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Chapter 03-Modes03 System Modes and SimpleBehaviorDescribe complex linear system using simple elements.
Lesson of the Course “Fondamenti di Controlli Automatici” of 27 Feb 2012
Cesare Fantuzzi , Università degli Studi di Modena e Reggio Emilia
03-Modes.1
1 Content and Learning Objectives
The learning objectives of this lesson and exercises.
1. Understand the concept of the “Modes of the Systems”, obtained from the inverse Laplace transformof the system.
2. Describe the behavior of elementar components (mode templates).3. Understand the concept of system stability and link the system stability with poles.
03-Modes.2
Contents
1 Content and Learning Objectives 1
2 Mathematical description of system 12.1 Input to output mathematical description using Laplace Transform . . . . . . . . . . . . . 12.2 System characteristic equation, poles and zeros . . . . . . . . . . . . . . . . . . . . . . . 72.3 Transfer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3 Stability 16
4 Examples 18
5 Step response of 1st and 2nd order systems 215.1 Step response of a first order system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.2 Step response of a Second order system . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
6 Notes on the Laplace transform 27 03-Modes.3
2 Mathematical description of system
2.1 Input to output mathematical description using Laplace Transform
Input-output mathematical model
• A generic input to output mathematical model:
andny(t)
dtn +an−1dn−1y(t)
dtn−1 + . . .+a1dy(t)
dt+a0y(t) =
= bmdmu(t)
dtm +bm−1dm−1u(t)
dtn−1 + . . .+b1du(t)
dt+b0u(t)
• In a compact form:
n
∑j=0
a jd jy(t)
dt j =m
∑j=0
b jd ju(t)
dt j
03-Modes.4
1
The Laplace TransformThe ability to obtain linear approximations of physical systems allows the analyst to consider the use of
the Laplace transformation.The Laplace transform exists for linear differential equations for which the transformation integral
converges.Therefore, for f (t) to be transformable, it is sufficient that∫
∞
0−| f (t)|e−σ1tdt < ∞
for some real, positive σ1.The 0− indicates that the integral should include any discontinuity, such as a deltafunction at t = 0. If the magnitude of f (t) is | f (t) < Meαt for all positive t, the integral will converge forσ1 > α . The region of convergence is therefore given by ∞ > σ1 > α and σ1 is known as the abscissa ofabsolute convergence.
• Definition of a mathematical tool for analyzing differential equations to identify structural properties• Signals that are physically realizable always have a Laplace transform.• The Laplace transformation for a function of time, f (t) : t ∈ℜ 7→ℜ, is
F(s) =∫
∞
0f (t)e−stdt
• The inverse Laplace transform is written as
f (t) =1
2π j
∫σ+ j∞
σ− j∞F(s)e+stds
• being s = σ + jω a variable in the complex field.03-Modes.5
Laplace transform pairs
Impulse function f0(t) = δ (t)L ( fo(t)) = F0(s) = 1
Heaviside step function
f1(t) ={
1 if t ≥ 00 else t < 0
L ( f1(t)) = F1(s) = 1s
The delta function is a generalized function that can be defined as the limit of a class of delta sequences.The delta function is sometimes called Dirac’s delta function or the impulse symbol.
Formally, delta is a linear functional from a space (commonly taken as a Schwartz space S or thespace of all smooth functions of compact support D) of test functions f (t). The action of δ on f (t),commonly denoted δ ( f (t)) then gives the value at 0 of f (t) for any function f (t). In engineering contexts,the functional nature of the delta function is often suppressed.
Having defined the Heaviside step function as:
f1(t) =
0, if t < 012 , if t = 01, if t > 0
The delta function can be viewed as the derivative of f1(t):
d f1(t)dt
= δ (t)
and it can be defined as:
δ (t) ={
+∞, if t = 00, if t 6= 0
since:
2
δ (t) = limτ 7→0
1τ( f1(t)− f1(t− τ))
The delta function has the fundamental property that:∫ +∞
−∞
f (t) ·δ (t−a)dt = f (a)03-Modes.6
Laplace transform pairs
Ramp function.
f2(t) ={
t if t ≥ 00 else t < 0
L ( f2(t)) = F2(s) = 1s2
Quadratic function:
f3(t) ={
t2 if t ≥ 00 else t < 0
L ( f3(t)) = F3(s) = 1s3
03-Modes.7
Laplace transform pairs
Exponential function.
f4(t) ={
eat if t ≥ 00 else t < 0
F(s) = 1(s−a)
Cosine function.
f5(t) ={
cos(ωt) if t ≥ 00 else t < 0
L ( f5(t)) = F5(s) = s(s2+ω2)
03-Modes.8
Laplace transform pairs
Sine function.
f6(t) ={
sin(ωt) if t ≥ 00 else t < 0
L ( f6(t)) = F6(s) = ω
(s2+ω2)
Generalized functionf7(t) = tneat L ( f7(t)) = F7(s) = n!
(s−a)n+1
03-Modes.9
Properties 03-Modes.10
3
Linearity: L (c1 f1(t)+ c2 f2(t)) = c1L ( f1(t))+ c2L ( f2(t))
Differential operatorL(
d f (t)dt
)= sL ( f (t))− f (0) = sF(s)− f (0)
2nd oder differentialoperator
L(
d2 f (t)dt2
)= s2L ( f (t))− s f (0)− d f (0)
dt = s2F(s)− s f (0)− d f (0)dt
Higher oder differen-tial operator
L(
dn f (t)dtn
)= snF(s)−∑
n−1i=0 si dn−i−1 f (t)
dtn−i−1
∣∣∣t=0
Integral Operator L(∫ t
0 f (τ)dτ)= 1
s L ( f (t)) = 1s F(s)
Laplace transform applied to a system model
• To illustrate the usefulness of the Laplace transformation and the steps involved in the system anal-ysis, consider again the spring-mass-damper system described by the following linear differentialequation:
r(t) = Md2y(t)
dt2 +Bdy(t)
dt+Ky(t)
• Let’s consider the homogeneous equation (i.e. r(t) = 0):
Md2y(t)
dt2 +Bdy(t)
dt+Ky(t) = 0
• with initial contitions:
y(t = 0) = y0,dy(t)
dt
∣∣∣∣t=0
= y′(t = 0) = 0
• Form physical point of view, the equation describe the dynamical behavior of a system with zeroinput and initial velocity and non zero initial position.
