What is the step function of the RC circuit?
2 Answers
The step function response of an RC (Resistor-Capacitor) circuit refers to how the circuit responds to a sudden change in input, typically a voltage applied at \( t = 0 \), often called a "step" input. This analysis is important because RC circuits are fundamental building blocks in electrical engineering, used in filters, integrators, differentiators, and timing circuits. Let’s break it down by examining both types of RC circuits:
### 1. **RC Charging Circuit (Series RC Circuit with a Step Input)**
This is the most common configuration. It involves a resistor \( R \) and a capacitor \( C \) connected in series with a voltage source. We are interested in how the capacitor charges over time when a step voltage \( V_{\text{in}} \) is applied at \( t = 0 \).
#### Governing Equation
For a series RC circuit, the voltage across the capacitor at any time \( t \) can be found using Kirchhoff's Voltage Law (KVL):
\[
V_{\text{in}}(t) = V_R(t) + V_C(t)
\]
Where:
- \( V_{\text{in}}(t) \) is the input step voltage applied at \( t = 0 \),
- \( V_R(t) = i(t)R \) is the voltage across the resistor,
- \( V_C(t) = \frac{q(t)}{C} \) is the voltage across the capacitor, with \( q(t) \) being the charge stored on the capacitor.
From Ohm’s law, \( i(t) = \frac{dq(t)}{dt} \), so the governing differential equation becomes:
\[
V_{\text{in}} = RC \frac{dV_C(t)}{dt} + V_C(t)
\]
#### Solution to the Equation (Step Response)
The solution for \( V_C(t) \), which is the voltage across the capacitor, when the step voltage \( V_{\text{in}} \) is applied at \( t = 0 \) is:
\[
V_C(t) = V_{\text{in}} \left( 1 - e^{-\frac{t}{RC}} \right)
\]
Where \( \tau = RC \) is the time constant of the circuit.
This equation describes how the capacitor charges over time:
- At \( t = 0 \), \( V_C(0) = 0 \), meaning no charge has yet built up on the capacitor.
- As \( t \to \infty \), \( V_C(t) \to V_{\text{in}} \), meaning the capacitor charges up to the input voltage.
- The time constant \( \tau = RC \) represents the time it takes for the capacitor to charge to approximately 63% of the input voltage.
#### Current in the Circuit
The current in the circuit \( i(t) \) is given by:
\[
i(t) = \frac{V_{\text{in}}}{R} e^{-\frac{t}{RC}}
\]
This shows that the current is maximum at \( t = 0 \) and decays exponentially to zero as the capacitor becomes fully charged.
### 2. **RC Discharging Circuit**
In this case, the capacitor is initially charged, and at \( t = 0 \), it is allowed to discharge through the resistor.
#### Governing Equation
For the discharging case, the voltage across the capacitor decreases as it loses its charge. The differential equation is:
\[
V_C(t) = V_C(0) e^{-\frac{t}{RC}}
\]
Where \( V_C(0) \) is the initial voltage across the capacitor at \( t = 0 \).
#### Solution to the Equation (Step Response)
The voltage across the capacitor at time \( t \) is:
\[
V_C(t) = V_C(0) e^{-\frac{t}{RC}}
\]
- At \( t = 0 \), \( V_C(0) \) is the initial charge.
- As \( t \to \infty \), \( V_C(t) \to 0 \), meaning the capacitor fully discharges.
#### Current in the Circuit
The current in this discharging circuit is:
\[
i(t) = -\frac{V_C(0)}{R} e^{-\frac{t}{RC}}
\]
This shows that the current is negative, as it flows in the opposite direction during discharge, and also decays exponentially over time.
### 3. **Step Function Representation in the Laplace Domain**
In the Laplace domain, we can represent the step input \( V_{\text{in}}(t) \) as:
\[
V_{\text{in}}(s) = \frac{V_{\text{in}}}{s}
\]
This simplifies the analysis of RC circuits in control theory or signal processing. Using Laplace transforms, we can easily derive the transfer functions and analyze the system’s response to various inputs.
### Summary of RC Step Response:
- **Charging:** Voltage across the capacitor increases exponentially according to \( V_C(t) = V_{\text{in}} \left( 1 - e^{-\frac{t}{RC}} \right) \).
- **Discharging:** Voltage across the capacitor decreases exponentially according to \( V_C(t) = V_C(0) e^{-\frac{t}{RC}} \).
- **Time Constant \( \tau = RC \):** Determines how fast the capacitor charges or discharges. After a time equal to \( \tau \), the voltage is about 63% of its final value (for charging) or 37% of its initial value (for discharging).
This step function response is critical in understanding the dynamic behavior of RC circuits in various applications, such as filters, oscillators, and timers.
### 1. **RC Charging Circuit (Series RC Circuit with a Step Input)**
This is the most common configuration. It involves a resistor \( R \) and a capacitor \( C \) connected in series with a voltage source. We are interested in how the capacitor charges over time when a step voltage \( V_{\text{in}} \) is applied at \( t = 0 \).
