# Laplace Transform Properties in Signal and Systems

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The Laplace transform fulfills a number of properties that are quite valuable in various applications. In particular, by using these properties, it is possible to derive many new transform pairs from a basic set of pairs. In this tutorial, we state most fundamental properties of the Laplace transform.

1. Linearity

If

$x(t)\leftrightarrow X(s)\text{ and v}(t)\leftrightarrow V(s)$

Then for any real or complex numbers a, b,

$ax(t)+bv(t)\leftrightarrow aX(s)+bV(s)$

This property says that the Laplace transform is a linear operation.

1. ### Right Shift in Time

If

$x(t)\leftrightarrow X(s)$

Then for any positive real number c,

$x(t-c)\leftrightarrow {{e}^{-cs}}X(s)$

The function x(t-c) is a c-second right shift of x(t). we can see that a c-second right shift in the time domain corresponds to multiplication by e-cs in the Laplace transform domain.

1. ### Time Scaling

If

$x(t)\leftrightarrow X(s)$

For any positive real number a,

$x(at)\leftrightarrow \frac{1}{a}X(\frac{s}{a})$

The function x(at) is a time-scaled version of the given function x(t). We can see that time scaling corresponds to scaling by the factor 1/a in the Laplace transform domain (plus multiplication of the transform by 1/a).

1. ### Multiplication by a power of t

If

$x(t)\leftrightarrow X(s)$

Then for any positive integer n,

${{t}^{n}}x(t)\leftrightarrow {{(-1)}^{n}}\frac{{{d}^{n}}X(s)}{d{{s}^{n}}}$

1. ### Multiplication by an exponential

If

$x(t)\leftrightarrow X(s)$

Then for any real or complex number a,

${{e}^{at}}x(t)\leftrightarrow X(s-a)$

By this property, multiplication by an exponential function in the time domain corresponds to a shift of the s variable in the Laplace transform domain.

1. ### Multiplication by sinwt or coswt

If

$x(t)\leftrightarrow X(s)$

Then for any real number ω,

$x(t)sin\omega t\leftrightarrow \frac{j}{2}[X(s+j\omega )-X(s-j\omega )]$

$x(t)cos\omega t\leftrightarrow \frac{j}{2}[X(s+j\omega )+X(s-j\omega )]$

1. ### Convolution

Given two functions x(t) and v(t) with x(t) and v(t) equal to zero for t<0, we define the convolution by

$x(t)*v(t)\leftrightarrow \int\limits_{0}^{\infty }{x(\tau )}v(t-\tau )d\tau \text{ t}\ge \text{0}$

Now, letting X(s) denote the Laplace transform of x(t) and v(s) denote the Laplace transform of v(t). we have the transform pair

$x(t)*v(t)\leftrightarrow X(s)V(s)$

By this property, convolution in the time domain corresponds to a product in the Laplace transform domain.

1. ### Integration

Given a function x(t) with x(t)=0 for all t<0, we define the integral of x(t) to be the function

$\int\limits_{0}^{\infty }{x(t)dt}\leftrightarrow \frac{1}{s}X(s)$

By this property, the Laplace transform of the integral of x(t) is equal to X(s) divided by s.

1. ### Differentiation in the time domain

If

$x(t)\leftrightarrow X(s)$

Then

$\overset{.}{\mathop{x}}\,(t)\leftrightarrow sX(s)-x(0)$

1. ### Initial-value theorem

Given a signal x(t) with transform X(s), we have

$x(0)\leftrightarrow \underset{s\to \infty }{\mathop{\lim }}\,\text{ }sX(s)$

$\overset{.}{\mathop{x}}\,(0)\leftrightarrow \underset{s\to \infty }{\mathop{\lim }}\,\text{ }\!\![\!\!\text{ }{{s}^{2}}X(s)-sx(0)]$

The initial-value theorem is useful since it allows for computation of the initial values of a function x(t) and its derivatives directly from the Laplace transform X(s) of x(t).Hence, if we know X(s) but not x(t), it is possible to compute these initial values without having to compute the inverse Laplace transform of x(t).

1. ### Final-value theorem

Given the signal x(t) with transform X(s), suppose that x(t) has a limit t→∞. Then the final-value theorem states that

$\underset{t\to \infty }{\mathop{\lim }}\,x(t)\leftrightarrow \underset{s\to 0}{\mathop{\lim }}\,\text{ }sX(s)$

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