\omega\tau_3}$$

  
  

  The log-magnitude of the Bode plot of this circuit is
  

  $$20\log \bigg\vert \frac{Z_1(\omega)}{Z_2(\omega)} \bigg\ver... ...{Z_1(\omega)} \bigg\vert =-Lm_{BP}(j\omega)=Lm_{BS}(j\omega)$$
  
  
  We see that this is band-stop filter, and its log-magnitude is a vertically flipped version of that of the band-pass filter considered previously.