What is the transfer function in the RL circuit?
2 Answers
### Transfer Function of an RL Circuit
The transfer function of an RL circuit describes the relationship between the input and output in the frequency domain. This circuit typically consists of a resistor (R) and an inductor (L) connected in series or parallel. Here, we'll derive the transfer function for a **series RL circuit**, which is a common configuration.
### 1. Series RL Circuit Configuration
A typical series RL circuit is composed of:
- Resistor \( R \)
- Inductor \( L \)
- An input voltage \( V_{in}(t) \)
- An output voltage across either the resistor or the inductor. For this example, let's find the transfer function with the output taken across the resistor.
### 2. Deriving the Transfer Function
#### Step 1: Circuit Analysis in the Time Domain
Using Kirchhoff's Voltage Law (KVL) for the series RL circuit, the sum of the voltages across the resistor and inductor is equal to the input voltage:
\[
V_{in}(t) = V_R(t) + V_L(t)
\]
Where:
- \( V_R(t) = i(t) R \) (Ohm's Law)
- \( V_L(t) = L \frac{di(t)}{dt} \) (Inductor voltage-current relationship)
The equation becomes:
\[
V_{in}(t) = i(t) R + L \frac{di(t)}{dt}
\]
#### Step 2: Apply Laplace Transform
To find the transfer function, we transform this equation into the frequency domain using the Laplace Transform. Assume zero initial conditions:
\[
V_{in}(s) = I(s) R + s L I(s)
\]
Where:
- \( V_{in}(s) \) is the Laplace transform of \( V_{in}(t) \)
- \( I(s) \) is the Laplace transform of \( i(t) \)
- \( s \) is the complex frequency variable
Factor out \( I(s) \):
\[
V_{in}(s) = I(s) (R + sL)
\]
#### Step 3: Determine the Transfer Function
The transfer function \( H(s) \) is the ratio of the output to the input in the Laplace domain. In this case, if we consider the output as the voltage across the resistor \( V_R(s) \):
\[
V_R(s) = I(s) R
\]
The input \( V_{in}(s) \) is given by:
\[
V_{in}(s) = I(s) (R + sL)
\]
Now, solve for the current \( I(s) \):
\[
I(s) = \frac{V_{in}(s)}{R + sL}
\]
Substitute this into the expression for \( V_R(s) \):
\[
V_R(s) = \frac{V_{in}(s)}{R + sL} \times R
\]
The transfer function \( H(s) \) is:
\[
H(s) = \frac{V_R(s)}{V_{in}(s)} = \frac{R}{R + sL}
\]
### 3. Interpretation of the Transfer Function
The transfer function \( H(s) = \frac{R}{R + sL} \) gives insights into the frequency response of the RL circuit:
- **Low-frequency behavior**: When \( s \) (which corresponds to \( j\omega \) in the frequency domain) is small (low frequencies), \( H(s) \approx 1 \). This means that at low frequencies, the output across the resistor approximately equals the input.
- **High-frequency behavior**: When \( s \) is large (high frequencies), \( H(s) \approx \frac{R}{sL} \), indicating that the output voltage across the resistor decreases with frequency.
### 4. Bode Plot Analysis
The transfer function can also be analyzed using Bode plots to visualize the magnitude and phase of the output concerning frequency.
1. **Magnitude**:
\[
|H(j\omega)| = \left|\frac{R}{R + j\omega L}\right|
\]
2. **Phase**:
\[
\angle H(j\omega) = \tan^{-1}\left(-\frac{\omega L}{R}\right)
\]
These expressions show how the circuit behaves as a low-pass filter, allowing low-frequency signals to pass through while attenuating higher frequencies.
### 5. Summary
For a series RL circuit with the output across the resistor, the transfer function is:
\[
H(s) = \frac{R}{R + sL}
\]
This function characterizes the circuit's response in the frequency domain, showing that it acts as a low-pass filter. The cut-off frequency (\( \omega_c \)) where the output power drops to half its maximum value is given by:
\[
\omega_c = \frac{R}{L}
\]
This transfer function is fundamental in understanding how RL circuits respond to different input signals in both the time and frequency domains.
The transfer function of an RL circuit describes the relationship between the input and output in the frequency domain. This circuit typically consists of a resistor (R) and an inductor (L) connected in series or parallel. Here, we'll derive the transfer function for a **series RL circuit**, which is a common configuration.
