3. Laplace Tarnsform

38

3. LAPLACE TRANSFORM

A wide range of engineering systems are modeled mathematically by using differential

equations. In general, the differential equation of n th

order system is written:

f(t)y(t)adt

dy(t)a

td

y(t)da

dt

y(t)d011n

1n

1nn

n

(3.1)

Which is also known as a linear differential equation, if the coefficient a0 , a1 , …, an are

not function of y(t). Solution of the differential equation with discontinuous input or of higher

than second order is laborious by the classical method. To simplify or systematize the solution

of differential equation, the Laplace Transform (LT) method is used extensively.

3.1. Definition of the Laplace Transform

The direct Laplace transform F(s) of a function of time (t) is given by.

0

stF(s)dtf(t)ef(t)L (3.2)

where L(t) is a shorthand notation for the Laplace integral.

A complex variable s is refereed to as the Laplace operator and has two components

s=j, a real and imaginary components. Figure 3.1 illustrates the complex s-plane, in

which any arbitrary point s=s1 is defined by the s1=1j1 .

The reverse process of finding the time function (t) from the F(s) is called the inverse

Laplace transform

L-1F(s) = (t) (3.3)

3.2. Derivation of Laplace Transforms of Simple Functions

Example 3.1. Unit-step function (see Figure 3.2)

0tif10tif0

(t)U0

(3.4)

(3.4) t

Figure 3.2

1

U0(t)

j

s1

1

Figure 3.1

1

0

3. Laplace Tarnsform

39

t

s

1

s

e

dtedte)t(U)t(UL

0

st

st

0

st

000

(3.5)

Example 3.2 Exponential function (Figure 3.3)

0tife0tif0

(t)Uat

1 (3.6)

as

1

as

e

dteedte)t(U)t(UL

0

t)sa(

st

0

atst

011

(3.7)

Example 3.3. Ramp function (Figure 3.4)

0tifAt0tif0

(t)r

U (3.8)

0;s

1A

s

e0Adt

s

e

s

teAdte tA(t)U

s

evanddtduhavewe,edvandtufor

vduuvudvusingByparts.byintegratedisThisdt.e At(t)UL

2

0

2

st

0

st

0

stst

0r

tsst

b

a

ba

b

a

st

0r

Example 3.4. Consider the sinusoidal function f(t) = cost

Laplace transform for sinusoidal functions are determined, in terms of Euler's transform

2

eexcos;

j2

eexsin

jxjxjxjx

(3.10)

Ur(t)

t

Figure 3.4

Figure 3.3

1

U1(t)

(3.9)

3. Laplace Tarnsform

40

22

22

0

t)sj(

0

t)sj(

0

t)sj(

0

t)sj(st

0

s

stcosL

s

s

sj

1

sj

1

2

1

sj

e

sj

e

2

1

dtedte2

1dtetcostcosL

3.3 Laplace Transform Theorems

Applications of the Laplace transform in many instances are simplified by utilization of the

properties of the transform. These properties are presented by the following theorems:

Theorem 1. Multiplication by a Constant

Let k be a constant and F(s) be the Laplace transform of f(t). The

L[kf(t)] = kF(s)

Theorem 2. Sum and Difference

Let F1(s) and F2(s) be the Laplace transform of f1(t) and f2(t), respectively. Then

L[f1(t) ± f2(t)] = F1(S) ± F2(S)

Theorem 3. Differentiation

Let F(s) be the Laplace transform of f(t), and let f(0) be the limit of f(t) as t approaches 0.

The Laplace transform of the time derivative of f(t) is

L[df(t)/dt] = sF(S) - )t(flim 0t = sF(s) - f(0)

In general, for higher-order derivation of f(t),

L

1n

1n

2

22n1n

0t

n

n

n

dt

)t(fd

dt

)t(fds)t(fslim)s(Fs

dt

)t(fd

= )0(f)o(fs)o(fs)s(Fs 1n)1(2n1nn

Where )o(f n denotes the nth

-order derivative of f t( ) with respect to t , evaluated at t o .

