Twin-T Active Filter

• The RCR network is is a T (or Y) network that can be converted to a $\pi$ (or $\Delta$) network (see here) of three branches:
    $$Z'_1=Z'_2=R+\frac{1}{2j\omega C}+\frac{R/2j\omega C}{R} =R+\frac{1}{j\omega C}=\frac{j\omega RC+1}{j\omega C}$$ (1)
    $$Z'_3=R+R+\frac{R^2}{1/2j\omega C}=2R+2R^2j\omega C=2R(1+j\omega RC)$$ (2)
  The frequency response function of the voltage divider formed by $Z'_3$ and $Z'_2$ is:
  

  $$H'(j\omega) = \frac{Z'_2}{Z'_2+Z'_3} =\frac{(1+j\omega RC)/j\omega C}{(1+j\omega RC)/j\omega C+2R(1+j\omega RC)}$$

  $$= \frac{1}{1+2j\omega RC}=\frac{1}{1+2j\omega\tau}$$

  where $\tau=RC$. This is a first-order low-pass filter with cut-off frequency at $\omega_c=1/2\tau$:
    $$\vert H'(j\omega)\vert=\left\{\begin{array}{ll}1 & \omega=0\\ 1/\sqrt{2}& \omega=1/2\tau\\ 0&\omega\rightarrow\infty\end{array}\right.$$ (3)

• The CRC network is a T (or Y) network that can be converted to a $\pi$ (or $\Delta$) network of three branches:
    $$Z''_1=Z''_2=\frac{R}{2}+\frac{1}{j\omega C}+\frac{R/2j\omega C}{1/j\omega C} =R+\frac{1}{j\omega C} =\frac{j\omega RC+1}{j\omega C}$$ (4)
    $$Z''_3=\frac{1}{j\omega C}+\frac{1}{j\omega C}+\frac{1/(j\omega C)... ...{j\omega C}+\frac{2}{R(j\omega C)^2} =\frac{2(1+j\omega RC)}{R(j\omega C)^2}$$ (5)
  The frequency response function of the voltage divider formed by $Z''_3$ and $Z''_2$ is:
  

  $$H''(j\omega) = \frac{Z''_2}{Z''_2+Z''_3} =\frac{(1+j\omega RC)/j\omega C} {(1+j\omega RC)/j\omega C+2(1+j\omega RC)/R(j\omega C)^2}$$

  $$= \frac{j\omega RC}{j\omega RC+2}=\frac{j\omega\tau}{j\omega\tau+2} =\frac{1}{1-j2/\omega\tau}$$

  This is a first-order high-pass filter with cut-off frequency at $\omega_c=2/\tau$:
    $$\vert H'(j\omega)\vert=\left\{\begin{array}{ll}0 & \omega=0\\ 1/\sqrt{2}& \omega=2/\tau\\ 1&\omega\rightarrow\infty\end{array}\right.$$ (6)

As these two $\pi$-networks are combined in parallel, they form a single $\pi$-network with three branches $Z_1=Z'_1\vert\vert Z''_1$, $Z_2=Z'_2\vert\vert Z''_2$, and $Z_3=Z'_3\vert\vert Z''_3$:

  $$Z_1=Z'_1\vert\vert Z''_1=Z_2=Z'_2\vert\vert Z''_2=\frac{1}{2}\left(R+\frac{1}{j\omega C}\right)$$ (7)

  $$Z_3=Z'_3\vert\vert Z''_3=\frac{Z'_3 Z''_3}{Z'_3+Z''_3} =\frac{2R(1+j\omega RC)}{1+(j\omega RC)^2}$$ (8)