### 1. Series RL Circuit Configuration
A typical series RL circuit is composed of:
- Resistor \( R \)
- Inductor \( L \)
- An input voltage \( V_{in}(t) \)
- An output voltage across either the resistor or the inductor. For this example, let's find the transfer function with the output taken across the resistor.
### 2. Deriving the Transfer Function
#### Step 1: Circuit Analysis in the Time Domain
Using Kirchhoff's Voltage Law (KVL) for the series RL circuit, the sum of the voltages across the resistor and inductor is equal to the input voltage:
\[
V_{in}(t) = V_R(t) + V_L(t)
\]
Where:
- \( V_R(t) = i(t) R \) (Ohm's Law)
- \( V_L(t) = L \frac{di(t)}{dt} \) (Inductor voltage-current relationship)
The equation becomes:
\[
V_{in}(t) = i(t) R + L \frac{di(t)}{dt}
\]
#### Step 2: Apply Laplace Transform
To find the transfer function, we transform this equation into the frequency domain using the Laplace Transform. Assume zero initial conditions:
\[
V_{in}(s) = I(s) R + s L I(s)
\]
Where:
- \( V_{in}(s) \) is the Laplace transform of \( V_{in}(t) \)
- \( I(s) \) is the Laplace transform of \( i(t) \)
- \( s \) is the complex frequency variable
Factor out \( I(s) \):
\[
V_{in}(s) = I(s) (R + sL)
\]
#### Step 3: Determine the Transfer Function
The transfer function \( H(s) \) is the ratio of the output to the input in the Laplace domain. In this case, if we consider the output as the voltage across the resistor \( V_R(s) \):
\[
V_R(s) = I(s) R
\]
The input \( V_{in}(s) \) is given by:
\[
V_{in}(s) = I(s) (R + sL)
\]
Now, solve for the current \( I(s) \):
\[
I(s) = \frac{V_{in}(s)}{R + sL}
\]
Substitute this into the expression for \( V_R(s) \):
\[
V_R(s) = \frac{V_{in}(s)}{R + sL} \times R
\]
The transfer function \( H(s) \) is:
\[
H(s) = \frac{V_R(s)}{V_{in}(s)} = \frac{R}{R + sL}
\]
### 3. Interpretation of the Transfer Function
The transfer function \( H(s) = \frac{R}{R + sL} \) gives insights into the frequency response of the RL circuit:
- **Low-frequency behavior**: When \( s \) (which corresponds to \( j\omega \) in the frequency domain) is small (low frequencies), \( H(s) \approx 1 \). This means that at low frequencies, the output across the resistor approximately equals the input.
- **High-frequency behavior**: When \( s \) is large (high frequencies), \( H(s) \approx \frac{R}{sL} \), indicating that the output voltage across the resistor decreases with frequency.
### 4. Bode Plot Analysis
The transfer function can also be analyzed using Bode plots to visualize the magnitude and phase of the output concerning frequency.
1. **Magnitude**:
\[
|H(j\omega)| = \left|\frac{R}{R + j\omega L}\right|
\]
2. **Phase**:
\[
\angle H(j\omega) = \tan^{-1}\left(-\frac{\omega L}{R}\right)
\]
These expressions show how the circuit behaves as a low-pass filter, allowing low-frequency signals to pass through while attenuating higher frequencies.
### 5. Summary
For a series RL circuit with the output across the resistor, the transfer function is:
\[
H(s) = \frac{R}{R + sL}
\]
This function characterizes the circuit's response in the frequency domain, showing that it acts as a low-pass filter. The cut-off frequency (\( \omega_c \)) where the output power drops to half its maximum value is given by:
\[
\omega_c = \frac{R}{L}
\]
This transfer function is fundamental in understanding how RL circuits respond to different input signals in both the time and frequency domains.
In an RL circuit, which consists of a resistor (R) and an inductor (L) connected in series, the transfer function describes the relationship between the input and output of the circuit in the frequency domain.
### 1. **Basic RL Circuit**
For a basic RL series circuit:
- **Resistor (R)**
- **Inductor (L)**
The input voltage is \( V_{in}(t) \), and the output voltage across the inductor is \( V_{L}(t) \).
### 2. **Impedances**
In the frequency domain, the impedance of the resistor \( R \) is \( R \), and the impedance of the inductor \( L \) is \( j\omega L \), where \( j \) is the imaginary unit and \( \omega \) is the angular frequency (\( \omega = 2 \pi f \), where \( f \) is the frequency in Hz).