Theorem 4. Integration

The Laplace transform of the first integral of f t( ) with respect to time is the Laplace

transform of f t( ) divided by s; that is,

s

F(s)d)(fL

For nth-order integration,

n1n21

t

o

t

0

t

0s

F(s)ddd)dτf(L 1 2 n

(311)

(3.14)

(3.15)

(3.16)

(3.12)

(3.13)

3. Laplace Tarnsform

41

Theorem 5. Shift in Time

The Laplace transform of f t( ) delayed by time T is equal to the Laplace transform f t( )

multiplied by Tse ; that is,

)s(FeTtu)Tt(fL Tss

Where u t Ts

denotes the unit-step function that is shifted to the right by T.

Theorem 6. Final-Value Theorem

If the Laplace transform of f (t) is F(s), and if sF(s) is analytic on the imaginary axis and in

the right half of the s-plane, then

sF(s) lim f(t) lim0st

(3.18)

The final-value theorem is very useful for the analysis and design of control systems,

since it gives the final value of a time function by knowing the behavior of its La lace

transform at s = 0. The final-value theorem is not valid if sF(s) contains any pole whose real

part is zero or positive, which is equivalent to the analytic requirement of sF(s) in the right-

half plane as stated in the theorem. The following examples illustrate the care that must be

taken in applying this theorem.

Example 3.5. Consider the function F(s)=5/s(s2

+ s + 2)

Since sF(s) is analytic on the imaginary axis and in the right-half s-plane, the final-value

theorem may by applied. Using Eq. (3.17), we have

5/2 2s5/s2 lim sF(s) lim f(t) lim 0s0st

Example 3.6. Consider the function

F(s)=/s2+2

Which is the Laplace transform of f(t)= sint. Since the function sF(s) has two poles on the

imaginary axis of the s-plane, the final value theorem cannot be applied in this case.

Theorem 7. Initial value theorem

sF(s) lim f(t) lim s0t (3.19)

Theorem 8. Complex Shifting

The Laplace transform of f (t) multiplied by et

, where is a constant, is equal to the

Laplace transform F(s), with s replaced by s; that is

Let

f(t) = F(s ) (3.20)

Theorem 9. Real Convolution (Complex Multiplication)

Let F1(s) and F2(s) be the Laplace transforms of f1 (t) and f2 (t), respectively, and f1 (t) = 0,

f2 (t) = 0, for t < 0; then

t

0212121 dtffLtf*tfLsFsF (3.21)

Where the symbol "*" denotes convolution in the time domain.

Equation (2.21) shows that multiplication of two transformed functions in the complex s-

domain is equivalent to the convolution of two corresponding real functions f1 (t) and f2 (t) in

the t-domain. The inverse Laplace transform of the product of two functions in the s-domain is

not equal to the product of the two corresponding real functions in the t-domain; that is, in

general, L-1F1(s)F2(s) f1 (t) f2 (t)

(3.17)

3. Laplace Tarnsform

42

Laplace Transform Table

Time function f(t) Laplace transform F(s)

2

2t

22

221

2ζωt

2

22

22ζωt

2

22

2

22

22

2

αt

2

tα

2

2

αtαt

2

αt

βtαt

1n

αtn

αt

1n

n

2

s

ω)s(s

ω)t1(e1.17

s)ω2s(

ω1ζ;)

ζ1tan

tζ1(sineζ1

1(1.16

ω2s

ω1ζ;tζ1ωsine

ζ1

ω15.

)ωs(s

ωtωcos114.

ωs

sωtcos13.

ωs

ωtωsin12.

α)(s

sαt)e(111.

α)(ss

1)e1t(α

α

110.

α)s(s

1)αtee(1

α

19.

α)s(s

1)e(1

α

18.

β)α)(s(s

1β)(α;)e(e

αβ

17.

α)(s

n!et6.

α)(s

1e5.

s

n!t4.

s

1t3.

s

1nfunctiostepunit(t)u2.

1functiondeltaδ(t)1.

2

2

22

2

22

2

22

2

22

22

2

2

2

1n

1n

2

ω)s(s

ω

s)ωs2s(

ω

ωs2s

ω

)ωs(s

ω

ωs

s

ωs

ω

α)(s

s

α)(ss

1

α)s(s

1

α)s(s

1

β)α)(s(s

1

α)(s

n!

