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

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

## Step Response of Series RC Circuit Using Laplace Transform

To obtain the step response of the series RC circuit, the applied input is given by,

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

By applying KVL to the circuit, the following equation describing the series RC circuit is obtained −

$$\mathrm{\mathit{Vu\left ( t \right )\mathrm{=}Ri\left ( t \right )\mathrm{\: +\: }\frac{\mathrm{1}}{C}\int_{-\infty }^{t}i\left ( t \right )dt}}$$

This equation can be written as,

$$\mathrm{\mathit{Vu\left ( t \right )\mathrm{=}Ri\left ( t \right )\mathrm{\: +\: }\frac{\mathrm{1}}{C}\int_{-\infty }^{\mathrm{0}}i\left ( t \right )dt\mathrm{\: +\: }\frac{\mathrm{1}}{C}\int_{\mathrm{0} }^{t}i\left ( t \right )dt}}$$

Taking the Laplace transform on both sides, i.e.,

$$\mathrm{\mathit{L\left [ Vu\left ( t \right ) \right ]\mathrm{=}L\left [ Ri\left ( t \right ) \right ]\mathrm{\: +\: }L\left [ \frac{\mathrm{1}}{C}\int_{-\infty }^{\mathrm{0}}i\left ( t \right )dt \right ]\mathrm{\: +\: }L\left [ \frac{\mathrm{1}}{C}\int_{\mathrm{0} }^{t}i\left ( t \right )dt \right ]}}$$

$$\mathrm{\mathit{\Rightarrow \frac{V}{s}\mathrm{=}RI\left ( s \right )\mathrm{\: +\: }\frac{\mathrm{1}}{C}\left [ \frac{I\left ( s \right )}{s} \right ]\mathrm{\: +\: }\frac{\mathrm{1}}{C}\left [ \frac{q\left ( \mathrm{0^{\mathrm{\: +\: }}} \right )}{s} \right ] }}$$

Where, $\mathrm{\mathit{q\left ( \mathrm{0^{\mathrm{\: +\: }}} \right )}}$ is the charge on the capacitor at $\mathrm{\mathit{t\mathrm{=}\left ( \mathrm{0^{\mathrm{\: +\: }}} \right )}}$, i.e., it is the initial charge. If the initial conditions are neglected, then,

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

$$\mathrm{\Rightarrow \mathit{\frac{V}{s}\mathrm{=}I\left ( s \right )\left ( R\mathrm{\: +\: }\frac{\mathrm{1}}{sC} \right )}}$$

Therefore, the current in the circuit is given by,

$$\mathrm{\mathit{I\left ( s \right )\mathrm{=}\frac{V}{s\left ( R\mathrm{\: +\: }\frac{\mathrm{1}}{sC} \right )}\mathrm{=}\frac{V}{s}\left ( \frac{sC}{sRC\mathrm{\: +\: }\mathrm{1}} \right )\mathrm{=}\frac{VC}{RC\left ( s\mathrm{\: +\: }\frac{\mathrm{1}}{RC} \right )}}}$$

$$\mathrm{\therefore \mathit{I\left ( s \right )\mathrm{=}\frac{V}{R}\left [ \frac{\mathrm{1}}{\left ( s \mathrm{\: +\: } \frac{\mathrm{1}}{RC} \right )} \right ]}}$$

Taking the inverse Laplace transform on both sides, we have,

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

This is the step response of the series RC circuit.

## Impulse Response of Series RC Circuit Using Laplace Transform

To obtain the impulse response of the series RC circuit (shown in Figure-1), the applied input is given by,

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

Thus, the equation describing the system is given by,

$$\mathrm{\mathit{\delta \left ( t \right )\mathrm{=}Ri\left ( t \right )\mathrm{\: +\: }\frac{\mathrm{1}}{C}\int_{-\infty }^{t}i\left ( t \right )dt\mathrm{=}Ri\left ( t \right )\mathrm{\: +\: }\frac{\mathrm{1}}{C}\int_{-\infty }^{\mathrm{0}}i\left ( t \right )dt\mathrm{\: +\: }\frac{\mathrm{1}}{C}\int_{\mathrm{0} }^{t}i\left ( t \right )dt}}$$

Taking the Laplace transform on both sides, we get,

$$\mathrm{\mathit{\mathrm{1}\mathrm{=}RI\left ( s \right )\mathrm{\: +\: }\frac{\mathrm{1}}{C}\left [ \frac{I\left ( s \right )}{s} \right ]\mathrm{\: +\: }\frac{\mathrm{1}}{C}\left [ \frac{q\left ( \mathrm{0^{\mathrm{\: +\: }}} \right )}{s} \right ] }}$$

Where, $\mathrm{\mathit{q\left ( \mathrm{0^{\mathrm{\: +\: }}} \right )}}$ is the initial charge on the capacitor and by neglecting the initial conditions, we get,

$$\mathrm{\mathit{\mathrm{1}\mathrm{=}RI\left ( s \right )\mathrm{\: +\: }\frac{\mathrm{1}}{C}\left [ \frac{I\left ( s \right )}{s} \right ]}}$$

$$\mathrm{\Rightarrow \mathit{\left ( R\mathrm{\: +\: }\frac{\mathrm{1}}{sC} \right )I\left ( s \right )\mathrm{=}\mathrm{1}}}$$

Therefore, the current flowing in the circuit is

$$\mathrm{ \mathit{I\left ( s \right )\mathrm{=}\frac{\mathrm{1}}{\left ( R\mathrm{\: +\: }\frac{\mathrm{1}}{sC} \right )}\mathrm{=}\frac{\mathrm{1}}{R\left ( \mathrm{1}\mathrm{\: +\: }\frac{\mathrm{1}}{sRC} \right )}\mathrm{=}\frac{sRC}{R\left ( sRC\mathrm{\: +\: }\mathrm{1} \right )}}}$$

$$\mathrm{\Rightarrow \mathit{I\left ( s \right )\mathrm{=}\frac{sC}{\left ( sRC\mathrm{\: +\: }\mathrm{1} \right )}\mathrm{=}\frac{sC}{RC\left ( s\mathrm{\: +\: }\frac{\mathrm{1}}{RC} \right )}\mathrm{=}\frac{s}{R\left ( s\mathrm{\: +\: }\frac{\mathrm{1}}{RC} \right )}}}$$

Adding and subtracting $\mathrm{\mathit{\left ( \mathrm{1}/Rc \right )}}$ in the numerator on the RHS of the above equation, we get,

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

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

Now taking the inverse Laplace transform on both sides, we get,

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

This is the impulse response of the series RC circuit.

Updated on: 05-Jan-2022

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