Skip to content Skip to navigation Skip to collection information

OpenStax-CNX

You are here: Home » Content » Signals, Systems, and Society » Properties of the Laplace Transform

Navigation

Recently Viewed

This feature requires Javascript to be enabled.

Tags

(What is a tag?)

These tags come from the endorsement, affiliation, and other lenses that include this content.
 

Properties of the Laplace Transform

Module by: Carlos E. Davila. E-mail the author

Summary: Derives basic properties of the Laplace transform.

Properties of the Laplace Transform

The properties associated with the Laplace transform are similar to those of the Fourier transform. First, let's set define some notation, we will use the notation L{}L{} to denote the Laplace transform operation. Therefore we can write X(s)=Lx(t)X(s)=Lx(t) and x(t)=L-1X(s)x(t)=L-1X(s) for the forward and inverse Laplace transforms, respectively. We can also use the transform pair notation used earlier:

x ( t ) X ( s ) x ( t ) X ( s )
(1)

With this notation defined, lets now look at some properties.

Linearity

Given that x1(t)X1(s)x1(t)X1(s) and x2(t)X2(s)x2(t)X2(s) then for any constants αα and ββ, we have

α x 1 ( t ) + β x 2 ( t ) α X 1 ( s ) + β X 2 ( s ) α x 1 ( t ) + β x 2 ( t ) α X 1 ( s ) + β X 2 ( s )
(2)

The linearity property follows easily using the definition of the Laplace transform.

Time Delay

The reason we call this the time delay property rather than the time shift property is that the time shift must be positive, i.e. if τ>0τ>0, then x(t-τ)x(t-τ) corresponds to a delay. If τ<0τ<0 then we would not be able to use the single-sided Laplace transform because we would have a lower integration limit of ττ, which is less than zero. To derive the property, lets evaluate the Laplace transform of the time-delayed signal

L x ( t - τ ) = 0 x ( t - τ ) e - s t d t L x ( t - τ ) = 0 x ( t - τ ) e - s t d t
(3)

Letting γ=t-τγ=t-τ leads to t=γ+τt=γ+τ and dt=dγdt=dγ. Substituting these quantities into Equation 3 gives

L x ( t - τ ) = - τ x ( γ ) e - s ( γ + τ ) d γ = e - s τ - τ x ( γ ) e - s γ d γ = e - s τ - τ 0 x ( γ ) e - s γ d γ + e - s τ 0 x ( γ ) e - s γ d γ L x ( t - τ ) = - τ x ( γ ) e - s ( γ + τ ) d γ = e - s τ - τ x ( γ ) e - s γ d γ = e - s τ - τ 0 x ( γ ) e - s γ d γ + e - s τ 0 x ( γ ) e - s γ d γ
(4)

where we note that the first integral in the last line is zero since x(t)=0,t<0x(t)=0,t<0. Therefore the time delay property is given by

L x ( t - τ ) = e - s τ X ( s ) L x ( t - τ ) = e - s τ X ( s )
(5)

s-Shift

This property is the Laplace transform corresponds to the frequency shift property of the Fourier transform. In fact, the derivation of the ss-shift property is virtually identical to that of the frequency shift property.

L e - a t x ( t ) = 0 e - a t x ( t ) e - s t d t = 0 x ( t ) e - ( a + s ) t d t = 0 x ( t ) e - ( a + σ + j Ω ) t d t = X ( s + a ) L e - a t x ( t ) = 0 e - a t x ( t ) e - s t d t = 0 x ( t ) e - ( a + s ) t d t = 0 x ( t ) e - ( a + σ + j Ω ) t d t = X ( s + a )
(6)

The ss-shift property also alters the region of convergence of the Laplace transform. If the region of convergence for X(s)X(s) is σ>σminσ>σmin, then the region of convergence for Le-atx(t)Le-atx(t) is σ>σmin-Re(a)σ>σmin-Re(a).

Multiplication by tt

Let's begin by taking the derivative of the Laplace transform:

d X ( s ) d s = d d s 0 x ( t ) e - s t d t = 0 x ( t ) d d s e - s t d t = - 0 t x ( t ) e - s t d t d X ( s ) d s = d d s 0 x ( t ) e - s t d t = 0 x ( t ) d d s e - s t d t = - 0 t x ( t ) e - s t d t
(7)

So we can write

L t x ( t ) = - d X ( s ) d s L t x ( t ) = - d X ( s ) d s
(8)

This idea can be extended to multiplication by tntn. Letting y(t)=tx(t)y(t)=tx(t), if follows that

t y ( t ) - d Y ( s ) d s d 2 X ( s ) d s 2 t y ( t ) - d Y ( s ) d s d 2 X ( s ) d s 2
(9)

Proceeding in this manner, we find that

t n x ( t ) ( - 1 ) n d n X ( s ) d s n t n x ( t ) ( - 1 ) n d n X ( s ) d s n
(10)

Time Scaling

The time scaling property for the Laplace transform is similar to that of the Fourier transform:

L x ( α t ) = 0 x ( α t ) e - s t d t = 1 α 0 x ( γ ) e - s α γ d γ = 1 α X s α L x ( α t ) = 0 x ( α t ) e - s t d t = 1 α 0 x ( γ ) e - s α γ d γ = 1 α X s α
(11)

where in the second equality, we made the substitution t=γαt=γα and dt=dγαdt=dγα.

