Chapter 9, Problem 9.9EP

An integrator with input and output voltages that are zero at $t=0$ is drivenby the input signal shown in Figure 9.32. (a) For circuit parameters $R_1=10k\Omega$ and $C_2=0.1\mu F$ , determine the output voltage at $t=$ (i) 1 ms, (ii) 2 ms, (iii) 3 ms,and (iv) 4 ms. (b) Repeat part (a) for circuit parameters $R_1=10k\Omega$ and $C_2=1\mu F$ . (Ans. (a) (i) $-1V$ , (ii) 0, (iii) $-1V$ , (iv) 0; (b) (i) $-0.1V$ , (ii) 0,(iii) $-0.1V$ , (iv) 0)

To determine

The value of the output voltage for different time interval.

Answer to Problem 9.9EP

Thevalue of the output voltage for the different time interval is $v_O=\{\begin{array}{l}-1\text{V}\text{for}\text{0}\text{ms}<t\le 1\text{ms}\\ \text{0}\text{for}1\text{ms}<t\le 2\text{ms}\\ -1\text{for}2\text{ms}<t\le 3\text{ms}\\ 0\text{for}3\text{ms}<t\le 4\text{ms}\end{array}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

From the above waveform the expression for the voltage is given by,

  $v_I\left(t\right)=\{\begin{array}{l}+1\text{for}\text{0}\text{ms}<t\le 1\text{ms}\\ -1\text{for}1\text{ms}<t\le 2\text{ms}\\ +1\text{for}2\text{ms}<t\le 3\text{ms}\\ -1\text{for}3\text{ms}<t\le 4\text{ms}\end{array}$

Mark the values and draw the integrated op-amp circuit.

The required diagram is shown in Figure 2

  

Apply KVL at the feedback loop.

  $v_C=v_O-v_1$

Substitute $0\text{V}$ for $v_1$ in the above equation.

  $\begin{array}{l}{v}_C=v_O-0\text{V}\\ v_C=v_O\end{array}$

The conversion from $1\text{ms}$ into $\text{s}$ is given by,

  $1\text{ms}={10}^{-3}\text{s}$

The conversion from $2\text{ms}$ into $\text{s}$ is given by,

  $2\text{ms}=2\times {10}^{-3}\text{s}$

The conversion from $3\text{ms}$ into $\text{s}$ is given by,

  $3\text{ms}=3\times {10}^{-3}\text{s}$

The conversion from $4\text{ms}$ into $\text{s}$ is given by,

  $4\text{ms}=4\times {10}^{-3}\text{s}$

The conversion from $1\text{μF}$ into $\text{F}$ is given by,

  $1\text{μF}={\text{10}}^{-6}\text{F}$

The conversion from $0.1\text{μF}$ into $\text{F}$ is given by,

  $0.1\text{μF}=0.1\times {\text{10}}^{-6}\text{F}$

The conversion from $1\text{k}\Omega$ into $\Omega$ is given by,

  $1\text{k}\Omega ={10}^3\Omega$

The conversion from $10\text{k}\Omega$ into $\Omega$ is given by,

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

Apply KCL at the inverting terminal.

  $\frac{v_I-v_1}{R_1}=-C_2\frac{d{v}_C}{dt}$

Substitute $0\text{V}$ for $v_1$ in the above equation.

  $\begin{array}{l}\frac{v_I}{R_1}=-C_2\frac{d{v}_C}{dt}\\ d{v}_C=\frac{v_I}{C_2{R}_1}dt\\ v_C\left(t\right)=-{\displaystyle {\int }_0^t\frac{v_I}{C_2{R}_1}dt}+v_C\left(0\right)\end{array}$

Substitute $0$ for $v_C\left(0\right)$ in the above equation.

  $v_O\left(t\right)=-{\displaystyle {\int }_0^t\frac{v_I}{C_2{R}_1}dt}+0$

The expression for the output voltage for the time $1\text{ms}$ is evaluated as,

  $\begin{array}{c}{v}_O\left({10}^{-3}\text{s}\right)=-{\displaystyle {\int }_0^{{10}^{-3}\text{s}}\frac{1}{C_2{R}_1}dt}+0\\ =-\frac{1}{C_2{R}_1}\left({10}^{-3}\right)\end{array}$

Substitute $10\times {10}^3\Omega$ for $R_1$ and $0.1\times {\text{10}}^{-6}\text{F}$ for $C_2$ in the above equation.

  $\begin{array}{c}{v}_O\left({10}^{-3}\text{s}\right)=-\frac{1}{\left(0.1\times {\text{10}}^{-6}\text{F}\right)\left(10\times {10}^3\Omega \right)}\left({10}^{-3}\right)\\ =-1\text{V}\end{array}$

The expression for the output voltage for the time $2\text{ms}$ is evaluated as,

  $\begin{array}{c}{v}_O\left(2\times {10}^{-3}\text{s}\right)=\frac{1}{C_2{R}_1}\left[{\displaystyle {\int }_0^{1\times {10}^{-3}\text{s}}\left(1\right)dt}+{\displaystyle {\int }_{1\times {10}^{-3}\text{s}}^{2\times {10}^{-3}\text{s}}\left(-1\right)dt}\right]\\ =-\frac{1}{C_2{R}_1}\left[\left(1\times {10}^{-3}\text{s}\right)-\left(1\times {10}^{-3}\text{s}\right)\right]\\ =0\end{array}$

The expression for the output voltage for the time $3\text{ms}$ is evaluated as,

  $\begin{array}{c}{v}_O\left(3\times {10}^{-3}\text{s}\right)=\frac{1}{C_2{R}_1}\left[{\displaystyle {\int }_0^{1\times {10}^{-3}\text{s}}\left(1\right)dt}+{\displaystyle {\int }_{1\times {10}^{-3}\text{s}}^{2\times {10}^{-3}\text{s}}\left(-1\right)dt+{\displaystyle {\int }_{2\times {10}^{-3}\text{s}}^{3\times {10}^{-3}\text{s}}\left(1\right)dt}}\right]\\ =-\frac{1}{C_2{R}_1}\left[\left(1\times {10}^{-3}\text{s}\right)-\left(1\times {10}^{-3}\text{s}\right)+\left(1\times {10}^{-3}\text{s}\right)\right]\\ =-\frac{1\times {10}^{-3}\text{s}}{C_2{R}_1}\end{array}$

Substitute $10\times {10}^3\Omega$ for $R_1$ and $0.1\times {\text{10}}^{-6}\text{F}$ for $C_2$ in the above equation.

  $\begin{array}{c}{v}_O\left(3\times {10}^{-3}\text{s}\right)=-\frac{\left(1\times {10}^{-3}\text{s}\right)}{\left(0.1\times {\text{10}}^{-6}\text{F}\right)\left(10\times {10}^3\Omega \right)}\\ =-1\text{V}\end{array}$

The expression for the output voltage for the time $4\text{ms}$ is evaluated as,

  $\begin{array}{c}{v}_O\left(4\times {10}^{-3}\text{s}\right)=\frac{1}{C_2{R}_1}\left[{\displaystyle {\int }_0^{1\times {10}^{-3}\text{s}}\left(1\right)dt}+{\displaystyle {\int }_{1\times {10}^{-3}\text{s}}^{2\times {10}^{-3}\text{s}}\left(-1\right)dt+{\displaystyle {\int }_{2\times {10}^{-3}\text{s}}^{3\times {10}^{-3}\text{s}}\left(1\right)dt}+{\displaystyle {\int }_{3\times {10}^{-3}\text{s}}^{4\times {10}^{-3}\text{s}}\left(-1\right)dt}}\right]\\ =-\frac{1}{C_2{R}_1}\left[\left(1\times {10}^{-3}\text{s}\right)-\left(1\times {10}^{-3}\text{s}\right)+\left(1\times {10}^{-3}\text{s}\right)-\left(1\times {10}^{-3}\text{s}\right)\right]\\ =0\end{array}$