α)(s

1

s

n!

s

1

s

1

1

3. Laplace Tarnsform

43

3.4. Inverse Laplace Transform by Partial-Fraction Expansion

When the Laplace Transform solution of a differential function is a rational function

it can be written as

)()()( sPsQsG (3.22)

Where Q(s) and P(s) are polynomial of s. It is assumed that the order of P(s) is greater than

that of Q(s). The polynomial P(s) my be written

P(s) = sn+an-1sn-1+an-2sn-2+…+a1s+a0

Where a0, a1,. . . an-1 are real coefficients. The roots of P(s) = 0 are called the poles of

systems. The roots of Q(s) = 0 are the zero of system. The methods of partial-fraction

expansion will now be given for the cases of simple poles, multiple-order poles, and complex

conjugate poles of G(s).

1. G(s) has simple poles. If all the poles of G(s) are simple and real, Eq. ( 3.22)

can be written as

)ss)...(ss)....(ss)(ss)(ss(

)s(Q

)s(P

)s(QsG

nk321 (3.23)

where s1 s2 . . . sn. Applying the partial-fraction expansion, is written

nk

k

2

2

1

1

ss

An

ss

A

ss

A

ss

A)s(G

(3.24)

The coefficient Ak (k= 1,2,. . . n) is determined as

)).......()()((

)(

)(

)()(

321

1

knkkkss

kssssssss

sQ

sP

sQssA

k

(3.25)

tns-

n

ts-

2

ts-

1

n

1k

tsk eAeAeAeAg(t) 21k

(3.26)

Example 3.7 Consider the function

G ss

s s s( )

( )( )( )

5 3

1 2 3 (3.27)

which is written in the partial-fraction expanded form:

3

3

2

2

1

1

ss

A

ss

A

ss

A)s(G

(3.28)

The coefficients A1, A2, and A3 are determined as follows:

;3s

62s

71s1

G(s)

becomes (3.28) Eq. Thus

63)3)(2(1

33)5(3)G(s)(sA

72)2)(3(1

32)5(2)G(s)(sA

11)1)(3(2

31)5(1)G(s)(sA

3s3

2s2

1s1

s1 = -1; s2 = -2; s3 = -3 ; g(t)= -e-t +7e-2t -6e-3t

3. Laplace Tarnsform

44

(r terms of repeated poles)

2. G(s) has multiple - order poles. If r of the n poles of G(s) are identical , G(s) is

written

r

irn21 )ss)(ss)......(ss)(ss(

)s(Q

)s(P

)s(QsG

; i =1, 2, . . . , n-r) (3.29)

s =-si is a multiple pole order-r. G(s) can be expended as

2ri

)2r(i

1ri

)1r(i

ri

ir

rn

rn

3

3

2

2

1

1

)ss(

A

)ss(

A

)ss(

A

ss

A

ss

A

ss

A

ss

A)s(G

The (n-r) coefficients, A1,A2, . . . ,A(n-r) which correspond simple poles ,may be evaluated by

the method described by Eq.(3.25). The coefficients that correspond to the multiple order

poles are defined as follows:

sisiir )G(s)s(sA

iss

ri1)i(r )s(Gss

ds

dA

(3.31)

iss

r

2

2

2)i(r )s(Gissds

d

2!

1A

iss

r

kr

kr

k)-i(r )s(Gissds

d

k)!(r

1A

In this case, Inverse Laplace transform is determined as

tseA

tseAeAteA

eA2!

teA

2)!(r

teA

1)!-(r

t(t)g

2111i

iii

21ts

11ts

12

ts13

2ts

1)i(r

2rts

ir

1-r

(3.32)

Example 3.8 Find inverse Laplace transform of the function

)3s()2s(

1)s(G

3

By using Eq. 3.30)

(n-r terms of simple poles)

(3.30)

3. Laplace Tarnsform

45

)2s(

A

)2s(

A

)2s(

A

3s

A)s(G 11

2

12

3

131

The coefficients are

1)s(G)2s(ds

d

!2

1A;1)s(G)2s(

ds

dA

1)s(G)2s(A1)s(G)3s(A

2s

3

2

2

11

2s

3

12

2s

3

133s

1

From Eq.(3.31) the Inverse Laplace transform

g(t) = t2

11

t2

12

t22

13

t3

1 eAteAe2

tAeA ;

2t2tt22

3t etee2

teg(t)

3. G(s) has simple complex-conjugate poles

The partial-fraction expansion of Eq. (3.24) is also valid for simple complex-conjugate poles.

Since complex-conjugate poles are more difficult to handle and are of special interest in

control system studies, they deserve special treatment here.