The frequency response function of this $\pi$-network (a voltage divider) is:

$$H(j\omega) = \frac{Z_2}{Z_2+Z_3}=\frac{R+1/j\omega C} {R+1/j\omega C+4R(1+j\omega RC)/(1+(j\omega RC)^2)}$$

$$= \frac{(1+j\omega RC)/j\omega C}{(1+j\omega RC)/j\omega C+4R(1+j\omega RC)/(1+(j\omega RC)^2)}$$

$$= \frac{1/j\omega C}{1/j\omega C+4R/(1+(j\omega RC)^2)} =\frac{1}{1+4j\omega RC/(1+(j\omega RC)^2)}$$

$$= \frac{1+(j\omega\tau)^2}{1+(j\omega\tau)^2+4j\omega\tau} =\frac{(j\omega)^2+(1/\tau)^2}{(j\omega)^2+4j\omega/\tau+(1/\tau)^2}$$

Alternatively, we apply KCL at the middle points of the RCR and CRC networks with $v_a$ and $v_b$:

  $$\frac{v_{in}-v_a}{R}+\frac{v_{out}-v_a}{R}=\frac{v_a}{1/j\omega 2... ...\frac{v_{in}-v_b}{1/j\omega C}+\frac{v_{out}-v_b}{1/j\omega C}=\frac{v_b}{R/2}$$ (9)

Solving for $v_a$ and $v_b$, we get

  $$v_a=\frac{v_{in}+v_{out}}{2(1+j\omega\tau)},\;\;\;\;\;\;\; v_b=\frac{(v_{in}+v_{out})j\omega\tau}{2(1+j\omega\tau)}$$ (10)

Also, apply KCL to the output to get

  $$\frac{v_{out}-v_a}{R}+\frac{v_{out}-v_b}{1/j\omega C}=0,\;\;\;\;\;\; v_{out}=\frac{v_a+j\omega\tau v_b}{1+j\omega\tau}$$ (11)

Substituting $v_a$ and $v_b$ into this equation we get:

  $$v_{out}=\frac{v_{in}(1+j\omega\tau)+v_{out}(1-(\omega\tau)^2)}{2(1+j\omega\tau)^2}$$ (12)

Solving this for $v_{out}$

  $$H(j\omega)=\frac{v_{out}}{v_{in}}=\frac{1+(j\omega\tau)^2}{1+4j\o... ...au)^2} =\frac{(j\omega)^2+\omega_n^2}{(j\omega)^2+4j\omega\omega_n+\omega_n^2}$$ (13)

where $\omega_n=1/\tau=1/RC$.

Active twin-T filter

  $$v_1=\frac{R_5}{R_4+R_5}v_{out}=\alpha v_{out}$$ (14)

where $\alpha=R_5/(R_4+R_5)$, i.e., $1-\alpha=R_4/(R_4+R_5)$.

  $$\frac{v_{in}-v_a}{R}+\frac{v_{out}-v_a}{R}=\frac{v_a-v_1}{1/j\ome... ...c{v_{in}-v_b}{1/j\omega C}+\frac{v_{out}-v_b}{1/j\omega C}=\frac{v_b-v_1}{R/2}$$ (15)

Solving these for $v_a$ and $v_b$, we get

  $$v_a=\frac{v_{in}+(1+2\alpha j\omega\tau)v_{out}}{2(1+j\omega\tau)... ...; v_b=\frac{j\omega\tau v_{in}+(j\omega\tau+2\alpha)v_{out}}{2(1+j\omega\tau)}$$ (16)

Substituting these into the expression for $v_{out}$, we get

  $$v_{out}=\frac{v_{in}+v_{out}(1+2\alpha j\omege\tau) +(j\omega\tau)^2v_{in}+j\omega\tau(j\omega\tau+2\alpha)v_{out}}{2(1+j\omega\tau)^2}$$ (17)

$$H(j\omega) = \frac{v_{out}}{v_{in}}=\frac{1+(j\omega\tau)^2}{1+4j\omega\tau(1-\alpha)+(j\omega\tau)^2}$$

$$= \frac{1+(j\omega\tau)^2}{1+4j\omega\tau R_4/(R_4+R_5)+(j\omega\ta... ...=\frac{\omega_n^2-\omega^2}{\omega_n^2+4j\omega\omega_n R_4/(R_4+R_5)-\omega^2}$$