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Control Systems
System responseL. Lanari
prelim
inary
versio
n
Monday, November 3, 2014
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Lanari: CS - System response 2
Outline
• study a particular response, the step response, to a unit step as input
• characterize the asymptotic and transient response on the step response
• define the long term, asymptotic, behavior (steady-state) of a response and understand when it exists
• compute the steady-state response for different test inputs
Monday, November 3, 2014
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Lanari: CS - System response 3
Step response
The forced response to a step input is referred as step response
u(t) = δ−1(t) U(s) =1
s
Ys(s) = W (s)U(s) = W (s)1
s
Xs(s) = H(s)U(s) = H(s)1
s
x0 = 0
integral property of the Laplace transform
xs(t) =
� t
0h(τ)dτ ys(t) =
� t
0w(τ)dτ
the step response is equal to the integral of the impulse response
Monday, November 3, 2014
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Lanari: CS - System response 4
Step response
Let S be asymptotically stable W (s) =N(s)�n
i=1(s− pi)
Ys(s) = W (s)U(s) =N(s)�n
i=1(s− pi)1
sYs(s) =
R0s
+n�
i=1
Ris− pi
these terms decay sinceS be asymptotically stable Re[pi] < 0
distinct poles
ys(t) =
�R0 +
n�
i=1
Riepit
�δ−1(t)
Monday, November 3, 2014
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Lanari: CS - System response 5
Step response
stable system (A.S.)
W (s)
δ−1(t)
1t
x0 = 0
constant value
t
1
steady-statetransient (asymptotic behavior)
ys(t)
ys(t) =
�R0 +
n�
i=1
Riepit
�δ−1(t)
R0
yZS(t) = ys(t)
conceptualfrontier
Monday, November 3, 2014
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Lanari: CS - System response 6
Step response
ys(t) =
� t
0CeA(t−τ)Bu(τ)dτ +Du(t)
= C
� t
0eA(t−τ)Bdτ +D
= C
� t
0eAσBdσ +D
= C(�A−1eAσB
�σ=tσ=0
+D
= CA−1eAtB + D − CA−1B
u(t) = 1
¾ = t - ¿
� �� � � �� �steady-statetransient
forced response (convolution integral + direct term)
Monday, November 3, 2014
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Lanari: CS - System response 7
Laplace transform
Final value theorem
limt→∞
f(t) = lims→0
sF (s)provided the limit on the left exists
if sF(s) is analytic (all the roots of the denominator of sF(s) in the open left half-plane)
example W (s) =1
s+ 1
u1(t) = δ−1(t) U1(s) =1
s
u2(t) = tδ−1(t) U2(s) =1
s2
u3(t) = sinωt U3(s) =ω
s2 + ω2sY3(s) = s
1
(s+ 1)
ω
(s2 + ω2)
sY2(s) = s1
(s+ 1)
1
s2
sY1(s) = s1
(s+ 1)
1
sok
no
no
Monday, November 3, 2014
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Lanari: CS - System response 8
Step response: steady-state
W (s) =N(s)�n
i=1(s− pi)
Ys(s) = W (s)U(s) =N(s)�n
i=1(s− pi)1
sYs(s) =
R0s
+n�
i=1
Ris− pi
S be asymptotically stable Re[pi] < 0
these termsdecay
steady-state
by the definition of residue R0
R0 = [sYs(s)]s=0 =
�sW (s)
1
s
�
s=0
= W (0)
or the application of the final value theorem
yss(t) = lims→0
sYs(s) = lims→0
sW (s)1
s= W (0) = R0 dc-gain
distinct poles
note that W (0) = D − CA−1B
ys(t) = R0δ−1(t) +n�
i=1
Riepitδ−1(t)
Monday, November 3, 2014
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Lanari: CS - System response 9
Step response: steady-state
The output of a linear asymptotically stable system to a unit step
input tends to a constant value given by the system’s gain F(0)
Since the output (total) response to a step input is given by
y(t) = yzi(t) + ys(t)
zero-inputresponse
stepresponse
and the zero-input response tends to zero for an asymptotically stable system, we can state that:
Monday, November 3, 2014
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Lanari: CS - System response 10
Step response: steady-state
0 2 4 6 8 10 122
1.5
1
0.5
0
0.5
1
1.5
2
2.5
time (sec)
Total output step response with different initial conditions
P (s) =3(s + 1)
(s + 0.5)(s + 2)(s + 3)
ySSZSR (x0 = 0)
0 1 2 3 4 5 6 7 8 9 102
1.5
1
0.5
0
0.5
1
1.5
2
time (sec)
Step response
P1(s) =2
(s + 1)(s + 2)
P2(s) =2s
(s + 1)(s + 2)
Input u(t) = 1(t)
step response y1step response y2
