Laplace Transform (Notes)

Chapter 2

Laplace Transform Review

Dr. Franjo Cecelja1

Process and Information Systems Engineering Research CentreFaculty of Engineering and Physical Sciences

University of Surrey

2.1 The Need for Laplace Transform

The use the Laplace Transform is strongly motivated by real engineering problems and especiallyby problems in the area of Control. In control, we are usually interested how the output of asystems depends on its input. Wast majority of control systems are dynamic systems, whichmeans that the output not only depends on the value of the input, but also on the change ofvalue of input.

Example 2.1: Lets take the control systems in Figure 2.1 represented in a block diagramform and which has input u(t) and the output y(t) which depends on the derivative of theinput value.

Figure 2.1: Example of a simple dynamic control system

It is evident from Figure 2.1 that the output is calculated as derivative of the input as

y(t) =d

dtu(t) (2.1)

In Example 2.1, the output y(t) of the system entirely depends on the change in the value of theinput u(t), not actual value; if the input is constant, the output is 0. Such a system is called adynamic system.

1These lecture notes have been compiled from the literature stated in the Bibliography Section

15

16 CHAPTER 2. LAPLACE TRANSFORM REVIEW

Example 2.2: Lets take an example of a moving car. Assume that the engine force umoves the car with mass m in x direction at a speed v, as shown in Figure 2.2, and that therotation inertia of the wheels is negligible [2]. We are interested to see how the velocity of thecar v changes with the engine force u. The car is then modeled using free body diagram asshown in Figure 2.3, with the force u generated by the engine being input (or manipulated)variable and the car speed v is obviously output variable of the system.

Figure 2.2: A car in motion

Note that the friction bx acts in opposite direction from the force u as much as the inertiaof the car mx, that the velocity is the first derivation of the position v = dx/dt = x andthat the acceleration is the first derivative of the velocity, hence the second derivative of theposition a = dv/dt = v = d2x/dt2 = x. Here, b [Ns/m] is the viscous friction constantbetween the car and the surface of the road. As a consequence, the friction is entered asnegative force and using the second Newtons law the force balance on the car is:

u bx = mx (2.2)

or

x+b

mx =

u

m(2.3)

Figure 2.3: Free body diagram for cruise control system

For the reason of the cruise control, it is the car speed v (= x) that is of interest here,hence the equation (2.3) can be expressed in terms of velocity and written as:

v +b

mv =

u

m(2.4)

This is obviously the first order differential equation, the solution of which, as you havealready experienced, is not easy to calculate. Yet, for the input in the form of step functionstarting at t = 0

u(t) =

{1000 t 00 t < 0

2.1. THE NEED FOR LAPLACE TRANSFORM 17

and zero initial conditions (v(0) = 0 [m/s]) the solution is

v(t) = 1000

(1

b 1

bet

b

m

)(2.5)

As you recognise, in equation (2.5) the term 1bis the particular integral and the term

1bet

b

m is the homogeneous solution. Numerical solution2 for two cars weighting 1000 [kg]

and 2000 [kg], with the same friction of b = 26 [Ns/m] (this data is obtained experimentally

as shown in ref. [2], chapter 2) is shown in Figure 2.4. At the beginning, the car is stationary,

and then at the moment of observation (t = 0 [s]) the engine provides step force of 1000 [N ].

The results clearly show that the velocity changes in an exponential manner and also that

it takes longer for a heavier car to achieve stationary speed than for a lighter one, both for

the same force of the engine u = 1000 [N ].

Figure 2.4: Computer solution of differential equation

The things become even more complicated if the order of the differential equation is higheror if we want to combine more systems together. Yes, there is a significant number of computertools to help you with this, but then we loose inside view into the system, the part which isso important in designing an appropriate control system. To compromise between the ease offinding the solution and still heaving good understanding of the system, numerous mathematicalmethods have been developed; the Laplace transform is probably the most commonly used.

So, the Laplace transform is one of the mathematical tools used to solve linear ordinarydifferential equations, and has the following features:

1. The homogeneous equation and the particular integral of the solution of differential equa-tions are obtained in one operation;

2. The Laplace transform converts the differential equation into an algebraic equation in anew domain, the s-domain. It is then possible to manipulate the algebraic equation by

2MatLab example Ex1.


simple algebraic rules to obtain the solution in time or simply t-domain. The final solutionis then obtained by taking the inverse Laplace transform.

Before we go to the definition of Laplace transform, it is important to note that the author ofthese notes assumes that you are fully familiar with the complex numbers and complex variables.If not, you will have to make an effort to familiarise yourself with that mathematical area fully.

