Question

Consider the circuit shown in the figure below:

Then,

• A. $\text{the output is}-\frac{t}{R^2{C}^2}u\left(t\right)\text{for}{V}_{\text{in}}\left(t\right)=\delta \left(t\right)$
• B. $\text{the output is}-\frac{t}{2R^2{C}^2}u\left(t\right)\text{for}{V}_{\text{in}}\left(t\right)=u\left(t\right)$
• C. $\text{the output is}-\frac{1}{R^{2}C}u\left(t\right)\text{for}{V}_{\text{in}}\left(t\right)=tu\left(t\right)$
• D. $\text{the output is}-\frac{t^2}{R^2{C}^2}u\left(t\right)\text{for}{V}_{\text{in}}\left(t\right)=u\left(t\right)$

Answer 1

Answer: (A), (B)

$\text{the output is}-\frac{t}{2R^2{C}^2}u\left(t\right)\text{for}{V}_{\text{in}}\left(t\right)=u\left(t\right)$

Taking Laplace transform, we have,


$I_1=\frac{V_{\text{in}}}{\frac{2R}{1+sRC}}\text{and}{I}_f\left(s\right)=V_{\text{out}}\left(s\right)\times \frac{sR{C}^2}{2(1+sRC)}$


$\text{Now,}{I}_1\left(s\right)=-I_f\left(s\right)$

$\therefore \frac{V_{\text{in}}\left(s\right)}{2R(sRC+1)}=-\frac{R{C}^2{s}^2}{2(sRC+1)}{V}_{V_{\text{out}}}\left(s\right)$

$V_{\text{out}}\left(s\right)=\frac{1}{s^2{R}^2{C}^2}{V}_{\text{in}}\left(s\right)=-\frac{\frac{1}{R_1^2{C}^2}}{s^3}=-\frac{\frac{1}{R_1^2{C}^2}}{s^3}$

$V_{\text{out}}\left(s\right)=-\frac{t^2}{2R^2{C}^2}u\left(t\right)$