Example 3.9

Consider the function that describes a second-order system

22

2n

nsn2ζsG(s)

(3.33)

Let us assume that the values of (damping ratio) and wn (natural frequency) are such that

the poles of G(s) are complex.

For < 1 the poles of system are determined as

2nn2,1 1js

Suppose

na and 2n 1

The complete partial- fraction expansion of Eq. (3.33) is

jas

K

jas

K)s(G 21 (3.34)

The coefficient in Eq. (3.34) are determined as

2jG(s))ja(sK

2n

jas1

2jG(s))ja(sK

2n

jas2

jas

1

jas

1

j2)s(G

2n (3.35)

3. Laplace Tarnsform

46

Taking the inverse Laplace transform on both sides of Eq. (3.35) gives

tζ1sineζ1

)e(ee2j

(t)g 2n

tζ

2

ntjtjat2n n

t 0 (3.36)

3.5. Application of Laplace Transform to the Solution of Linear

Differential Equations

Linear ordinary differential equations can be solved by the Laplace transform. The

procedure is outlined as follows:

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

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

3. Perform partial-fraction expansion

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

Let us illustrate the method by several examples.

Example 3.10. Consider the differential equation

)t(u5)t(y2dt

)t(dy3

dt

)t(yds2

2

(3.37)

Where u ts ( ) is the unit-step function. The initial condition are 1)0(y and

2dt/)t(dy)0(y0t

)1(

. To solve the differential equation, we first take the Laplace

transform on both sides of Eq. (3.37):

5/s2Y(s)3Y(0)3sY(s)(o)ysy(0)Y(s)s (1)2

Substituting the values of the initial conditions into Eq. (3.37) and solving for Y(s), we get

Y ss s

s s s

s s

s s s( )

( ) ( )( )

2

2

25

3 2

5

1 2 (3.38)

Equation (3.38) is expended by partial fraction expansion to give

Y ss s s

( )( )

5

2

5

1

3

2 1

Taking the inverse Laplace transform, we get the complete solution

Y(t) = 2.5 - 5e-t + 1.5e

-2t

3.5.1 Matlab files of Laplace transform

a) Direct Laplace Transform

L = LAPLACE(F) is the Laplace transform of the scalar sym F with default independent

variable t. The default return is a function of s. If F = F(s), then LAPLACE returns a

function of t: L = L(t).

L = LAPLACE(F,w,z) makes L a function of z instead of the default s (integration with

respect to w).

Examples:

syms a s t w x

laplace(t^5) returns 120/s^6

3. Laplace Tarnsform

47

laplace(exp(a*s)) returns 1/(t-a)

laplace(sin(w*x),t) returns w/(t^2+w^2)

laplace(cos(x*w),w,t) returns t/(t^2+x^2)

laplace(x^sym(3/2),t) returns 3/4*pi^(1/2)/t^(5/2)

laplace(diff(sym('F(t)'))) returns laplace(F(t),t,s)*s-F(0)

b) Inverse Laplace transform

F = ILAPLACE(L) is the inverse Laplace transform of the scalar sym L with default

independent variable s. The default return is a function of t. If L = L(t), then ILAPLACE

returns a function of x: F = F(x).

F = ILAPLACE(L,y) makes F a function of y instead of the default t:

F = ILAPLACE(L,y,x) makes F a function of x instead of the default t:

Examples:

syms s t w x y

ilaplace(1/(s-1)) returns exp(t)

ilaplace(1/(t^2+1)) returns sin(x)

ilaplace(t^(-sym(5/2)),x) returns 4/3/pi^(1/2)*x^(3/2)

ilaplace(y/(y^2 + w^2),y,x) returns cos(w*x)

ilaplace(sym('laplace(F(x),x,s)'),s,x) returns F(x)

3.6 Impulse Response and Transfer Function of Systems

An important and first step in the analysis and design control system is the

mathematical modeling. The classical way of modeling linear time-invariant system is to use

transfer function to represent input-output relation between the variables. One way to define

the transfer function is to use impulse response, which is defined in the following.

Consider linear time-invariant system (see Figure 3.5).

