What is the general equation of step response?
2 Answers
The step response of a system refers to how the system reacts over time when subjected to a step input, usually a unit step function. This is a crucial aspect in control systems and signal processing to understand how a system will behave in response to a sudden change or command.
Here’s a general overview of how the step response can be described for different types of systems:
### 1. **First-Order Systems**
For a first-order linear time-invariant (LTI) system, the general form of the differential equation is:
\[ \tau \frac{dy(t)}{dt} + y(t) = K u(t) \]
where:
- \( \tau \) is the time constant of the system,
- \( K \) is the system gain,
- \( u(t) \) is the step input (often \( u(t) = 1 \) for a unit step function),
- \( y(t) \) is the output of the system.
The step response \( y(t) \) for a first-order system is:
\[ y(t) = K \left(1 - e^{-\frac{t}{\tau}}\right) \]
Here, \( \tau \) determines how quickly the system responds to the step input. The term \( e^{-\frac{t}{\tau}} \) describes the exponential approach to the final value \( K \).
### 2. **Second-Order Systems**
For a second-order LTI system, the differential equation is:
\[ \frac{d^2y(t)}{dt^2} + 2\zeta \omega_n \frac{dy(t)}{dt} + \omega_n^2 y(t) = \omega_n^2 u(t) \]
where:
- \( \omega_n \) is the natural frequency of the system,
- \( \zeta \) is the damping ratio,
- \( u(t) \) is the step input (often \( u(t) = 1 \)),
- \( y(t) \) is the output of the system.
The step response \( y(t) \) depends on the damping ratio \( \zeta \) and is categorized into different cases based on \( \zeta \):
- **Underdamped (\( \zeta < 1 \))**: The response oscillates before settling to the final value.
\[ y(t) = 1 - \frac{1}{\sqrt{1 - \zeta^2}} e^{-\zeta \omega_n t} \left( \sin \left( \omega_d t + \phi \right) \right) \]
where \( \omega_d = \omega_n \sqrt{1 - \zeta^2} \) is the damped natural frequency and \( \phi \) is a phase angle.
- **Critically damped (\( \zeta = 1 \))**: The system returns to equilibrium as quickly as possible without oscillating.
\[ y(t) = 1 - \left(1 + \omega_n t \right) e^{-\omega_n t} \]
- **Overdamped (\( \zeta > 1 \))**: The response returns to equilibrium without oscillating, but more slowly compared to the critically damped case.
\[ y(t) = 1 - \frac{e^{-\lambda_1 t} - e^{-\lambda_2 t}}{\lambda_2 - \lambda_1} \]
where \( \lambda_1 \) and \( \lambda_2 \) are the roots of the characteristic equation.
### 3. **Higher-Order Systems**
For higher-order systems, the step response can be more complex and generally requires solving the corresponding higher-order differential equations. In practice, these systems are often analyzed using techniques such as Laplace transforms or numerical simulation.
### Summary
- **First-Order System**: \( y(t) = K \left(1 - e^{-\frac{t}{\tau}}\right) \)
- **Second-Order Underdamped**: \( y(t) = 1 - \frac{1}{\sqrt{1 - \zeta^2}} e^{-\zeta \omega_n t} \left( \sin \left( \omega_d t + \phi \right) \right) \)
- **Second-Order Critically Damped**: \( y(t) = 1 - \left(1 + \omega_n t \right) e^{-\omega_n t} \)
- **Second-Order Overdamped**: \( y(t) = 1 - \frac{e^{-\lambda_1 t} - e^{-\lambda_2 t}}{\lambda_2 - \lambda_1} \)
Understanding the step response of a system helps in designing and analyzing systems to ensure they meet desired performance criteria.
Here’s a general overview of how the step response can be described for different types of systems:
### 1. **First-Order Systems**
For a first-order linear time-invariant (LTI) system, the general form of the differential equation is:
\[ \tau \frac{dy(t)}{dt} + y(t) = K u(t) \]
where:
- \( \tau \) is the time constant of the system,
- \( K \) is the system gain,
- \( u(t) \) is the step input (often \( u(t) = 1 \) for a unit step function),
- \( y(t) \) is the output of the system.
