The Laplace transform is utilized by engineers and mathematicians as a method of solving linear differential and integral equations. The transform itself has a very close relation to essential economic ideas, and is an ordinary tool to delight a variety of economic problems concerning stocks, flows, and time, comprising reduction patterns, lagged adjustments, adaptive outlook, and dynamic manufacture issues. The Laplace transform plays a function in the theory of probability distributions, where its derivatives can be understand as the instant of a probability distribution over the nonnegative real. In statistical workings the Laplace transform of the pointer function for the state space is the partition function, which reviews the properties of the highest entropy pattern of the system.
This is due the large part of the fact that broad classes of signals can be represented as linear combinations of periodic complex exponentials and that complex exponentials are eigen functions of LTI system. The continuous-time Fourier transform provides us with a representation for signals as linear combinations of complex exponentials of the form
â„®st with s=јω. Many of its consequences apply as well for arbitrary value of s and not only that value that are purely imaginary. Laplaces and z-transform can be applied to the analysis of many unstable systems and consequently play an important role in the investigation of the stability or instability of systems.
Definition of Laplace transform
Therefore the Laplace transform is as the integral:
∞
â„“ (f (t)) = F (S) = ∫ f (t) e-st dt
0
where s = σ + πω, and the inverse Laplace transform is given by:
σ + i∞
â„“ -1 (F(s)) = f (t) = 1 / 2πп ∫ F (s) e st ds
σ - i∞
Theory of Laplace Transform
The real-valued functions f (t) have a Fourier transform and that the basic
∞
g (w) = ∫ f (t) e -iwt dt
-∞
g(w) which represents as the complex function of the real variable w. If we multiply the integrand f (t) e -iwt by e-σt, then we generate a complex function G (σ + πω) of the complex variable σ + πω
∞ ∞
G (σ + πω) = ∫ f (t) e-σt e -iwt dt = ∫ f (t) e -( σ + iω)t dt
-∞ -∞
The function G (σ + πω) is called the two-sided Laplace transform of f (t) or bilateral Laplace transform of f (t) and it subsist when the Fourier transform of the function f (t) e-σt exists. From Fourier transform theory, an adequate condition for G (σ + πω) to subsist is that
∞
∫ | f (t) | e-σt dt < ∞
-∞
For a function f (t), this integral is finite for values of σ that lie in some interval a < σ < b
The bilateral LT has the inferior limit of integration and consequently requires an information of the past history of the function f (t) that is when t < 0. For most physical function, we are involved in the deeds of a structure only for 0 ≤ t. The initial conditions f (0), f '(0), f '' (0) …are a result of the past history of the structure and are frequently all that we recognize. For this rationale, it is helpful to identify the one-sided Laplace transform of f (t), which is normally referred to basically as the Laplace transform of f (t), which is also defined as an integral:
∞
â„“ (f (t)) = F (S) = ∫ f (t) e-st dt
0
where s = σ + πω. If the integral in Equation, for the Laplace transform exists for s0 = σ0 + πω0, then values of σ with σ > σ0 involve that e-σt < e - σ0 t and so that
∞ ∞
∫ | f (t) | e-σt dt < ∫ | f (t) | e-σot dt < ∞
-∞ -∞
from which it follows that F(s) exists for s = σ + πω. Therefore the Laplace transform â„“ (f (t)) is definite for all position s in the right half-plane Re (s) > σ0. Another way to view the association between the FT and the LT is to regard as the function U (t) given by
U (t) = {f (t), for 0 ≤ t; 0, for t < 0
Then the Fourier transform theory illustrate that
∞ ∞
U (t) = 1 / 2π ∫ [ ∫ U (t) e -iwt dt ] e iwt dw
- ∞ -∞
and, since the integrand U (t) is zero for t < 0, we can write this equation as
∞ ∞
f (t) = 1 / 2π ∫ [ ∫ f (t) e -iwt dt ] e iwt dw
- ∞ -∞
Now use the change of variable s = σ + πω and embrace σ > σ0 fixed. We have ds = 0 + πω and dω = ds / π. Then the new limits of integration are from s = σ - πω to s = σ + πω. The resulting equation is
σ + i∞ ∞
f (t) = 1 / 2πп ∫ [ ∫ f (t) e -st dt ] e st ds
σ - i∞ 0
Definition of the Inverse Laplace Transform.
