Signals and Systems – Solving Differential Equations with Laplace Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

Laplace Transform

The Laplace transform is a mathematical tool which is used to convert the differential equation in time domain into the algebraic equations in the frequency domain or sdomain.

Mathematically, if $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is a time domain function, then its Laplace transform is defined as −

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\int_{-\infty }^{\infty }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{st}}\:\mathit{dt}}}$$

Solution of Differential Equations Using Laplace Transform

A linear time invariant (LTI) system is described by constant coefficient differential equations which are relating the input and output of the system. The response of the LTI system is obtained by solving these differential equations.

The Laplace transformation technique can be used for solving the differential equation describing the LTI system. Using the Laplace transform, the differential equations in time domain are converted into algebraic equations in s-domain. The solution is obtained in s-domain by manipulating the algebraic equations. The result is then converted back into time domain by taking the inverse Laplace transform of the response.

Explanation

Consider an LTI system with input $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ and output $\mathit{y}\mathrm{\left(\mathit{t}\right)}$ and is described by a differential equation given as follows −

$$\mathrm{\mathit{a_{n}}\frac{\mathrm{\mathit{d}} ^{\mathit{n}}}{\mathrm{\mathit{d}}\mathit{t}^{\mathit{n}}}\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:\mathit{a_{n-\mathrm{1}}}\frac{\mathrm{\mathit{d}} ^{\mathit{n-\mathrm{1}}}}{\mathrm{\mathit{d}}t^{\mathit{n-\mathrm{1}}}}\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:...\:\mathrm{+}\mathit{a_{\mathrm{0}}}\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathit{b_{m}}\frac{\mathrm{\mathit{d}} ^{\mathit{m}}}{\mathrm{\mathit{d}}\mathit{t}^{\mathit{m}}}\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:\mathit{b_{m-\mathrm{1}}}\frac{\mathrm{\mathit{d}} ^{\mathit{m-\mathrm{1}}}}{\mathrm{\mathit{d}}t^{\mathit{m-\mathrm{1}}}}\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:...\:\mathrm{+}\mathit{b_{\mathrm{0}}}\:\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\:\:\:\:\:...(1)}$$

When the initial conditions of the system are zero, i.e.,

$$\mathrm{\mathit{y}\mathrm{\left(\mathrm{\mathrm{0}^{-}}\right)}\:\mathrm{=}\:\frac{\mathrm{\mathit{d}\mathit{y}\mathrm{\left(\mathrm{0}^{-}\right)}} }{\mathrm{\mathit{d}} \mathit{t}}\:\mathrm{=}\frac{\mathrm{\mathit{d}^{\mathrm{2}}\mathit{y}\mathrm{\left(\mathrm{0}^{-}\right)}} }{\mathrm{\mathit{d}} \mathit{t}^{\mathrm{2}}}\:...\:\mathrm{=}\:\frac{\mathit{d^{\mathrm{\left ( \mathit{n-\mathrm{1}}\right)}}\mathit{y}\mathrm{\left(\mathrm{0}^{-}\right)}}}{\mathit{dt^{\mathrm{\left(\mathit{n-\mathrm{1}}\right )}}}}\:\mathrm{=}\:\mathrm{0}}$$

And

$$\mathrm{\mathit{x}\mathrm{\left(\mathrm{\mathrm{0}^{-}}\right)}\:\mathrm{=}\:\frac{\mathrm{\mathit{d}\mathit{x}\mathrm{\left(\mathrm{0}^{-}\right)}} }{\mathrm{\mathit{d}} \mathit{t}}\:\mathrm{=}\frac{\mathrm{\mathit{d}^{\mathrm{2}}\mathit{x}\mathrm{\left(\mathrm{0}^{-}\right)}} }{\mathrm{\mathit{d}} \mathit{t}^{\mathrm{2}}}\:...\:\mathrm{=}\:\frac{\mathit{d^{\mathrm{\left ( \mathit{m-\mathrm{1}}\right)}}\mathit{x}\mathrm{\left(\mathrm{0}^{-}\right)}}}{\mathit{dt^{\mathrm{\left(\mathit{m-\mathrm{1}}\right )}}}}\:\mathrm{=}\:\mathrm{0}}$$

Now applying the Laplace transform on both sides of the differential equation given in eq.(1) and neglecting the initial conditions, we get,

