Chapter 15, Problem 15.18P

(a)

To determine

To derive: the expression for the voltage transfer function

Explanation of Solution

Given:

  

Calculation:

Redraw the given circuit in s -domain as

  

From the circuit,

  $\begin{array}{l}\frac{0-V_i(s)}{R_1+\frac{1}{s{C}_1}}+\frac{0-V_o(s)}{R_2\frac{1}{s{C}_2}}=0\\ \frac{\left(s{C}_1\right)V_i(s)}{s{R}_1{C}_1+1}=\frac{-V_o(s)}{\frac{R_2\frac{1}{s{C}_2}}{R_2+\frac{1}{s{C}_2}}}\\ \frac{\left(s{C}_1\right)V_i(s)}{s{R}_1{C}_1+1}=\frac{-\left(s{R}_2{C}_2+1\right)V_o(s)}{R_2}\\ \frac{s{R}_2{C}_1}{\left(s{R}_1{C}_1+1\right)\left(s{R}_2{C}_2+1\right)}=\frac{-V_o(s)}{V_i(s)}\end{array}$

Now the transfer function is

  $T(s)=\frac{V_o(s)}{V_i(s)}$

Hence

  $\begin{array}{l}T(s)=\frac{-s{R}_2{C}_1}{s^2{R}_1{R}_2{C}_1{C}_2+s\left(R_1{C}_1+R_2{C}_2\right)+1}\\ =\frac{-s{R}_2{C}_1}{s{R}_1{C}_1\left[s{R}_2{C}_2+\left(1+\frac{R_2{C}_2}{R_1{C}_1}\right)+\frac{1}{s{R}_1{C}_1}\right]}\end{array}$

Thus, the required voltage transfer function is

  $T(s)=\frac{-R_2}{R_1}\left\{\frac{1}{s{R}_2{C}_2+\left(1+\frac{R_2{C}_2}{R_1{C}_1}\right)+\frac{1}{s{R}_1{C}_1}}\right\}$

Conclusion:

Thus, the required voltage transfer function is

  $T(s)=\frac{-R_2}{R_1}\left\{\frac{1}{s{R}_2{C}_2+\left(1+\frac{R_2{C}_2}{R_1{C}_1}\right)+\frac{1}{s{R}_1{C}_1}}\right\}$

(b)

To determine

To find: The value of $C_1,C_2,R_2$

Answer to Problem 15.18P

The required values are $C_2=64.92\text{nF}$ , $C_1=156.3\text{nF}$ and $R_2=510.6\text{k}\Omega$

Explanation of Solution

Given:

  $R_1=10\text{k}\Omega$

Magnitude of the mid band gain is 50

Cut-off frequency $f_{3dB1}=200\text{Hz}$

Cut-off frequency $f_{3dB2}=5\text{kHz}$

  

Calculation:

Consider $s=j\omega$

The magnitude of the voltage transfer function is given by

  $|T(s)|=-\frac{R_2}{R_1}\left|\frac{1}{j\omega R_2{C}_2+\left(1+\frac{R_2{C}_2}{R_1{C}_1}\right)+\frac{1}{j\omega R_1{C}_1}}\right|$

  $|T(s)|=-\frac{R_2}{R_1}\left|\frac{1}{\left(1+\frac{R_2{C}_2}{R_1{C}_1}\right)+j\left(\omega R_2{C}_2-\frac{1}{\omega R_1{C}_1}\right)}\right|$

  $|T(s)|=-\frac{R_2}{R_1}\left\{\frac{1}{\sqrt{{\left(1+\frac{R_2{C}_2}{R_1{C}_1}\right)}^2+{\left(\omega R_2{C}_2-\frac{1}{\omega R_1{C}_1}\right)}^2}}\right\}$

  $|T(s)|=-\frac{R_2}{R_1}{\left\{{\left(1+\frac{R_2{C}_2}{R_1{C}_1}\right)}^2+{\left(\omega R_2{C}_2-\frac{1}{\omega R_1{C}_1}\right)}^2\right\}}^{-\frac{1}{2}}$

  $\begin{array}{l}\text{ When }\omega R_2{C}_2-\frac{1}{\omega R_1{C}_1}=0\hfill \\ |T(s)|=-\frac{R_2}{R_1}\frac{1}{\left(1+\frac{R_2{C}_2}{R_1{C}_1}\right)}\hfill \end{array}$

Given $|T(s)|=50$

To determine the cut-off frequency $\frac{|T(s)|}{\sqrt{2}}$ assume the equality function as

  $\omega R_2{C}_2-\frac{1}{\omega R_1{C}_1}=\pm \left(1+\frac{R_2{C}_2}{R_1{C}_1}\right)$

For $f=200\text{Hz}$

  $\begin{array}{l}{\omega }_1=2\pi (200)\\ =1257\end{array}$

  $=1.257\text{kradlsec}$

For $f=5\text{kHz}$

  $\begin{array}{l}{\omega }_2=2\pi (5\text{k})\\ =31415.93\end{array}$

  $=31.42\text{kradlsec}$

Let us consider ${\tau }_1=R_1{C}_1$ and ${\tau }_2=R_2{C}_2.$ now, substitute ${\tau }_1,{\tau }_2$ in equation (1) then

  $\begin{array}{l}|T(s)|=+\frac{R_2}{R_1}\left(\frac{1}{1+\frac{{\tau }_2}{{\tau }_1}}\right)\\ =50\end{array}$

The equation (1) becomes,

  $\begin{array}{l}{\omega }_2{\tau }_2-\frac{1}{{\omega }_2{\tau }_1}=+\left(1+\frac{{\tau }_2}{{\tau }_1}\right)\dots \dots (3)\hfill \\ \text{ And also }{\omega }_1{\tau }_2-\frac{1}{{\omega }_1{\tau }_1}=-\left(1+\frac{{\tau }_2}{{\tau }_1}\right)\cdots \dots (4)\hfill \end{array}$