The value of the voltage output voltage for different time interval is given by,

  $v_O=\{\begin{array}{l}-1\text{V}\text{for}\text{0}\text{ms}<t\le 1\text{ms}\\ \text{0}\text{for}1\text{ms}<t\le 2\text{ms}\\ -1\text{for}2\text{ms}<t\le 3\text{ms}\\ 0\text{for}3\text{ms}<t\le 4\text{ms}\end{array}$

Conclusion:

Therefore, the value of the output voltage for the different time interval is $v_O=\{\begin{array}{l}-1\text{V}\text{for}\text{0}\text{ms}<t\le 1\text{ms}\\ \text{0}\text{for}1\text{ms}<t\le 2\text{ms}\\ -1\text{for}2\text{ms}<t\le 3\text{ms}\\ 0\text{for}3\text{ms}<t\le 4\text{ms}\end{array}$ .

To determine

The value of the output voltage for different time interval.

Answer to Problem 9.9EP

Thevalue of the output voltage for the different time interval is $v_O=\{\begin{array}{l}-0.1\text{V}\text{for}\text{0}\text{ms}<t\le 1\text{ms}\\ \text{0}\text{for}1\text{ms}<t\le 2\text{ms}\\ -0.1\text{for}2\text{ms}<t\le 3\text{ms}\\ 0\text{for}3\text{ms}<t\le 4\text{ms}\end{array}$ .

Explanation of Solution

Calculation:

The expression for the output voltage for the time $1\text{ms}$ is evaluated as,

  $v_O\left({10}^{-3}\text{s}\right)=-\frac{1}{C_2{R}_1}\left({10}^{-3}\right)$

Substitute $10\times {10}^3\Omega$ for $R_1$ and $1\times {\text{10}}^{-6}\text{F}$ for $C_2$ in the above equation.

  $\begin{array}{c}{v}_O\left({10}^{-3}\text{s}\right)=-\frac{1}{\left(1\times {\text{10}}^{-6}\text{F}\right)\left(10\times {10}^3\Omega \right)}\left({10}^{-3}\right)\\ =-0.1\text{V}\end{array}$

The expression for the output voltage for the time $2\text{ms}$ is evaluated as,

  $v_O\left(2\times {10}^{-3}\text{s}\right)=0$

The expression for the output voltage for the time $3\text{ms}$ is evaluated as,

  $v_O\left(3\times {10}^{-3}\text{s}\right)=-\frac{1\times {10}^{-3}\text{s}}{C_2{R}_1}$

Substitute $10\times {10}^3\Omega$ for $R_1$ and $0.1\times {\text{10}}^{-6}\text{F}$ for $C_2$ in the above equation.

  $\begin{array}{c}{v}_O\left(3\times {10}^{-3}\text{s}\right)=-\frac{1}{\left(1\times {\text{10}}^{-6}\text{F}\right)\left(10\times {10}^3\Omega \right)}\left({10}^{-3}\right)\\ =-0.1\text{V}\end{array}$

The expression for the output voltage for the time $4\text{ms}$ is evaluated as,

  $v_O\left(4\times {10}^{-3}\text{s}\right)=0$

The value of the voltage output voltage for different time interval is given by,

  $v_O=\{\begin{array}{l}-0.1\text{V}\text{for}\text{0}\text{ms}<t\le 1\text{ms}\\ \text{0}\text{for}1\text{ms}<t\le 2\text{ms}\\ -0.1\text{for}2\text{ms}<t\le 3\text{ms}\\ 0\text{for}3\text{ms}<t\le 4\text{ms}\end{array}$

Conclusion:

Therefore, the value of the output voltage for the different time interval is $v_O=\{\begin{array}{l}-0.1\text{V}\text{for}\text{0}\text{ms}<t\le 1\text{ms}\\ \text{0}\text{for}1\text{ms}<t\le 2\text{ms}\\ -0.1\text{for}2\text{ms}<t\le 3\text{ms}\\ 0\text{for}3\text{ms}<t\le 4\text{ms}\end{array}$ .