#### Governing Equation
For a series RC circuit, the voltage across the capacitor at any time \( t \) can be found using Kirchhoff's Voltage Law (KVL):
\[
V_{\text{in}}(t) = V_R(t) + V_C(t)
\]
Where:
- \( V_{\text{in}}(t) \) is the input step voltage applied at \( t = 0 \),
- \( V_R(t) = i(t)R \) is the voltage across the resistor,
- \( V_C(t) = \frac{q(t)}{C} \) is the voltage across the capacitor, with \( q(t) \) being the charge stored on the capacitor.
From Ohm’s law, \( i(t) = \frac{dq(t)}{dt} \), so the governing differential equation becomes:
\[
V_{\text{in}} = RC \frac{dV_C(t)}{dt} + V_C(t)
\]
#### Solution to the Equation (Step Response)
The solution for \( V_C(t) \), which is the voltage across the capacitor, when the step voltage \( V_{\text{in}} \) is applied at \( t = 0 \) is:
\[
V_C(t) = V_{\text{in}} \left( 1 - e^{-\frac{t}{RC}} \right)
\]
Where \( \tau = RC \) is the time constant of the circuit.
This equation describes how the capacitor charges over time:
- At \( t = 0 \), \( V_C(0) = 0 \), meaning no charge has yet built up on the capacitor.
- As \( t \to \infty \), \( V_C(t) \to V_{\text{in}} \), meaning the capacitor charges up to the input voltage.
- The time constant \( \tau = RC \) represents the time it takes for the capacitor to charge to approximately 63% of the input voltage.
#### Current in the Circuit
The current in the circuit \( i(t) \) is given by:
\[
i(t) = \frac{V_{\text{in}}}{R} e^{-\frac{t}{RC}}
\]
This shows that the current is maximum at \( t = 0 \) and decays exponentially to zero as the capacitor becomes fully charged.
### 2. **RC Discharging Circuit**
In this case, the capacitor is initially charged, and at \( t = 0 \), it is allowed to discharge through the resistor.
#### Governing Equation
For the discharging case, the voltage across the capacitor decreases as it loses its charge. The differential equation is:
\[
V_C(t) = V_C(0) e^{-\frac{t}{RC}}
\]
Where \( V_C(0) \) is the initial voltage across the capacitor at \( t = 0 \).
#### Solution to the Equation (Step Response)
The voltage across the capacitor at time \( t \) is:
\[
V_C(t) = V_C(0) e^{-\frac{t}{RC}}
\]
- At \( t = 0 \), \( V_C(0) \) is the initial charge.
- As \( t \to \infty \), \( V_C(t) \to 0 \), meaning the capacitor fully discharges.
#### Current in the Circuit
The current in this discharging circuit is:
\[
i(t) = -\frac{V_C(0)}{R} e^{-\frac{t}{RC}}
\]
This shows that the current is negative, as it flows in the opposite direction during discharge, and also decays exponentially over time.
### 3. **Step Function Representation in the Laplace Domain**
In the Laplace domain, we can represent the step input \( V_{\text{in}}(t) \) as:
\[
V_{\text{in}}(s) = \frac{V_{\text{in}}}{s}
\]
This simplifies the analysis of RC circuits in control theory or signal processing. Using Laplace transforms, we can easily derive the transfer functions and analyze the system’s response to various inputs.
### Summary of RC Step Response:
- **Charging:** Voltage across the capacitor increases exponentially according to \( V_C(t) = V_{\text{in}} \left( 1 - e^{-\frac{t}{RC}} \right) \).
- **Discharging:** Voltage across the capacitor decreases exponentially according to \( V_C(t) = V_C(0) e^{-\frac{t}{RC}} \).
- **Time Constant \( \tau = RC \):** Determines how fast the capacitor charges or discharges. After a time equal to \( \tau \), the voltage is about 63% of its final value (for charging) or 37% of its initial value (for discharging).
This step function response is critical in understanding the dynamic behavior of RC circuits in various applications, such as filters, oscillators, and timers.
The step response of an RC (Resistor-Capacitor) circuit is a fundamental concept in electrical engineering that describes how the circuit reacts when a step input voltage is applied. Here’s a detailed explanation:
### RC Circuit Overview
An RC circuit is composed of a resistor \( R \) and a capacitor \( C \) connected in series. When a voltage is applied across the series combination, the circuit exhibits distinct behavior due to the capacitor’s ability to store and release charge.
### Step Input
A step input is a sudden change in voltage, often represented as a step function. Mathematically, a step function \( u(t) \) can be defined as:
- \( u(t) = 0 \) for \( t < 0 \)
- \( u(t) = 1 \) for \( t \geq 0 \)
When a step input is applied, the voltage \( V_{in}(t) \) across the RC circuit suddenly changes from 0 to a constant value \( V_0 \) at \( t = 0 \).