03-Modes.11
Solution of Homogeneous Differential Equation
• The solution of the homogeneous differential equation is formed by a linear combination of functions:
y(t) = eλ t
• by substitution y(t) into the previous differential equation::
Mλ2eλ t +Bλeλ t +Keλ t = 0
• that is, by collecting eλ t :
eλ t(Mλ2 +Bλ +K) = 0
• Division by eλ t gives the second-order polynomial, the characteristic polynomial.:
P(λ ) = Mλ2 +Bλ +K = 0
• This algebraic equation P(λ ) = 0, is the characteristic equation, which solutions are:
λ1 =−√
B2−4KM−B2M
,λ2 =
√B2−4KM−B
2MFormally, the terms
y(k) (k = 1,2).
of the original differential equation are replaced by λ k.Solving the polynomial gives the 2 values of λ1 and λ2.Substitution of any of those values for λ into eλ t gives a solution eλit .Since homogeneous linear differential equations obey the superposition principle, any linear combi-nation of these functions also satisfies the differential equation.
03-Modes.12
4
Solution of Homogeneous Differential Equation (cont.)
• If the roots of P(λ ) are both real non zero (i.e. λ1 6= λ2 ∈ℜ):
y(t) = c1eλ1t + c2eλ2t
• if they are real and coincident (i.e. λ1 = λ2), then se solution is:
y(t) = (c1 + c2 · t)eλ1t
• then, if they come in conjugate pairs (i.e. λ1,2 = α ± iβ ), then it’s possibile to conisder separatelythe real part and the imaginary part:
y(t) = eαt (c1 cosβ t + ic2 sinβ t)
• where the c1 and c2 constant are calculated by applying the initital conditions:
y(t = 0) = c1eλ1t + c2eλ2t = y0y′(t = 0) = c1λ1eλ1t + c2λ2eλ2t = 0
03-Modes.13
Solution of Homogeneous Differential Equation (cont.)The solution of the differential equation is:
y(t) =y0e−
t(√
B2−4KM+B)2M
(B
(e
t√
B2−4KMM −1
)+√
B2−4KM
(e
t√
B2−4KMM +1
))2√
B2−4KM03-Modes.14
An overdamped system
20 40 60 80 100t
0.2
0.4
0.6
0.8
1.0
yHtL
Figure 1: When the friction is high, the energy in the system is dissipated in a short time, and the systemoutput doesn’t oscillate. The figure shows the system output y(t) for: M→ 0.5,B→ 10,K→ 0.5,y0→ 1.
λ1 =−19.95, λ2 =−0.05 y(t) = 0.05(−0.05e−19.95t +19.95e−0.05t)
03-Modes.15
5
A critically damped system
2 4 6 8 10t
0.2
0.4
0.6
0.8
1.0
yHtL
Figure 2: When the friction is decreasing, the system output tends to increase the oscillation behavior. The
figure shows the system output y(t) for: M → 0.5,B→ 1,K → 0.5,y0 → 1, λ1 = −1, λ2 = −1. y(t) =te−t + e−t
03-Modes.16
Underdamped system
2 4 6 8 10t
0.2
0.4
0.6
0.8
1.0
yHtL
Figure 3: When the friction decreases, the system output starts to oscillatate. The figure shows the
system output y(t) for: M → 0.5,B → 0.5,K → 0.5,y0 → 1. λ1 = −0.5− 0.86I,λ2 = −0.5 + 0.86I.y(t) = e−0.5t(0.58sin(0.86t)+ cos(0.86t))
03-Modes.17
6
Undamped system
10 20 30 40t
-1.0
-0.5
0.5
1.0
yHtL
Figure 4: When the friction goes down to zero, the energy in the system is not dissipated any longer,therefore the system output oscillates steadily. The figure shows the system output y(t) for: M→ 0.5,B→0,K→ 0.5,y0→ 1. lambda = 0− I, λ = 0+ I. y(t) = cos(t)
03-Modes.18
Laplace trasform
• Now, we analyize the same system using the Laplace Transform.• The Laplace transform of the function y(t) is:a
R(s) = M
(s2Y (s)− sy(0)− dy(t)
dt
∣∣∣∣y=0
)+B(sY (s)− y(0))
+KY (s)
(1)
• collecting terms in Y (s) (transform of output signal) and R(s) (transform of input signal).
R(s) = Ms2Y (s)+BsY (s)+KY (s)
−(M(sy(0)+ f ′(0))−By(0)
)where: dy(t)
dt
∣∣∣y=0
= f ′(0)
03-Modes.19
• obtaining the relationship between input, initial conditions and output:
@@I
Action of initial conditions y(0), y′(0) to Output Y (s)
+ (M(sy(0)+ f ′(0))+By(0)Ms2+Bs+K
Action of input R(s) to Ouptut Y (s)��������9
Y (s) = 1Ms2+Bs+K R(s)+
03-Modes.20
2.2 System characteristic equation, poles and zeros
System characteristic equation, poles and zeros
• The denominator polynomia p(s), when set equal to zero, is called the characteristic equation,• The roots of this equation determine the character of the time response.• The roots of this characteristic equation are also called the poles of the system.
7
• The roots of the numerator polynomial q(s) are called the zeros of the system;• Poles and zeros are critical values. At the poles, the function Y (s) becomes infinite, whereas at the
zeros, the function becomes zero.• The complex frequency s-plane plot of the poles and zeros graphically portrays the character of the
natural transient response of the system.03-Modes.21
Characteristic equation
• Let’s consider the trasnfer function (1) with null input R(s) = 0 and intial velocity y′(0) = 0 and notnull initial position (y(0) = y0 )
Y (s) =1
Ms2 +Bs+K(Ms+B)y0
where:
y(t) output variable (position)M system massB friction coefficientK Spring stiffnessR(s) system input (external force)
• Let’s us rearrange the equation as follows:
Y (s) = Kpω2
ns2 +2ζ ωns+ω2
n(Ms+B)y0
whereKp =
1K is the Steady State gain (or Static Sensitivity)
ζ = B2√
K·M Damping coefficient
ωn =√
KM Natural Frequency
• in the following we discusse the behavior of the system output y(t) with respect the poles p1 and p2.03-Modes.22
System poles
• System poles are:
p1,2 =−ζ ωn± jωn
√1−ζ 2
................................
................................
-ZZZ
ZZZ
ZZ}
.....................................
......
7
6
QQs
p1
p2
−ζ ωn
+ jωn√
1−ζ 2
− jωn√
1−ζ 2
θ = arccos(ζ )
03-Modes.23
Modes and system behaviour characterisationThe poles defines the behaviour of the system output, we can define four different behaviors:
• Overdamped (ζ > 1): The system returns (exponentially decays) to equilibrium without oscillating.Larger values of the damping ratio ζ return to equilibrium more slowly.
• Critically damped (ζ = 1): The system returns to equilibrium as quickly as possible without oscil-lating.
• Underdamped (0 < ζ < 1): The system oscillates (at reduced frequency compared to the undampedcase) with the amplitude gradually decreasing to zero.
• Undamped (ζ = 0): The system oscillates at its natural resonant frequency (ωn).03-Modes.24
8
Overdamped system (ζ > 1)
....................
.................... -
6
p1p2
−ζ ωn
ℜ(s)
ℑ(s)
In such a case the poles are both real:
p1 =−ζ ωn +ωn
√ζ 2−1,
p2 =−ζ ωn−ωn
√ζ 2−1
The following simulation shows the behviour of the systen with parameters are: B = 10,M = 0.5,K =0.5,ζ = 10,ωn = 1rad/sec
0 2 4 6 8 10 12 14 16 18 20−4
−2
0
2
4
6
8
B=10 M=0.5 K=0.5 damp=10 freq=1 rad/sec
oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
03-Modes.25
Critically damped system (ζ = 1)
....................
.................... -
6
p1
p2
−ζ ωn
ℜ(s)
ℑ(s)
In such a case the poles are both real and coincident:
p1 = p2 =−ζ ωn
The following simulation shows the behviour of the systen with parameters are: B = 1,M = 0.5,K =0.5,ζ = 1,ωn = 1rad/sec
0 2 4 6 8 10 12 14 16 18 20−4
−2
0
2
4
6
8
B=1 M=0.5 K=0.5 damp=1 freq=1 rad/sec
oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
03-Modes.26
9
Underdamped system (0 < ζ < 1)
....................
....................
-
6p1
p2
−ζ ωn
+ jωn√
1−ζ 2
− jωn√
1−ζ 2
In such a case the poles are complex:
p1 =−ζ ωn + jωn
√1−ζ 2,
p2 =−ζ ωn− jωn
√1−ζ 2
The following simulation shows the behviour of the systen with parameters are: B = 0.5,M = 0.5,K =0.5,ζ = 0.5,ωn = 1rad/sec
0 2 4 6 8 10 12 14 16 18 20−4
−2
0
2
4
6
8
B=0.5 M=0.5 K=0.5 damp=0.5 freq=1 rad/sec
oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
03-Modes.27
Undamped system (ζ = 0)
..........
....................
-
6p1
p2..........
In such a case the poles are both imaginary:
p1 =+ jωn
√1−ζ 2,
p2 =− jωn
√1−ζ 2
The following simulation shows the behviour of the systen with parameters are: B = 0,M = 0.5,K =0.5,ζ = 0ωn = 1rad/sec
10
0 2 4 6 8 10 12 14 16 18 20−4
−2
0
2
4
6
8
B=0 M=0.5 K=0.5 damp=0 freq=1 rad/sec
ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
ooooooooooooooooooooo
oooooooooooooooo
03-Modes.28
2.3 Transfer Function
Transfer function
• For a generic differential equation:
n
∑j=0
a jd jy(t)
dt j =m
∑j=0
b jd ju(t)
dt j
• we can write the correspondent Laplace transform:
n
∑j=0
a js jY (s)−n
∑j=1
a j
j−1
∑i=0
si d j−i−1y(t)dt j−i−1
∣∣∣∣t=0
=m
∑j=0
b js jU(s)
• it’s noteworthy saying that the input signal is zero as along as its derivatives for t = 0:
u(0) = 0,du(t)
dt
∣∣∣∣t=0
= 0, . . . ,dmu(t)
dtm
∣∣∣∣t=0
= 0
03-Modes.29
Transfer FunctionThe Laplace transform of system model is the sum of H1(s) and H0(s)
Forced response H1(t)The output of the system from the input signal, assuming zero intial conditions:
Y1(s) =∑
mj=0 b js j
∑nj=0 a js j U(s) = H1(s)U(s)
Natural response H0(t)The output of the system from intial conditions, assuming null input signal:
Y0(s) =∑
nj=1 a j ∑
j−1i=0 si d j−i−1y(t)
dt j−i−1
∣∣∣t=0
∑nj=0 a js j = H0(s)
03-Modes.30
Transfer Function
• It’s noteworthy saying that both H0(s) and H1(s) share the same characteristic polynomial ∑nj=0 a js j,
therefore they share the same poles.• in summary the Laplace transform of the system response can be written as:
Y (s) = Y1(s)+Y0(s) = H1(s)U(s)+H0(s)03-Modes.31
11
Transfer Function A formal definition
• The transfer function of a linear system is defined as the ratio of the Laplace transform of the outputvariable to the Laplace transform of the input variable, with all initial conditions assumed to be zero.Thus:
Y1(s) = H1(s)U(s)
• H1(s) is the Transfer Function of the system.• The transfer function of a system (or element) represents the relationship describing the dynamics of
the system under consideration.A transfer function may be defined only for a linear, stationary (constant parameter) system. Anonstationary system, also called a time–varying system, has one or more time–varying parameters,and the Laplace transformation may not be utilized. Furthermore, a transfer function is an input–output description of the behavior of a system. Thus, the transfer function description does notinclude any information concerning the internal structure of the system and its behavior.
03-Modes.32
Inverse Laplace transform of rational functions
• For Laplace transform defined using rational functions, or ratios of polynomials in s, the Laplacetransform can be computed using algebraic techniques:
H1(s) =∑
mj=0 b js j
∑nj=0 a js j =
q(s)p(s)
• we assume that bm,an 6= 0 and q(s) and p(s) have no common factors.• Let p1, . . . , pn the roots of equation p(s) = 0, i.e. the poles of H1(s):
p(s) = an(s− pn)(s− pn−1) . . .(s− p1)
• Let z1, . . . ,zn the roots of equation q(s) = 0, i.e. the zeros of H1(s):
q(s) = bm(s− zm)(s− zn−1) . . .(s− z1)03-Modes.33
Inverse Laplace transforms via partial fraction expansions
• Let us write H1(s) in the form
H1(s) =q(s)
an(s− pn)(s− pn−1) . . .(s− p1)
• We will use this factorization to decompose H1(s) into partial fractions and then use known Laplacetransform pairs to compute the inverse Laplace transform L −1(H1(s))
• We assume for now that the rational function H1(s) is proper, i.e, m < n.• Depending on the structure of the set of poles of H1(s), the methodology for finding the partial
fraction expansion will differ. We will consider several cases:
– Distinct poles, all real
– Distinct poles, two or more complex
– Repeated poles03-Modes.34
Distinct poles, all real
• We first consider the case when all the poles p1, . . . , pn are real and distinct (i.e. pi 6= p j , when i 6= j)• then H1(s) has the partial fraction expansion:
H1(s) =c1
(s− p1)+
c2
(s− p2)+ · · ·+ cm
(s− pm)
• where the constant c1, . . . ,cm (called the residues of H1(s)) are determined via the formula:
ci = [(s− pi)H1(s)]|s=pi, i = 1, . . . ,n
03-Modes.35
12
• We can now compute L −1(H1(s)) using linearity property and the transform pair:
ciepi·tstep(t) = L −1(
ci
(s− pi)
)being step(t) =
{1, if t ≥ 00, if t < 0 the step function
• thus:
h1(t) = L −1(H1(t)) = step(t)n
∑i=1
cie−pi·t =
= step(t)(c1e−p1·t + . . .+ cne−pn·t)
03-Modes.36
Transfer function as a sum of partial fraction expansion
Transfer function can be seen a the sum of partial fraction expansion os simplier elements (Laplacetransform)
03-Modes.37
System output as a sum of single modes of the system
Each siple fraction can be transformed back into the time domain function using the laplace pairs table, i.e.