Impulse response g(t) is defined as the output when the input is unit-impulse function

(t). The response Y(t) of system is determined by the use of convolution integral

t

0d)t(g)(U)t(Y (3.39)

Using the convolution property of the Laplace transform, we have

)s(G)s(Ud)t(g)(ULt

0

=Y(s) (3.40)

Transfer function G(s) of the system is determined by

means of Laplace transform Lg(t)= G(s). In general

transfer function G(s) is related to the Laplace

transforms of the input U(s) and the output Y(s)

through the following relation: G(s) = Y(s)/U(s) with all the

initial conditions are set to zero.

Let us consider that the input-output relation of LTIS is described by the following nth

order differential equation with constant real coefficients:

g(t) Y(t) U(t)

Figure 3.5

3. Laplace Tarnsform

48

)t(ubdt

)t(dub

dt

)t(udb

dt

)t(udb

)t(yadt

)t(dya

dt

)t(dya

dt

)t(dy

011m

1m

1mm

m

m

01n

1n

1nn

n

To obtain the transfer function of a linear system we simplify take the Laplace transform on

both sides of the equation and assume zero initial condition. The result is:

(sn

+ an-1sn-1

+ . . .

a1s +a0) Y(s) = (bmsm-1

+ bm-1sm-1

+. . . .

+b1s+ b0)U(s) (3.42)

The transfer function G(s) between U(t) and Y(t) is given by

)s(P

)s(Q

asasas

bsbsbsb)s(G

01

1n

1n

n

01

1m

1m

m

m

(3.43)

The properties of the transfer function are summarized as follows:

1. The transfer function is defined only for a linear-invariant system.

2. The transfer function between an input and output variables of a system is defined as a

Laplace transform of the impulse response. Alternatively, the transfer function is the ratio of

the Laplace transform of the output to the Laplace transform of the input.

3. All initial conditions of the system are set to zero.

4. The transfer function is independent of the input of the system.

Characteristic equation. The characteristic equation of a LTIS is defined as the equation

obtained by setting the denominator polynomial of the transfer function to zero. Thus, from

Eq. (3.43), the characteristic equation of the system is:

P(s) = sn

+ a n -1sn-1

+ . . . + a1s + a0 = 0 (3.44)

The Laplace transform of the output is given by Y(s) = H(s) U(s). Here

Evaluation the inverse transform of the partial fraction expansion of Y(s) gives

Example 3.12 Evaluate the step response of a system whose transfer function is given

(3.41)

2s

1

s

1

)2s(s

)1s(2)s(Y,

s

1)s(U,

2s

)1s(2)s(H

t2e1)t(y

)4s)(5.0s(

)2s()s(G

The roots of the obtained from P(s) = 0 are called the poles of the system. The roots

obtained from Q(s) = 0 are called the zero of system.

Example 3.11 What is the step response of a system whose transfer function has a

zero at –1, a pole at –2, and a gain factor of 2 ?

3. Laplace Tarnsform

49

The pole-zero map of the output is obtained by adding the poles and zeros of the input to the

pole-zero map of the transfer function. The output pole-zero map therefore has poles at 0, -0,5

and –4 and a zero at –2 as shown below.

Partial fraction expansion for the output gives 4s

C

5.0s

B

s

A)s(Y

Where

Example 3.13 Find the transfer function and evaluate the step response of the system which

has a gain factor of 3 and the pole-zero configuration shown in Figure 3.7. The time response

is therefore

The transfer function has zero at -2 j and poles at -3 and at -1 j. The transfer function are

therefore

)j1s)(j1s)(3s(

)j2s)(j2s(3)s(G

Expanding Y(s) into partial fractions yields

Figure 3.6

j

Pole due to the input

* * *

Zero of P(s)

Poles of P(s) S-plane

-0.5 -2 -4

-3 -2

Im

m

Re

-1

-j

j

*

*

*

Figure 3.7

1)4(5.0

2A 857.0

)5.3(5.0

5.1B

143.0

)5.3(4

2C

t4t5.0t4t5.0 e143.0e857.01CeBeA)t(y

)j1s)(j1s)(3s(s

)j2s)(j2s(3)s(U)s(G)s(Y

3. Laplace Tarnsform

50

o1 13.8]71[tan

Where

Evaluating the inverse Laplace transform, we have

Example 3.14. Find the output of the system

shown in Figure 3.8. U(t) = sin4t ; g(t) = 0.5 U0(t)

From the Table of Laplace transform we have

)4s(s

2)s(G)s(U)s(Y;

s

15.0)t(U5.0L)s(G;

4s

4t4sinL)s(U

22022

Taking the Inverse Laplace transform from Y(s) we obtain

y(t) = t4cos18

1

3.7. Modeling of Electrical Systems Using Laplace Transform

3.7.1. First-Order Electrical Circuits

The mathematical model of an electrical circuit can be obtained by applying Ohm's,

one or both of Kirshhoff’s, and Faraday’s laws.

Basic laws governing electrical circuits are Kirshhoff's laws, which states:

1. The algebraic sum of the currents entering a node is equal to zero. In other word, the sum

of the current entering a node is equal of current leaving the same node.

2. The algebraic sum of the voltages around any loop in an electrical circuit is zero.

The voltage drop across a resistor is given by Ohm’s law, which states that the voltage VR

drop across a resistor R is equal to the product of the current I and its resistance R

y(t) U(t) g(t)

Figure 3.8

j1s

A

j1s

A

3s

A

s

A)s(Y 4321

2

5

)j1)(j1(3

)j2)(j2(3A1

)j7(20

3

)j2)(j2)(j1(

)j21)(1(3A3

)tcos(e4

23e

5

2

2

5]ee[e

4

23e

5

2

2

5y tt3)t(j)t(jtt3

)j7(20

3

)j2)(j1)(j2(

)1)(j21(3A4

;

R

I VR

3. Laplace Tarnsform

51

L

I VL

t

VL(t)

1

b)

L

R

E

I

Fig. 3.9

VL

V

a)

VR = RI

The voltage drop across an inductor is given by Faraday’s law, which states that the

voltage VL drop across L an inductor is equal to the product of the inductance L and the time

rate of increase of current

dt

dILVL

The voltage across a capacitor is is determined as follows

IdtC

1VC

Consider the different electrical circuits.

Example. 3.5. Find the transfer function serial R-L circuit

Assuming the circuit is not loaded, i.e. no current flows through the output terminal.

Applying Kirshhof’s voltage law to the system (see Figure 3.9 a), we obtain the following

equations:

)47.3()s(E)s(LsI)s(RI

obtainweconditionsinitialzeroassumingand,(3.45)equationof

transformaplaceLtheTakingcircuit.theofmodelalmathematicofgivequationseThes

)46.3(Edt

dILRI

)45.3(EVV LR

The output voltage across the inductor is VL= LdI/dt. The current through the inductor is

therefore I(s)=VL/Ls.

Substituting this value into the Eq. (3.46)

givesR

LsV s V s E s V s

R

LsE sL L L( ) ( ) ( ) ; ( )( ) ( ) 1

The transfer function of system of system is found to be

R

LsV s V s E s V s

R

LsE sL L L( ) ( ) ( ) ; ( )( ) ( ) 1

The transfer function of system of system is found to be

C

I Vc

3. Laplace Tarnsform

52

)s(E)1Ls

R)(s(V;)s(E)s(V)s(V

Ls

RLLL

The transfer function of system of system is found to be

G sV s

E s

Ls

R Ls

L( )

( )

( )

For unit-step input, we have response

R

Lτwhere;t/e(t)

LV(s)

LV1-L

;L/Rs

1

LsR

L

LsR

Ls

s

1(s)V

L

Unit step response of system shown in Fig.3.5 (b)

Example.3.6. Serial R-C circuit

Applying Kirshhof’s voltage law to the system, we obtain the following equations:

)s(EV)s(RI 0 ; IdtC

1V0 ; )s(I

sC

1)s(V0 ; I(s) = sV0(s)C

1RCs

1

)s(E

)s(V)s(Gand

)s(EsRC1)s(V;)s(E)s(sRCV)s(V

0

000

Impulse response of system

;/1s

11

1s

1

1RCs

1)s(G)s()s(V0

where = RC

Taking the Inverse Laplace transform we define impulse response g(t) (see Fig.3.10 b)

For step impulse we have

(3.48)

Figure 3.10

t

I

V0 C

R

VR E

g(t) 1/

a)

V0(t)

t

c) b)

(3.49) ;t/e

1)t(Vg(t) 0

3. Laplace Tarnsform

53

(3.15)

(3.16)

(3.51)

(3.52)

Taking the Inverse Laplace transform we define step response shown in Figure 3.10 (b)

3. 7. 2 Second-Order Electrical Circuits

Applying Kirchhof’s voltage law to the system shown in Figure 3.11, we obtain the

following equations:

dt)s(IC

1)t(V

)t(E)t(V)t(RIdt

)t(dIL

C

c

Equations (3.51) and (3.52) give a mathematical model of the system. Taking the Laplace

transforms of equations (3.51) and (3.52), assuming zero initial conditions, we obtain

1RCsLCs

1

)s(E

)s(V)s(G

)s(E)s(V)s(sRCV)s(LCVs);s(sCV)s(I

)s(V)s(Is

1

C

1;)s(E)s(I

s

1

C

1)s(RI)s(LsI

2

C

CCC

2

C

C

Putting

RC

LCand

LC2

10

we have G ss s

( )

0

2

2

0 0

22 (3.53)

(3.16)

The poles of this system are s1 2 0 0

2 1,

Where is called a damping ratio and 0 - natural frequency of system.

3.8 Time-Response Characteristic of Second Order Systems

The solution of the Eq.(3.53)

G ss s

( )

0

2

2

0 0

22 (3.54)

Where s1 2 0 0

2 1,

depends on whether is smaller than unity, greater than unity, or equal to unity

;t/e1)t(V0

Fig.3.11

L R

E C

VC

(3.50)

1ss

1

)1s(s

1)s(V;

s

1)s(E 0

3. Laplace Tarnsform

54

1. When 1, the response of system is said to be underdamped. For a unit-step input

U(s)=1/s, output Y(s) of system

)s2s(s)s(Y

2

00

2

2

0

The Inverse Laplace transform (see Table of Laplace Transform)

)1

tant1sin(1

e1)t(y

212

02

t0

(3.55)

2. When =1, the roots are real and equal s1 = s2 = -0, and system is said to be critically.

For a unit-step input

2

0

2

0

)s(s)s(Y

0ttω1e1)t(y0

0t

(3.56)

3. When 1, the response is called overdamped. For a unit-step input

0te1)t(yt0)12(

(3.57)

A family of curve corresponding to this equation is shown in Figure 3.12, where

abscissa is the dimensionless variable 0t

This curves show that the amount of overshoot depends on the damping ratio . For the

overdamped and critically damped case ≥1, there is no overshoot. For the underdamped

case, <1, the system oscillates around the final value. The oscillation decrease with time, and

the system response approaches the final value. The peak overshoot for the underdamped

system is first overshoot. The time tp at which the peak overshoot occurs, can be determined

by differentiating Y(t) from Eq.(3.55) with respect to time and setting this derivative to zero:

y(t)

0t

Figure 3.12

=0.1

=2

=1

2.0

1.5

1.0

0.5

0 5 10

=0

3. Laplace Tarnsform

55

0t1sin(1

e

dt

dy 20

2

t0

0

This derivative is zero at 0

21 t = 0, , 2, . . . The peak overshoot occurs at the

first value after zero, provided there are zero initial conditions; therefore

t p

0

21

Inserting this value of time in Eq. (3.20) gives the peak overshoot as

Mp Cp

11 2

exp( )

The transient-response characteristic for the second order system shows a number of

significant parameters of system.

The overdamped system is slow acting and does not oscillate about the final position.

For some applications the absence of oscillation may be necessary. For example, an elevator

cannot be allowed to oscillate at each stop. But for systems where a fast response is necessary,

the slow response can not be tolerated.

The underdamped system reaches the final value faster than the overdamped system,

but the response oscillates about this final value. The amount the permissible overshoot

determines the desirable value of the damping ratio.

The performance of a system may be evaluated in term of the following

quantities, as shown, in Figure 3.13.

1. Peak overshoot Cp is the magnitude of the largest overshoot and often occurs at the first

overshoot. This may be expressed in percent of the final value

2. Time of maximum overshoot tp is the time required reaching the maximum overshoot.

3. Time of first zero t0 is the time required reaching a final value the first time. It is often

referred to as duplicating time

4. Setting time ts is the time required for the output response first to reach and thereafter

remain within a prescribed of the final value. Common values used for setting time are 2

and 5%. The setting time for a 2% error criterion is approximately 4 time constants; for a

5% error criterion, it is 3 time constants

t0 tp ts

t

1

Cp Allowable

tolerance -(2 or 5)%

Figure 3.13

y(t)