The step response \( y(t) \) for a first-order system is:
\[ y(t) = K \left(1 - e^{-\frac{t}{\tau}}\right) \]
Here, \( \tau \) determines how quickly the system responds to the step input. The term \( e^{-\frac{t}{\tau}} \) describes the exponential approach to the final value \( K \).
### 2. **Second-Order Systems**
For a second-order LTI system, the differential equation is:
\[ \frac{d^2y(t)}{dt^2} + 2\zeta \omega_n \frac{dy(t)}{dt} + \omega_n^2 y(t) = \omega_n^2 u(t) \]
where:
- \( \omega_n \) is the natural frequency of the system,
- \( \zeta \) is the damping ratio,
- \( u(t) \) is the step input (often \( u(t) = 1 \)),
- \( y(t) \) is the output of the system.
The step response \( y(t) \) depends on the damping ratio \( \zeta \) and is categorized into different cases based on \( \zeta \):
- **Underdamped (\( \zeta < 1 \))**: The response oscillates before settling to the final value.
\[ y(t) = 1 - \frac{1}{\sqrt{1 - \zeta^2}} e^{-\zeta \omega_n t} \left( \sin \left( \omega_d t + \phi \right) \right) \]
where \( \omega_d = \omega_n \sqrt{1 - \zeta^2} \) is the damped natural frequency and \( \phi \) is a phase angle.
- **Critically damped (\( \zeta = 1 \))**: The system returns to equilibrium as quickly as possible without oscillating.
\[ y(t) = 1 - \left(1 + \omega_n t \right) e^{-\omega_n t} \]
- **Overdamped (\( \zeta > 1 \))**: The response returns to equilibrium without oscillating, but more slowly compared to the critically damped case.
\[ y(t) = 1 - \frac{e^{-\lambda_1 t} - e^{-\lambda_2 t}}{\lambda_2 - \lambda_1} \]
where \( \lambda_1 \) and \( \lambda_2 \) are the roots of the characteristic equation.
### 3. **Higher-Order Systems**
For higher-order systems, the step response can be more complex and generally requires solving the corresponding higher-order differential equations. In practice, these systems are often analyzed using techniques such as Laplace transforms or numerical simulation.
### Summary
- **First-Order System**: \( y(t) = K \left(1 - e^{-\frac{t}{\tau}}\right) \)
- **Second-Order Underdamped**: \( y(t) = 1 - \frac{1}{\sqrt{1 - \zeta^2}} e^{-\zeta \omega_n t} \left( \sin \left( \omega_d t + \phi \right) \right) \)
- **Second-Order Critically Damped**: \( y(t) = 1 - \left(1 + \omega_n t \right) e^{-\omega_n t} \)
- **Second-Order Overdamped**: \( y(t) = 1 - \frac{e^{-\lambda_1 t} - e^{-\lambda_2 t}}{\lambda_2 - \lambda_1} \)
Understanding the step response of a system helps in designing and analyzing systems to ensure they meet desired performance criteria.
The **step response** of a system refers to how the system behaves when subjected to a step input, typically a sudden change from zero to a constant value. In electrical engineering, especially in control systems and signal processing, analyzing the step response helps understand a system's dynamics, such as stability, rise time, and settling time.
The general equation for the step response of a linear time-invariant (LTI) system depends on its **transfer function** \( H(s) \). Here's how the step response is derived:
### 1. **Transfer Function** \( H(s) \):
The transfer function of an LTI system is given by:
\[
H(s) = \frac{Y(s)}{X(s)}
\]
Where:
- \( Y(s) \) is the Laplace transform of the output.
- \( X(s) \) is the Laplace transform of the input.
For a step input, \( x(t) = u(t) \), where \( u(t) \) is the unit step function, its Laplace transform is:
\[
X(s) = \frac{1}{s}
\]
### 2. **Step Response in the Laplace Domain**:
Using the transfer function and the step input, the output in the Laplace domain \( Y(s) \) is:
\[
Y(s) = H(s) \cdot \frac{1}{s}
\]
### 3. **Inverse Laplace Transform**:
To get the time-domain step response \( y(t) \), we take the inverse Laplace transform of \( Y(s) \).
Thus:
\[
y(t) = \mathcal{L}^{-1} \left[ H(s) \cdot \frac{1}{s} \right]
\]
The specific form of \( y(t) \) depends on the system's transfer function. Below are typical examples for first-order and second-order systems.
