# Step Response and Impulse Response of Series RL Circuit using Laplace Transform

An electric circuit consisting of a resistance (R) and an inductor (L), connected in series, is shown in Figure-1. Consider the switch (S) is closed at time $\mathrm{\mathit{ t=\mathrm{0}}}$.

## Step Response of Series RL Circuit

To obtain the step response of the series RL circuit, the input $\mathrm{\mathit{x\left ( t \right )}}$ applied to the circuit is given by,

$$\mathrm{\mathit{x\left ( t \right )\mathrm{=}Vu\left ( t \right )}}$$

Now by applying KVL in the loop, we obtain the following differential equation,

$$\mathrm{\mathit{Vu\left ( t \right )\mathrm{=}Ri\left ( t \right )\mathrm{+}L\frac{di\left ( t \right )}{dt}}}$$

Taking the Laplace transform on both sides, we get,

$$\mathrm{\mathit{\frac{V}{s}\mathrm{=}RI\left ( s \right )\mathrm{+}L\left [ sI\left ( s \right )-i\left ( \mathrm{0}^{-} \right ) \right ]}}$$

The current in an inductor cannot change abruptly. Therefore, the current in the circuit before $\mathrm{\mathit{ t\mathrm{=}\mathrm{0}}}$ is zero, i.e., initial current in the inductor is $\mathrm{\mathit{i\left ( \mathrm{0}^{-} \right )\mathrm{=}\mathrm{0}}}$. Thus,

$$\mathrm{\mathit{\frac{V}{s}\mathrm{=}RI\left ( s \right )\mathrm{+}sLI\left ( s \right )}}$$

$$\mathrm{\Rightarrow \mathit{\frac{V}{s}\mathrm{=}\left ( R\mathrm{+}sL \right )I\left ( s \right )}}$$

Hence, the circuit current is then given by,

$$\mathrm{\Rightarrow \mathit{I\left ( s \right )\mathrm{=}\frac{V}{s\left ( R\mathrm{+}sL \right )}}}$$

$$\mathrm{\Rightarrow \mathit{I\left ( s \right )\mathrm{=}\frac{V}{L}\left [ \frac{\mathrm{1}}{s\left ( s\mathrm{+}\frac{R}{L} \right )} \right ]\mathrm{=}\frac{V}{L}\cdot \frac{L}{R}\left [ \frac{\mathrm{1}}{s}-\frac{\mathrm{1}}{\left ( s\mathrm{+}\frac{R}{L} \right )} \right ]}}$$

$$\mathrm{\Rightarrow \mathit{I\left ( s \right )\mathrm{=}\frac{V}{R}\left [ \frac{\mathrm{1}}{s}-\frac{\mathrm{1}}{\left ( s\mathrm{+}\frac{R}{L} \right )} \right ]}}$$

Taking the inverse Laplace transform of the above equation, we get,

$$\mathrm{\mathit{i\left ( t \right )\mathrm{=}\frac{V}{R}\left [ \mathrm{1}-e^{-\left (R/L \right )t} \right ]}}$$

This is the step response of the series RL circuit.

## Impulse Response of Series RL Circuit

To obtain the impulse response of the series RL circuit (shown in Figure-1), the input $\mathrm{\mathit{x\left ( t \right )}}$ to the circuit is given by,

$$\mathrm{\mathit{x\left ( t \right )\mathrm{=}\delta \left ( t \right )}}$$

By applying KVL, the following differential equation of the circuit is obtained −

$$\mathrm{\mathit{\delta \left ( t \right )\mathrm{=}Ri\left ( t \right )\mathrm{+}L\frac{di\left ( t \right )}{dt}}}$$

Taking Laplace transform on both sides, we get,

$$\mathrm{\mathit{\mathrm{\mathbf{L}}\left [ \delta \left ( t \right ) \right ]\mathrm{=}\mathrm{\mathbf{L}}\left [ Ri\left ( t \right ) \right ]\mathrm{+}\mathrm{\mathbf{L}}\left [ L\frac{di\left ( t \right )}{dt} \right ]}}$$

(L is the Laplace transform operator and $\mathrm{\mathit{L}}$ is the inductance of the inductor)

$$\mathrm{\mathit{\Rightarrow \mathrm{1}\mathrm{=}RI\left ( s \right )\mathrm{+}L\left [ sI\left ( s \right )-i\left ( \mathrm{0}^{-} \right ) \right ]}}$$

Neglecting the initial conditions i.e. $\mathrm{\mathit{i\left ( \mathrm{0}^{-} \right )\mathrm{=}\mathrm{0}}}$. Then,

$$\mathrm{\mathit{\left ( R\mathrm{+}sL \right )I\left ( s \right )\mathrm{=}\mathrm{1}}}$$

Therefore, for the impulse input, the current in the RL series circuit is,

$$\mathrm{\mathit{I\left ( s \right )\mathrm{=}\frac{\mathrm{1}}{\left ( R\mathrm{+}sL \right )}\mathrm{=}\frac{\mathrm{1}}{L\left ( s\mathrm{+}\frac{R}{L} \right )}}}$$

Taking the inverse Laplace transform, we have,

$$\mathrm{\mathit{L^{-\mathrm{1}}\left [ I\left ( s \right ) \right ]\mathrm{=}L^{-\mathrm{1}}\left [ \frac{\mathrm{1}}{L\left ( s\mathrm{+}\frac{R}{L} \right )} \right ]}}$$

$$\mathrm{\mathit{\therefore i\left ( t \right )\mathrm{=}\frac{\mathrm{1}}{L}e^{-\left ( R/L \right )t}u\left ( t \right )}}$$

This is the impulse response of the series RL circuit.