2.1.1 The Concept of the Laplace Transform

Normal form of differential equations is in so called input-output form as demonstrated byeq. (2.4). Lets break the system input u(t) into simpler inputs consisting of sum of impulsesand then determine the system response, the output y(t), to each of these impulses. By usingsuperposition principle, it is possible then to obtain the system response to the actual input.

For a typical system input as shown in Figure 2.5 a), we approximate this input by thesquare pulse train so that

u(t)

k=

u(t, tk,t) =

k=

u(tk)p(t tk,t)t (2.6)

where p(t tk,t) is the pulse of height 1/t and width t and centered at t = tk. This pulseis called unit pulse since its area is unity.

Lets define the response of initially unexcited linear and time invariant system to a unitpulse p(t ,t) as h(t, ,t) where is the time the centre of the unit pulse occurs and t isits width. Now, if we let the width of the pulse approach zero, than the unit impulse is definedas

(t ) = limt0

p(t ,t) (2.7)which is called the Dirac delta function , or simply function. The corresponding system

Figure 2.5: Decomposition of input signal u(t)

output for this unit impulse is then

h(t ) = limt0

h(t, ,t) (2.8)

2.2. DEFINITION OF THE LAPLACE TRANSFORM 19

The actual system input time is not just the unit pulse but the unit pulse multiplied by u(tk)t.The response due to the complete pulse train, by applying the superposition principle and fort 0, is then the integral expressed as:

y(t) =

h(t, )u()d (2.9)

The integral given by eq. (2.9) is called the convolution integral. Note that for t 0, theimpulse function (t) has infinitely short duration and infinitely high amplitude to maintain theunity area.

The interpretation of the integral 2.9 is that the response of the system to the stringof impulses is just the sum of responses to individual impulses. In eq. (2.9), the functionh(t, ) = h(t ) represent a stream of delayed impulses (delayed by ).

One of the uses of the convolution integral (2.9) is to find the response of the system asa sum of individual responses to individual impulse functions. If we take a real system whichcan not respond to the input before it occurs (there is no prediction built in the system), thenthe convolution integral (2.9) can be written in so called restricted form

y(t) =

0h(t, )u()d (2.10)

which assumes that the system excitation starts at t = 0 and which serves as an idea for theLaplace transform which has more practical use.

2.2 Definition of the Laplace Transform

Given the real function f(t) that satisfies the condition

0

f(t)et dt < (condition ofcontinuity) for some finite and real , the Laplace transform of a time dependent function f(t)is defined as

F (s) = L {f(t)} =

0f(t)estdt (2.11)

The variable s is referred to as Laplace operator, and it is a complex variable

s = + j (2.12)

where is the real component, is the imaginary component and j =1. In control terms,

(= 2pif) is the frequency and that is the reason why it is commonly said that the Laplacetransform converts the time domain function into the frequency domain. Here, f (= 1/T )denotes the actual frequency which is measured in Hertz [Hz] (1 [Hz] is a value of one periodper second) as shown in Figure 2.6. Note in eq. (2.11) that the integration goes between 0 and. That means that all information in f(t) that date before t = 0 is ignored. As explained forthe convolution integral, this limitation does not impose any limitation for real systems includingcontrol systems. Control systems are always considered from t = 0. Even if the excitation startsat t = 0, the function response can never starts before t = 0, so the equation (2.11) is alwaysfully valid. Also note that in equation (2.11) we use low case letters for time domain function(f(t)) and capital letters for functions in s-domain (F (s)), which is an established conventionin control (and mathematics).


Figure 2.6: Definition of frequency

Example 2.3: Find Laplace transform for the unit step function defined as:

f(t) =

{1 t 00 t < 0

and which is shown in Figure 2.7.

Figure 2.7: Unity step function

The Laplace transform is the obtained as:

F (s) = L {f(t)} =

0

estdt = 1sest

0

= 0 +1

s=

1

s

Example 2.4: Find the Laplace transform for a ramp function of the form

f(t) =

{bt t 00 t < 0

where b defines the slope and which is shown in Figure 2.8.The Laplace transform is obtained by applying the partial integration3 of the form

u dvdxdx = uv v dv

dxwhere both functions u and v are functions of t, we have

F (s) =

0

btestdt =

(bte

st

s be

st

s2

)

0

=b

s2

3If we have a function y = uv, then by differentiating both sides we get dydx

= u dvdx

+ v dudx. Now, by integrating

this equation we get

dy

dxdx =

u dv

dxdx +

v du

dxdx. The left hand side just returns to y, which writes uv =

u dvdx

dx +

v dudx

dx. By rearranging we get

u dvdx

dx = uv

v dudx

dx

2.2. DEFINITION OF THE LAPLACE TRANSFORM 21

Figure 2.8: Ramp function

Example 2.5: Find the Laplace transform of an exponential function of the form

f(t) =

{eat t 00 t < 0

The Laplace transform is

F (s) =

0

eatestdt =

0

e(s+a)tdt = 0 +1

s+ a=

1

s+ a

In literature you can find derived Laplace transform for numerous functions, especiallythose most commonly in use today. Some of those are summarised in Table 2.1, at least thosewhich will be used in the course of this module4.