∞
â„“ (f (t)) = F (S) = ∫ f (t) e-st dt
0
where s = σ + пτ. If the integral in Equation for the Laplace transform exists for S0 = σ0 + пτ, then values of σ with σ > σ0 imply that e-σt > e-σ0t and therefore
∞ ∞
∫ | f (t) | e-σt dt < ∫ | f (t) | e-σot dt < ∞
-∞ -∞
from which it follows that F (s) exists for s = σ + пτ.
Therefore the Laplace transform F (s) = â„“ (f (t)) is definite for all position S in the right half-plane Re (s) > σ0. For many practical principle, the function f (t) will have a Laplace transform F(s) is defined at all points in the compound plane except at a finite number of remarkable points s1, 2, ..sn where F(s) has poles. This is the state we will think. The inverse Laplace Transform is definite with a contour integral
σ0 + i∞
â„“ -1 (F(s)) = f (t) = 1 / 2πп ∫ F (s) e st ds = 1 / 2πп ∫ F (s) e st ds
σ0- i∞ C
the Bromwich contour C = {s = σ0 + i∞: - ∞ < τ <∞} is a perpendicular line in the complex plane where all singularities of F(s) lie in the left half-plane {s = σ + пτ : σ < σ0. This integral is called the Bromwich integral
The singularities s1, s2, s3 … sn of F(s) lie to the left of the Bromwich contour. We can utilize the Residue Calculus to estimate the Bromwich integral. We shall presume that the singularities of F(s) lie inside the easy closed contour consisting of the section of the Bromwich contour CR = { s = σ0 + пτ : -R ≤ τ ≤R}and a semicircle ЃR of radius R
The singularities S1, S2 ……sn of F(s) lie inside the contour CR + ЃR. The Cauchy Residue Theorem can be employed to estimate the contour integral along CR + ЃR
n
∫ F (s) e st ds = 2πп Σ Res [f, sk]
CR + ЃR k = 1
Then
n
∫ F (s) e st ds + ∫ F (s) e st ds = 2πп Σ Res [f, sk]
CR ЃR k = 1
Taking limits we have
∫ F (s) e st ds = lim ∫ F (s) e st ds
C R à CR
If adequate conditions are imposed on then it can be shown that
lim ∫ F (s) e st ds
R à ЃR
we will examine functions of the form F (s) = P (s) / Q (s) , where P (s) and Q (s) are polynomials of degree m and n, correspondingly, and n >m. This will assure that
lim ∫ F (s) e st ds = 0
R à ЃR
For this condition we can utilize the complex function f (s) = F(s) e st and write
f (t) = â„“ -1 (F(s))
= 1 / 2πп ∫ F (s) e st ds
C
= 1 / 2πп lim ∫ F (s) e st ds + 1 / 2πп lim ∫ F (s) e st ds
R à CR R à ЃR
= 1 / 2πп lim [∫ F (s) e st ds + ∫ F (s) e st ds]
R à CR ЃR
= 1 / 2πп lim [∫ F (s) e st ds
R à CR+ ЃR
n
= 2πп / 2πп Σ Res [f, sk]
K = 1
n
= Σ Res [f, sk]
K = 1
Theorem (Inverse Laplace Transform)
Let F (s) = P (s) / Q (s), where P(s) and Q (s) are polynomials of degree m and n respectively, and n>m. The inverse Laplace transform of F(s) can be calculate using residues, and is given by
f (t) = â„“ -1 (F(s)) = Σ Res [F (s) e st, sk]
where the sum is taken whole of the singularities s1,s2 …. Sn of F(s)
Properties of the Laplace transform
We consider the corresponding se of properties for the Laplace transform. The derivations of any these result are analogous to those of the corresponding properties for the Fourier transform. Consequently, we will not present the derivations in detail,
Linearity of the laplace transform
If £
x1 (t) X1(s) with a region convergence
that will be denoted as R1
and
£
x2(t) X2(s) with a region of convergence that will be
Denoted as R2
Then
ax1(t) + bx2(t) aX1(s) + bX2(s), with ROC containing R1∩R2
As indicated, the region of convergence of X(s) is at least the intersection of R1 and R2, which could be empty, in which case X(s) has no region of convergence (i.e.), x(t) has no laplace transform. The ROC for X(s) is the intersection of the ROCs for the two terms in the sum, If b < 0, there are no common points in R1 and R2, that is the intersection is empty, and thus, x(t) has no laplace transform. The ROC can also be larger than the intersection, (i.e. ) for x1(t) =x2(t) and a = -b , x(t) = 0, and thus(s) = 0, The ROC of X(s) is then the entire s-plane.
