How to find transient response?
2 Answers
The transient response of a system refers to how the system behaves in response to a change in its operating conditions, particularly after a sudden disturbance or input. Finding the transient response involves analyzing how the system's output evolves over time before reaching a steady state.
Here's a detailed process for finding the transient response of a system:
### 1. **Define the System and Its Dynamics**
- **System Representation**: Determine if the system is represented by a differential equation, transfer function, or state-space model. For many systems, a transfer function or state-space representation is used to describe the system's dynamics.
- **Inputs and Outputs**: Identify the input signal (e.g., step input, impulse input) and the corresponding output of the system.
### 2. **Obtain the System's Transfer Function**
- **Transfer Function**: For linear time-invariant (LTI) systems, the transfer function \( H(s) \) is often used. It is a ratio of the Laplace Transform of the output \( Y(s) \) to the Laplace Transform of the input \( U(s) \):
\[
H(s) = \frac{Y(s)}{U(s)}
\]
- **Differential Equation**: If the system is described by a differential equation, you can derive the transfer function by taking the Laplace Transform of the equation, assuming initial conditions are zero.
### 3. **Determine the Input Signal**
- **Types of Inputs**: Common inputs include step functions, impulse functions, or sinusoidal functions. Each type will produce a different transient response.
- **Laplace Transform**: Apply the Laplace Transform to the input signal to get its Laplace domain representation \( U(s) \).
### 4. **Find the System’s Response in the Laplace Domain**
- **Compute Output**: Multiply the transfer function \( H(s) \) by the Laplace Transform of the input \( U(s) \) to find the Laplace Transform of the output \( Y(s) \):
\[
Y(s) = H(s) \cdot U(s)
\]
### 5. **Inverse Laplace Transform**
- **Inverse Transform**: To find the transient response in the time domain, take the Inverse Laplace Transform of \( Y(s) \). This process converts the solution from the Laplace domain back to the time domain.
### 6. **Analyze the Transient Response**
- **Transient Characteristics**: Examine the time-domain response for characteristics such as rise time, settling time, overshoot, and oscillatory behavior.
- **System Stability**: Check if the system is stable by analyzing the poles of the transfer function. The location of poles in the Laplace domain (i.e., the roots of the denominator polynomial) determines the stability and nature of the transient response.
### Example
Let's go through an example with a simple first-order system:
1. **System**: Consider a first-order system with the transfer function:
\[
H(s) = \frac{1}{\tau s + 1}
\]
where \( \tau \) is the time constant of the system.
2. **Input**: Suppose the input is a unit step function \( u(t) \). The Laplace Transform of \( u(t) \) is \( \frac{1}{s} \).
3. **Output in Laplace Domain**: Compute the output in the Laplace domain:
\[
Y(s) = H(s) \cdot \frac{1}{s} = \frac{1}{\tau s + 1} \cdot \frac{1}{s} = \frac{1}{s(\tau s + 1)}
\]
4. **Inverse Laplace Transform**: To find the time-domain response, perform the Inverse Laplace Transform of \( Y(s) \):
\[
y(t) = \mathcal{L}^{-1}\left\{\frac{1}{s(\tau s + 1)}\right\}
\]
This results in:
\[
y(t) = 1 - e^{-\frac{t}{\tau}}
\]
5. **Analysis**: The transient response \( y(t) \) shows that the system output starts at 0 and asymptotically approaches 1. The exponential term \( e^{-\frac{t}{\tau}} \) represents the transient part of the response, which decays over time.
### Summary
Finding the transient response involves determining the system’s dynamics, applying the input signal, calculating the system’s output in the Laplace domain, and then converting this output back to the time domain to analyze how the system responds over time. This process helps in understanding how quickly and effectively a system reacts to changes or disturbances.
Here's a detailed process for finding the transient response of a system:
### 1. **Define the System and Its Dynamics**
- **System Representation**: Determine if the system is represented by a differential equation, transfer function, or state-space model. For many systems, a transfer function or state-space representation is used to describe the system's dynamics.
- **Inputs and Outputs**: Identify the input signal (e.g., step input, impulse input) and the corresponding output of the system.
### 2. **Obtain the System's Transfer Function**
- **Transfer Function**: For linear time-invariant (LTI) systems, the transfer function \( H(s) \) is often used. It is a ratio of the Laplace Transform of the output \( Y(s) \) to the Laplace Transform of the input \( U(s) \):
\[
H(s) = \frac{Y(s)}{U(s)}
\]
- **Differential Equation**: If the system is described by a differential equation, you can derive the transfer function by taking the Laplace Transform of the equation, assuming initial conditions are zero.
### 3. **Determine the Input Signal**
- **Types of Inputs**: Common inputs include step functions, impulse functions, or sinusoidal functions. Each type will produce a different transient response.
- **Laplace Transform**: Apply the Laplace Transform to the input signal to get its Laplace domain representation \( U(s) \).
### 4. **Find the System’s Response in the Laplace Domain**
- **Compute Output**: Multiply the transfer function \( H(s) \) by the Laplace Transform of the input \( U(s) \) to find the Laplace Transform of the output \( Y(s) \):
\[
Y(s) = H(s) \cdot U(s)
\]
### 5. **Inverse Laplace Transform**
- **Inverse Transform**: To find the transient response in the time domain, take the Inverse Laplace Transform of \( Y(s) \). This process converts the solution from the Laplace domain back to the time domain.
