GATE 2024 IN Q.59: solution and answer

Official answer: -2.6 to -2.4Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. Assumptions: the op-amp is ideal (infinite gain, zero input current, output inside the supply rails, so not saturated) and the capacitor starts uncharged, so $V_{out}(0)=0$.

2. Virtual short: the + input is grounded, so the − input sits at $0\text{V}$ (virtual ground). The resistor current is therefore $i=\frac{V_{in}}{R}$.

3. KCL at the inverting node: no current enters the op-amp, so all of $i$ flows through the capacitor. The input side of $C$ is at 0 and the output side at $V_{out}$, so $C\frac{d(0-V_{out})}{dt}=\frac{V_{in}}{R}$, which gives $$V_{out}(t)=V_{out}(0)-\frac{1}{RC}{\int }_0^t{V}_{in}dt=-\frac{1}{RC}{\int }_0^t{V}_{in}dt$$

4. Time constant: $RC=(1\text{k}\Omega )(2\mu \text{F})=2\text{ms}$.

5. For $0\le t\le 1\text{ms}$, $V_{in}=5\text{V}$ is constant, so the output ramps down linearly: $$V_{out}(1\text{ms})=-\frac{5\text{V}\times 1\text{ms}}{2\text{ms}}=-2.5\text{V}$$

6. For $1\text{ms}<t<2\text{ms}$, $V_{in}=0$. No current flows through $R$, so none flows through $C$ either. The capacitor keeps its charge and $V_{out}$ stays at $-2.5\text{V}$.

7. At $t=2\text{ms}$ the second pulse is only just starting. The integral is a continuous function of $t$, so the step in $V_{in}$ has not added any area yet: $$V_{out}(2\text{ms})=-2.5\text{V}$$