The ROC associated with a linear combination of terms can always be constructed by using the properties of the ROC developed. From the intersection of the ROCs for the individual terms, we can find a line or strip that is in the ROC of the linear combination. We then extend this to the right increasing and to the lift decreasing to the nearest poles.
Time shifting
If
£
X(t) X(s) with ROC = R,
Then
£
X(t-t0) â„® -st0 X(s), with ROC = R,
Shifting in the s- Domain
If
£
X(t) X(s), with ROC = R,
Then
£
â„® s0t x(t) X(s - s0), with ROC = R + Râ„®{s0},
that is, the ROC associated with X(s - s0)is that of X(s), shifted by Râ„®{s0}, thus for value s that is in R, the value s + Râ„®{s0} will be in R1. that if X(s) has a pole or zero at s = a, then X(s - s0) has a pole or zero at s - s0 = a- i.e s = a + s0.
An important when s0 = јω0 ie when a signal x(t) is used to modulated a periodic complex exponential â„® јω0t, in this becomes
£
â„® јω0tx (t) X(s - јω0), with ROC = R
The right hand side of e q., can be interpreted as a shift in the s-plane parallel to the јω-axis. That is, if the laplace transform of x(t) has a pole or zero at s = a, then the laplace transform of â„® јω0tx(t) has a pole or zero at s= a + јω0.
Time scaling
If
x(t) X(s), with ROC = R,
Then
£
x (at) 1/ |a| X( s/a), with ROC R1 = R/a,
That is, for value s in R , the value s/a will be in R1, For a positive value of a>1. note that for a>1. there is a compression in OC is expanded by a factor of 1/a. implies that if a is negative, the ROC undergoes a reversal plus a scaling. In particular, the ROC of 1/|a|X(s/a) for 0>a>-1 involve
A reversal about the јω- axis, together with a change in the size of the ROC by a factor of 1/|a|. Thus, time reversal of x(t) result in a reversal of the ROC. That is,
£
X(-t) X(-s), with ROC = R.
Conjugation
If £
x(t) X(s), with ROC = R,
Then
£
x*(t) X*(s*), with ROC = R.
Therefore,
X(s) = X*(s*) when x(t) is real.
Consequently, if x(t) is real and if X9s) has a pole or zero at s = s0 then X9s) also has a pole or zero at the complex conjugate point s = s*0 the transform X(s) for real signal x(t) has poles at s = 1 ±3 ј and zero at s = (-5± ј√71)/2.
Convolution property
If
£
x1(t) X1(s), with ROC = R1,
and
£
x2(t) X2(s), with ROC = R2,
then
£
x1(t) *x2(t) X1(s)X2(s), with ROC containing R1 ∩ R2,
the ROC of X1(s)X2(s) includes the intersection of the ROCs of X1(s) and X2(s) and may be larger if pole zero cancellation occurs in the product.
X1(s) = s + 1/s + 2, Râ„®{s}>-2,
And
X2(s) = s+2/s + 1, Râ„®{s}>-1,
Then X1(s) X2(s) = 1, and its ROC is the entire s-plane,
Convolution property in the context of the Fourier transform plays an important role in the analysis of linear time invariant systems.
Differentiation in the time domain
If
£
x(t) X(s), with ROC = R,
then
£
dx(t)/dt sX(s), with ROC containing R,
This property follows by differentiating both sides of the inverse laplace transform.
σ + ј∞
x(t) = 1/2πј∫ X(s)â„® st ds.
σ - ј∞
then
σ + ј∞
dx(t)/dt = 1/2πј ∫ sX(s) )â„® st ds.
σ - ј∞
dx(t)/dt is the inverse laplace transform of sX(s). The ROC of sX(s) includes the ROC of X(s) and many be larger if X(s) has a first order pole at s = 0 that is canceled b the multiplication by s. The derivation of x(t) is an impulse with an associated laplace transform that is unity and an ROC that is the entire s-plane.
Differentiation in the s-domain
Differentiating both side of the Laplace transform
+∞
X(s) = ∫ x(t) â„® -st dt,
-∞
We obtain
+∞
dX(s)/ds = ∫ (-t)x(t) )â„® -st dt
-∞
Consequently, if
£
x(t) X(s), with ROC = R,
Then
£
-tx (t) dX(s)/ds, with ROC = R,
Integration in the time domain
If £
x(t) X(s), with ROC = R,
Then
t £
∫ x (τ) dτ 1/s X(s), with ROC containing R ∩ {Râ„®{s}>0},
-∞
This property id the inverse of the differentiation property .It can be derived using the convolution property presented.
t
∫ x (τ)dτ = u(t) * x(t).