From equation (2),

  $\begin{array}{l}\frac{{\omega }_2^2{\tau }_1{\tau }_2-1}{{\omega }_2{\tau }_1}=\frac{{\tau }_1+{\tau }_2}{{\tau }_1}\hfill \\ {\omega }_2{\tau }_1{\tau }_2-\frac{1}{{\omega }_2}={\tau }_1+{\tau }_2\hfill \\ {\tau }_1\left({\omega }_2{\tau }_2-1\right)=\frac{1}{{\omega }_2}+{\tau }_2\hfill \\ {\tau }_1=\frac{\frac{1}{{\omega }_2}+{\tau }_2}{{\omega }_2{\tau }_2-1}\hfill \end{array}$

Substituting equation (5) in equation (4),

  $\begin{array}{l}{\omega }_1{\tau }_2-\frac{1}{{\omega }_1\left[\frac{\frac{1}{{\omega }_2}+{\tau }_2}{{\omega }_2{\tau }_2-1}\right]}=-\left[1+\frac{{\tau }_2}{\frac{\frac{1}{{\omega }_2}+{\tau }_2}{{\omega }_2{\tau }_2-1}}\right]\\ {\omega }_1{\tau }_2-\frac{{\omega }_2{\tau }_2-1}{{\omega }_1\left[\frac{1}{{\omega }_2}+{\tau }_2\right]}=-\left[1+\frac{{\tau }_2\left({\omega }_2{\tau }_2-1\right)}{\frac{1}{{\omega }_2}+{\tau }_2}\right]\end{array}$

Substitute the values of ${\omega }_1,{\omega }_2$ in the above expression as it is in the form of polynomial equation as

  $\begin{array}{l}\left[\begin{array}{c}{\tau }_2^2\{{\left(1.257\times {10}^3\right)}^2\left(31.42\times {10}^3\right)\\ +\left(1.257\times {10}^3\right){\left(31.42\times {10}^3\right)}^2\\ +\tau \left\{{\left(1.257\times {10}^3\right)}^2-{\left(31.42\times {10}^3\right)}^2\right\}\\ +\left(1.257\times {10}^3+31.42\times {10}^3\right)\\ \end{array}\right]=0\\ {\tau }_2^2\left(1.291\times {10}^{12}\right)-\left(985.64\times {10}^6\right){\tau }_2+32.677\times {10}^3=0\end{array}$

Hence the roots are obtained as

  $\begin{array}{l}{\tau }_2=\frac{\left\{-\left(-985.64\times {10}^6\right)\pm \sqrt{{\left(-985.64\times {10}^6\right)}^2-4\left(1.291\times {10}^{12}\right)\left(32.677\times {10}^3\right)}\right\}}{2\left(32.677\times {10}^3\right)}\\ =7.287\times {10}^{-4},3.315\times {10}^{-5}\end{array}$

Since ${\omega }_2$ is larger, ${\tau }_2$ should be small then consider

  ${\tau }_2=0.3315\times {10}^{-4}\text{s}$

Now, substitute the value of ${\tau }_2$ in equation (5) then

  $\begin{array}{l}{\tau }_1=\frac{\frac{1}{{\omega }_2}+{\tau }_2}{{\omega }_2{\tau }_2-1}\\ =\frac{\frac{1}{31.42\times {10}^3}+0.3315\times {10}^{-4}}{\left(31.42\times {10}^3\right)\left(0.3315\times {10}^{-4}\right)-1}\\ =\frac{64.98\times {10}^{-6}}{0.04157}\\ =1.563\times {10}^{-3}s\end{array}$

  $\begin{array}{l}\text{Mid band }\mathrm{gain}=\frac{R_2}{R_1}\cdot \frac{1}{1+\frac{{\tau }_2}{{\tau }_1}}\\ 50=\frac{R_2}{R_1}\left(\frac{1}{1+\frac{{\tau }_2}{{\tau }_1}}\right)\end{array}$

Hence, by substituting the values

  $\begin{array}{l}50=\frac{R_2}{R_1}\left(\frac{1}{1+\frac{0.3315\times {10}^{-4}}{1.563\times {10}^{-3}}}\right)\\ =\frac{R_2}{R_1}\left(\frac{1}{1.0212}\right)\\ =\frac{R_2}{R_1}(0.979)\end{array}$

Further simplify as

  $\begin{array}{l}{R}_2=R_1\left(\frac{50}{0.979}\right)\hfill \\ R_2=51.06R_1\hfill \end{array}$

Given $R_1=10\text{k}\Omega$ then

  $\begin{array}{l}{R}_2=51.06\left(10\times {10}^3\right)\\ =510.6\times {10}^3\end{array}$

Therefore, the required resistor value is $R_2=510.6\text{k}\Omega$

Now

  $\begin{array}{l}{\tau }_1=R_1{C}_1\\ C_1=\frac{{\tau }_1}{R_1}\\ =\frac{1.563\times {10}^{-3}}{10\times {10}^3}\\ =1.56\times {10}^{-7}\end{array}$

Therefore, the required capacitor $C_1$ value is $C_1=156.3\text{nF}$

  $\begin{array}{l}{\tau }_2=R_2{C}_2\\ C_2=\frac{{\tau }_2}{R_2}\\ =\frac{0.3315\times {10}^{-4}}{510.6\times {10}^3}\\ =6.492\times {10}^{-11}\end{array}$

Conclusion:

Therefore, the required values are $C_2=64.92\text{nF}$ , $C_1=156.3\text{nF}$ and $R,=510.6\text{k}\Omega$