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What is the step response of a RC circuit?

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The step response of a **RC (Resistor-Capacitor)** circuit refers to how the voltage across the capacitor changes when a step input voltage (a sudden change from 0 to a fixed value, such as a DC voltage) is applied to the circuit.

### Types of RC Circuits:
1. **Series RC Circuit**: A resistor and capacitor are connected in series with a voltage source.
2. **Parallel RC Circuit**: A resistor and capacitor are connected in parallel to a voltage source.

Here, we focus on the **series RC circuit**, as it's more commonly analyzed in step response scenarios.

### Step Response of a Series RC Circuit

When a step voltage \( V_{\text{in}}(t) \) is applied to a series RC circuit at \( t = 0 \), the capacitor voltage \( V_C(t) \) changes according to the charging or discharging behavior of the capacitor.

#### 1. **Charging the Capacitor** (Applying a Step Input):
Consider a step input voltage \( V_{\text{in}}(t) = V_0 u(t) \), where \( u(t) \) is the unit step function. For \( t \geq 0 \), the input suddenly jumps from 0 to \( V_0 \).

- **Governing Equation**: The voltage across the capacitor, \( V_C(t) \), and the current through the resistor \( R \) are governed by Kirchhoff’s Voltage Law (KVL):
  
  \[
  V_{\text{in}}(t) = V_R(t) + V_C(t)
  \]
  where \( V_R(t) = i(t)R \) and \( i(t) = C \frac{dV_C(t)}{dt} \).

  Substituting into KVL:
  \[
  V_0 = RC \frac{dV_C(t)}{dt} + V_C(t)
  \]
  This is a first-order linear differential equation.

- **Solution**:
  Solving this equation for \( V_C(t) \):
  \[
  V_C(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right)
  \]
  where \( RC \) is the time constant \( \tau \) of the circuit. The time constant represents the time it takes for the voltage to rise to approximately 63% of its final value.

- **Interpretation**:
  - At \( t = 0 \), the capacitor voltage \( V_C(0) = 0 \) (initially uncharged).
  - As \( t \rightarrow \infty \), the capacitor voltage asymptotically approaches the input voltage, i.e., \( V_C(\infty) = V_0 \).

#### 2. **Discharging the Capacitor** (Removing the Step Input):
If the capacitor is charged initially to \( V_0 \) and then the input is removed (or set to zero), the capacitor discharges through the resistor.

- **Governing Equation**:
  \[
  0 = RC \frac{dV_C(t)}{dt} + V_C(t)
  \]

- **Solution**:
  Solving this differential equation, we get:
  \[
  V_C(t) = V_0 e^{-\frac{t}{RC}}
  \]
  This describes how the voltage across the capacitor decreases exponentially with time as it discharges.

- **Interpretation**:
  - At \( t = 0 \), \( V_C(0) = V_0 \).
  - As \( t \rightarrow \infty \), \( V_C(\infty) = 0 \).

### Key Concepts:

- **Time Constant (\( \tau = RC \))**: It determines the rate at which the capacitor charges or discharges. After \( t = \tau \), the voltage has reached about 63% of its final value (for charging) or has decreased to about 37% of its initial value (for discharging).
  
- **Steady-State and Transient Response**:
  - The **transient response** refers to the period during which the voltage is changing, i.e., the exponential growth or decay.
  - The **steady-state response** is the final value of the capacitor voltage after a long period, which is either \( V_0 \) (charging) or 0 (discharging).

### Practical Examples:
1. **Charging**: Suppose \( V_{\text{in}}(t) = 10 \, \text{V} \), \( R = 1 \, \text{k}\Omega \), and \( C = 1 \, \mu F \). The time constant is:
   \[
   \tau = RC = 1 \times 10^3 \times 1 \times 10^{-6} = 1 \, \text{ms}
   \]
   So, after 1 ms, the capacitor voltage will be approximately 63% of 10V, i.e., 6.3V. After 5 ms (5τ), it will be almost fully charged.

2. **Discharging**: If the capacitor starts with 10V, the voltage will decrease to 3.7V after 1 ms and will approach 0V after a longer time.

### Conclusion:
The step response of an RC circuit shows how the capacitor voltage reacts over time to a sudden change in input voltage. It involves exponential charging and discharging behavior, governed by the circuit's time constant \( RC \).
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The step response of an RC circuit describes how the circuit reacts over time when a sudden change (step input) is applied to it. Let’s break down the behavior of an RC (Resistor-Capacitor) circuit, focusing on its response to a step input.

### RC Circuit Overview

An RC circuit consists of a resistor \( R \) and a capacitor \( C \) connected in series. When a voltage step is applied to this circuit, the capacitor initially behaves like a short circuit (if it was uncharged) and eventually behaves like an open circuit once fully charged.

### Step Response Analysis

#### 1. **Initial Conditions**

At \( t = 0 \), when the step input is first applied, the capacitor has no charge and behaves as if it were a short circuit. Thus, the initial voltage across the capacitor \( V_C(0) \) is zero.

#### 2. **Applying a Step Input**

Suppose the step input is a voltage source \( V_{in} \) that switches from 0V to \( V_{0} \) at \( t = 0 \).

#### 3. **Governing Differential Equation**

The voltage across the capacitor \( V_C(t) \) and the current through the resistor \( I(t) \) are governed by the following differential equation derived from Kirchhoff’s Voltage Law (KVL):

\[ V_{in} = V_R(t) + V_C(t) \]

Since \( V_R(t) = I(t) \cdot R \) and \( I(t) = C \frac{dV_C(t)}{dt} \), we can rewrite the equation as:

\[ V_{in} = R \cdot C \frac{dV_C(t)}{dt} + V_C(t) \]

#### 4. **Solving the Differential Equation**

To solve for \( V_C(t) \), we need to solve this first-order linear differential equation. Rearranging, we get:

\[ \frac{dV_C(t)}{dt} + \frac{1}{RC} V_C(t) = \frac{V_{in}}{RC} \]

This is a standard form for a first-order linear differential equation. The general solution can be found using methods for solving such equations. The solution consists of a homogeneous part and a particular part.

The homogeneous solution (when \( V_{in} = 0 \)) is:

\[ V_C(t) = A e^{-\frac{t}{RC}} \]

The particular solution (when \( V_{in} \) is a constant \( V_{0} \)) is:

\[ V_{C_{p}}(t) = V_{0} \]

Combining these, the total solution is:

\[ V_C(t) = V_{0} \left(1 - e^{-\frac{t}{RC}}\right) \]

#### 5. **Time Constant**

The term \( \tau = RC \) is called the time constant of the circuit. It represents the time required for the capacitor to charge up to approximately 63.2% of the step input voltage \( V_{0} \). The voltage across the capacitor \( V_C(t) \) approaches \( V_{0} \) asymptotically as \( t \) becomes large.

#### 6. **Key Points of the Step Response**

- **Initial Voltage Across Capacitor:** At \( t = 0 \), \( V_C(0) = 0 \) V.
- **Final Voltage Across Capacitor:** As \( t \rightarrow \infty \), \( V_C(t) \rightarrow V_{0} \) V.
- **Time Constant \( \tau \):** \( \tau = RC \). It characterizes the rate of charging of the capacitor.

### Summary

In summary, the step response of an RC circuit shows that the capacitor voltage increases exponentially from 0V to the step input voltage \( V_{0} \) with a time constant \( \tau = RC \). The response curve follows the equation:

\[ V_C(t) = V_{0} \left(1 - e^{-\frac{t}{RC}}\right) \]

This exponential behavior is characteristic of first-order linear circuits.
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