L −1(
ci
s− pi
)= ciepi·t
These elementary functions are called the “system modes”.
03-Modes.38
Example Distinct poles, all real
H1(s) =s+1
s2 +7s+12
• We have q(s) = (s+1), and p(s) = (s2 +7s+12)
13
• We can factor p(s)as = (s+4)(s+3),• so poles are p1 =−4 and p2 =−3• we compute the residues:
c1 = [(s+4)H1(s)]|s=−4 = [���(s+4)(s+1)
���(s+4)(s+3)]
∣∣∣∣s=−4
= 3
c2 = [(s+3)H1(s)]|s=−3 = [���(s+3)(s+1)
(s+4)���(s+3)]
∣∣∣∣s=−3
=−2
• hence:h1(t) = step(t)(3e−4t −2e−3t)
03-Modes.39
Distinct poles, two or more complex• Now suppose that the poles are still all distinct, but two or more of them are complex.• Since complex roots of polynomials occur always in conjugate pairs, for each complex pole pi =
σi + jωi there will be another complex coniugate pole p∗i = σi− jωi.• The residual of the poles are also complex coniugate, in fact:
ci = [(s− pi)H1(s)]|s=pi, [(s− p∗i )H1(s)]|s=p∗i
= c∗iIn fact for a complex function f (s) = f (σ + jω), if we substitute the independent variable with it’sconjuagate s = σ − jω , we obtain the complex function conjugate, that is:
f (s∗) = f (σ − jω) = f ∗(sigma+ jω) = ℜ( f (s))− iℑ( f (s))• Let us now assume that the rest of the poles are all real• Hence, the partial fraction expansion will be
H1(s) =c1
(s− p1)+ · · · ci
(s− pi)+
c∗i(s− p∗i )
+ . . .cm
(s− pm) 03-Modes.40
• Inverting we get:
h1(t) =step(t)(c1e−p1·t + . . .
. . .+ cie(pi·t)+ c∗i e(p∗i ·y)+ . . .cne−pn·t)
• It can be verified that the two terms corresponding to conjugate pair can be combined as follows:
cie(pi·t)+ c∗i e(p∗i ·t) = 2|ci|eσ ·t cos(ω · t + arg(ci))
where |ci| and arg(ci) are the magnitude and the phase of the residue ci, and σ and ω are the real andthe imaginary parts of the pole pi
• Thus, we can write:
h1(t) =step(t)(c1e−p1·t + . . .
+2|ci|eσ ·t cos(ω · t + arg(ci))+ . . .
+ cne−pn·t)03-Modes.41
Example Distinct poles, two or more complex
• Let’s consider the following example: H1(s) = −s2+1s3+9s
• q(s) =−s2 +1, p(s) = s3 +9s = s(s2 +9) = s(s+3 j)(s−3 j)• poles: p1 = 3 j, p2 =−3 j, p3 = 0• compute the residues:
c1 = [(s−3 j)H1(s)](s=3 j) =
[ −s2 +1s(s+3 j)
](s=3 j)
=10
3 j ·6 j=−5
9
c2 = c∗1 =−59
c3 = [sH1(s)](s=0) =
[ −s2 +1(s+3 j)(s−3 j)
](s=3 j)
=1
−3 j ·3 j=
19
• and thus:
h(t) =[
109
cos(3t +π)+19
]step(t)
03-Modes.42
14
Repeated poles• We assumed that n poles of the rational function H1(s) can be divided into h groups, each formed by
ri, (i = 1, ...,h) repeated poles.
h
∑i=1
ri = n
• The development in fractional simple in this case is given by:
H1(s) =q(s)p(s)
=q(s)
(s− pi)r1(s− p2)r2 · · ·(s− ph)rh=
=h
∑i=1
ri
∑k=1
ci,k
(s− pi)ri−k+1
• where the residuals ci,k are computed as:
ci,k =1
(k−1)!dk−1
dsk−1 (s− pi)ri
q(s)p(s)
∣∣∣∣s=pi
, (i = 1, . . . ,h;k = 1, . . . ,ri)
03-Modes.43
Modes of a system• by considering the Laplace–time function pair:
L (tnepi) =n!
(s− pi)n+1
• Finally, one can obtain the inverse Laplace transform as
h1(t) =n
∑i=1
mi(t) =h
∑i=1
ri
∑k=1
ci,k
(ri− k)!tri−kepit
Also in this case the residuals ci,k are complex conjugates in correspondence of complex conju-gate poles, whereby the complex exponential can be replaced with products of real exponential andtrigonometric functions, with similar procedure to that followed in the case of poly distinct.