f(t) F (s) = L {f(t)} f(t) F (s) = L {f(t)}1 1

s1 cos(at) a2

s(s2+a2)

t 1s2

at sin(at) a3s2(s2+a2)

tn n!sn+1

tcos(at) a2a2

(s2+a2)2

eat 1sa

tsin(at) 2ass(s2+a2)

teat 1(sa)2

eatcos(bt) s+a(s+a)2+b2

1(n1)!

tn1eat 1(sa)n

eatsin(bt) b(s+a)2+b2

eatebt

ab(a 6= b) 1

(sa)(sb)u(t) 1

saeatbebt

ab(a 6= b) s

(sa)(sb)u(t a) eas

s

cos(at) ss2+a2

(t) 1sin(at) a

s2+a2(t a) eas

cosh(at) ss2a2

sinh(at) as2a2

Table 2.1: Laplace transform of the most commonly used functions

4Data prepared by Dr. N. Rockliff


2.3 Properties of Laplace Transforms

2.3.1 Superposition (linearity)

The Laplace transform is linear: this is one of the most important properties. Given two time-domain functions f1(t) and f2(t) and two real coefficients and , the linearity property of theLaplace transform can be written as

L {f1(t) + f2(t)} =

0(f1(t) + f2(t)) e

stdt =

0f1(t)e

stdt+

0f2(t)e

stdt

(2.13)Hence

L {f1(t) + f2(t)} = F1(s) + F2(s) (2.14)The scaling property follows naturally from equation (2.14) as a special case, that is

L {f(t)} = F (s) (2.15)

2.3.2 Transforms of Derivatives

The Laplace transform of derivative of a time domain function f(t)

L {df(t)/dt} =

0

df(t)

dtestdt

can be shown in the following way: applying the integration by parts on function5 f(t), withu = f(t) and dv/dt = est we can show that from the definition of the Laplace transform (2.11):

F (s) =

0f(t)estdt = f(t) e

st

s

0

0

df(t)

dt

est

s dt (2.16)

Hence

F (s) =f(0)

s+

1

sL

{df(t)

dt

}(2.17)

which then gives

L

{df(t)

dt

}= sF (s) f(0) (2.18)

Here, f(0) is the initial condition of the function f(t), which means the value it takes at t = 0.Similarly, we obtain a general formula for transform of derivatives of second order as

L

{d2f(t)

dt2

}= s2F (s) sf(0) df(0)

dt(2.19)

Or, in general for derivative of any order the Laplace transform is

L

{dnf(t)

dtn

}= snF (s) sn1f(0) sn2df(0)

dt d

n1f(0)

dtn1(2.20)

5If we have a function y = uv, then by differentiating both sides we get dydx

= u dvdx

+ v dudx. Now, by integrating

this equation we get

dy

dxdx =

u dv

dxdx +

v du

dxdx. The left hand side just returns to y, which writes uv =

u dvdx

dx +

v dudx

dx. By rearranging we get

u dvdx

dx = uv

v dudx

dx

2.3. PROPERTIES OF LAPLACE TRANSFORMS 23

Example 2.6: Find the Laplace transform of the differential equation

d2y(t)

dt2+ 3

dy(t)

dt+ 2y(t) = 5u(t)

with u(t) as input function and with initial conditions y(0) = 1 and y(0) = dy(t)dt

t=0

= 2.

Using the derivative property of the Laplace transform, as per eq. (2.20), we get

s2Y (s) sy(0) y(0) + 3sY (s) 3y(0) + 2Y (s) = 5U(s)or

Y (s)(s2 + 3s+ 2

)+ s+ 1 = 5U(s)

2.3.3 Transform of the Integral of a Function

For the Laplace transform of an integral we have

F1(s) = L

{ t0f()d

}=

0

[ t0f()d

]estdt (2.21)

and employing the integral by parts where u = t0 f()d and dv = e

stdt we get

F1(s) = est

s

t0f()d

0

0e

st

sf(t)dt =

F (s)

s(2.22)

Example 2.7: Calculate the Laplace transform of the integral of a time domain functionf(t) =

tdt assuming the initial condition f(0) = 0.

