1 The Laplace transform

If f ( t ) is a causal function then the Laplace transform of f ( t ) is written L { f ( t ) } and defined by:

L { f ( t ) } = ∫ 0 ∞ e − s t f ( t ) d t .

Clearly, once the integral is performed and the limits substituted the resulting expression will involve the s parameter alone since the dependence upon t is removed in the integration process. This resulting expression in s is denoted by F ( s ) ; its precise form is dependent upon the form taken by f ( t ) . We now refine Key Point 1 (page 4).

Key Point 3

The Laplace Transform of a Causal Function

L { f ( t ) u ( t ) } ≡ ∫ 0 ∞ e − s t f ( t ) u ( t ) d t ≡ F ( s )

To begin, we determine the Laplace transform of some simple causal functions. For example, if we consider the ramp function f ( t ) = t . u ( t ) with graph

Figure 11

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we find:

L { t u ( t ) } = ∫ 0 ∞ e − s t t u ( t ) d t = ∫ 0 ∞ e − s t t d t  since in the range of the integral u ( t ) = 1 = t e − s t ( − s ) 0 ∞ − ∫ 0 ∞ e − s t ( − s ) d t  using integration by parts = t e − s t ( − s ) 0 ∞ − e − s t ( − s ) 2 0 ∞

Now we have the difficulty of substituting in the limits of integration. The only problem arises with the upper limit ( t = ∞ ). We shall always assume that the parameter s is so chosen that no contribution ever arises from the upper limit ( t = ∞ ). In this particular case we need only demand that s is real and positive. Using this ‘rule of thumb’:

L { t u ( t ) } = [ 0 − 0 ] − 0 − 1 ( − s ) 2 = 1 s 2

Thus, if f ( t ) = t u ( t ) then F ( s ) = 1 ∕ s 2 .

A similar, but more tedious, calculation yields the result that if f ( t ) = t n u ( t ) in which n is a positive integer then:

L { t n u ( t ) } = n ! s n + 1

[We remember n ! ≡ n ( n − 1 ) ( n − 2 ) … ( 3 ) ( 2 ) ( 1 ) . ]

Task!

Find the Laplace transform of the step function u ( t ) .

Begin by obtaining the Laplace integral:

You should obtain ∫ 0 ∞ e − s t d t since in the range of integration, t > 0 and so u ( t ) = 1 leading to

L { u ( t ) } = ∫ 0 ∞ e − s t u ( t ) d t = ∫ 0 ∞ e − s t d t

Now complete the integration:

You should have obtained:

L { u ( t ) } = ∫ 0 ∞ e − s t d t = e − s t ( − s ) 0 ∞ = 0 − 1 ( − s ) = 1 s

where, again, we have assumed the contribution from the upper limit is zero.

As a second example, we consider the decaying exponential f ( t ) = e − a t u ( t ) where a is a positive constant. This function has graph:

Figure 12

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In this case,

L { e − a t u ( t ) } = ∫ 0 ∞ e − s t e − a t d t = ∫ 0 ∞ e − ( s + a ) t d t = e − ( s + a ) t − ( s + a ) 0 ∞ = 1 s + a  (zero contribution from the upper limit)

Therefore, if f ( t ) = e − a t u ( t ) then F ( s ) = 1 s + a .

Following this approach we can develop a table of Laplace transforms which records, for each causal function f ( t ) listed, its corresponding transform function F ( s ) . Table 1 gives a limited table of transforms.


Table 1: Table of Laplace Transforms
Rule   Causal function Laplace transform
1 f ( t ) F ( s )
     
2 u ( t ) 1 s
     
3 t n u ( t ) n ! s n + 1
     
4 e − a t u ( t ) 1 s + a
     
5 sin a t . u ( t ) a s 2 + a 2
     
6 cos a t . u ( t ) s s 2 + a 2
     
7 e − a t sin b t . u ( t ) b ( s + a ) 2 + b 2
     
8 e − a t cos b t u ( t ) s + a ( s + a ) 2 + b 2

Note: For convenience, this table is repeated at the end of the Workbook.

1.1 The linearity property of the Laplace transformation

If f ( t ) and g ( t ) are causal functions and c 1 , c 2 are constants then

L { c 1 f ( t ) + c 2 g ( t ) } = ∫ 0 ∞ e − s t [ c 1 f ( t ) + c 2 g ( t ) ] d t = c 1 ∫ 0 ∞ e − s t f ( t ) d t + c 2 ∫ 0 ∞ e − s t g ( t ) d t = c 1 L { f ( t ) } + c 2 L { g ( t ) }
Key Point 4

Linearity Property of the Laplace Transform

L { c 1 f ( t ) + c 2 g ( t ) } = c 1 L { f ( t ) } + c 2 L { g ( t ) }

That is, the Laplace transform of a linear sum of causal functions is a linear sum of Laplace transforms. For example,

L { 2 cos t . u ( t ) − 3 t 2 u ( t ) } = 2 L { cos t . u ( t ) } − 3 L { t 2 u ( t ) } = 2 s s 2 + 1 − 3 2 s 3
Task!

Obtain the Laplace transform of the hyperbolic function sinh a t .

Begin by expressing sinh a t in terms of exponential functions:

sinh a t = 1 2 ( e a t − e − a t ) Now use the linearity property (Key Point 4) to obtain the Laplace transform of the causal function sinh a t . u ( t ) :

You should obtain a ∕ ( s 2 − a 2 ) since

L { sinh a t . u ( t ) } = L e a t − e − a t 2 . u ( t ) = 1 2 L { e a t . u ( t ) } − 1 2 L { e − a t . u ( t ) } = 1 2 1 s − a − 1 2 1 s + a  (Table 1, Rule 4) = 1 2 2 a ( s − a ) ( s + a ) = a s 2 − a 2
Task!

Obtain the Laplace transform of the hyperbolic function cosh a t .

You should obtain s s 2 − a 2 since

L { cosh a t . u ( t ) } = L e a t + e − a t 2 . u ( t ) = 1 2 L { e a t . u ( t ) } + 1 2 L { e − a t . u ( t ) } = 1 2 1 s − a + 1 2 1 s + a  (Table 1, Rule 4) = 1 2 2 s ( s − a ) ( s + a ) = s s 2 − a 2
Task!

Find the Laplace transform of the delayed step-function u ( t − a ) , a > 0 .

Write the delayed step-function here in terms of an integral:

You should obtain L { u ( t − a ) } = ∫ a ∞ e − s t d t (note the lower limit is a ) since:

L { u ( t − a ) } = ∫ 0 ∞ e − s t u ( t − a ) d t = ∫ 0 a e − s t u ( t − a ) d t + ∫ a ∞ e − s t u ( t − a ) d t

In the first integral 0 < t < a and so ( t − a ) < 0 , therefore u ( t − a ) = 0 .

In the second integral a < t < ∞ and so ( t − a ) > 0 , therefore u ( t − a ) = 1 . Hence

L { u ( t − a ) } = 0 + ∫ a ∞ e − s t d t .

Now complete the integration:

L { u ( t − a ) } = ∫ a ∞ e − s t d t = e − s t ( − s ) a ∞ = e − s a s

Exercise

Determine the Laplace transform of the following functions.

  1. e − 3 t u ( t )
  2. u ( t − 3 )
  3. e − t sin 3 t . u ( t )
  4. ( 5 cos 3 t − 6 t 3 ) . u ( t )
  1. 1 s + 3
  2. e − 3 s s
  3. 3 ( s + 1 ) 2 + 9
  4. 5 s s 2 + 9 − 36 s 4