Chapter 15, Problem 6E

For the circuit in Fig. 15.56, (a) determine the transfer function H(jω) = V_out/V_in in terms of circuit parameters R₁, R₂, and C; (b) determine the magnitude and phase of the transfer function at ω = 0, 3 × 10⁴ rad/s, and as ω → ∞ for the case where circuit values are R₁ = 500 Ω, R₂ = 40 kΩ, and C = 10 nF.

FIGURE 15.56

To determine

The transfer function $H\left(j\omega \right)$.

Answer to Problem 6E

The transfer function $H\left(j\omega \right)$ is $\frac{-j\omega C{R}_2}{2-{\omega }^2{C}^2{R}_1{R}_2+j\omega C{R}_1}$.

Explanation of Solution

Given data:

The required diagram is shown in Figure 1.

Calculation:

The equivalent impedance of the capacitor is given as,

$X_C=\frac{1}{j\omega C}$

The Kirchhoff’s current law equation at node ${\text{v}}_{\text{x}}$ is written as,

$\frac{{\text{v}}_{\text{x}}-{\text{v}}_{\text{in}}}{R_1}+\frac{{\text{v}}_{\text{x}}}{R_1}+\frac{{\text{v}}_{\text{x}}-{\text{v}}_{\text{out}}}{\frac{1}{j\omega C}}+\frac{{\text{v}}_{\text{x}}}{\frac{1}{j\omega C}}=0$        (1)

Here,

$R_1$ is the resistor.

$C$ is the capacitor.

$\omega$ is the angular frequency.

${\text{v}}_{\text{out}}$ is the output voltage.

${\text{v}}_{\text{in}}$ is the input current.

The Kirchhoff’s current law equation at ${\text{v}}_{\text{out}}$ is written as,

$\frac{{\text{v}}_{\text{out}}-{\text{v}}_{\text{x}}}{\frac{1}{j\omega C}}+\frac{{\text{v}}_{\text{out}}}{R_2}=0$

${\text{v}}_{\text{out}}-{\text{v}}_{\text{x}}=-\frac{1}{j\omega C}\frac{{\text{v}}_{\text{out}}}{R_2}=0$

${\text{v}}_{\text{x}}=\left(\text{1}+\frac{1}{j\omega R_{2}C}\right){\text{v}}_{\text{out}}$ (2)\

The transfer function $H\left(j\omega \right)$ is written as,

$H\left(j\omega \right)=\frac{{\text{v}}_{\text{out}}}{{\text{v}}_{\text{in}}}$        (3)

Substitute $\left(1+\frac{1}{j\omega C{R}_2}\right){\text{v}}_{\text{out}}$ for ${\text{v}}_{\text{x}}$ in equation (1).

$\begin{array}{l}\left[\begin{array}{l}\frac{\left(\left(1+\frac{1}{j\omega C{R}_2}\right){\text{v}}_{\text{out}}\right){\text{-v}}_{\text{in}}}{R_1}+\frac{\left(\left(1+\frac{1}{j\omega C{R}_2}\right){\text{v}}_{\text{out}}\right)}{R_1}+\frac{\left(\left(1+\frac{1}{j\omega C{R}_2}\right){\text{v}}_{\text{out}}\right){\text{-v}}_{\text{out}}}{\frac{1}{j\omega C}}+\\ \frac{\left(\left(1+\frac{1}{j\omega C{R}_2}\right){\text{v}}_{\text{out}}\right)}{\frac{1}{j\omega C}}=0\end{array}\right]\\ {\text{v}}_{\text{out}}\left(\frac{1+\frac{1}{j\omega C{R}_2}}{R_1}+\frac{1+\frac{1}{j\omega C{R}_2}}{R_1}+\frac{1+\frac{1}{j\omega C{R}_2}}{\frac{1}{j\omega C}}-\frac{1}{\frac{1}{j\omega C}}+\frac{1+\frac{1}{j\omega C{R}_2}}{\frac{1}{j\omega C}}\right)={\text{v}}_{\text{in}}\left(\frac{1}{R_1}\right)\\ {\text{v}}_{\text{in}}=\frac{2-{\omega }^2{C}^2{R}_1{R}_2+j\omega C{R}_1}{-j\omega C{R}_2}{\text{v}}_{\text{out}}\end{array}$

Substitute $\frac{2-{\omega }^2{C}^2{R}_1{R}_2+j\omega C{R}_1}{-j\omega C{R}_2}{\text{v}}_{\text{out}}$ for ${\text{v}}_{\text{in}}$ in equation (3).