even with non-zero initial condition,due to the asymptotic stability of Sall the responses tend to the sameconstant value
the final constant value coincides with the system gain which can also be zero (due to the presence of a zero in s = 0)
steady-state is independentfrom the initial conditions
P2(0) = 0
Monday, November 3, 2014
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Lanari: CS - System response 11
Step response: transient
transient is defined as the forced response minus the steady-state
yss(t) = R0
yt(t) =n�
i=1
Riepit
distinct poles case
alternative
W (s) =n�
i=1
R�is− pi
w(t) =n�
i=1
R�iepit
ys(t) =
� t
0w(τ)dτ =
n�
i=1
R�ipi
�epit − 1
�=
n�
i=1
R�ipi
epit −n�
i=1
R�ipi
W (0)
ys(t) = R0δ−1(t) +n�
i=1
Riepitδ−1(t)
Monday, November 3, 2014
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Lanari: CS - System response 12
Transient behavior of a system
Defined on a particular time response: step response
t
y(t)
yss1
rise time
overshoot
settling time
peak time
0.1
0.9
Monday, November 3, 2014
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Lanari: CS - System response 13
Transient
tr rise time: amount of time required for the signal to go from
10% to 90% of its final value
yss steady-state value: asymptotic output value
Mp overshoot: maximum excess of the output w.r.t. the final value (can be
defined as a percentage of the final value). In a normalized y(t)/y ss plot the
overshoot is given by the maximum of the normalized output minus one.
tp peak time: time required for the step response to reach the overshoot
ts settling time: amount of time required for the step response to stay within
2% of its final value for all future times
Monday, November 3, 2014
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Lanari: CS - System response 14
Transient
−|α|
ω
Re
Im
ωnϑ
ζ = sinϑ
Quantities related to complex plane position of the poles
Monday, November 3, 2014
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Lanari: CS - System response 15
Transient
Re
Im
Quantities related to complex plane position of the poles (real poles)
0 1 2 3 4 5 6 7 8
0
0.5
1
1.5
t
p1 = −1
p2 = −5
Comparison between two systems
-1-5Re
Im
-1-5
0 1 2 3 4 5 6 7 8
0
0.5
1
1.5
t
p1 = −1
(p1,p
2) = (−1,−5)
� �� �
Single system
P (s) =p1p2
(s− p1)(s− p2)Pi(s) =
−pis− pi
we need to “wait” for the slower one
Monday, November 3, 2014
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Lanari: CS - System response 16
Transient
Re
Im
Quantities related to complex plane position of the poles (complex poles)
Comparison between four systems with pair of complex poles and unit gain
-1-5
1
5
0 1 2 3 4 5 6
0
0.5
1
1.5
t
A
B
C
D
P (s) =ω2n
s2 + 2ζωns+ ω2n
Monday, November 3, 2014
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Lanari: CS - System response 17
Steady state
Existence
we require
1. the existence of the steady-state also for the state (not only for the output)
2. and the independence, of the asymptotic behavior, from the initial conditions x(0)
Beingx(t) = eAtx(0) +
� t
0eA(t−τ)Bu(τ)dτ
x(t) will be independent for from the initial condition iff all the modes are
converging (all the eigenvalues have negative real part)
i.e. the system is asymptotically stable
t → ∞
Monday, November 3, 2014
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Lanari: CS - System response 18
Steady state
Exists for asymptotically stable systems
Depends on the particular input applied to the system
polynomial inputs polynomial steady state
sinusoidal inputs sinusoidal steady state
(canonical test signals)
Is independent from the initial state
Monday, November 3, 2014
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Lanari: CS - System response 19
Polynomial input
Let the canonical input (signal of order k) be u(t) =tk
k!δ−1(t) with transform U(s) =
1
sk+1
For an asymptotically stable system with distinct poles, the output is
Y (s) = P (s)U(s) =N(s)�n
i=1(s− pi)1
sk+1
and expandingthese give in t contributions that tend to 0 as t increases
Y (s) =R11s
+R12s2
+ · · ·+ R1,k+1sk+1
+n�
i=1
Ris− pi
Yss(s) =R11s
+R12s2
+ · · ·+ R1,k+1sk+1
yss(t) =
�R11 +R12t+ · · ·+R1,k+1
tk
k!