-∞
With a = 0,
£
u(t) 1/s, Râ„®{s}>0,
and thus, from the convolution property,
£
u(t) *x(t) 1/sX(s),
With an ROC that contains the intersection of the ROC of X(s) and the ROC of the laplace transform of u(t).
The initial and final value theorems
Under the specific constraints that x(t) = 0 for t<0 and x(t) contains no impulses or higher order singularities at the origin, one can directly calculate. From the laplace transform, the initial value x(0+) i.e. x(t) as t approaches zero from positive value of t- and the final value i.e. the limit as t ∞ of x(t) . The initial value theorem states that
x(0+) = lim sX(s),
s ∞
While the final value theorem says that
lim x(t) = lim sX(s)
t ∞ s 0
Laplace transforms pairs
The inverse Laplace transform can often be easily evaluated by decomposing X(s) into a linear combination of simpler terms, the inverse transform of each of which can be recognized.
Transform pair
signal
Transform
ROC
1
2
3
4
5
6
7
8
9
10
11
12
13
14
δ(t)
u (t)
-u (-t)
tn-1/(n-1)! u(t)
- tn-1/(n-1)!u(-t)
â„® -atu(t)
-â„® -atu(-t)
tn-1/(n-1)! â„® -atu(t)
-tn-1/(n-1)!â„® -atu(-t)
δ(t - Ñ‚)
[cos ω0t] u(t)
[sin ω0t] u(t)
[â„® -atcos ω0t] u(t)
[â„® -atsin ω0t] u(t)
1
1/s
1/s
1/sn
1/sn
1/(s + α)
1/(s + α)
1/(s + α)n
1/(s + α)n
℮ -sт
s/s2+ ω02
ωu/s2+ ω02
s + α / (s + α)2+ ω02
ωu/ (s + α)2+ ω02
Alls
Râ„®{s}>0
Râ„®{s}>0
Râ„®{s}>0
Râ„®{s}<0
Râ„®{s}>-α
Râ„®{s}<-α
Râ„®{s}>- α
Râ„®{s}<-α
All s
Râ„®{s}>0
Râ„®{s}>0
Râ„®{s}>- α
Râ„®{s}>-α
15
16
Un(t) = dnδ(t) / dtn
u-n(t) = u(t)….u(t)
n times
sn
1/sn
All s
Râ„®{s}>0
Application of Laplace transform
One of the important applications of the Laplace transform is in the analysis and characterization of LTI systems. Its role for this class of systems directly from the convolution property. The Laplace transform of the input and output of an LTI systems are related through multiplication by the Laplace transform of the impulse response of the systems.
Y(s) = H(s) X(s),
Where X(s),Y(s) and H(s) are the laplace transform of the input, output and impulse response of the systems. In the context of the laplace transform in eq., for the Fourier transform for s = јω, each of the laplace transform the eq., reduces to the respective Fourier transform, From our discussion on the response of LTI systems to complex exponentials, if the input to as LTI systems is x (t) = â„® st, then the output will be H(s)â„®st i.e. is an eigen function of the systems with eigen value equal to the laplace transform of the impulse response.
For s = јω, H(s) is the frequency response of the LTI systems. In the broader context of the Laplace transform, H(s) is commonly referred to as the system function or, alternatively, the transform function. Many properties of LTI systems can be closely associated with the characteristics of the system function in the s-plane.
Causality
For a causal LTI system, the impulse is zero for t< 0and thus is right sided. The ROC associated with the system function for a causal system is a right- half plane.
It should be stressed that the converse of this statement is not necessarily true. An ROC to the rightmost pole dose nit guarantee that a system is causal rather it guarantees only that the impulse response is right sided. If H(s) is rational, we can determine whether the system is causal simply by checking to see if its ROC is a right-half plane. for the system with a rational system function causality of the system is equivalent to the ROC being the right-half plane to the right of the rightmost pole.
Stability
The ROC of H(s) can also be related to the stability of a system. The stability of the LTI system is equivalent to its impulse response being absolutely integrable, in which case the Fourier transform of the impulse response converges. The Fourier transform of the signal equals the laplace transform evaluated along the јω-axis, An LTI system is stable if and only if the ROC of its system function H(s) includes the јω-aixs[ i.e. Râ„®{s}=0]