Convolution

The derivation of the convolution property for the Laplace transform is virtually identical to that of the Fourier transform. We begin with

L - x ( τ ) h ( t - τ ) d τ = - x ( τ ) L h ( t - τ ) d τ L - x ( τ ) h ( t - τ ) d τ = - x ( τ ) L h ( t - τ ) d τ
(12)

Applying the time-delay property of the Laplace transform gives

- x ( τ ) L h ( t - τ ) d τ = H ( s ) - x ( τ ) e - s τ d τ = H ( s ) X ( s ) - x ( τ ) L h ( t - τ ) d τ = H ( s ) - x ( τ ) e - s τ d τ = H ( s ) X ( s )
(13)

If h(t)h(t) is the the impulse response of a linear time-invariant system, then we call H(s)H(s) the system function of the system. The frequency response results by setting s=jΩs=jΩ in H(s)H(s). The system function provides us with a very powerful means of determining the output of a linear time-invariant filter given the input signal. It will also enable us to determine a means of establishing the stability1 of a linear-time invariant filter, something which was not possible with the frequency response.

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 L d d t x ( t ) = 0 - x ' ( t ) e - s t d t
(14)

this can be integrated by parts:

u = e - s t v ' = x ' ( t ) u ' = - s e - s t v = x ( t ) u = e - s t v ' = x ' ( t ) u ' = - s e - s t v = x ( t )
(15)

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 ) 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 )
(16)

therefore we have,

d d t x ( t ) s X ( s ) - x ( 0 - ) d d t x ( t ) s X ( s ) - x ( 0 - )
(17)

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 - ) y ( t ) = d x ( t ) d t s X ( s ) - x ( 0 - )
(18)

it follows that

d y ( t ) d t s Y ( s ) - y ( 0 - ) d y ( t ) d t s Y ( s ) - y ( 0 - )
(19)

which leads to

d 2 d t 2 x ( t ) s 2 X ( s ) - s x ( 0 - ) - d x ( 0 - ) d t d 2 d t 2 x ( t ) s 2 X ( s ) - s x ( 0 - ) - d x ( 0 - ) d t
(20)

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 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
(21)

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 d m x ( 0 - ) d t m d m x ( t ) d t m t = 0 - , m = 1 , ... , n - 1
(22)

Integration

Let

g ( t ) = 0 - t x ( τ ) d τ g ( t ) = 0 - t x ( τ ) d τ
(23)

it follows that

d g ( t ) d t = x ( t ) d g ( t ) d t = x ( t )
(24)

and g0-=0g0-=0. Moreover, we have

X ( s ) = L d g ( t ) d t = s G ( s ) - g 0 - = s G ( s ) X ( s ) = L d g ( t ) d t = s G ( s ) - g 0 - = s G ( s )
(25)

therefore

G ( s ) = X ( s ) s G ( s ) = X ( s ) s
(26)

but since

G ( s ) = L 0 - t x ( τ ) d τ G ( s ) = L 0 - t x ( τ ) d τ
(27)

we have

0 - t x ( τ ) d τ X ( s ) s 0 - t x ( τ ) d τ X ( s ) s
(28)

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

t x ( τ ) d τ = - 0 - x ( τ ) d τ + 0 - t x ( τ ) d τ t x ( τ ) d τ = - 0 - x ( τ ) d τ + 0 - t x ( τ ) d τ
(29)

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

u ( t ) - 0 - x ( τ ) d τ u ( t ) - 0 - x ( τ ) d τ
(30)

it follows that

t x ( τ ) d τ - 0 - x ( τ ) d τ s + X ( s ) s t x ( τ ) d τ - 0 - x ( τ ) d τ s + X ( s ) s
(31)

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

The Initial Value Theorem

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

L d x ( t ) d t = s X ( s ) - x 0 - L d x ( t ) d t = s X ( s ) - x 0 -
(32)

Taking the limit ss

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 - 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 -
(33)

There are two cases, the first is when x(t)x(t) is continuous at t=0t=0. In this case it is clear that dx(t)dte-st0dx(t)dte-st0 as ss, so Equation 33 can be written as

0 = lim s s X ( s ) - x 0 - 0 = lim s s X ( s ) - x 0 -
(34)

Since x(t)x(t) is continuous at t=0t=0, x0-=x0+x0-=x0+, the Initial Value Theorem follows,

x 0 + = lim s s X ( s ) x 0 + = lim s s X ( s )
(35)

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

d x ( t ) d t t = 0 = x 0 + - x 0 - δ ( t ) d x ( t ) d t t = 0 = x 0 + - x 0 - δ ( t )
(36)

For example, if we integrate the right-hand side of Equation 36 with x0-=0x0-=0 and x0+=1x0+=1, we get the unit step function, u(t)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 - lim s 0 - d x ( t ) d t e - s t d t = lim s s X ( s ) - x 0 -
(37)

The left-hand side of Equation 37 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 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
(38)

From the sifting property of the unit impulse, the first term in Equation 38 is

x 0 + - x 0 - x 0 + - x 0 -
(39)

while the second term is zero since in the limit, the real part of ss goes to infinity. Substituting these results into the left-hand side of Equation 37 again leads to the initial value theorem, in Equation 35.

The Final Value Theorem

The Final Value Theorem allows us to determine

lim t x ( t ) lim t x ( t )
(40)

from X(s)X(s). Taking the limit as ss 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 - lim s 0 0 - d x ( t ) d t e - s t d t = lim s 0 s X ( s ) - x 0 -
(41)

The left-hand-side of Equation 41 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 - 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 -
(42)

Substituting this result back into Equation 41 leads to the Final Value Theorem

x ( ) = lim s 0 s X ( s ) x ( ) = lim s 0 s X ( s )
(43)

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

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

Footnotes

  1. We will discuss stability shortly

Collection Navigation

Content actions

Download:

Collection as:

PDF | EPUB (?)

What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

Downloading to a reading device

For detailed instructions on how to download this content's EPUB to your specific device, click the "(?)" link.

| More downloads ...

Module as:

PDF | More downloads ...

Add:

Collection to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

Module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks