Upload
others
View
14
Download
0
Embed Size (px)
Citation preview
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)