$\begin{array}{c}H\left(j\omega \right)=\frac{{\text{v}}_{\text{out}}}{\frac{2-{\omega }^2{C}^2{R}_1{R}_2+j\omega C{R}_1}{-j\omega C{R}_2}{\text{v}}_{\text{out}}}\\ =\frac{-j\omega C{R}_2}{2-{\omega }^2{C}^2{R}_1{R}_2+j\omega C{R}_1}\end{array}$

Conclusion:

Therefore, the transfer function $H\left(j\omega \right)$ is $\frac{-j\omega C{R}_2}{2-{\omega }^2{C}^2{R}_1{R}_2+j\omega C{R}_1}$.

To determine

The magnitude of transfer function $\left|H\left(j\omega \right)\right|$ and phase of transfer function $\angle H\left(j\omega \right)$ for the given condition.

Answer to Problem 6E

The magnitude of transfer function $\left|H\left(j\omega \right)\right|$ at $0\text{rad/s}$ is $0$, at $3\times {10}^4\text{rad/s}$ is $48$ and at $\infty \text{rad/s}$ is $0$ the phase of transfer function $\angle H\left(j\omega \right)$ at $0\text{rad/s}$ is $-90°$, at $3\times {10}^4\text{rad/s}$ is $-126.9°$ and at $\infty \text{rad/s}$ is $-90°$.

Explanation of Solution

Given data:

The resistance $R_1$ is $500\Omega$.

The resistance $R_2$ is $40\text{k}\Omega$.

The capacitor $C$ is $10\text{nF}$

The angular frequency ${\omega }_1$ is $0$.

The angular frequency ${\omega }_2$ is $0.3\times {10}^4\text{rad/s}$.

The angular frequency ${\omega }_3$ is $\infty$.

Calculation:

The conversion of $\text{k}\Omega$ to $\Omega$ is written as,

$1\text{k}\Omega =1\times {10}^3\Omega$

The conversion of $\text{40}\text{k}\Omega$ to $\Omega$ is written as,

$40\text{k}\Omega =40\times {10}^3\Omega$

The conversion of $\text{nF}$ to $\text{F}$ is written as,

$1\text{nF}=1\times {10}^{-9}\text{F}$

The conversion of $\text{10}\text{nF}$ to $\text{F}$ is written as,

$10\text{nF}=10\times {10}^{-9}\text{F}$

The magnitude of transfer function $\left|H\left(j{\omega }_1\right)\right|$ is written as,

$\left|H\left(j{\omega }_1\right)\right|=\frac{{\omega }_{1}C{R}_2}{\sqrt{{\left(2-{\omega }_1{}^2{C}^2{R}_1{R}_2\right)}^2+{\left({\omega }_{1}C{R}_1\right)}^2}}$        (4)

The phase of transfer function $\angle H\left(j{\omega }_1\right)$ is written as,

$\angle H\left(j{\omega }_1\right)=-90°-{\tan }^{-1}\frac{{\omega }_{1}C{R}_1}{2-{\omega }_1{}^2{C}^2{R}_1{R}_2}$        (5)

Substitute $0$ for ${\omega }_1$, $500\Omega$ for $R_1$, $40\times {10}^3\Omega$ for $R_2$, and $10\times {10}^{-9}\text{F}$ for $C$ in equation (4).

