1. No category

Analysis of Linear Continuous Systems
Paula Raica
Department of Automation
Dorobantilor Str., room C21, tel: 0264 - 401267
Baritiu Str., room C14, tel: 0264 - 202368
email: Paula.Raica@aut.utcluj.ro
http://rrg.utcluj.ro/st
Technical University of Cluj-Napoca
Analysis of Linear Continuous Systems
System Analysis
First step: derive a mathematical model
Various methods are available for analysis
Performance is analyzed based on various test signals
The aim of analysis: study the system behavior in transient
and steady-state when the system model and input are known.
Typical test signals: step, ramp, impulse, sinusoidal
t
transient state
steady-state
Analysis of Linear Continuous Systems
System Analysis
Predict the dynamic behavior of the system from a knowledge
of the system model.
Important characteristics: absolute stability, transient
response, steady-state error
A stable LTI system: the output comes back to its equilibrium
state (system is subjected to a disturbance)
An unstable LTI system: either sustained oscillation of the
output or the output diverges from equilibrium (system is
subjected to a disturbance)
If the output of a system at steady state does not exactly agree
with the input, the system is said to have steady-state error
Analysis of Linear Continuous Systems
System Analysis
The system is broken down in simple elements of at most
second order and the effects of each element are analyzed
The behavior of simple elements can be studied using some
characteristic parameters:
Time constants, T
Time delay constant, Tm
Damping factor ζ
Natural frequency ωn
Gain constant, K
Analysis of Linear Continuous Systems
First-order systems
The input-output relationship is given by:
C (s)
K
=
R(s)
Ts + 1
K - the gain constant
T - the time constant
We will analyze the system responses to inputs as the unit step,
unit ramp and unit impulse functions. The initial conditions are
assumed to be zero.
Analysis of Linear Continuous Systems
Unit-step response
1
K 1
r (t) = 1, R(s) = , C (s) =
s
Ts + 1 s
K
KT
c(t) = L−1 [C (s)] = L−1
−
= K (1−e −t/T ),
s
Ts + 1
(t ≥ 0)
Property: At t = T the value of c(t) is 0.632K , or the response
has reached 63.2% of its total change:
c(T ) = K (1 − e −1 ) = 0.632K
Property: The slope of the tangent at t = 0 is 1/T :
dc(t)
K
K
= e −t/T |t=0 =
dt
T
T
Analysis of Linear Continuous Systems
Unit-step response
For t ≥ 4T the response remains within 2% of the final value.
The response time is about 4 time constants.
Analysis of Linear Continuous Systems
Unit-ramp response
r (t) = t, R(s) =
1
,
s2
C (s) =
1
K
Ts + 1 s 2
Expanding C (s) into partial fraction gives
1
T
T2
−1
−1
c(t) = L [C (s)] = L
K
− +
s2
s
Ts + 1
= K (t − T + Te −t/T ),
(t ≥ 0)
For K = 1, the error signal e(t):
e(t) = r (t) − c(t) = T (1 − e −t/T ),
e(∞) = T
Analysis of Linear Continuous Systems
Unit-ramp response
r(t) c(t)
steady-state
error
T
T
r(t)=t
c(t)
t
0
2T
4T
6T
Analysis of Linear Continuous Systems
Unit-impulse response
r (t) = δ(t), R(s) = 1, C (s) =
K
K
, c(t) = e −t/T , (t ≥ 0)
Ts + 1
T
Analysis of Linear Continuous Systems
Second-order systems
R(s)
H(s) =
C (s)
=
R(s)
1
1 2
s
ωn2
C(s)
H(s)
+
2ζ
ωn s
+1
=
ωn2
s 2 + 2ζωn s + ωn2
ωn - the natural frequency, ζ - the damping factor
Example.
H(s) =
1
1
,
= 1;
s 2 + s + 1 ωn2
2ζ
= 1; ⇒ ωn = 1;
ωn
ζ=
Analysis of Linear Continuous Systems
1
2
Second-order systems
The roots of the characteristic equation (System poles)
s 2 + 2ζωn s + ωn2 = 0
are:
s1,2 = −ζωn ± ωn
p
ζ2 − 1
Poles are complex for 0 < ζ < 1 and real for ζ ≥ 1.