### 6. **Analyze the Transient Response**
- **Transient Characteristics**: Examine the time-domain response for characteristics such as rise time, settling time, overshoot, and oscillatory behavior.
- **System Stability**: Check if the system is stable by analyzing the poles of the transfer function. The location of poles in the Laplace domain (i.e., the roots of the denominator polynomial) determines the stability and nature of the transient response.
### Example
Let's go through an example with a simple first-order system:
1. **System**: Consider a first-order system with the transfer function:
\[
H(s) = \frac{1}{\tau s + 1}
\]
where \( \tau \) is the time constant of the system.
2. **Input**: Suppose the input is a unit step function \( u(t) \). The Laplace Transform of \( u(t) \) is \( \frac{1}{s} \).
3. **Output in Laplace Domain**: Compute the output in the Laplace domain:
\[
Y(s) = H(s) \cdot \frac{1}{s} = \frac{1}{\tau s + 1} \cdot \frac{1}{s} = \frac{1}{s(\tau s + 1)}
\]
4. **Inverse Laplace Transform**: To find the time-domain response, perform the Inverse Laplace Transform of \( Y(s) \):
\[
y(t) = \mathcal{L}^{-1}\left\{\frac{1}{s(\tau s + 1)}\right\}
\]
This results in:
\[
y(t) = 1 - e^{-\frac{t}{\tau}}
\]
5. **Analysis**: The transient response \( y(t) \) shows that the system output starts at 0 and asymptotically approaches 1. The exponential term \( e^{-\frac{t}{\tau}} \) represents the transient part of the response, which decays over time.
### Summary
Finding the transient response involves determining the system’s dynamics, applying the input signal, calculating the system’s output in the Laplace domain, and then converting this output back to the time domain to analyze how the system responds over time. This process helps in understanding how quickly and effectively a system reacts to changes or disturbances.
Let's break down the problem step by step:
1. **Determine Vph (Phase Voltage):**
In a star (wye) connection, the phase voltage \( V_{ph} \) is related to the line voltage \( V_L \) by:
\[
V_{ph} = \frac{V_L}{\sqrt{3}}
\]
Given \( V_L = 415 \text{ V} \):
\[
V_{ph} = \frac{415}{\sqrt{3}} \approx 240 \text{ V}
\]
2. **Determine Iph (Phase Current):**
Each coil has an impedance of \( Z = 4 + j5 \text{ ohms} \). The phase current \( I_{ph} \) is given by:
\[
I_{ph} = \frac{V_{ph}}{Z}
\]
First, calculate the magnitude of \( Z \):
\[
|Z| = \sqrt{4^2 + 5^2} = \sqrt{41} \approx 6.4 \text{ ohms}
\]
Thus:
\[
I_{ph} = \frac{240}{6.4} \approx 37.5 \text{ A}
\]
3. **Wattmeter Readings (W1, W2):**
In a balanced star-connected load, each wattmeter reads:
\[
W = V_{ph} \cdot I_{ph} \cdot \cos \phi
\]
Where \( \cos \phi \) is the power factor of the coil. For \( Z = 4 + j5 \):
\[
\cos \phi = \frac{R}{|Z|} = \frac{4}{6.4} \approx 0.625
\]
Thus:
\[
W_{1} = W_{2} = 240 \cdot 37.5 \cdot 0.625 \approx 5,625 \text{ W}
\]
Each wattmeter reads approximately 5,625 W.
1. **Determine Vph (Phase Voltage):**
In a star (wye) connection, the phase voltage \( V_{ph} \) is related to the line voltage \( V_L \) by:
\[
V_{ph} = \frac{V_L}{\sqrt{3}}
\]
Given \( V_L = 415 \text{ V} \):
\[
V_{ph} = \frac{415}{\sqrt{3}} \approx 240 \text{ V}
\]
2. **Determine Iph (Phase Current):**
Each coil has an impedance of \( Z = 4 + j5 \text{ ohms} \). The phase current \( I_{ph} \) is given by:
\[
I_{ph} = \frac{V_{ph}}{Z}
\]
First, calculate the magnitude of \( Z \):
\[
|Z| = \sqrt{4^2 + 5^2} = \sqrt{41} \approx 6.4 \text{ ohms}
\]
Thus:
\[
I_{ph} = \frac{240}{6.4} \approx 37.5 \text{ A}
\]
3. **Wattmeter Readings (W1, W2):**
In a balanced star-connected load, each wattmeter reads:
\[
W = V_{ph} \cdot I_{ph} \cdot \cos \phi
\]
Where \( \cos \phi \) is the power factor of the coil. For \( Z = 4 + j5 \):
\[
\cos \phi = \frac{R}{|Z|} = \frac{4}{6.4} \approx 0.625
\]
Thus:
\[
W_{1} = W_{2} = 240 \cdot 37.5 \cdot 0.625 \approx 5,625 \text{ W}
\]
Each wattmeter reads approximately 5,625 W.