Chapter 10, Problem 78P

Determine v_o(t) in the op amp circuit in Fig. 10.121 below.

To determine

Calculate the output voltage $v_o\left(t\right)$ of the op amp circuit in Figure 10.121 using MATLAB.

Answer to Problem 78P

The value of output voltage $v_o\left(t\right)$ for the given circuit is $19.735\sin \left(400t-9.48°\right)\text{V}$.

Explanation of Solution

Given data:

Refer to Figure 10.121 in the textbook for op amp circuit.

Formula used:

Write the expression to calculate impedance of the capacitor.

$Z_C=\frac{1}{j\omega C}$ (1)

Here,

$\omega$ is the angular frequency, and

$C$ is the value of capacitor.

Write the general representation of sinusoidal sine function.

$v_s=V_m\sin \left(\omega t+\varphi \right)$ (2)

Here,

$\varphi$ is the phase angle, and

$V_m$ is the magnitude of source voltage.

Write the general expression to phasor transform of sinusoidal function from time domain to frequency domain.

$v_s=P\left\{V_m\sin \left(\omega t+\varphi \right)\right\}=V_m{e}^{i\varphi }$

Here,

$V_m\sin \left(\omega t+\varphi \right)$ is the time domain representation of source voltage, and

$V_m{e}^{i\varphi }$ is the frequency domain representation of source voltage.

Write the polar form representation of frequency domain.

$v_s=V_m\angle \varphi$ (3)

Calculation:

Comparing $\left[10\sin \left(400t+0°\right)\text{V}\right]$ source voltage with equation (2), the magnitude, angular frequency, and phase angle of source voltage are $10\text{V}$, $400\frac{\text{rad}}{\text{s}}$ and $0°$ respectively.

Substitute $10\text{V}$ for $V_m$ and $0°$ for $\varphi$ in equation (3).

$v_s=10\angle 0°\text{V}$

Substitute $400\frac{\text{rad}}{\text{s}}$ for $\omega$ and $0.25\mu \text{F}$ for $C$ in equation (1) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(400\frac{\text{rad}}{\text{s}}\right)\left(0.25\mu \text{F}\right)}\\ =\frac{1}{j\left(400\frac{\text{rad}}{\text{s}}\right)\left(0.25\times {10}^{-6}\frac{\text{s}}{\Omega }\right)}\left\{\begin{array}{l}\because 1\mu \text{F}=1\times {10}^{-6}\text{F}\\ 1\text{F}=1\frac{\text{s}}{\Omega }\end{array}\right\}\\ =-j10\text{k}\Omega \end{array}$

Substitute $400\frac{\text{rad}}{\text{s}}$ for $\omega$ and $0.5\mu \text{F}$ for $C$ in equation (1) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(400\frac{\text{rad}}{\text{s}}\right)\left(0.5\mu \text{F}\right)}\\ =\frac{1}{j\left(400\frac{\text{rad}}{\text{s}}\right)\left(0.5\times {10}^{-6}\frac{\text{s}}{\Omega }\right)}\left\{\begin{array}{l}\because 1\mu \text{F}=1\times {10}^{-6}\text{F}\\ 1\text{F}=1\frac{\text{s}}{\Omega }\end{array}\right\}\\ =-j5\text{k}\Omega \end{array}$

The frequency domain representation of given figure is shown in Figure 1.

Apply Kirchhoff’s current law at node $V_1$.

$\begin{array}{l}\frac{V_1-10}{10\text{k}}+\frac{V_1-V_o}{20\text{k}}+\frac{V_1-V_2}{-j5\text{k}}+\frac{V_1}{-j10\text{k}}=0\\ \frac{V_1-10}{10}+\frac{V_1-V_o}{20}+\frac{V_1-V_2}{-j5}+\frac{V_1}{-j10}=0\\ \frac{V_1}{10}-1+\frac{V_1}{20}-\frac{V_o}{20}-\frac{V_1}{j5}+\frac{V_2}{j5}-\frac{V_1}{j10}=0\\ 0.1V_1+0.05V_1-0.05V_o+j0.2V_1-j0.2V_2+j0.1V_1=1\end{array}$

Simplify the equation as follows.

$\begin{array}{l}\left(0.1+0.05+j0.2+j0.1\right)V_1-j0.2V_2-0.05V_o=1\\ \left(0.15+j0.3\right)V_1-j0.2V_2-0.05V_o=1\end{array}$

$\left(3+j6\right)V_1-j4V_2-V_o=20$ (1)

Apply Kirchhoff’s current law at node $V_2$.

$\begin{array}{l}\frac{V_2-V_1}{-j5\text{k}}+\frac{V_2-0}{10\text{k}}=0\\ \frac{V_2-V_1}{-j5}+\frac{V_2-0}{10}=0\\ j0.2V_2-j0.2V_1+0.1V_2=0\\ -j0.2V_1+\left(0.1+j0.2\right)V_2=0\end{array}$

$V_1=\left(1-j0.5\right)V_2$ (2)

Apply voltage division rule at node $V_b$.

$\begin{array}{c}{V}_b=\frac{20\text{kΩ}}{20\text{kΩ}+40\text{kΩ}}{V}_o\\ =\frac{20\text{kΩ}}{60\text{kΩ}}{V}_o\end{array}$

$V_b=\frac{V_o}{3}$ (3)

According to the properties of ideal op amp, the voltage at the input of the non-inverting terminal of the op amp is equal to the voltage at the input of the inverting terminal. Hence, $\left(V_2=V_b\right)$.

$V_2=\frac{V_o}{3}$ (4)

Substitute equation (5) in (1).

$\begin{array}{l}\left(3+j6\right)V_1-j4\left(\frac{V_o}{3}\right)-V_o=20\\ \left(3+j6\right)V_1-j1.333V_o-V_o=20\end{array}$

$\left(3+j6\right)V_1+\left(-1-j1.333\right)V_o=20$ (5)

Substitute equation (5) in (2).

$\begin{array}{l}{V}_1=\left(1-j0.5\right)\frac{V_o}{3}\\ V_1=\left(0.3333-j0.1666\right)V_o\\ V_1-\left(0.3333-j0.1666\right)V_o=0\end{array}$

$V_1+\left(-0.3333+j0.1666\right)V_o=0$ (7)

Represent the equations (6) and (7) in matrix form.

$\left[\begin{array}{cc}\left(3+j6\right)& \left(-1-j1.333\right)\\ 1& \left(-0.3333+j0.1666\right)\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_o\end{array}\right]=\left[\begin{array}{c}20\\ 0\end{array}\right]$ (8)

Write the MATLAB code to solve equation (8) to find $V_1$, $V_o$.

A = [3+6*1i -1-1.333*1i; 1 -0.3333+0.1666*1i];

b = [20; 0];

x = A\b

The MATLAB result is shown below.

x =

   5.9463 - 4.3272i   19.4666 - 3.2525i

The polar representation of obtained result is shown below.

$V_o=19.735\angle -9.48°\text{V}$

Represent the output voltage in time domain.

$v_o\left(t\right)=19.735\sin \left(400t-9.48°\right)\text{V}$

Conclusion:

Thus, the value of output voltage $v_o\left(t\right)$ for the given circuit is $19.735\sin \left(400t-9.48°\right)\text{V}$.