0 < ζ < 1 : the system is underdamped and the transient
response is oscillatory.
ζ = 1 : the system is critically damped
ζ > 1 : system is overdamped (response does not oscillate)
ζ = 0: the transient response does not die out.
Analysis of Linear Continuous Systems
Step-response
r (t) = 1, R(s) =
ωn2
1
, C (s) =
s
s(s 2 + 2ζωn2 s + ωn2 )
Underdamped case (0 < ζ < 1)
1
s + ζωn
ζωn
−
−
2
2
s
(s + ζωn ) + ωd
(s + ζωn )2 + ωd2
p
where ωd = ωn 1 − ζ 2 - damped natural frequency.
C (s) =
L
−1
e −ζωn t
[C (s)] = c(t) = 1 − p
· sin ωd t + arctan
1 − ζ2
p
1 − ζ2
ζ
Analysis of Linear Continuous Systems
!
Step-response
ζ = 0: the response becomes undamped and oscillations continue
indefinitely.
c(t) = 1 − cos ωn t, (t ≥ 0)
ωn = the undamped natural frequency of the system = frequency
at which the system would oscillate if the damping were decreased
to zero.
Critically damped case, (ζ = 1): poles are equal and real
C (s) =
ωn2
, ⇒, c(t) = 1 − e −ωn t (1 − ωn t), (t ≥ 0)
(s + ωn )2 s
Analysis of Linear Continuous Systems
Step-response
Overdamped case, (ζ > 1): poles are negative real
C (s) =
ωn2
p
ζ 2 − 1)(s + ζωn − ωn ζ 2 − 1)
−s1 t
ωn
e
e −s2 t
c(t) = 1 + p
−
s1
s2
2 ζ2 − 1
p
p
where s1 = (ζ + ζ 2 − 1)ωn and s2 = (ζ − ζ 2 − 1)ωn are the
system poles.
The response c(t) includes two decaying exponential terms.
s(s + ζωn + ωn
p
Analysis of Linear Continuous Systems
Step-response
c(t)
0.1
0.2
0.5
0.7
1
2
t
0
Analysis of Linear Continuous Systems
Transient-response specifications
Rise time, Peak time, Maximum overshoot, Settling time
c(t)
Allowable
tollerance
Mp
±0.05
or
±0.02
1
0.9
0.1
0
t
tr1
tr
tp
ts
Analysis of Linear Continuous Systems
Transient-response specifications
Rise time, tr - the time required for the response to rise from
10% to 90%, 5% to 95% or 0% to 100% of its final value.
Peak time, tp - the time required for the response to reach the
first peak of the overshoot.
Maximum (percent) overshoot Mp - the maximum peak value
of the response curve measured from the steady-state value of
the response
Settling time, ts - the time required for the response curve to
reach and stay within a range of 2% or 5% about the final
value.
Analysis of Linear Continuous Systems
Rise time
Obtain the rise time tr by letting c(tr ) = 1 or
c(tr ) = 1 = 1 − e −ζωn tr (cosωd tt + p
cosωd t + p
ζ
1 − ζ 2 sinωd tr
tr =
ζ
1 − ζ 2 sinωd tr
= 0, tanωd tr = − p
1
· arctan
ωd
β and σ are defined in next Figure.
ωd
−σ
=
ζ
1 − ζ2
)
=−
π−β
ωd
Analysis of Linear Continuous Systems
ωd
σ
Poles of 2nd order system
Important picture !!!
jw
jwd
wn
wd = w n 1 - z 2
b
s
zwn
- jwd
-s
Analysis of Linear Continuous Systems
Peak time
Obtain the peak time by differentiating c(t) with respect to time
and letting the derivative equal zero:
dc(t)
ωn
|t=tp = sin(ωd tp ) · p
· e −ζωn tp = 0
2
dt
1−ζ
c(t)
Allowable
tollerance
Mp
sin(ωd tp ) = 0
ωd tp = 0, π, 2π, 3π, ...,
π
tp =
ωd
±0.05
or
±0.02
1
0.9
0.1
0
t
tr1
tr
tp
ts
Analysis of Linear Continuous Systems
Percent overshoot
Mp occurs at the peak time or at t = tp =
π
ωd .
c(t)
Allowable
tollerance
Mp
±0.05
or
±0.02
1
0.9
0.1
0
Mp = c(tp )−1 = −e
t
tr1
tr
tp
ts
−ζωn π/ωd
cosπ + p
ζ
1 − ζ2
sinπ
!
