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Differentiation

The Laplace transform of the derivative of a signal will be used widely. Consider

L d d t x ( t ) = 0 - x ' ( t ) e - s t d t

this can be integrated by parts:

u = e - s t v ' = x ' ( t ) u ' = - s e - s t v = x ( t )

which gives

L d d t x ( t ) = u v 0 - - 0 - u ' v d t = e - s t x ( t ) 0 - + 0 - s x ( t ) e - s t d t = - x ( 0 - ) + s X ( s )

therefore we have,

d d t x ( t ) s X ( s ) - x ( 0 - )

Higher order derivatives

The previous derivation can be extended to higher order derivatives. Consider

y ( t ) = d x ( t ) d t s X ( s ) - x ( 0 - )

it follows that

d y ( t ) d t s Y ( s ) - y ( 0 - )

which leads to

d 2 d t 2 x ( t ) s 2 X ( s ) - s x ( 0 - ) - d x ( 0 - ) d t

This process can be iterated to get the Laplace transform of arbitrary higher order derivatives, giving

d n x ( t ) d t n s n X ( s ) - s n - 1 x ( 0 - ) - k = 2 n s n - k d k - 1 x ( 0 - ) d t k - 1

where it should be understood that

d m x ( 0 - ) d t m d m x ( t ) d t m t = 0 - , m = 1 , ... , n - 1

Integration

Let

g ( t ) = 0 - t x ( τ ) d τ

it follows that

d g ( t ) d t = x ( t )

and g 0 - = 0 . Moreover, we have

X ( s ) = L d g ( t ) d t = s G ( s ) - g 0 - = s G ( s )

therefore

G ( s ) = X ( s ) s

but since

G ( s ) = L 0 - t x ( τ ) d τ

we have

0 - t x ( τ ) d τ X ( s ) s

Now suppose x ( t ) has a non-zero integral over negative values of t . We have

t x ( τ ) d τ = - 0 - x ( τ ) d τ + 0 - t x ( τ ) d τ

The quantity - 0 - x ( τ ) d τ is a constant for positive values of t , and can be expressed as

u ( t ) - 0 - x ( τ ) d τ

it follows that

t x ( τ ) d τ - 0 - x ( τ ) d τ s + X ( s ) s

where we have used the fact that u ( t ) 1 s .

The initial value theorem

The initial value theorem makes it possible to determine x ( t ) at t = 0 + from X ( s ) . From the derivative property of the Laplace transform, we can write

L d x ( t ) d t = s X ( s ) - x 0 -

Taking the limit s

lim s 0 - d x ( t ) d t e - s t d t = lim s s X ( s ) - x 0 - 0 - lim s d x ( t ) d t e - s t d t = lim s s X ( s ) - x 0 -

There are two cases, the first is when x ( t ) is continuous at t = 0 . In this case it is clear that d x ( t ) d t e - s t 0 as s , so [link] can be written as

0 = lim s s X ( s ) - x 0 -

Since x ( t ) is continuous at t = 0 , x 0 - = x 0 + , the Initial Value Theorem follows,

x 0 + = lim s s X ( s )

The second case is when x ( t ) is discontinuous at t = 0 . In this case, we use the fact that

d x ( t ) d t t = 0 = x 0 + - x 0 - δ ( t )

For example, if we integrate the right-hand side of [link] with x 0 - = 0 and x 0 + = 1 , we get the unit step function, u ( t ) . Proceeding as before, we have

lim s 0 - d x ( t ) d t e - s t d t = lim s s X ( s ) - x 0 -

The left-hand side of [link] can be written as

lim s 0 + 0 - x 0 + - x 0 - δ ( t ) e - s t d t + lim s 0 + d x ( t ) d t e - s t d t

From the sifting property of the unit impulse, the first term in [link] is

x 0 + - x 0 -

while the second term is zero since in the limit, the real part of s goes to infinity. Substituting these results into the left-hand side of [link] again leads to the initial value theorem, in [link] .

The final value theorem

The Final Value Theorem allows us to determine

lim t x ( t )

from X ( s ) . Taking the limit as s approaches zero in the derivative property gives

lim s 0 0 - d x ( t ) d t e - s t d t = lim s 0 s X ( s ) - x 0 -

The left-hand-side of [link] can be written as

0 - lim s 0 d x ( t ) d t e - s t d t = 0 - d x ( t ) d t d t = x ( ) - x 0 -

Substituting this result back into [link] leads to the Final Value Theorem

x ( ) = lim s 0 s X ( s )

which is only valid as long as the limit x ( ) exists.

Laplace Transform properties.
Property y ( t ) Y ( s )
Linearity α x 1 ( t ) + β x 2 ( t ) α X 1 ( s ) + β X 2 ( s )
Time Delay x ( t - τ ) X ( s ) e - s τ
s-Shift x ( t ) e - a t X ( s + a ) )
Multiplication by t t x ( t ) - d X ( s ) d s
Multiplication by t n t n x ( t ) ( - 1 ) n d n X ( s ) d s n
Convolution x ( t ) * h ( t ) X ( s ) H ( s )
Differentiation d x ( t ) d t s X ( s ) - x 0 -
d 2 x ( t ) d t 2 s 2 X ( s ) - s x 0 - - d x 0 - d t
d n x ( t ) d t n s n X ( s ) - s n - 1 x 0 - - k = 2 n s n - k d k - 1 x 0 - d t k - 1
Integration t x ( τ ) d τ - 0 - x ( τ ) d τ s + X ( s ) s
Initial Value Theorem x 0 + = lim s s X ( s )
Final Value Theorem x ( ) = lim s 0 s X ( s )

Questions & Answers

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Source:  OpenStax, Signals, systems, and society. OpenStax CNX. Oct 07, 2012 Download for free at http://cnx.org/content/col10965/1.15
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