From eq. (2.22) the Laplace transform is:

F (s) = L

{tdt

}=

L {t}s

=1

s s2 =1

s3

Alternatively and to prove validity of the Laplace transform of integration, we derive theintegral of the given function as:

f(t) =

tdt =

t2

2+ f(0) =

t2

2

then the Laplace transform from the Table 2.1 is:

F (s) = L {f(t)} = L{t2

2

}=

2

2s3=

1

s3

which proves the validity.

2.3.4 Delayed (in time) functions: the second shift theorem

Suppose that the function f(t) is delayed by > 0 units of time as shown in Figure 2.9. Then,the Laplace transform is F (s) =

0 f(t )estdt. If we define a = t , then da = dt since is a constant, and f(t) = 0 for t < 0. Thus

F (s) =

0f(a)es(a+)da =

0f(a)esaesda (2.23)


Figure 2.9: Delayed unity step function

Since es is independent of time, it can be taken out of the integral, so

F1(s) = es

0f(a)esada = esF (s) (2.24)

So, the delay in the time-domain by units of time of a time domain function f(t) is equal tomultiplication by exponent function es in s-domain.

Example 2.8: Determine the Laplace transform of a ramp function f(t) = t delayed by 2sec, as shown in Figure 2.11.

Figure 2.10: Delayed ramp function

The delayed function in time domain is f(t 2) = t 2. Because F (s) = L {f(t)} = 1s2,

the Laplace transform of the delayed function according to the eq. (2.24) is

L {f(t 2)} = e2s

s2

Example 2.9: Represent the function in Figure 2.11 in analytical form and find its Laplacetransform.

The analytical format isf(t) = u(t 1) u(t 3)

where u(t) is the unity step function and u(t a) is the unity step function u(t) shifted bya. Therefore, the Laplace transform is, using the linearity theorem (2.14) and delay theorem

2.3. PROPERTIES OF LAPLACE TRANSFORMS 25

Figure 2.11: A pulse function

(2.24):

F (s) = L {f(t)} = L {u(t 1)} L {u(t 3)} = 1s

(es e3s)

2.3.5 The first shift theorem: frequency shift

Multiplication (modulation) of f(t) by a time dependent exponential eat expression in thetime-domain corresponds to a shift in frequency by a in the s-domain:

F1(s) =

0f(t)eatestdt =

0f(t)e(a+s)tdt = F (s+ a) (2.25)

Example 2.10: Calculate the Laplace transform of the function f(t) = t3e4t.It is obvious that the function f(t) can be decomposed to a function f1(t) = t

3 shiftedin frequency by -4 (e4t). Hence, the Laplace transform per Table 2.1 is

F1(s) =6

s4

Now, from eq. (2.25) the Laplace transform for the function f(t) is

F (s) = F1(s 4) = 6(s 4)4

2.3.6 Multiplication by time: differentiation of the Laplace transform

Multiplication by time in the time-domain corresponds to differentiation in the frequency do-main:

d

dsF (s) =

d

ds

0f(t)estdt =

0testf(t)dt =

0t(f(t))estdt = L {tf(t)} (2.26)

Then

L {tf(t)} = ddsF (s) (2.27)

Example 2.11: Find the Laplace transform of a time-domain function f(t) = te3t.Taking that f1(t) = e

3t, the the Laplace transform of it is

L {f1(t)} = F1(s) = 1s+ 3


The derivative of F1(s) is

d

dsF1(s) =

d

ds

[1

s+ 3

]=

0 1(s+ 3)

2 =1

(s+ 3)2

So, the solution is

L {f(t)} = L{te3t} = ddsF1(s) =

1

(s+ 3)2

2.3.7 Division by time: integration of the Laplace transform

The Laplace transform of a time function f(t) divided by time t is equivalent to the integrationis s-domain:

L

{f(t)

t

}=

s

F (s)ds (2.28)

To prove this, lets start from integral of the Laplace transform

sF (s)ds. This yields

s

F (s)ds =

0

(

0estf(t)dt

)ds =

0

(

0estds

)f(t)dt =

0

1

testf(t)dt =

= L

{f(t)

t

}(2.29)

2.3.8 The Laplace transform of convolution

For the convolution of two functions of time domain f1(t) and f2(t)

f1(t) f2(t) = t0f1()f2(t )d (2.30)

it can be shown that by permutation of variables, the convolution is a symmetric operation, sothat f1(t) f2(t) = f2(t) f1(t), hence

t0f1()f2(t )d =

t0f2()f1(t )d (2.31)

Then, the Laplace transform is

L {f1(t) f2(t)} = L{ t

0f1()f2(t )d

}=

t=0

t=0

sstf1()f2(t )ddt (2.32)

Substituting = t and d = dt and using the valid extension of the upper bounds ofintegration to yields