• Each function mi(t) defined in the time domain, obtained as a inverse transform of a term of a fractionexpansion, is callede mode of the system.
mode of the system: mi(t) =ci,k
(ri− k)!tri−kepit
• A mode mi(t) correspons univocally to a system pole pi. 03-Modes.44
Example Repeated poles
• Let’s consider the following example
H1(s) =1
s3 +4s2 +5s+2=
1(s+2)(s+1)2
with two poles h = 2, being p1 =−2,(r1 = 1) and p2 =−1,(r2 = 2).• The fraction expansion can be written as
H1(s) =h
∑i=1
ri
∑k=1
ci,k
(s− pi)ri−k+1 =c1,1
s+2+
c2,2
s+1+
c2,1
(s+1)203-Modes.45
• and residuals:
ci,k =1
(k−1)!dk−1
dsk−1 (s− pi)ri
q(s)p(s)
∣∣∣∣s=pi
c1,1 = (s+2) H1(s)|s=−2 = 1,
c2,2 =dds
(s+1)2 H1(s)|s=−1 =−1,
c2,1 = (s+1)2 H1(s)|s=−1 = 1• so that
H1(s) =1
s+2− 1
s+1+
1(s+1)2
• and then the inverse Laplace transform::
h1(t) =h
∑i=1
ri
∑k=1
ci,k
(ri− k)!tri−kepit = e−2t − e−t + te−t
03-Modes.46
15
Pole locations and the form of a signal
• When the signal h1(t) has a rational Laplace transform H1(s), we can tell the functional form of h1(t)without actually inverting H1(s),
• it’s possible simply by looking at the poles pi, i = 1, . . . ,n, n order of characteritic polynomial, assummarized here:
If H1(s) has: Then h1(t) contains a term: (multiplied by step(t))
a nonrepeated real pole pi ciepi·t
a real pole pi repeated twice ci,1epi·t + ci,2tepi·t
a single pair of complex conjugatepoles σ ±ω
2|ci|eσ ·t cos(ω · t + arg(ci))
a pair of complex conjugate polesσ ±ω repeated twice
2|ci,1|eσ ·t cos(ω · t + arg(ci,1)) + 2|ci,2|teσ ·t cos(ω ·t + arg(ci,2))
• Constants ci,ci+1 are determined by the zeros of H1(s).03-Modes.47
3 Stability
Stability
• The stability of control systems is an important property.• Considering any bounded input signal of a system, and if the output signal of the system to such a
signal is also bounded,• then the system is called bounded-input-bounded-output (BIBO) stable.• If the output signal does not show this property, the system is unstable.
03-Modes.48
Definition of StabilityWe use the impulse response of the system as the basis for definition of the stability properties.The impulse response is the time–response in output variable of the system due to an impulse on the
input. Using the impulse response makes it relatively simple to relate stability properties to the modes andthen the poles of the system.
The system defined by the Transfer function H1(s), and thus, the impulse response: h1(t)=L−1 {H1(s)},is said to be:
• Asymptotically stable system if the stationary impulse response h(t) is asymptotically zero:
limt 7→∞
hi(t) = 0
• Marginally stable system if the stationary impulse response h(t) is asymptotically different fromzero, but limited:
0 < limt 7→∞
hi(t)< ∞
• Unstable system if the stationary impulse response h(t) is asymptotically unlimited:
| limt 7→∞
hi(t)|= ∞
03-Modes.49
03-Modes.50Stability criteria
Since the impulse response h(t) is the sum of the modes hi(t), i = 1, . . . ,n, n being the number of polesof the transfer function of the system, we can say:
• Suppose that the real part of the pole pi is strictly negative (ℜ(pi) < 0), which means that the polelies in the left half plane.This implies hi(t)→ 0, as t→ ∞
• Suppose that the real part of the pole pi is zero, which means that the pole lies on the imaginary axis.This implies hi(t) goes towards a constant value different from zero as t→ ∞.
• Suppose that the real part of the pole pi is strictly positive (ℜ(pi) > 0), which means that the polelies in the left right plane. This implies |hi(t)| → ∞, as t→ ∞.
From the above analysis we can conclude as follows for transfer functions having pole multiplicity one:
1. If each of the poles lies inthe left half plane, the system isasymptotically stable, because then each ofthe partial impulse response terms, hi(t), goes towards zero as t→ ∞.
16
Figure 5: Definition of Stability
2. If one pole lies on the imaginary axis while the rest of the poles lies on the left half plane, the systemismarginally stable, because then one of the hi(t)-terms goes towards a constant value different fromzero as t→ ∞.
3. If at least one of the poles lies in the right half plane, the system isunstable, because then at least oneterm |hi(t)| goes to inf as t→ ∞.
Multiple poles:It would have been nice to conclude about stability and pole placement now, but wehave to look closer at the case ofmultiple poles of H(s).
In that case mode hi(t) will contain terms as tepit . By performing the asymptotic analysis as for simplepoles, we will find the following:
1. hi(t)→ 0 for a pole with negative real part (since tepit goes towards zero because epit decreases fasterthan t increases).
2. hi(t)→ ∞ for a pole on the imaginary axis (tepit equals t since ℜ(pi) = 0).3. hi(t)→ ∞ for a pole having positive real part. We will get the same results if the multiplicitymis
greater than two.
Characteristics Modes ModesAsymptotically stable sys-tem
limt 7→∞ hi(t) = 0, ∀ i ℜ(pi)< 0, ∀i
Marginally stable system ∃ i : 0 < limt 7→∞ hi(t)< ∞∃ i ℜ(pi) = 0, which isnonrepeating pole
Unstable system ∃ i : | limt 7→∞ hi(t)|= ∞ ∃ i ℜ(pi)> 0
Table 1:
03-Modes.51
In summary: stability, dynamic and poles 03-Modes.52
17
Figure 6: Stability, dynamic and poles
4 Examples
Transfer function of the DC motorThe DC motor is a power actuator device that delivers energy to a load, as shown in Figure 8; a sketch
of a DC motor is shown in Figure 7. The DC motor converts direct current (DC) electrical energy intorotational mechanical energy.
A major fraction of the torque generated in the rotor (armature) of the motor is available to drive anexternal load. Because of features such as high torque, speed controllability over a wide range, portabil-ity, well-behaved speed-torque characteristics, and adaptability to various types of control methods, DCmotors are widely used in numerous control applications, including robotic manipulators, tape transportmechanisms, disk drives, machine tools, and servovalve actuators.
The transfer function of the DC motor will be developed for a linear approximation to an actual motor,and second-order effects, such as hysteresis and the voltage drop across the brushes, will be neglected.
The input voltage may be applied to the field or armature terminals. The air-gap flux φ of the motor isproportional to the field current, provided the field is unsaturated, so that
φ = K f i f
The torque developed by the motor is assumed to be related linearly to φ and the armature current asfollows:
Tm(t) = Klφ ia(t) = KlK f i f (t)ia(t)
Figure 7: An electrical diagram of a DC motor
03-Modes.53
Analysis of DC motor mathematical model
• To have a linear system, one among ia or i f must be maintained constant while the other becomes theinput current.
18
Figure 8: A sketch of a DC motor
• First, we shall consider the field current controlled motor, which provides a substantial power ampli-fication. In Laplace transform notation:
Tm(s) = (KlK f Ia)I f (s) = KmI f (s)
where ia = Ia is a constant armature current, and Km is defined as the motor constant.• The field current is related to the field voltage as
V f (s) = (R f +L f (s))I f (s)03-Modes.54
• The motor torque Tm(s) is equal to the torque delivered to the load. This relation may be expressedas
Tm(s) = TL(s)+Td(s)
where TL(s) is the load torque and Td(s) is the disturbance torque,• Td(s) is often negligible (if motor is alone), but must be considered in systems subjected to external
forces.• The load torque for rotating inertia, is written as
TL(s) = Js2θ(s)+bsθ(s)
• Rearranging the previous equations, we have
TL(s) = Tm(s)−Td(s),
Tm(s) = KmI f (s),
I f (s) =V f (s)
R f +L f s03-Modes.55
• Therefore, the transfer function of the motor-load combination, with Td(s) = 0, is:
θ(s)V f (s)
=Km
s(Js+b)(L f s+R f )=
Km/(JL f )
s(s+b/J)(s+R f /L f )
• Alternatively, the transfer function may be written in terms of the time constants of the motor as:
θ(s)V f (s)
= H1(s) =Km/(bR f )
s(τ f s+1)(τLs+1)
where τ f = L f /R f and τL = J/b.• Typically is τL > τ f often the field time constant may be neglected.
03-Modes.56
θ(s)V f (s)
= H1(s) =Km/(bR f )
s(τ f s+1)(τLs+1)03-Modes.57
19
- - - - --?kVf (s)
Field
1R f +LF s
Km
Load
1b+Js
1s
Speed ω(s)
Position θ(s)
Output
Td(s)
Tm(s) TL(s)
+
−
Figure 9: The scheme of equations for the flux-controlled DC motor.
kk- - - - --?
�
-
6
Va(s)
Armature
1Ra+Las Km
Load
1b+Js
1s
Speed ω(s)
Position θ(s)Output
Td(s)
Tm(s) TL(s)
+
−
+−
Kb
Figure 10: The scheme of equations for the armature-controlled DC motor.
The Armature-Controlled DC motor
• The armature-controlled DC motor uses the armature current ia as the control variable. The statorfield can be established by a constant field coil:
Tm(s) = (KlK f I f )Ia(s) = KmIa(s)
• or a permanent magnet:
Tm(s) = KmIa(s),
where Km is a function of the permeahility of the magnetic material.03-Modes.58
• The armature current is related to the input voltage applied to the armature by
Va(s) = (Ra +Las)Ia(s)+Vb(s)
where Vb(s) is the back electromotive-force voltage proportional to the motor speed.• Therefore, we have:
Vb(s) = Kbω(s),
where ω(s) = sθ(s) is the transform of the angular speed• The armature current is
Ia(s) =Va(s)−Kbω(s)
Ra +Las• and the load torque is:
TL(s) = Js2θ(s)+bsθ(s) = Tm(s)−Td(s)
03-Modes.59
• We can obtain the motor transfer function
θ(s)Va(s)
=Km
s[(Js+b)(Las+Ra)+KbKm]=
Km
s(s2 +2ζ ωns+ωn)
However, for many DC motors, the time constant of the armature, τa = La/R, is negligible; therefore
G(s) =θ(s)Va(s)
=Km
s[(Js+b)Ra +KbKm]=
Km/(Rab+KbKm)
s(τls+1)
with the equivalent time constant τl = RaJ/(Rab+KbKm). Note that Km is equal to Kb. This equalitymay be shown by considering the steady-state motor operation and the power balance when the rotorresistance is neglected.
03-Modes.60
• The power input to the rotor is (Kbω)ia, and the power delivered to the shaft is T ω ,
20
• In the steady-state condition, the power input is equal to the power delivered to the shaft so that(Kbω)ia = T ω
• Since T = Kmia, we find that
Kb = Km
.03-Modes.61
5 Step response of 1st and 2nd order systems
5.1 Step response of a first order system
Step response of a first order system
• An first order elementary system is characterized by a transfer function as
Y (s) =Kp
τs+1• where τ is the time constant of the system, and Kp is the system static gain.• Appling a step function with amplitude A0, step(s) = A0
s , we obtain:
y(t) =L −1[
KP
τs+1A0
s
]=A0(1− e−
tτ )
Figure 11: Step response for a frist–order system
03-Modes.62
Settling time
• Settling time tais time needed for the output remains within 5% of final value.
ta ≈ 3 · τ
Figure 12: Definition of Settling time.
• For t = 5τ the output reaches 99.3% of the steady state value.• For t = 7τ the output reaches 99.91% of the steady-state value, i.e. the settling residue remains below
to one per thousand.03-Modes.63
21
Step response of a first order system
• Since the pole of the system is in |p|= 1τ
• the gretest in absolute value of p, then τ and ta are the shortest.• hence, fast system have pole on the far left side on Gauss plane.
Figure 13: Relationship between pole and settling time.
03-Modes.64
5.2 Step response of a Second order system
Step response of a second order system
• We consider now the problem of characterizing the step response of the second order system:
Y (s) = Kpω2
ns2 +2ζ ωns+ω2
nR(s)
• with R(s) = R01s (step input).
• The step response can be calculated using the Laplace inverse transform:
y(t) =L −1(
Kpω2
ns2 +2ζ ωns+ω2
nR0
1s
)=
=R0Kp
(1− e−ζ ωnt√
1−ζ 2sin((ωn
√1−ζ 2)t + arctan(
√1−ζ 2
ζ)
) (2)
03-Modes.65
step response of a second order system
Figure 14: Relationship between step response and pole location in the Gauss plane.
03-Modes.66
22
System response The most important parameters on which it can be based a measure of the quality ofthe transient of a second order system.
difference between the maximum value achieved fromthe output and the final value, usually expressed in % of final value.
S
S: Maximum overshoot (or maximum overtaking),
ta
a range of 5% of its final value.
ta: settling time, time neededfor the output remains within
10% of final val. (5+5).
tmax:time at which thre response has themaximum overshoot
tmax
Figure 15: Step response for a second order system.
03-Modes.67
System response and poles of a 2nd order system
• It’s important to understand the link between the maximum overshot S and the dumping coefficientζ .
