Documentation

### This is machine translation

Mouseover text to see original. Click the button below to return to the English version of the page.

To view all translated materials including this page, select Country from the country navigator on the bottom of this page.

## Analyzing the Response of an RLC Circuit

This example shows how to analyze the time and frequency responses of common RLC circuits as a function of their physical parameters using Control System Toolbox™ functions.

### Bandpass RLC Network

The following figure shows the parallel form of a bandpass RLC circuit: Figure 1: Bandpass RLC Network.

The transfer function from input to output voltage is:

` `

The product `LC` controls the bandpass frequency while `RC` controls how narrow the passing band is. To build a bandpass filter tuned to the frequency 1 rad/s, set `L=C=1` and use `R` to tune the filter band.

### Analyzing the Frequency Response of the Circuit

The Bode plot is a convenient tool for investigating the bandpass characteristics of the RLC network. Use `tf` to specify the circuit's transfer function for the values

```%|R=L=C=1|: R = 1; L = 1; C = 1; G = tf([1/(R*C) 0],[1 1/(R*C) 1/(L*C)]) ```
```G = s ----------- s^2 + s + 1 Continuous-time transfer function. ```

Next, use `bode` to plot the frequency response of the circuit:

```bode(G), grid ``` As expected, the RLC filter has maximum gain at the frequency 1 rad/s. However, the attenuation is only -10dB half a decade away from this frequency. To get a narrower passing band, try increasing values of R as follows:

```R1 = 5; G1 = tf([1/(R1*C) 0],[1 1/(R1*C) 1/(L*C)]); R2 = 20; G2 = tf([1/(R2*C) 0],[1 1/(R2*C) 1/(L*C)]); bode(G,'b',G1,'r',G2,'g'), grid legend('R = 1','R = 5','R = 20') ``` The resistor value `R=20` gives a filter narrowly tuned around the target frequency of 1 rad/s.

### Analyzing the Time Response of the Circuit

We can confirm the attenuation properties of the circuit `G2` (`R=20`) by simulating how this filter transforms sine waves with frequency 0.9, 1, and 1.1 rad/s:

```t = 0:0.05:250; opt = timeoptions; opt.Title.FontWeight = 'Bold'; subplot(311), lsim(G2,sin(t),t,opt), title('w = 1') subplot(312), lsim(G2,sin(0.9*t),t,opt), title('w = 0.9') subplot(313), lsim(G2,sin(1.1*t),t,opt), title('w = 1.1') ``` The waves at 0.9 and 1.1 rad/s are considerably attenuated. The wave at 1 rad/s comes out unchanged once the transients have died off. The long transient results from the poorly damped poles of the filters, which unfortunately are required for a narrow passing band:

```damp(pole(G2)) ```
``` Pole Damping Frequency Time Constant (rad/TimeUnit) (TimeUnit) -2.50e-02 + 1.00e+00i 2.50e-02 1.00e+00 4.00e+01 -2.50e-02 - 1.00e+00i 2.50e-02 1.00e+00 4.00e+01 ```

### Interactive GUI

To analyze other standard circuit configurations such as low-pass and high-pass RLC networks, click on the link below to launch an interactive GUI. In this GUI, you can change the R,L,C parameters and see the effect on the time and frequency responses in real time.

Open the RLC Circuit GUI

```rlc_gui ``` 