�δ−1(t)u(t) =
tk
k!δ−1(t)
steady-state
Monday, November 3, 2014
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Lanari: CS - System response 20
0 2 4 6 8 10 12 14 16 18 201.5
1
0.5
0
0.5
1
1.5
time (sec)
Zero State response and Input
P (s) =10
s + 10
InputZSR
0 1 2 3 4 5 6 7 8 9 102
1.5
1
0.5
0
0.5
1
1.5
2
2.5
3
time (sec)
Response to a ramp input
P1(s) =2
(s + 1)(s + 2)
P2(s) =2s
(s + 1)(s + 2)
Input u(t) = tOutput y1Output y2
0 1 2 3 4 5 6 7 8 9 108
6
4
2
0
2
4
6
8
10
12
time (sec)
Response to a ramp input: gain varies
P (s) =Kp
(s + 1)(0.5s + 1)
Input u(t) = t 1(t)
Kp = 1Kp = 2Kp = 0.5
Steady state
Monday, November 3, 2014
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Lanari: CS - System response 21
Sinusoidal input
u(t) = sin ω̄t =ejω̄t − e−jω̄t
2jU(s) = L{sin ω̄t} = ω̄
s2 + ω̄2=
ω̄
(s+ jω̄)(s− jω̄)
P (s) =N(s)�(s− pi)
Y (s) = P (s)U(s) =N �(s)�(s− pi)
+R1
s− jω̄ +R2
s+ jω̄
R1 = [(s− jω̄)Y (s)]s=jω̄ =�P (s)
ω̄
s+ jω̄
�
s=jω̄
=1
2jP (jω̄)
R2 = [(s+ jω̄)Y (s)]s=−jω̄ =
�P (s)
ω̄
s− jω̄
�
s=−jω̄= − 1
2jP (−jω̄) = R∗1
Input
System (asymptotically stable)
Output
with
Monday, November 3, 2014
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Lanari: CS - System response 22
Sinusoidal inputAsymptotic stability
L−1�
N �(s)�(s− pi)
�= L−1
��� Rik(s− pi)mi−k
�→ 0 when t → ∞
Asymptotic behavior (steady-state) is
yss(t) = L−1�
R1s− jω̄ +
R2s+ jω̄
�
but being P (jω̄) = |P (jω̄)|ej∠P (jω̄)
we have
|P (−jω̄)| = |P (jω̄)| ∠P (−jω̄) = −∠P (jω̄)with and
yss(t) = R1ejω̄t +R2e
−jω̄t
=1
2j
�P (jω̄)ejω̄t − P (−jω̄)e−jω̄t
�
=1
2j
�|P (jω̄)|ej∠P (jω̄)ejω̄t − |P (−jω̄)|e−j∠P (jω̄)e−jω̄t
�
=|P (jω̄)|
2j
�ej(ω̄t+∠P (jω̄)) − e−j(ω̄t+∠P (jω̄))
�
= |P (jω̄)| sin(ω̄t+ ∠P (jω̄))
Monday, November 3, 2014
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Lanari: CS - System response
• steady-state has same frequency than input
• can be amplified
attenuated
• also phase variation
• depends only on the frequency of the input and the system characteristics
23
Sinusoidal input
The steady-state response of an asymptotically stable system P(s)
to a sinusoidal input is given byu(t) = sin ω̄t
yss(t) = |P (jω̄)| sin(ω̄t+ ∠P (jω̄))
|P (jω̄)| > 1
|P (jω̄)| < 1
gain curve phase curve
|P (jω)| ∠P (jω)
Monday, November 3, 2014
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Lanari: CS - System response 24
Sinusoidal input
10 2 100 1020
0.2
0.4
0.6
0.8
1
frequency (rad/s)
Gain
Monday, November 3, 2014