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{a_{n}}\frac{\mathrm{\mathit{d}} ^{\mathit{n}}}{\mathrm{\mathit{d}}\mathit{t}^{\mathit{n}}}\mathit{y}\mathrm{\left(\mathit{t}\right)} \right]}\:\mathrm{+}\:\mathit{L}\mathrm{\left[\mathit{a_{n-\mathrm{1}}}\frac{\mathrm{\mathit{d}} ^{\mathit{n-\mathrm{1}}}}{\mathrm{\mathit{d}}t^{\mathit{n-\mathrm{1}}}}\mathit{y}\mathrm{\left(\mathit{t}\right)} \right]}\:\mathrm{+}\:...\mathit{L}\mathrm{\left[ \mathit{a_{\mathrm{0}}}\mathit{y}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{L}\mathrm{\left[ \mathit{b_{m}}\frac{\mathrm{\mathit{d}} ^{\mathit{m}}}{\mathrm{\mathit{d}}\mathit{t}^{\mathit{m}}}\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{+}\:\mathit{L}\mathrm{\left[\mathit{b_{m-\mathrm{1}}}\frac{\mathrm{\mathit{d}} ^{\mathit{m-\mathrm{1}}}}{\mathrm{\mathit{d}}t^{\mathit{m-\mathrm{1}}}}\mathit{x}\mathrm{\left(\mathit{t}\right)} \right]}\:\mathrm{+}\:...\mathit{L}\mathrm{\left[\mathit{b_{\mathrm{0}}}\:\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}}$$

$$\mathrm{\Rightarrow \mathit{a_{n}}\mathit{s^{n}}\mathit{Y}\mathrm{\left(\mathit{s}\right)}\:\mathrm{+}\:\mathit{a_{n-\mathrm{1}}}\mathit{s^{n-\mathrm{1}}}\mathit{Y}\mathrm{\left(\mathit{s}\right)}\:\mathit{+}\:...\mathrm{+}\mathit{a_{\mathrm{0}}\mathit{Y}\mathrm{\left(\mathit{s}\right)}}\:\mathrm{=}\:\mathit{b_{m}}\mathit{s^{m}}\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{+}\:\mathit{b_{m-\mathrm{1}}}\mathit{s^{m-\mathrm{1}}}\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{+}\:...\:\mathrm{+}\:\mathit{b_{\mathrm{0}}\mathit{X}\mathrm{\left(\mathit{s}\right)}}}$$

$$\mathrm{\Rightarrow \mathrm{\left(\mathit{a_{n}}\mathit{s^{n}}\:\mathrm{+}\:\mathit{a_{n-\mathrm{1}}}\mathit{s^{n-\mathrm{1}}}\:\mathrm{+}\:...\:\mathrm{+}\:\mathit{a}_{\mathrm{0}} \right)}\mathit{Y}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\mathrm{\left(\mathit{b_{m}}\mathit{s^{m}}\:\mathrm{+}\:\mathit{b_{m-\mathrm{1}}}\mathit{s^{m-\mathrm{1}}}\:\mathrm{+}\:...\:\mathrm{+}\:\mathit{b}_{\mathrm{0}} \right)}\mathit{X}\mathrm{\left(\mathit{s}\right)}}$$

$$\mathrm{\Rightarrow \frac{\mathit{\mathit{Y}\mathrm{\left(\mathit{s}\right)}}}{\mathit{X}\mathrm{\left(\mathit{s}\right)}}\:\mathrm{=}\:\frac{\mathrm{\left( \mathit{b_{m}}\mathit{s^{m}}\:\mathrm{+}\:\mathit{b_{m-\mathrm{1}}}\mathit{s^{m-\mathrm{1}}}\:\mathrm{+}\:...\:\mathrm{+}\:\mathit{b}_{\mathrm{0}}\right )}}{\mathrm{\left ( \mathit{a_{n}}\mathit{s^{n}}\:\mathrm{+}\:\mathit{a_{n-\mathrm{1}}}\mathit{s^{n-\mathrm{1}}}\:\mathrm{+}\:...\:\mathrm{+}\:\mathit{a}_{\mathrm{0}} \right )}}\:\:\:\:\:\:...(2)}$$

The equation (2) is known as the transfer function of the LTI system. Therefore, the transfer function of the LTI system is defined as the ratio of the Laplace transform of the output [Y(s)] to the Laplace transform of the input [X(s)] with the zero initial conditions.

The transfer function of an LTI system may also be defined as the Laplace transform of the impulse response of the system, i.e.,

$$\mathrm{\mathit{H}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\frac{\mathit{\mathit{Y}\mathrm{\left(\mathit{s}\right)}}}{\mathit{X}\mathrm{\left(\mathit{s}\right)}}\:\:\:\:\:\:...(3)}$$

The function H(s) depends only upon the coefficients of the differential equation describing the system and it does not depend upon the input signal or the initial energy stored by the system.