$\begin{array}{c}\left|H\left(j{\omega }_1\right)\right|=\frac{\left(0\right)\left(10\times {10}^{-9}\text{F}\right)\left(40\times {10}^3\Omega \right)}{\sqrt{\begin{array}{l}{\left[2-{\left(0\right)}^2{\left(10\times {10}^{-9}\text{F}\right)}^2\left(500\Omega \right)\left(40\times {10}^3\Omega \right)\right]}^2+\\ {\left[\left(0\right)\left(10\times {10}^{-9}\text{F}\right)\left(500\Omega \right)\right]}^2\end{array}}}\\ =0\end{array}$

Substitute $0$ for ${\omega }_1$, $500\Omega$ for $R_1$, $40\times {10}^3\Omega$ for $R_2$, and $10\times {10}^{-9}\text{F}$ for $C$ in equation (5).

$\begin{array}{c}\angle H\left(j{\omega }_1\right)=-90°-{\tan }^{-1}\frac{\left(0\right)\times \left(10\times {10}^{-9}\text{F}\right)\times \left(500\Omega \right)}{2-\left[{\left(0\right)}^2\times {\left(10\times {10}^{-9}\text{F}\right)}^2\times \left(500\Omega \right)\times \left(40\times {10}^3\Omega \right)\right]}\\ =-90°-{\tan }^{-1}0\\ =-90°-0\\ =-90°\end{array}$

The magnitude of transfer function $\left|H\left(j{\omega }_2\right)\right|$ is written as,

$\left|H\left(j{\omega }_2\right)\right|=\frac{{\omega }_{2}C{R}_2}{\sqrt{{\left(2-{\omega }_2{}^2{C}^2{R}_1{R}_2\right)}^2+{\left({\omega }_{2}C{R}_1\right)}^2}}$        (6)

The magnitude of transfer function $\angle H\left(j{\omega }_2\right)$ is written as,

$\angle H\left(j{\omega }_2\right)=-180°+90°-{\tan }^{-1}\frac{{\omega }_{2}C{R}_1}{2-{\omega }_2{}^2{C}^2{R}_1{R}_2}$        (7)

Substitute $3\times {10}^4\text{rad/s}$ for ${\omega }_2$, $500\Omega$ for $R_1$, $40\times {10}^3\Omega$ for $R_2$, and $10\times {10}^{-9}\text{F}$ for $C$ in equation (6).

$\begin{array}{c}\left|H\left(j{\omega }_2\right)\right|=\frac{\left(3\times {10}^4\text{rad/s}\right)\left(10\times {10}^{-9}\text{F}\right)\left(40\times {10}^3\Omega \right)}{\sqrt{\begin{array}{l}{\left[2-{\left(3\times {10}^4\text{rad/s}\right)}^2{\left(10\times {10}^{-9}\text{F}\right)}^2\left(500\Omega \right)\left(40\times {10}^3\Omega \right)\right]}^2+\\ {\left[\left(3\times {10}^4\text{rad/s}\right)\left(10\times {10}^{-9}\text{F}\right)\left(500\Omega \right)\right]}^2\end{array}}}\\ =\frac{12}{\sqrt{{\left[2-9\times {10}^8\times {10}^{-16}\times 5\times 4\times {10}^6\right]}^2+{\left[3\times {10}^4\times {10}^{-8}\times 500\right]}^2}}\\ =\frac{12}{\sqrt{{\left[2-1.8\right]}^2+{\left[0.15\right]}^2}}\\ =48\end{array}$

Substitute $3\times {10}^4\text{rad/s}$ for ${\omega }_2$, $500\Omega$ for $R_1$, $40\times {10}^3\Omega$ for $R_2$, and $10\times {10}^{-9}\text{F}$ for $C$ in equation (7).