=e
Analysis of Linear Continuous Systems
−π √
ζ
1−ζ 2
Settling time
c(t) = 1 − e −ζωn t · sin ωd t + arctan
envelope curves:
1 − ζ2
ζ
!
c(t)
p
c1,2 (t) = ±e −ζωn t / 1 − ζ 2
c1(t)
c1 , c2 , c(t) will reach 2% from the
final value approximately when
e −ζωn ts < 0.02, or ζωn ts ∼
=4
ts =
p
4
ζωn
c2(t)
0
Analysis of Linear Continuous Systems
t
Example
Consider the closed-loop system with the transfer function:
H(s) =
25
ωn2
=
s 2 + 6s + 25
s 2 + 2ζωn + ωn2
Calculate:
ωn = 5,
ζ = 0.6, ωd = ωn
The poles negative real part:
p
p
1 − ζ 2 = 5 1 − 0.62 = 4
σ = −ζωn = −3.
Analysis of Linear Continuous Systems
Example
ωd
= 0.93
σ
π−β
3.14 − 0.93
tr =
=
= 0.55sec
ωd
4
β = arctan
jw
jwd
wn
wd = w n 1 - z 2
π
3.14
= 0.78sec
=
ωd
4
√
2
Mp = e −πζ/ 1−ζ = 0.095
tp =
b
s
zwn
- jwd
-s
Mp (%) = 9.5%
ts =
4
4
4
=
= = 1.33sec
−σ
ζωn
3
Analysis of Linear Continuous Systems
Example
The step response of the system. The values of the system
parameters can be seen from the plot.
1.2
1
0.8
0.6
0.4
0.2
0
0
0.5
1
Time (sec.)
1.5
Analysis of Linear Continuous Systems
Note
Consider a 1st or 2nd order system with the transfer function
Hk (s) =
k
ωn2
= k·H(s), or Hk (s) = k· 2
= k·H(s)
Ts + 1
s + 2ζωn + ωn2
The system response when the input is a unit step R(s) = 1/s is:
c(t) = L−1 [Hk (s) · R(s)] = L−1 [
k · H(s)
H(s)
] = k · L−1 [
]
s
s
The time response is k · c(t) where c(t) is the response of the
system with a unity gain.
Analysis of Linear Continuous Systems
Steady-state error
The steady-state error = the error between the input (r (t)) and
the output (c(t)) at steady-state.
ess = lim (r (t) − c(t))
t→∞
The open-loop system error :
R(s)
C(s)
G(s)
E (s) = R(s) − C (s) = (1 − G (s))R(s)
Analysis of Linear Continuous Systems
Steady-state error
For a unity feedback closed-loop system, the error is:
E (s) = R(s) − C (s) = R(s) − R(s)
1
G (s)
=
R(s)
1 + G (s)
1 + G (s)
Final value theorem:
ess = lim e(t) = lim sE (s)
t→∞
s→0
Analysis of Linear Continuous Systems
Steady-state error
For a unit step input
Open-loop system:
1
ess = lim s(1 − G (s))( ) = lim (1 − G (s)) = 1 − G (0)
s→0
s→0
s
Unity feedback closed-loop system:
1
1
1
ess = lim s
=
s→0
1 + G (s)
s
1 + G (0)
Analysis of Linear Continuous Systems
Steady-state error - Example
Consider a first-order system:
G (s) =
k
Ts + 1
Input: R(s) = 1/s. The steady-state error of the open-loop
system:
ess = 1 − G (0) = 1 − k
If G (s) is the open-loop transfer function for a unity feedback
closed-loop system:
ess =
1
1
=
1 + G (0)
1+k
Analysis of Linear Continuous Systems