L {f1(t) f2(t)} =

=

=0es(+)f1()f2()dd (2.33)

2.4. INVERSE LAPLACE TRANSFORM 27

Because both functions f1(t) and f2(t) have zero values for t < 0, it follows with respect to lowerlimit of integration that

L {f1(t) f2(t)} =

t=0esf1()d

0ssf2()d = F1(s)F2(s) (2.34)

In consequence, the Laplace transform of the convolution of two functions is the productof the Laplace transforms of individual functions:

L {f1(t) f2(t)} = F1(s)F2(s) (2.35)

2.3.9 The Laplace transform of periodic functions

To find the Laplace transform of basic periodic functions sin(t) and cos(t), we can use theirproperties of periodicity. Lets take for example the function f(t) = sin(t). For this functionwe have f(t) = cos(t), f(t) = 2sin(t) where initial conditions are f(0) = 0 and f(0) = .So, we can now write

2L {f(t)} = L{f(t)

}= s2L {f(t)} sf(0) f(0)

By rearranging this equation we get

L {f(t)} (s2 + 2) = which finally gives the solution

L {f(t)} = L {sin(t)} = s2 + 2

In the similar way we can determine the Laplace transform for all other periodic functions.The summary of the Laplace transform properties6 is given in Table 2.2.

2.4 Inverse Laplace Transform

The inverse Laplace transform refers to the process of finding the time-domain function f(t) forthe corresponding Laplace transform (s-domain function) F (s). The inverse Laplace transformof a function F (s), denoted as L1 {F (s)}, is given by:

L1 {F (s)} = f(t) = c+jcj

F (s)estds (2.36)

where c is a real constant that is greater than the real parts of all singularities (poles) of F (s).Now, we will not be solving the integral given by the eq. (2.36) as it is a very difficult job. Instead,several easier methods are available for finding the inverse Laplace transforms. The simplest ofall, of course, is to use tables (Table 2.1) of Laplace transforms to find the corresponding functionf(t). However, given s-domain function F (s) is not always in a table form, so several methodsare available to prepare F (s) for direct table reading. One of the most useful, especially withrational functions, and hence the most commonly used is the Partial-fraction Expansion theuse of which is explained in the follow-on section.

6The data is prepared by Dr. N. Rockliff


1. Definition Laplace transform of f(t) is F (s) = L {f(t)} = 0 f(t)estdt2. Linearity L {af(t)bg(t)} = aL {f(t)}+ bL {g(t)}

3. Derivative 1st: L {f (t)} = sL {f(t)} f(0)2nd: L {f (t)} = s2L {f(t)} sf(0) f (0)nth: L

{f (n)(t)

}= snL {f(t)} s(n1)f(0) s(n2)f (0)

f (n1)(0)

4. Integral L{ t

0 f()d}= 1

sL {f(t)}

5. 1st shift If L {f(t)} = F (s) then L{eatf(t)} = F (s a)alternatively L1 {F (s a)} = eatf(t)

6. 2nd shift If L {f(t)} = F (s) then L {u(t a)f(t a)} = easF (s)alternatively L1 {easF (s)} = u(t a)f(t a)where u(t a) denotes the shift step function

7. Differentiation If L {f(t)} = F (s) then L {tf(t)} = F (s)

8. Integration If L {f(t)} = F (s) then L{f(t)t

}=

sF (s)ds

9. Convolution For f and g, con.is (f g) (t) = t0 f()g(t )dIf L {f(t)} = F (s) and L {g(t)} = G(s) then the Laplacetransform is L

{ t0 f()g(t )d

}= F (s)G(s)

10. Periodic func. The Laplace transform of a periodic function f withperiod p is L {f(t)} = 1

1eps

p0 f(t)e

stdt

Table 2.2: Properties of the Laplace transform


2.4.1 Partial-Fraction Expansion

When the Laplace transform solution of a differential equation is a rational function in s (fromTable 2.1 it is apparent that most of functions of the Laplace transform are rational functions),it can be written in a general form as

G(s) =Q(s)

P (s)=

bmsm + bm1s

m1 + + b0sn + an1sn1 + + a0 (2.37)

where Q(s) and P (s) are polynomials of s. It is assumed that the order n of P (s) is greaterthan the order m of Q(s), n > m, the assumption which holds for most of practical engineeringproblems. The polynomial P (s) (= sn+ an1s

n1+ + a1s+ a0), where an1, . . . , a1, a0 arereal coefficients, has solutions obtained by setting

P (s) = sn + an1sn1 + + a1s+ a0 = 0 (2.38)

The method of partial-fraction is based on manipulating solutions of the polynomial P (s), whichare in control normally called the poles: solutions of the polynomial P (s) are values for whichP (s) = 0 and hence G(s) =. The ultimate outcome of applying the partial-fraction expansionis to convert the function G(s) into the form which is available from Table 2.1. We will considerhere the use of partial-fraction expansion for cases of simple poles, multiple-order poles andcomplex-conjugate poles of G(s).

