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SIMPLE SYSTEM PROTOTYPES
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In order to give an idea about the various types of behav-
ior of dynamical systems we will next discuss a numberof simple system prototypes.
Recall that linear dynamical systems can be described by
its transfer function G(p),
y(t) = G(p)u(t), G(p) =
B(p)
A(p)
The system properties are uniquely determined by its
transfer function:
Poles of G(p)The zeros pi of the denominator polynomial A(p),
i.e., the solutions to
A(pi) = 0
determine how the autonomous behavior of a system
(i.e., when u(t) = 0). The poles pi thus determine
system stability and whether the output oscillates or
not.
Remark. pi are real or (often) complex-valued num-bers. The term pole is used in the theory of analyt-
ical functions to denote a point at which a function
is infinite; in this case G(p) is infinite at p = pi.
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Zeros of G(p)
The zeros zi of the numerator polynomial B(p), i.e.,the solutions to
B(zi) = 0
determine the response to the input u(t), such as
whether the system has an inverse response or not.
Just as the poles, the zeros of a system may be real
or complex.
The system behavior will be demonstrated by considering
the step response, i.e., the systems output when the input
is a step,
u(t) =
0 , for t < 0
usteg , for t
0
and y(t) = 0, t < 0.
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The systems we will consider are:
First-order systems.Cf. above.
Systems with two time constants.These are second-order systems consisting of two first-
order systems in series.
Underdamped system.Systems with an oscillating response. Second-order
systems with complex-conjugate poles can be under-
damped.
Systems with time delay.
Systems with integration.
Systems with inverse response.The response starts in the wrong direction before
changing direction. These systems can be described
(and are often caused) by two subsystems operating
in parallel which act in opposite directions.
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First-order systems
A first-order system is described a first-order differential
equationdy(t)
dt+ ay(t) = bu(t)
or
Tdy(t)
dt+ y(t) = Ku(t)
where T = 1/a and K = b/a.Transfer function:
G(p) =b
p + a=
K
T p + 1
Step response:
ystep
(t) = K(1
et/T)ustep
, t
0
Static gain = K: y(t) Kustep as t .Time constant = T: y(T) = K
1 e1ustep = 0.632 y()
Stability:
|y(t)| as t if (and only if) T 0 (a 0) the system is unstable if a 0Or:System is unstable if (and only if) the pole p1 = a ofG(p) is positive (or zero)
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Step response of first-order system with
a = 1, b = 1(T = 1, K = 1)
0
1
u(t)
1 0 1 2 3 4 5
0
1
y(t)
time
Note:
Output derivative reacts instantaneously to a discontin-
uous change in the input.
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System with two time constants
Two first-order systems in series:
G(p) =b2
p + a2 b1p + a1
=b1b2
p2 + (a1 + a2)p + a1a2
This is a second-order system with the real poles
p1 = a1 and p2 = a2.Differential equation:
d2y(t)
dt2+ (a1 + a2)
dy(t)
dt+ a1a2 y(t) = b1b2 u(t)
Response:
IfT1 = T2 the system can be represented in terms of twofirst-order systems in parallel:
Define the time constants T1
= 1/a1, T
2= 1/a
2and
static gains K1 = b1/a1, K1 = b2/a2. Then
G(p) =K
(T2p + 1)(T1p + 1), K = K1K2
Partial fraction expansion:
G(p) =KT1/(T1 T2)
T1p + 1
+KT2/(T2 T1)
T2p + 1ifT1 = T2.
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Then
y(t) = G(p)u(t) = y1(t) + y2(t)
where
y1(t) =KT1/(T1 + T2)
T1p + 1u(t), y2(t) =
KT2/(T1 + T2)
T2p + 1u(t)
Step response:
ystep(t) = y1,step(t) + y2,step(t)
=KT1
T1 T2
1 et/T1
ustep +KT2
T2 T1
1 et/T2
ustep
= K
1 T1
T1 T2et/T1 T2
T2 T1et/T2
ustep
The case T1
= T2: taking the limit as T
1 T2
0 gives
ystep(t) = K
1 (1 + t
T)et/T
ustep
Stability: |y(t)| if (and only if) a1 = 1/T1 0and/or a2 = 1/T2 0= system is stable if and only if the poles ofG(p) are
all negative; p1 = a1 < 0 and p2 = a2 < 0.
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Step response of system with two time constants
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0
0.2
0.4
0.6
0.8
1
1.2
u
2 0 2 4 6 8 10 12 14 16
0
0.2
0.4
0.6
0.8
1
1.2
y
tid
Note:
Response smoother than for first-order system,dy(t)dt is not affected immediately by discontinuous change
in input.
Instead second derivatived2y(t)dt2 reacts instantaneously to
the discontinuous change in the input.
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Underdamped systems
General second-order system:
d2y(t)
dt2+ a1
dy(t)
dt+ a2y(t) = b1
du(t)
dt+ b2u(t)
has transfer function:
G(p) =b1p + b2
p2
+ a1p + a2In the general case, G(p) may have complex conjugate
poles.