Since the response of the system in s-domain is given by,

$$\mathrm{\mathrm{Response\:,}\mathit{Y}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\mathit{H}\mathrm{\left(\mathit{s}\right)}\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\:\:\:\:\:...(4)}$$

Taking inverse Laplace of equation (4), we have,

$$\mathrm{\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathit{L}^{-\mathrm{1}}\mathrm{\left[\mathit{H}\mathrm{\left(\mathit{s}\right)}\mathit{X}\mathrm{\left(\mathit{s}\right)}\right]}\:\:\:\:\:\:...(5)}$$

Equation (5) is known as the zero state response of the system.

Now, if impulse function is the input to the system, i.e., $\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathit{\delta \mathrm{\left(\mathit{t}\right)}}$, then $\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\mathrm{1}$. Thus, eq. (5) is written as

$$\mathrm{\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathit{L}^{-\mathrm{1}}\mathrm{\left[\mathit{H}\mathrm{\left(\mathit{s}\right)}\right]}\:\mathrm{=}\:\mathit{h}\mathrm{\left(\mathit{t}\right)}\:\:\:\:\:\:...(6)}$$

Equation (6) is known as the impulse response of the system.

Numerical Example

Find the transfer function of the system which is described by the following differential equation −

$$\mathrm{\frac{\mathrm{\mathit{d}} ^{\mathrm{2}}}{\mathrm{\mathit{dt}}^{\mathrm{2}}}\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:\mathrm{5}\frac{\mathrm{\mathit{d}} }{\mathrm{\mathit{dt}}}\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:\mathrm{10}\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathrm{3}\frac{\mathrm{\mathit{d}} }{\mathrm{\mathit{dt}}}\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:\mathrm{8}\mathit{x}\mathrm{\left(\mathit{t}\right)}}$$

Solution

The given differential equation is,

$$\mathrm{\frac{\mathrm{\mathit{d}} ^{\mathrm{2}}}{\mathrm{\mathit{dt}}^{\mathrm{2}}}\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:\mathrm{5}\frac{\mathrm{\mathit{d}} }{\mathrm{\mathit{dt}}}\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:\mathrm{10}\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathrm{3}\frac{\mathrm{\mathit{d}} }{\mathrm{\mathit{dt}}}\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:\mathrm{8}\mathit{x}\mathrm{\left(\mathit{t}\right)}}$$

Neglecting the initial conditions and taking the inverse Laplace transform on both sides of the equation, we get,

$$\mathrm{\mathit{L}\mathrm{\left[\frac{\mathrm{\mathit{d}} ^{\mathrm{2}}}{\mathrm{\mathit{dt}}^{\mathrm{2}}}\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:\mathrm{5}\frac{\mathrm{\mathit{d}} }{\mathrm{\mathit{dt}}}\mathit{y}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:\mathrm{10}\mathit{y}\mathrm{\left(\mathit{t}\right)} \right]}\:\mathrm{=}\:\mathit{L}\mathrm{\left[\mathrm{3}\frac{\mathrm{\mathit{d}} }{\mathrm{\mathit{dt}}}\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{+}\:\mathrm{8}\mathit{x}\mathrm{\left(\mathit{t}\right)} \right ]}}$$

$$\mathrm{\Rightarrow \mathit{s}^{\mathrm{2}}\mathit{Y}\mathrm{\left(\mathit{s}\right)}\:\mathrm{+}\:\mathrm{5}\mathit{s}\mathit{Y}\mathrm{\left(\mathit{s}\right)}\:\mathrm{+}\:\mathrm{10}\mathit{Y}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\mathrm{3}\mathit{s}\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{+}\:\mathrm{8}\mathit{X}\mathrm{\left(\mathit{s}\right)}}$$

$$\mathrm{\Rightarrow \mathrm{\left( \mathit{s^{\mathrm{2}}\:\mathrm{+}\:\mathrm{5}\mathit{s}\:\mathrm{+}\:\mathrm{10}}\right)}\mathit{Y}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\mathrm{\left( \mathrm{3}\mathit{s}\:\mathrm{+}\:\mathrm{8}\right)}\mathit{X}\mathrm{\left(\mathit{s}\right)}}$$

Thus, the transfer function of the system is,

$$\mathrm{\mathit{H}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\frac{\mathit{\mathit{Y}\mathrm{\left(\mathit{s}\right)}}}{\mathit{X}\mathrm{\left(\mathit{s}\right)}}\:\mathrm{=}\:\mathrm{\left( \frac{\mathrm{3}\mathit{s}\:\mathrm{+}\:\mathrm{8}}{\mathit{s^{\mathrm{2}}\:\mathrm{+}\:\mathrm{5}\mathit{s}\:\mathrm{+}\:\mathrm{10}}} \right )}}$$

raja
Updated on 11-Jan-2022 11:15:59

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