Simple poles

If all poles of G(s) are simple and real, the equation (2.37) can be written as

G(s) =Q(s)

P (s)=

Q(s)

(s+ s1) (s+ s2) (s+ sn) (2.39)

where s1 6= s2 6= 6= sn. By applying the partial-fraction expansion, the eq. (2.39) can befurther expanded into the form

G(s) =Q(s)

(s+ s1) (s+ s2) (s+ sn) =Ks1s+ s1

+Ks2s+ s2

+ + Ksns+ sn

(2.40)

The coefficients Ksi (i = 1, 2, , n) are determined by multiplying both sides of the equation(2.39) by the factor (s+ si) and then setting s equal to si such as:

Ksi =

[(s+ si)

Q(s)

P (s)

]s=si

=Q(si)

(si + s1) (si + s2) (si + sn) (2.41)

For Ks1, for example, we have

Ks1 =

[(s+ s1)

Q(s)

P (s)

]s=s1

=Q(s1)

(s1 + s2) (s1 + s3) (s1 + sn) (2.42)


Example 2.12: Find the partial-fraction expansion of the function G(s) = 5s+3(s+1)(s+2)(s+3) .

Given function G(s) can be written in the partial form as

G(s) =K1s+ 1

+K2s+ 2

+K3s+ 3

(2.43)

the coefficients are:

K1 = [(s+ 1)G(s)]|s=1 =5 + 3

(1 + 2) (1 + 3) = 1

K2 = [(s+ 2)G(s)]|s=2 =10 + 3

(2 + 1) (2 + 3) = 7

K3 = [(s+ 3)G(s)]|s=3 =15 + 3

(3 + 1) (3 + 2) = 6 (2.44)

So, from equations (2.43) and (2.44) the solution is

G(s) =1s+ 1

+7

s+ 2 6s+ 3

(2.45)

The function (2.45) is now directly readable from the Laplace transform table (Table 2.1) as:

f(t) = et + 7e2t 6e3t (2.46)

To test the validity of derived partial-fraction expansion, we can simply sum fractions in eq.

(2.45) and the result should be the same as the original function.

Simple and Repeated Poles

Similarly to the simple poles procedure (eq. (2.41)), we can also use partial fraction expansionfor the system G(s) which has repeated poles [4]. The procedure will be explained on thefollowing example.

Example 2.11: Find the partial-fraction expansion of the s-domain function

G(s) =2

(s+ 1) (s+ 2)2 (2.47)

where evidently the poles of (s+ 2)2(the solutions of the denominator) are repeated since

the factor is raised to an integer power higher than 1. In this case the pole s = 2 is amultiple pole of multiplicity 2.

We can write the partial-fraction expansion as a sum of terms, where each factor of thedenominator forms the denominator of each terms. In addition, each multiple pole generatesadditional terms consisting of denominator factor of reduced multiplicity. So, for the rationalfunction given by eq. (2.47), we have

G(s) =2

(s+ 1) (s+ 2)2 =

K1(s+ 1)

+K2

(s+ 2)2 +

K3(s+ 2)

(2.48)

ThenK1 = 2, which can be found as described in the previous Section, whereas the coefficientK2 can be isolated by multiplying equation (2.48) by (s+ 2)

2, yielding

2

s+ 1= (s+ 2)

K1(s+ 1)

+K2 + (s+ 2)K3 (2.49)


Letting s = 2, we find K2 = 2. To find K3, we see that if we differentiate eq. (2.49) withrespect to s, we get

2(s+ 1)

2 =(s+ 2)

2

(s+ 1)2K1 +K3 (2.50)

K3 is isolated and we find it if we take s = 2. Hence, K3 = 2. Consequently, thepartial-fraction expansion of the rational function described by eq. (2.47) has the form

G(s) =2

s+ 1 2

(s+ 2)2

2

s+ 2(2.51)

Shortly, the above procedure of partial fraction expansion of function (2.47) could beshortened as follows:

G(s) =2

(s+ 1) (s+ 2)2 =

K1(s+ 1)

+K2

(s+ 2)2 +

K3(s+ 2)

which gives

K1 = [(s+ 1)G(s)]|s=1 =2

(1 + 2)2 = 2

K2 =[(s+ 2)

2G(s)

]s=2

=2

(2 + 1) = 2

K3 =

[d

ds(s+ 2)

2G(s)

]s=2

=2

(s+ 1)

s=2

=2

(2 + 1) = 2

Complex-Conjugate Poles

If the rational function (2.39) contains a pair of complex-conjugate poles s1 = + j ands2 = j (with a system which has a real-life implementation, complex poles always appearas a complex-conjugate pairs), then the corresponding coefficients of these poles can be foundby using the equation (2.41) as:

K+j = (s+ j)G(s)|s=+jKj = (s+ + j)G(s)|s=j (2.52)

Example 2.13: Write the function G(s) = 1s2+2s+5 in the fractional form.