IfG(p) has poles p1 and p2, we have
p2 + a1p + a2 = (p p1) (p p2)= p2 (p1 + p2)p + p1p2
= a1 = (p1 + p2) , a2 = p1p2System properties can be conveniently characterized in
terms of the parameters n, (zeta) and K defined by
2n = a2 = p1p2.n is called the natural frequency of the system
2n = a1 =
(p1 + p2).
is called the relative damping of the system.
K2n = b2K is the static gain of the system.
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Then (assuming for simplicity b1 = 0):
d2y(t)
dt2+ 2n
dy(t)
dt+ 2ny(t) = K
2nu(t)
and
G(p) =K2n
p2 + 2np + 2nThe system poles are given by
p2 + a1p + a2 = p2 + 2np +
2n = 0
=p1,2 = n
2 1 n
We see that the determines the nature of the poles:
assuming that > 0,
If 1, the system has real poles If = 1, the system has a double pole at
p1 = p2 = n If < 1, the system has complex conjugate poles
p1,2 = j.In this case, n =
|p1,2
|=
2 + 2
and = /|p1,2|.
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Step response:
The case 1 corresponds to a system with two timeconstants, which we have already considered.
For < 1, the step response is given by
ystep(t)=K
1 1
ent [sin(nt) + cos(nt)]
ustep
=K
1
1
ent sin(nt + )
ustep
where
=
1 2, = arccos()We observe:
The response oscillates with frequency n.For = 0, the oscillation frequency equals the natu-
ral frequency n. The amplitude of the oscillations decreases exponen-
tially according to ent. Hence, the larger (smaller)the relative damping is, the faster (slower) the ex-
ponential decay of the oscillation.
The relative damping reduces the oscillation frequen-cy n: from n = n for = 0 to n = 0 for = 1.
Stability: If n < 0, the output diverges as t in-creases, and the system is unstable.
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As n =
1
2
(p1 + p2) (cf. above), we have that the
system is unstable if it has a pole with a positive real
component.
The relative damping determines the amount of oscil-
lation of the system. We have the following classification: > 1: the system is overdamped.
It has two real negative poles, corresponding to sys-
tem with two time constants.
= 1: the system is critically dampled.It has a real negative double pole.
< 1: the system is underdamped.It has two complex conjugate poles with negative real
components.
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0 1 2 3 4 5 6 7 8 9 100.2
0
0.2
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1.2
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1.6
1.8
y
tid
Step responses of second-order systems with relative damp-
ings = 2, 1, 0.5 and 0.1 (from below), natural frequency
n = 1 and static gain K = 1.
Maximum overshoot M (in % of static value):
M = exp
100%
Time tM at which the maximum is obtained:
tM =
n
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Example
Mechanical system composed of spring and damper.
u(t)
(t)
m
bk
y(t): deviation from equilibrium
u(t): external force acting on mass
Newtons law:
md2y(t)dt2
= u(t) + Ff(t) + Fd(t)
where:
- Ff(t): spring force, proportional and directed in the
opposite direction to deviation y(t),
Ff(t) = ky(t)- Fd(t): damping force, proportional and directed in theopposite direction to rate of change dy(t)/dt,
Fd(t) = b dy(t)dt
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=
m
d2y(t)
dt2= u(t) ky(t) b dy(t)
dt
ord2y(t)
dt2+
b
m
dy(t)
dt+
k
my(t) =
1
mu(t)
This is a second-order system with transfer function
y(t) = 1/mp2 + bmp +
km
u(t)
We see that the system has:
- natural frequency n =
km
- static gain K = 1/m2n
= 1k
- relative damping = b/m2n =12
bmk
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Example - Harmonic oscillator.
Case with zero relative damping; = 0:
d2y(t)
dt2+ 2ny(t) = K
2nu(t)
When left to itself (u(t) = 0), the output is sinusoidally
oscillating with the natural frequency frequency n.
Indeed, the signal
y(t) = A sin(nt)
has second derivative
d2y(t)
dt2= A2n sin(nt) = 2ny(t)
so that
d2y(t)
dt2+ 2ny(t) = 0
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Remark
Response of system with b1 = 0,
G(p) =b1p + b2
p2 + a1p + a2
We can write G(p) as a parallel coupling of two second-
order systems:
G(p) = b1pp2 + a1p + a2
+ b2p2 + a1p + a2
Hence
y(t) = G(p)u(t) = y1(t) + y2(t)
where
y1(t) = G1(p)u(t), G1(p) =b1p
p2
+ a1p + a2u(t)
and
y2(t) = G2(p)u(t), G2(p) =b2
p2 + a1p + a2u(t)
We already know how to compute step response y2(t) of
G2(p).
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The first system G1(p) can be written as a series connec-
tion
G1(p) = pb1
p2 + a1p + a2= p G1(p)
where
G1(p) =b1
p2 + a1p + a2
Hence we obtain the output by first computing the output
ofG1(p),y1(t) = G1(p)u(t)
followed by the differential operator p,
y1(t) = p y1(t) =dy1(t)
dt
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Systems with inverse response
Inverse response systems typically arise when two dynam-
ical processes act in parallel, but in opposite directions
and with different time constants. Examples of inverse
0.2
0
0.2
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u
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0.5
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y
tid
response systems:
Response of water level in boiler to changes in waterflow to boiler.