$\begin{array}{c}\angle H\left(j{\omega }_2\right)=-90°-{\tan }^{-1}\frac{\left(3\times {10}^4\text{rad/s}\right)\times \left(10\times {10}^{-9}\text{F}\right)\times \left(500\Omega \right)}{2-\left[{\left(3\times {10}^4\text{rad/s}\right)}^2\times {\left(10\times {10}^{-9}\text{F}\right)}^2\times \left(500\Omega \right)\times \left(40\times {10}^3\Omega \right)\right]}\\ =-90°-{\tan }^{-1}\frac{15\times {10}^{-2}}{2-\left(18\times {10}^{-1}\right)}\\ =-90°-36.869°\\ =-126.9°\end{array}$

The magnitude of transfer function $\left|H\left(j{\omega }_3\right)\right|$ is written as,

$\begin{array}{c}\left|H\left(j{\omega }_3\right)\right|=\frac{{\omega }_{3}C{R}_2}{\sqrt{{\left(2-{\omega }_3{}^2{C}^2{R}_1{R}_2\right)}^2+{\left({\omega }_{3}C{R}_1\right)}^2}}\\ =\frac{{\omega }_{3}C{R}_2}{{\omega }_3{}^2\sqrt{{\left(\frac{2}{{\omega }_3{}^2}-C^2{R}_1{R}_2\right)}^2+{\left(\frac{C{R}_1}{{\omega }_3}\right)}^2}}\\ =\frac{C{R}_2}{{\omega }_3\sqrt{{\left(\frac{2}{{\omega }_3{}^2}-C^2{R}_1{R}_2\right)}^2+{\left(\frac{C{R}_1}{{\omega }_3}\right)}^2}}\end{array}$        (8)

The magnitude of transfer function $\angle H\left(j{\omega }_3\right)$ is written as,

$\begin{array}{c}\angle H\left(j{\omega }_3\right)=-90°-{\tan }^{-1}\frac{{\omega }_{3}C{R}_1}{2-{\omega }_3{}^2{C}^2{R}_1{R}_2}\\ =-90°-{\tan }^{-1}\frac{{\omega }_{3}C{R}_1}{{\omega }_3{}^2\left(\frac{2}{{\omega }_3{}^2}-C^2{R}_1{R}_2\right)}\\ =-90°-{\tan }^{-1}\frac{C{R}_1}{{\omega }_3\left(\frac{2}{{\omega }_3{}^2}-C^2{R}_1{R}_2\right)}\end{array}$        (9)

Substitute $\infty$ for ${\omega }_3$, $500\Omega$ for $R_1$, $40\times {10}^3\Omega$ for $R_2$, and $10\times {10}^{-9}\text{F}$ for $C$ in equation (8).

$\begin{array}{c}\left|H\left(j{\omega }_3\right)\right|=\frac{\left(10\times {10}^{-9}\text{F}\right)\left(40\times {10}^3\Omega \right)}{\infty \sqrt{\begin{array}{l}{\left[\frac{2}{{\infty }^2}-{\left(10\times {10}^{-9}\text{F}\right)}^2\left(500\Omega \right)\left(40\times {10}^3\Omega \right)\right]}^2+\\ {\left[\frac{\left(10\times {10}^{-9}\text{F}\right)\left(500\Omega \right)}{\infty }\right]}^2\end{array}}}\\ =0\end{array}$

Substitute $\infty$ for ${\omega }_3$, $500\Omega$ for $R_1$, $40\times {10}^3\Omega$ for $R_2$, and $10\times {10}^{-9}\text{F}$ for $C$ in equation (9).

$\begin{array}{c}\angle H\left(j{\omega }_3\right)=-90°-{\tan }^{-1}\frac{\left(10\times {10}^{-9}\text{F}\right)\times {\left(500\Omega \right)}_1}{\infty \left(\frac{2}{{\infty }^2}-{\left(10\times {10}^{-9}\text{F}\right)}^2\times \left(500\Omega \right)\times \left(40\times {10}^3\Omega \right)\right)}\\ =-90°-{\tan }^{-1}0\\ =-90°-0\\ =-90°\end{array}$

Conclusion:

Therefore, the magnitude of transfer function $\left|H\left(j\omega \right)\right|$ at $0\text{rad/s}$ is $0$, at $3\times {10}^4\text{rad/s}$ is $48$ and at $\infty \text{rad/s}$ is $0$ the phase of transfer function $\angle H\left(j\omega \right)$ at $0\text{rad/s}$ is $-90°$, at $3\times {10}^4\text{rad/s}$ is $-126.9°$ and at $\infty \text{rad/s}$ is $-90°$.