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What is the Laplace transform of u(t-2)?

$ ( a )  \dfrac { 1 } { s } + 2 $

$ ( b ) \dfrac { 1 } { s } \: – \: 2 $

$ ( c )  \dfrac { e ^ { 2 s } } { s } $

$ ( d ) \dfrac {e ^ { – 2 s } } { s } $

This article aims to find the Laplace transform of a given function. The article uses the concept of how to find the Laplace transform of the step function. The reader should know the basics of Laplace transform.

In mathematics, Laplace transform, named after its discoverer Pierre-Simon Laplace, is an integral transformation that converts function of a real variable (usually $ t $, in the time domain) to a part of a complex variable $ s $ (in the complex frequency domain, also known as $ s $-domain or s-plane).

The transformation has many applications in science and engineering because it is a tool for solving differential equations. In particular, it converts ordinary differential equations to algebraic equations and convolution to multiplication.

For any given function $ f $, the Laplace transform is given as

\[F ( s ) = \int  _ { 0 } ^ { \infty } f ( t ) e ^ { – s  t } dt\]

Expert Answer

We know that

\[ L ( u ( t ) )  =  \dfrac { 1 } { s } \]

By $ t $ shifting theorem

\[ L ( u ( t – 2 ) ) = e ^ { – 2 s } L ( u ( t ) )  = \dfrac  { e ^ { – 2 s } } { s }  \]

Option $ d $ is correct.

Numerical Result

The Laplace transform of $ u( t – 2 ) $ is $ \dfrac { e ^ { – 2 s } } { s } $.

Option $ d $ is correct.

Example

What is the Laplace transform of $ u ( t – 4 ) $?

$ ( a )  \dfrac { 1 } { s } + 4 $

$ ( b ) \dfrac { 1 } { s } \: – \: 4 $

$ ( c )  \dfrac { e ^ { 4 s } } { s } $

$ ( d ) \dfrac {e ^ { – 4 s } } { s } $

Solution

\[ L ( u ( t ) )  =  \dfrac { 1 } { s } \]

By $ t $ shifting theorem

\[ L ( u ( t – 4 ) ) = e ^ { – 4 s } L ( u ( t ) )  = \dfrac  { e ^ { – 4 s } } { s }  \]

\[ L (  u (  t – 4 ) ) = \dfrac  { e ^ { – 4 s } } { s } \]

Option $ d $ is correct.

The Laplace transform of $ u( t – 4 ) $ is $ \dfrac  { e ^ { – 4 s } } { s }$.

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