### 3. **Transfer Function Derivation**
The transfer function \( H(s) \) is defined as the ratio of the Laplace transform of the output voltage to the Laplace transform of the input voltage. For the RL circuit:
- **Input Voltage (\( V_{in}(s) \))**: Laplace transform of the input voltage.
- **Output Voltage (\( V_{L}(s) \))**: Laplace transform of the voltage across the inductor.
The impedance of the RL series circuit is:
\[ Z_{RL}(s) = R + sL \]
where \( s \) is the complex frequency variable in the Laplace domain.
The voltage across the inductor \( V_{L}(s) \) can be found using the voltage divider rule:
\[ V_{L}(s) = V_{in}(s) \cdot \frac{Z_{L}}{Z_{R} + Z_{L}} \]
where \( Z_{L} = sL \) and \( Z_{R} = R \). So,
\[ V_{L}(s) = V_{in}(s) \cdot \frac{sL}{R + sL} \]
The transfer function \( H(s) \) is given by:
\[ H(s) = \frac{V_{L}(s)}{V_{in}(s)} \]
Substituting the voltage divider result:
\[ H(s) = \frac{\frac{sL}{R + sL}}{1} \]
Simplify to:
\[ H(s) = \frac{sL}{R + sL} \]
### 4. **Frequency Response**
In the frequency domain (\( s = j\omega \)):
\[ H(j\omega) = \frac{j\omega L}{R + j\omega L} \]
This transfer function describes how the RL circuit responds to different frequencies. At low frequencies, the impedance of the inductor is small, so the output voltage \( V_{L} \) is small compared to \( V_{in} \). At high frequencies, the impedance of the inductor becomes large, and \( V_{L} \) approaches \( V_{in} \).
### Summary
The transfer function for an RL circuit is:
\[ H(s) = \frac{sL}{R + sL} \]
or, in the frequency domain:
\[ H(j\omega) = \frac{j\omega L}{R + j\omega L} \]
This function provides insight into how the circuit filters signals based on frequency.
### 1. **Basic RL Circuit**
For a basic RL series circuit:
- **Resistor (R)**
- **Inductor (L)**
The input voltage is \( V_{in}(t) \), and the output voltage across the inductor is \( V_{L}(t) \).
### 2. **Impedances**
In the frequency domain, the impedance of the resistor \( R \) is \( R \), and the impedance of the inductor \( L \) is \( j\omega L \), where \( j \) is the imaginary unit and \( \omega \) is the angular frequency (\( \omega = 2 \pi f \), where \( f \) is the frequency in Hz).
### 3. **Transfer Function Derivation**
The transfer function \( H(s) \) is defined as the ratio of the Laplace transform of the output voltage to the Laplace transform of the input voltage. For the RL circuit:
- **Input Voltage (\( V_{in}(s) \))**: Laplace transform of the input voltage.
- **Output Voltage (\( V_{L}(s) \))**: Laplace transform of the voltage across the inductor.
The impedance of the RL series circuit is:
\[ Z_{RL}(s) = R + sL \]
where \( s \) is the complex frequency variable in the Laplace domain.
The voltage across the inductor \( V_{L}(s) \) can be found using the voltage divider rule:
\[ V_{L}(s) = V_{in}(s) \cdot \frac{Z_{L}}{Z_{R} + Z_{L}} \]
where \( Z_{L} = sL \) and \( Z_{R} = R \). So,
\[ V_{L}(s) = V_{in}(s) \cdot \frac{sL}{R + sL} \]
The transfer function \( H(s) \) is given by:
\[ H(s) = \frac{V_{L}(s)}{V_{in}(s)} \]
Substituting the voltage divider result:
\[ H(s) = \frac{\frac{sL}{R + sL}}{1} \]
Simplify to:
\[ H(s) = \frac{sL}{R + sL} \]
### 4. **Frequency Response**
In the frequency domain (\( s = j\omega \)):
\[ H(j\omega) = \frac{j\omega L}{R + j\omega L} \]
This transfer function describes how the RL circuit responds to different frequencies. At low frequencies, the impedance of the inductor is small, so the output voltage \( V_{L} \) is small compared to \( V_{in} \). At high frequencies, the impedance of the inductor becomes large, and \( V_{L} \) approaches \( V_{in} \).
### Summary
The transfer function for an RL circuit is:
\[ H(s) = \frac{sL}{R + sL} \]
or, in the frequency domain:
\[ H(j\omega) = \frac{j\omega L}{R + j\omega L} \]
This function provides insight into how the circuit filters signals based on frequency.