The given function can be written in the following format:

G(s) =1

(s+ 1 + 2j) (s+ 1 2j) =K1

s+ 1 + 2j+

K2s+ 1 2j (2.53)

where coefficients K1,2, according to the equation (2.52), are:

K1 = (s+ 1 + 2j)G(s)|s=12j =1

1 2j + 1 2j = 1

4j

K2 = (s+ 1 2j)G(s)|s=1+2j =1

1 + 2j + 1 + 2j =1

4j(2.54)

Now, the original function G(s) in a fractional form is:

G(s) = 14j

s+ 1 + 2j+

14j

s+ 1 2j (2.55)


The imaginary form of equation (2.40) does not make any difference. Taking from theLaplace transform table (Table 2.1), the one for F (s) = b

sain s-domain with the time

domain response is f(t) = beat, the time-domain response for equation (2.55) is:

f(t) = 14je(1+2j)t +

1

4je(12j)t =

1

4j

(e(12j)t e(1+2j)t

)(2.56)

Using the identity e(a+jb)t = eat (cos(bt) + jsin(bt)), the equation (2.56) becomes

f(t) =et

4j(cos(2t) + jsin(2t) cos(2t) sin(2t)) =

=et

4j(cos(2t) + jsin(2t) cos(2t) + jsin(2t)) =

=et

2sin(2t) (2.57)

2.4.2 Inverse Laplace Transform Process

Once the partial-fraction form of the function is obtained, it is very likely that correspondingtime-domain function can be found directly from the Laplace transform table (Table 2.1), asshown by eq. (2.46) and (2.57).

2.4.3 Solving Linear, Time-Invariant Differential Equations

Linear and time-invariant differential equations can be solved by using the Laplace transformmethod with the aid of the theorems of Laplace transform, the partial-fraction expansion andthe Table 2.1 of Laplace transforms. The procedure is as follows:

1. Transform the differential equation to the s-domain using Laplace transform;

2. Manipulate the transformed algebraic equation and solve for output variable;

3. If the obtained relationship is not directly readable from the table, perform the partial-fraction expansion to the transformed algebraic equation;

4. Obtain inverse Laplace transform from the Laplace transform table.

Example 2.14: Lets take the differential equation

d2y(t)

dt2+ 3

dy(t)

dt+ 2y(t) = 5u(t) (2.58)

where u(t) is the unit step function, and initial conditions are y(0) = 1 and y(0) =dy(t)

dt

t=0

= 2. Now, the first step is to find the Laplace transform of equation (2.58).

Using the derivative property of the Laplace transform, as given in the Section 2.3.2, and

the Laplace transform of the unity step input function u(t) ={

1 t00 t

2.5. SKILL-ASSESSMENT EXERCISE 33

which, solving it for the output Y (s) gives

Y (s)(s2 + 3s+ 2

) sy(0) y(0) 3y(0) = 5s

(2.60)

Y (s)(s2 + 3s+ 2

)=

5

s s+ 2 3 = s

2 s+ 5s

(2.61)

Thus

Y (s) =s2 s+ 5

s (s2 + 3s+ 2)=

s2 s+ 5s (s+ 1) (s+ 2)

=K1s

+K2s+ 1

+K3s+ 2

(2.62)

Now, applying partial-fraction expansion we get K1 = 5/2, K2 = 5 and K3 = 3/2,

K1 = sY (s)|s=0 =5

(0 + 1)(0 + 2)=

5

2

K2 = (s+ 1)Y (s)|s=0 =1 + 1 + 51(1) = 5

K3 = (s+ 2)Y (s)|s=0 =4 + 2 + 52(1)) =

3

2

hence the equation (2.62) becomes

Y (s) =5

2s 5s+ 1

+3

2(s+ 2)(2.63)

Applying the inverse Laplace transform, the equation (2.63) takes the time-domain form as:

y(t) =5

2 5et + 3

2e2t (2.64)

The first term ( 52 ) in (2.64) is the steady-state solution or the particular integral; the last

two terms (5et + 32e2t) are transient, or homogeneous solution.