---
### **First-Order System**:
For a first-order system, the transfer function is:
\[
H(s) = \frac{K}{\tau s + 1}
\]
The step response is:
\[
y(t) = K \left( 1 - e^{-\frac{t}{\tau}} \right)
\]
Where:
- \( K \) is the system gain.
- \( \tau \) is the time constant, representing how fast the system responds.
---
### **Second-Order System**:
For a second-order system, the transfer function is:
\[
H(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}
\]
The step response varies depending on the damping ratio \( \zeta \):
- **Overdamped** (\( \zeta > 1 \))
- **Critically damped** (\( \zeta = 1 \))
- **Underdamped** (\( \zeta < 1 \))
For the underdamped case (\( 0 < \zeta < 1 \)), the step response is:
\[
y(t) = 1 - e^{-\zeta \omega_n t} \left( \cos(\omega_d t) + \frac{\zeta \omega_n}{\omega_d} \sin(\omega_d t) \right)
\]
Where:
- \( \omega_n \) is the natural frequency.
- \( \zeta \) is the damping ratio.
- \( \omega_d = \omega_n \sqrt{1 - \zeta^2} \) is the damped natural frequency.
---
### Summary:
The **general equation** of the step response involves taking the inverse Laplace transform of the product of the system's transfer function \( H(s) \) and the step input \( \frac{1}{s} \). For specific systems, such as first-order or second-order systems, the step response can be expressed as a time-domain function involving exponential and sinusoidal terms depending on the system's parameters (e.g., gain, time constant, natural frequency, damping ratio).
The general equation for the step response of a linear time-invariant (LTI) system depends on its **transfer function** \( H(s) \). Here's how the step response is derived:
### 1. **Transfer Function** \( H(s) \):
The transfer function of an LTI system is given by:
\[
H(s) = \frac{Y(s)}{X(s)}
\]
Where:
- \( Y(s) \) is the Laplace transform of the output.
- \( X(s) \) is the Laplace transform of the input.
For a step input, \( x(t) = u(t) \), where \( u(t) \) is the unit step function, its Laplace transform is:
\[
X(s) = \frac{1}{s}
\]
### 2. **Step Response in the Laplace Domain**:
Using the transfer function and the step input, the output in the Laplace domain \( Y(s) \) is:
\[
Y(s) = H(s) \cdot \frac{1}{s}
\]
### 3. **Inverse Laplace Transform**:
To get the time-domain step response \( y(t) \), we take the inverse Laplace transform of \( Y(s) \).
Thus:
\[
y(t) = \mathcal{L}^{-1} \left[ H(s) \cdot \frac{1}{s} \right]
\]
The specific form of \( y(t) \) depends on the system's transfer function. Below are typical examples for first-order and second-order systems.
---
### **First-Order System**:
For a first-order system, the transfer function is:
\[
H(s) = \frac{K}{\tau s + 1}
\]
The step response is:
\[
y(t) = K \left( 1 - e^{-\frac{t}{\tau}} \right)
\]
Where:
- \( K \) is the system gain.
- \( \tau \) is the time constant, representing how fast the system responds.
---
### **Second-Order System**:
For a second-order system, the transfer function is:
\[
H(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}
\]
The step response varies depending on the damping ratio \( \zeta \):
- **Overdamped** (\( \zeta > 1 \))
- **Critically damped** (\( \zeta = 1 \))
- **Underdamped** (\( \zeta < 1 \))
For the underdamped case (\( 0 < \zeta < 1 \)), the step response is:
\[
y(t) = 1 - e^{-\zeta \omega_n t} \left( \cos(\omega_d t) + \frac{\zeta \omega_n}{\omega_d} \sin(\omega_d t) \right)
\]
Where:
- \( \omega_n \) is the natural frequency.
- \( \zeta \) is the damping ratio.
- \( \omega_d = \omega_n \sqrt{1 - \zeta^2} \) is the damped natural frequency.
---
### Summary:
The **general equation** of the step response involves taking the inverse Laplace transform of the product of the system's transfer function \( H(s) \) and the step input \( \frac{1}{s} \). For specific systems, such as first-order or second-order systems, the step response can be expressed as a time-domain function involving exponential and sinusoidal terms depending on the system's parameters (e.g., gain, time constant, natural frequency, damping ratio).