## Numerical Example

Find the step response and impulse response of the series RL circuit shown in Figure-2.

### Solution

The differential equation describing the series RL circuit of Figure-2 is,

$$\mathrm{\mathit{x\left ( t \right )\mathrm{=}\mathrm{20}i\left ( t \right )\mathrm{+}\mathrm{5}\frac{di\left ( t \right )}{dt}}}$$

Taking inverse Laplace transform on both sides, we get,

$$\mathrm{\mathit{X\left ( s \right )\mathrm{=}\mathrm{20}I\left ( s \right )\mathrm{+}\mathrm{5}\left [ sI\left ( s \right )-i\left ( \mathrm{0}^{-} \right ) \right ]}}$$

Neglecting the initial conditions of the circuit, we have,

$$\mathrm{\mathit{X\left ( s \right )\mathrm{=}\mathrm{20}I\left ( s \right )\mathrm{\: +\: }\mathrm{5} sI\left ( s \right )\mathrm{=}\mathrm{5}I\left ( s \right )\left ( s\mathrm{\: +\: }\mathrm{4} \right )}}$$

$$\mathrm{\mathit{\therefore I\left ( s \right )\mathrm{=}\frac{X\left ( s \right )}{\mathrm{5}\left ( s\mathrm{\: +\: }\mathrm{4} \right )}}}$$

Refer the circuit of Figure-2, the output of the circuit is,

$$\mathrm{\mathit{Y\left ( s \right )\mathrm{=}\mathrm{20}I\left ( s \right )\mathrm{=}\mathrm{20}\left [ \frac{X\left ( s \right )}{\mathrm{5}\left ( s\mathrm{\: +\: }\mathrm{4} \right )} \right ]}}$$

$$\mathrm{\mathit{\Rightarrow Y\left ( s \right )\mathrm{=}\left [ \frac{\mathrm{4}}{\left ( s\mathrm{\: +\: }\mathrm{4} \right )} \right ]X\left ( s \right )}}$$

### For unit step response

$$\mathrm{Input,\mathit{x\left ( t \right )\mathrm{=}u\left ( t \right )}}$$

$$\mathrm{\mathit{\therefore X\left ( s \right )\mathrm{=}\frac{\mathrm{1}}{s}}}$$

Therefore, the output of the circuit is,

$$\mathrm{\mathit{Y\left ( s \right )\mathrm{=}\left [ \frac{\mathrm{4}}{\left ( s\mathrm{\: +\: }\mathrm{4} \right )} \right ]\left ( \frac{\mathrm{1}}{s} \right )\mathrm{=}\frac{\mathrm{4}}{s\left ( s\mathrm{\: +\: }\mathrm{4} \right )}}}$$

$$\mathrm{\mathit{\Rightarrow Y\left ( s \right )\mathrm{=}\frac{\mathrm{1}}{s}-\frac{\mathrm{1}}{\left ( s\mathrm{\: +\: }\mathrm{4} \right )}}}$$

Taking the inverse Laplace transform, the step response of the circuit is given by,

$$\mathrm{\mathit{L^{-\mathrm{1}}\left [ Y\left ( s \right ) \right ]\mathrm{=}L^{-\mathrm{1}}\left [ \frac{\mathrm{1}}{s} \right ]-L^{-\mathrm{1}}\left [ \frac{\mathrm{1}}{\left ( s\mathrm{\, +\,}\mathrm{4} \right )} \right ]}}$$

$$\mathrm{\mathit{\Rightarrow y\left ( t \right )\mathrm{=}u\left ( t \right )-e^{-\mathrm{4}t}u\left ( t \right )}}$$

$$\mathrm{\mathit{\therefore y\left ( t \right )\mathrm{=}u\left ( t \right )\left [\mathrm{1} -e^{-\mathrm{4}t} \right ]}}$$

### For Impulse Response

$$\mathrm{\mathit{x\left ( t \right )\mathrm{=}\delta \left ( t \right )}}$$

$$\mathrm{\mathit{\therefore X\left ( s \right )\mathrm{=}\mathrm{1}}}$$

$$\mathrm{Output,\mathit{ Y\left ( s \right )\mathrm{=}\left [ \frac{\mathrm{4}}{\left ( s\mathrm{\, +\,}\mathrm{4} \right )} \right ]\left ( \mathrm{1} \right )\mathrm{=}\frac{\mathrm{4}}{\left ( s\mathrm{\, +\,}\mathrm{4} \right )}}}$$

Taking inverse Laplace transform, the impulse response of the circuit is given by,

$$\mathrm{\mathit{ y\left ( t \right )\mathrm{=}\mathrm{4}e^{-\mathrm{4}t}u\left ( t \right )}}$$

Updated on: 05-Jan-2022

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