2.5 Skill-Assessment Exercise

2.5.1 Review Questions

1. Give the Laplace transform definition and explain its importance for control theory. Whatis the inverse Laplace transform?;

2. Explain how the fact that the Laplace transform operates in the range between t = 0 andt = affects the process of solving control problems;

3. Several properties of the Laplace transform are very useful for solving the control problem.Show that for a time-domain function f(t) delayed by seconds such that f1(t ) theLaplace transform is F1(s) = e

sF (s);

4. Show that multiplication of a time domain function f(t) by eat corresponds to frequencyshift in frequency domain as F (s+ a);

5. Show that multiplication by time t in time domain (tf(t)) corresponds to differentiationin frequency domain d

dsF (s).


2.5.2 Solving Problems

Task 2.1:

Derive the Laplace transform for time domain function f(t) = cos(t).

Hint: use the derivative property of Laplace transform of the form

L

{dnf(t)

dtn

}= snF (s) sn1f(0) sn2df(0)

dt d

n1f(0)

dtn1

Task 2.2:

Find inverse Laplace transform, that is function g(t), for the following s-domain function:

G(s) =s (s+ + 5) + 3

s3 + 5s2 + 4s

where and are real coefficients.

Task 2.3:

For the system in Figure 2.12 described by the differential equation

y(t) + 5y(t) + 4y(t) = u(t)

find the time domain solution (the system output y(t)) for the initial conditions y(0) = 0,y(0) = 0, and the input function u(t) = 2e2t. Sketch the response y(t) graphically.

Figure 2.12: A time-domain system

Task 2.4:

Given the dynamic model of a system as:

d2y(t)

dt2+ 2

dy(t)

dt+ 2y(t) = u(t)

where y(t) is the output of the system and u(t) is the input of the system.

Find the time-domain response for zero initial conditions and impulse input u(t) = (t) and

sketch it in the time diagram. Derive the transfer function of the system in the form G(s) = Y (s)U(s) .

2.5. SKILL-ASSESSMENT EXERCISE 35

Task 2.5:

For the system described by the transfer function

G(s) =Y (s)

U(s)=

1

s (s+ 2)

with Y (s) being the output and U(s) being the input, find the time domain response y(t) forthe step input function of the form

u(t) =

{1 t 00 t < 0

Solution: The time domain response is

y(t) =1

2 1

2e2t

Task 2.6:

Find the transfer function7 G(s) = Y (s)U(s) corresponding to the differential equation

d3y(t)

dt3+ 3

d2y(t)

dt2+ 7

dy(t)

dt+ 5y(t) =

d2u(t)

dt2+ 4

du(t)

dt+ 3u(t)

Solution: The transfer function is

G(s) =Y (s)

U(s)=

s2 + 4s+ 3

s3 + 3s2 + 7s+ 5

Task 2.7:

Find the response8 y(t) of the system described by the transfer function

G(s) =Y (s)

U(s)=

s

(s+ 4) (s+ 8)

for the ramp input of the form

u(t) =

{t t 00 t < 0

Solution: the time-domain response is

y(t) =1

32 1

16e4t +

1

32e8t

7This example is taken from the reference [4]8This example is taken from the reference [5]


Task 2.8:

Find the Laplace transform9 of the function

f(t) = te5t

Solution: The Laplace transform of t is 1s2

(directly from the Table ??). Applying the 1st shifttheorem (Table 2.2) we get the solution as:

F (s) =1

(s+ 5)2

Task 2.9:

Find the inverse Laplace transform for the function10

F (s) =10

s (s+ 2) (s+ 3)2

Solution: The inverse Laplace transform is

f(t) =5

9 5e2t + 10

3te3t +

40

9e3t

Task 2.10:

Find the inverse Laplace transform for the s-domain function

F (s) =3

s (s2 + 2s+ 5)


f(t) =3

5 3

5et

(cos(2t) +

1

2sin(2t)

)

Task 2.11:

Find the inverse Laplace transform for the s-domain function

G(s) =s+ 1

s3 + 3s2 + 2s


g(t) =1

2

(1 e2t)

9This example is taken from the reference [5]10This example is taken from reference [4]

Bibliography

[1] Fraklin GF, Powell JD, Emami-Naeini A; Feedback Control of Dynamic Systems - fourtedition, Prentince Hall 2002 - Chapter 3.

[2] Coughanowr DR; Process Systems Analysis and Controlm- second edition, McGraw-Hill In-ternational 1991 - Chapter 3

[3] Katsuhiko, O; System Dynamics - third edition, Prentince-Hall inc. 1998 - Chapter 2

[4] Nise, NS; Control Systems Engineering - third edition, John Wiley & Sons, Inc., 2007

[5] Website www.wiley.com/college/nise

37

Laplace Transform (Notes)

Documents

Transcript of Laplace Transform (Notes)