### Step Response Analysis
To find the step response of the RC circuit, we can analyze the circuit’s behavior over time. Let's denote:
- \( V_{in}(t) \) as the input voltage (step function)
- \( V_{out}(t) \) as the voltage across the capacitor
- \( I(t) \) as the current through the circuit
**1. Differential Equation:**
The voltage across the capacitor \( V_{out}(t) \) and the current through the resistor \( I(t) \) are related by:
\[ V_{in}(t) = V_{out}(t) + I(t) \cdot R \]
Since \( I(t) = C \frac{dV_{out}(t)}{dt} \), we can substitute this into the equation:
\[ V_{in}(t) = V_{out}(t) + R \cdot C \frac{dV_{out}(t)}{dt} \]
Rearranging gives us the differential equation:
\[ \frac{dV_{out}(t)}{dt} + \frac{1}{R \cdot C} V_{out}(t) = \frac{V_{in}(t)}{R \cdot C} \]
**2. Step Response Solution:**
For a step input \( V_{in}(t) = V_0 \cdot u(t) \):
Substitute \( V_{in}(t) = V_0 \) into the differential equation:
\[ \frac{dV_{out}(t)}{dt} + \frac{1}{R \cdot C} V_{out}(t) = \frac{V_0}{R \cdot C} \]
To solve this, we use the method of integrating factors or the Laplace transform. The general solution for \( V_{out}(t) \) is:
\[ V_{out}(t) = V_0 \left(1 - e^{-\frac{t}{R \cdot C}}\right) \]
**3. Interpretation:**
- **Initial Condition (t = 0):** At \( t = 0 \), \( V_{out}(0) = 0 \). This is because the capacitor initially behaves like an open circuit, and the full input voltage appears across the resistor.
- **Long-Term Behavior (t → ∞):** As \( t \) approaches infinity, \( V_{out}(t) \) approaches \( V_0 \). The capacitor eventually charges up to the input voltage \( V_0 \), and the voltage across the resistor becomes zero.
### Summary
The step response of an RC circuit shows that the voltage across the capacitor gradually increases from 0 to \( V_0 \) following an exponential curve. The time constant \( \tau = R \cdot C \) characterizes how quickly the capacitor charges, with the voltage reaching approximately 63.2% of \( V_0 \) after one time constant \( \tau \).
### RC Circuit Overview
An RC circuit is composed of a resistor \( R \) and a capacitor \( C \) connected in series. When a voltage is applied across the series combination, the circuit exhibits distinct behavior due to the capacitor’s ability to store and release charge.
### Step Input
A step input is a sudden change in voltage, often represented as a step function. Mathematically, a step function \( u(t) \) can be defined as:
- \( u(t) = 0 \) for \( t < 0 \)
- \( u(t) = 1 \) for \( t \geq 0 \)
When a step input is applied, the voltage \( V_{in}(t) \) across the RC circuit suddenly changes from 0 to a constant value \( V_0 \) at \( t = 0 \).
### Step Response Analysis
To find the step response of the RC circuit, we can analyze the circuit’s behavior over time. Let's denote:
- \( V_{in}(t) \) as the input voltage (step function)
- \( V_{out}(t) \) as the voltage across the capacitor
- \( I(t) \) as the current through the circuit
**1. Differential Equation:**
The voltage across the capacitor \( V_{out}(t) \) and the current through the resistor \( I(t) \) are related by:
\[ V_{in}(t) = V_{out}(t) + I(t) \cdot R \]
Since \( I(t) = C \frac{dV_{out}(t)}{dt} \), we can substitute this into the equation:
\[ V_{in}(t) = V_{out}(t) + R \cdot C \frac{dV_{out}(t)}{dt} \]
Rearranging gives us the differential equation:
\[ \frac{dV_{out}(t)}{dt} + \frac{1}{R \cdot C} V_{out}(t) = \frac{V_{in}(t)}{R \cdot C} \]
**2. Step Response Solution:**
For a step input \( V_{in}(t) = V_0 \cdot u(t) \):
Substitute \( V_{in}(t) = V_0 \) into the differential equation:
\[ \frac{dV_{out}(t)}{dt} + \frac{1}{R \cdot C} V_{out}(t) = \frac{V_0}{R \cdot C} \]
To solve this, we use the method of integrating factors or the Laplace transform. The general solution for \( V_{out}(t) \) is:
\[ V_{out}(t) = V_0 \left(1 - e^{-\frac{t}{R \cdot C}}\right) \]
**3. Interpretation:**
- **Initial Condition (t = 0):** At \( t = 0 \), \( V_{out}(0) = 0 \). This is because the capacitor initially behaves like an open circuit, and the full input voltage appears across the resistor.
- **Long-Term Behavior (t → ∞):** As \( t \) approaches infinity, \( V_{out}(t) \) approaches \( V_0 \). The capacitor eventually charges up to the input voltage \( V_0 \), and the voltage across the resistor becomes zero.
### Summary
The step response of an RC circuit shows that the voltage across the capacitor gradually increases from 0 to \( V_0 \) following an exponential curve. The time constant \( \tau = R \cdot C \) characterizes how quickly the capacitor charges, with the voltage reaching approximately 63.2% of \( V_0 \) after one time constant \( \tau \).