• To obtain this link we should find the time in which the system respons is maximum, i.e. by derivatingthe step response eq. (2):
dy(t)dt
=
=R0Kp
(−Ae−ζ ωnt
ω cos(ωt +ψ)+Aζ ωne−ζ ωnt sin(ωt +ψ))
with ψ = arctan(√
1−ζ 2
ζ) = arcos(ζ ), A = 1√
1−ζ 2, ω = ωn
√1−ζ 2
• posing dy(t)dt = 0, we find the maximun and minimum for the function:
−ω cos(ωt +ψ)+ζ ωn sin(ωt +ψ) = 003-Modes.68
• obtaining:
ω cos(ωt +ψ) = ζ ωn sin(ωt +ψ)
sin(ωt +ψ)
cos(ωt +ψ)= tan(ωt +ψ) =
√1−ζ 2
ζ
• and then:
ωt +ψ = arctan(
√1−ζ 2
ζ) ⇒ ωt = nπ
because ψ = arctan(√
1−ζ 2
ζ)
• and then, finally:t =
nπ
ωn√
1−ζ 2, (n = 0,1, . . .)
• substituting values for t into equation (2), we obtain:
y(t)| maxmin
= 1− e−nπζ√
1−ζ 2√1−ζ 2
sin(nπ +ψ)⇒ y(t)| maxmin
= 1− (−1)ne−nπζ√
1−ζ 2
03-Modes.69
23
Maximun and minimum values
y(t)
ωnt
1 + e−πζ√1−ζ2
1− e−2πζ√1−ζ2
π√1−ζ2
2π√1−ζ2
3π√1−ζ2
4π√1−ζ2
1 + e−ζωnt
1− e−ζωnt
Figure 16: Maximun and minimum values.
03-Modes.70
Maximum overshoot S
• In a second order system the maximum overshoot is a function solely of the damping coefficient ζ
and is equal to 100% when this coefficient is zero:
S% = 100 · (ymax−1) = 100e−πζ√
1−ζ
Figure 17: The relationship between S% and ζ
03-Modes.71
secon dorder system specification Settling time ta
• To obtain a specification on output settling time ta
y(t)|max 1− e−nπζ√
1−ζ 2 ⇒ e−ζ ωn ta = 0.05
ζ ωnta ≈ 3,⇒ ta ≈3
ζ ωn
24
• The product ζ ωn is equal in magnitude, with opposite sign, to the real part σ of the poles of thesystem.
• this constraint is equivalent to restrict the position of the poles to the left of a vertical straight lineparallel to imaginary axis that crosses the real axes in
σ =− 3ta
• to satisfy a specified system settling time it should be necessary that the real part of the poles, σ̄ :
σ̄ <− 3ta
03-Modes.72
secon dorder system specification Maximum overshoot
• A specification on Maximum overshoot S̄% corresponds to a specification on damping coefficient fothe complex conjugate poles, that can be solved numerically:
S̄% < 100e−πζ√
1−ζ
• or graphically on figure 17.• The damping coefficient ζ depends on the position of the complex conjugate poles.• If the value of maximum overshoot must not exceed a certain maximum given, the poles of the system
must be included in the area bounded by the lines b and b′.03-Modes.73
Sector admissible in the Gauss plane For a desired t̄a and S̄%
• Considering specification on settling time t̄a and percentual overshoot S̄%,• they turned out in specification on σ̄ =−ζ̄ ω̄n and ζ̄ .• enforcing these specification means identify an allowable region for poles location into Gauss plane.
σ̄ = −ω̄nζ̄
θ = arcos(ζ̄)
Figure 18: Allowable region on gaus plane on the basis of step response specifications.
03-Modes.74
Poles with constant overshoot ζ = const
Figure 19: Poles with constant overshoot
03-Modes.75
25
Poles with constant oscillation period ω = ωn√
1−ζ 2 = const
Figure 20: Poles with constant oscillation period
03-Modes.76
Poles with constant ωn ωn = const
Figure 21: Poles with constant
03-Modes.77
26
6 Notes on the Laplace transform
Our aim is to develop a method for solving initial-value problems
y(n)+a1y(n−1)+ · · ·+an−1y′+any = g,
y(t0) = y0, y′(t0) = y′0, . . . , y(n−1)(t0) = y(n−1)0 ,
(3)
where the coefficients ai are constant, although the forcing function g need not be constant. To give theconditions that the forcing function will have to satisfy, we need some definitions:
A function is piecewise continuous on an interval [a,b] if the interval has a partition
a = t0 < t1 < · · ·< tn−1 < tn = b (4)
such that the function is continuous on each sub-interval (tk, tk+1) and has finite limits at the endpointsof these intervals. The value of the function at the points tk is irrelevant. Indeed, we may say that f ispiecewise continuous on [a,b] even if it is undefined at the points tk.
Theorem and definition. Suppose f is piecewise continuous on [0,b] for each positive b, and
limt→∞
e−at f (t) = 0 (5)
for some a. Then f has a Laplace transform, namely the function F on (a,∞) given by
F(s) =∫
∞
0e−st f (t)d t. (6)
Proof. By (5), there is some (positive) R such that∣∣e−at f (t)
∣∣ < 1 when R < t. But then f is piecewisecontinuous on [0,R], so
∫ R0 e−st f (t)d t exists. Also, when R < t, we have∣∣e−st f (t)
∣∣6 e(a−s)t ; (7)
and when a < s, the integral∫
∞
R e(a−s)t d t converges; so∫
∞
R e−st f (t)d t converges. Therefore the integralin (6) converges when a < s.
The property expressed by equation (5) is that f has exponential order. If f has the Laplace transformF , then we may write
F = L{ f} . (8)
Often F is expressed in terms of its values F(s), and f in terms of f (t); then we may write
F(s) = L{ f (t)} . (9)
The solution of problems (3) by means of Laplace transforms involves three essential lemmas:
Lemma 1. The operation f 7→ L{ f} is linear, that is,
L{c0 f0 + c1 f1}= c0 ·L{ f0}+ c1 ·L{ f1} . (10)
Proof. Multiplication by e−st and integration are linear.
Remembering the proof of the following may be an aid in remembering the lemma itself:
Lemma 2. If the derivatives f (k) are of exponential order and are continuous on [0,∞) when k < n, and if
f (n) is piecewise continuous on [0,∞), then L{
f (n)}
exists and is given by
L{
f (n)}= sn ·L{ f}− (sn−1 · f (0)+ sn−2 · f ′(0)+ · · ·+ f (n−1)(0)). (11)
In particular,
L{
f ′}= s ·L{ f}− f (0), (12)
L{
f ′′}= s2 ·L{ f}− s · f (0)− f ′(0). (13)
Proof. Use integration by parts and induction. Equation (11) holds trivially when n = 0. By integration byparts,
L{
f (m+1)}=∫
∞
0e−st f (m)(t)d t (14)
=[e−st f (m)(t)
]∞
0+ s ·
∫∞
0e−st f (m)(t)d t (15)
=− f (m)(0)+ s ·L{
f (m)}, (16)
assuming f (m) has exponential order. Hence, if (11) holds when n = m, then it holds when n = m+1 (underthe stated conditions).