Response of temperature to changes in fuel input ratein combustion of solid fuels.
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Altitude control of airplanes.
Economic systems: response of tax revenues to changesin tax level(?).
To see how inverse response behavior may arise, consider
two first-order system in parallel, so that
y(t) = y1(t) + y2(t)
where
y1(t) =b1
p + a1u(t)
and
y2(t) =b2
p + a2u(t)
Inverse response as shown in the figure is obtained if andonly if the following conditions hold:
Final condition:y(t) = y1(t) + y2(t) > 0, as t
Initial condition:dy(t)
dt
t=0
= dy1(t)dt
t=0
+ dy2(t)dt
t=0
< 0
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But
y1(t) b1a1
ustep as t and
y2(t) b2a2
ustep as t Final condition =
b1
a1+
b2
a2> 0
which implies (as a1, a2 > 0; stable systems assumed)
a1b2 + a2b1 > 0
Moreover,dy1(t)
dt
t=0
= b1 ustep
anddy2(t)
dt
t=0
= b2 ustep
Initial condition =b1 + b2 < 0
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Relation to transfer function of the parallel connected
system:
Gpar(p) =b1
p + a1+
b2p + a2
=(b1 + b2)p + a1b2 + a2b1
(p + a1)(p + a2)
Zero:
(b1 + b2)p + a1b2 + a2b1 = 0
= pzero = a1b2 + a2b1
b1 + b2Inverse response conditions
a1b2 + a2b1 > 0
and
b1 + b2 < 0
imply that pzero > 0.
General property of inverse response systems:
Transfer function of an inverse response system has at
least one zero which is positive (if real) or has a positive
real component (if complex valued)
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Systems with integration
These are systems consisting of pure integration:
y(t) = kt0
[u() u0] dHere the output keeps changing as long as u(t) u0 = 0,cf. figure.
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0
0.2
0.4
0.6
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1.2
u
2 0 2 4 6 8 100
1
2
3
4
5
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y
tid
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Examples of integrators:
Inventory models, where y(t) is the amount of ma-terial in stock, and u(t) is inflow and u0 the outflow
of material. For example, y(t) could be the liquid
volume of a container and u(t) and u0 the in- and
outflows, respectively.
Servo motors used for position control. The posi-
tion is controlled by a voltage u(t), where the rate
of change of position is proportional to the voltage,
dy(t)/dt = ku(t).
Differential equation and transfer function
Differentiation gives
dy(t)dt
= k [u(t) u0]=
y(t) =k
p[u(t) u0]
Transfer function:
G(p) =k
pThe system has a pole atp = 0. Hence it is on the stability
boundary.
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Systems with time delay
A time delay means that it takes a time L before the
input affects the system. The figure shows the response
of a first-order system with time delay L = 1:
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0.2
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0.6
0.8
1
1.2
u
2 0 2 4 6 8 100.2
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1.2
y
tid
The control of time delay systems is difficult as the effect
of a control action is seen only after a certain time. Forgood control, one should predict the future system output
y(t + L), which depends on the past control actions
u(s), s [t L, t].
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Examples of time-delay systems:
In the process industry time delays arise typicallydue to transport delays (for example, when material
is transported in pipes etc.).
In a paper machine it is the time it takes for the webto travel from the point where control actions are
taken to the point where measurements are made.
In web services, time delays may arise due to the timeit takes to start a new server.
Transfer function:
A pure time-delay system is described by
y(t) = u(t
L)
We can express this relation with the differential operator
p = ddt by introducing the Taylor-series expansion
u(t + h) = u(t) + hdu(t)
dt+
h2
2!
d2u(t)
dt2+ + h
k
k!
dku(t)
dtk+
=
1 + hp +
(hp)2
2! + +(hp)k
k! +
u(t)
= ehpu(t)
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With p =
L we have
y(t) = u(t L) = eLpu(t)or
y(t) = GL(p)u(t)
where the transfer function of the time delay is
GL(p) = eLp
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Higher-order systems
In general, dynamical systems are not restricted to first-
order or second-order systems.
However, by partial fraction expansion a transfer func-
tion G(p) with simple poles can be written as a parallel
connection of first-order and second-order systems:
G(p) =B(p)
A(p)
=k
KkTkp + 1
+
l
bl1p + bl2p2 + al1p + al2
When G(p) has multiple poles, there are also terms con-
sisting of series connections of first-order or second-order
systems.
The decomposition above implies:
The output of a higher-order system can be comput-ed as a sum of the outputs of first-order and second-
order systems.
The system G(p) is stable if and only if all the first-order and second-order systems in the parallel con-
nection are stable. Equivalently, the system is stableif and only if all the poles of G(p) (i.e., the zeros of
A(p)) have negative real parts.
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