The circuit shown is an inverting integrator using an operational amplifier. In this configuration, the op-amp integrates the input voltage over time.

Step-by-Step Solution:

1. Identify Circuit Components and Inputs:

   • Input Voltage, $V_{in}$: A step input that changes from 0 V to 5 V at $t = 0$.
   • Resistor, $R$: $R = 56 \, \text{k}\Omega$
   • Capacitor, $C$: $C = 0.022 \, \mu\text{F}$

2. Inverting Integrator Formula: For an inverting integrator, the output voltage $V_{out}(t)$ is given by: $V_{out}(t) = -\frac{1}{RC} \int V_{in}(t) \, dt$

3. Calculate the Integrator Constant: Substitute $R$ and $C$ values: $\frac{1}{RC} = \frac{1}{56 \times 10^3 \, \Omega \times 0.022 \times 10^{-6} \, \text{F}} \approx 0.812 \, \text{s}^{-1}$

4. Compute $V_{out}(t)$ for $V_{in} = 5 \, \text{V}$ (Step Input): When $V_{in}$ steps from 0 to 5 V at $t = 0$, the integral of a constant $V_{in}$ over time becomes: $V_{out}(t) = -0.812 \times 5 \times t = -4.06t$

   Thus, the output voltage $V_{out}(t)$ is: $V_{out}(t) = -4.06t \quad \text{(for } t \geq 0\text{)}$

Final Answer:

For $t \geq 0$: $V_{out}(t) = -4.06t \, \text{V}$

Would you like more details, or do you have any questions?

Here are some related questions to further explore this topic:

1. How does the output waveform change if the input is a sinusoidal signal instead of a step function?
2. What happens to $V_{out}$ if the resistor $R$ value is increased?
3. How would adding a feedback resistor change the circuit’s behavior?
4. What is the effect of changing the capacitor value on the integrator output?
5. How would you design a differentiator circuit using an op-amp?

Tip: For an integrator circuit, the output slope is directly related to the input voltage and the $RC$ time constant.