27
Lemma 3. Suppose f and g are piecewise continuous on each interval [0,a]. Then L{ f} = L{g} if andonly if f (t) = g(t) for all but finitely many t in each interval [0,a].
Proof. Part of this is easy, since the integral of a function is not changed if the value of the function ischanged at finitely many points. We shall not try to prove the hard part.
Lemma 3 means that, if L{ f}= F , then we can write
L−1 {F}= f : (17)
in a word, the operation f 7→ L{ f} is invertible.To simplify matters, let us say we want to solve the intermediate-value problem
y′′+ay′+by = g, y(0) = y0, y′(0) = y′0. (18)
(What is important here is that the differential equation is linear with constant coefficients.) By Lemma 1we get
L{
y′′}+a ·L
{y′}+b ·L{y}= L{g} . (19)
Writing Y for L{y}, by Lemma 2 we get
s2 ·Y − s · y0− y′0 +as ·Y −ay0 +b ·Y = L{g} , (20)
which we can solve for Y :
Y =y0s+ay0 + y′0
s2 +as+b+
L{g}s2 +as+b
. (21)
So, to be able to solve (18) by means of Laplace transforms, we need:
1. to be able to find L{g} easily;2. to be able to recover y as L−1 {Y}.
If g is a reasonable function, then (because of Lemma 3) we can accomplish these tasks by knowing somestandard Laplace transforms.
First, we find the transforms of polynomials:
Theorem 4. For all non-negative integers n,
L{tn}= n!sn+1 . (22)
Proof. Just compute:
L{1}=∫
∞
0e−st d t =
[e−st
−s
]∞
0=
1s
; (23)
L{
tk+1}=∫
∞
0e−st tk+1 d t (24)
=
[e−st tk+1
−s
]∞
0+
k+1s
∫∞
0e−st tk d t (25)
=k+1
s·L{
tk}
; (26)
Now the claim follows by induction.
If we know that L{ f}= F , then we can recover the function whose Laplace transform is F(s− c):
Theorem 5.L{
ect f (t)}= F(s− c). (27)
Proof. Apply definition (6) twice:
L{
ect f (t)}=∫
∞
0e−(s−c)t f (t)d t, (28)
which is F(s− c).
Theorem 6.
L{coshct}= ss2− c2 , L{cosct}= s
s2 + c2 ,
L{sinhct}= cs2− c2 , L{sinct}= c
s2 + c2 .
28
Proof. By Theorem 5 and (23),
L{
ect}= 1s− c
. (29)
Hence, by using the definitions
cosh t =et + e−t
2, cos t = cosh it,
sinh t =et − e−t
2, sin t =
sinh iti
,
(30)
we can compute the desired transforms.
An alternative proof of Theorem 6 uses the following rule.
Theorem 7.L{ f (ct)}= 1
c·F( s
c
). (31)
Proof. The computation is
L{ f (ct)}=∫
∞
0e−st f (ct)d t =
1c
∫∞
0e−su/c f (u)du, (32)
which is (1/c) ·F(s/c).
A special case of (29) is
L{exp}= 1s−1
. (33)
Then, from (30), we get
L{cosh}= ss2−1
, L{cos}= ss2 +1
,
L{sinh}= 1s2−1
, L{sin}= 1s2 +1
.
Now Theorem (6) follows.
Theorem 8. If F is a rational function, then L−1 {F} exists and is a linear combination of ect and ect cosbtand ect sinbt (for various b and c).
Proof. If F is a rational function, then we can write F(s) as a sum of ‘partial’ fractions of the forms
As−a
andAs+B
s2 +as+b, (34)
where a2 < 4b in the latter. Hence we can write F as a linear combination of fractions of the forms
1s− c
ands− c
(s− c)2 +b2 andb
(s− c)2 +b2 . (35)
These are respectively the transforms of
ect and ect cosbt and ect sinbt; (36)
so, by Lemma 1, we can compute L−1 {F}.
Now we can use Laplace transforms to solve (18) (and higher-order problems) in case g is a linearcombination of polynomial and exponential and trigonometric functions. The solution does not requiresolving a homogeneous equation first, or finding a general solution before using the initial conditions.
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Piecewise-continuous forcing functions
If g is merely piecewise continuous, will still may be able to compute L{g} by expressing g in terms ofstep-functions. Let uc be the function on (−∞,∞) given by
uc(t) =
{0, if t < c;1, if c 6 t.
(37)
If, for example, g is given by
g(t) =
g0(t), if t 6 t1;g1(t), if t1 < t < t2;g2(t), if t2 6 t;
(38)
theng(t) = g0(t)+ut1(t) · (g1(t)−g0(t))+ut2(t) · (g2(t)−g1(t)), (39)
unless t = t1; this one difference won’t affect computation of the Laplace transform, by Lemma 3. We canperform this computation by means of the following.
Theorem 9.L{uc(t) · f (t− c)}= e−csF(s). (40)
Proof. Compute:
L{uc(t) · f (t− c)}=∫
∞
ce−st f (t− c)d t =
∫∞
0e−s(u+c) f (u)du, (41)
which is e−csF(s).
Impulse functions
We have encountered no function whose Laplace transform is a non-zero constant. In calculus, when wenote that 1/x is not the derivative of a rational function, we define lnx =
∫ x1 d t/t. Similarly, we look for a
function δ (the Dirac delta) satisfyingL{δ}= 1 (42)
and—so that Theorem 9 is satisfied—L{δ (t− c)}= e−cs. (43)
These properties are achieved if δ (t) = 0 when t 6= 0, but∫
∞
−∞δ = 1. There is no such function, literally;
but the equation (21) still makes sense if g is a multiple of δ (t− c), and then we can find y by means ofTheorem 9.
Convolution
Theorem and definition. If L{ f}= F and L{g}= G, then
F ·G = L{ f ∗g} , (44)
where the convolution f ∗g is defined by
f ∗g(t) =∫ t
0f (t− x) ·g(x)dx. (45)
In particular, Equation (21) has the solution y = φ +g∗h, where
L{φ}= y0s+ay0 + y′0s2 +as+b
and L{h}= 1s2 +as+b
. (46)
Note that φ and h are independent of g; also φ = 0 if y0 = 0 and y′0 = 0. Once φ and h are found, thensolutions to (21) can be obtained by integration.
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