B1
TRANSIENT RESPONSE ANALYSIS
Test signals:
•
•
•
•
Impulse
Step
Ramp
Sin and/or cos
Transient Response:
for t between 0 and T
Steady-state Response:
for t
:\
System Characteristics:
• Stability
• Relative stability
• Steady-state error
Æ
Æ
Æ
transient
transient
steady-state
B2
First order systems
C (s)
1
=
R ( s ) Ts + 1
R(s)
E(s)
C(s)
1
Ts
+
-
Unit step response:
C (s) =
1
1
⋅
Ts + 1 s
c (t ) = 1 − e
1
T
−
s sT + 1
=
− t T
t •
−t T
e(t ) = r(t) − c(t ) = e
e(’
c(T) = 1 – e-1 = 0.632
dc ( t )
dt
t= 0
=
1
T
e
− t T
t= 0
=
1
T
B3
Unit ramp response
T2
1
1 1 T
⋅ 2= 2− +
C(s) =
Ts +1 s s
s Ts + 1
c(t ) = t − T + Te
−t
T
t≥0
−t
e(t ) = r (t ) − c (t ) = T 1 − e T 


t_
e(’) =T
Unit-ramp response of the system
B4
Impulse response:
r(t) = /(t)
R(s) = 1
C (s) =
c (t ) =
1
sT + 1
e
− t / T
t_
T
4T
Unit-impulse response of the system
Output
Input
Ramp
Step
r(t) = t
r(t) = 1
Impulse r(t) = /(t)
t•
t•
−t T
c(t ) = t − T + Te
c (t ) = 1 − e
e −t /T
c (t ) =
T
− t T
t_
t_
t_
B5
Observation:
Response to the derivative of an input equals to derivative of
the response to the original signal.
Y(s) = G(s) U(s)
U(s): input
U1(s) = s U(s) Y1(s) = s Y(s)
Y(s): output
G(s) U1(s) = G(s) s U(s) = s Y(s) = Y1(s)
How can we recognize if a system is 1st order ?
System
r(t)
c(t)
r(t) = step
Plot log |c(t) – c(’_
If the plot is linear,
then the system is 1st order
Explanation:
c (t ) = 1 − e
− t T
log |c(t) – c(’_
c(\
ORJ _H-t/T |=
t
T
B6
Second Order Systems
Block Diagram
F
+
K
J
-
R(s)
+
-
K
s (Js + F
Position signal C (s)
)
Transfer function:
C(s)
K
= 2
R(s) Js + Fs + K
K
J
=
2
2
 F
F  K  F
F  K


s + +   −  s + −   − 
 2J
 2J  J   2J
 2J  J 

B7
Substitute in the transfer function:
n²
K
J
F
J
= 2 n
ζ =
n:
-
F
2 JK
damping ratio
undamped natural frequency
stability ratio
to obtain
ωn
C ( s)
= 2
R ( s ) s + 2ζω n s + ω n 2
2
• Underdamped FDVH
F² - 4 J K < 0
two complex conjugate poles
• Critically damped FDVH F² - 4 J K = 0
two equal real poles
• Overdamped FDVH !
F² - 4 J K > 0
two real poles
B8
Under damped case (0 < < 1):
ωn
C (s)
=
R ( s ) (s + ζω n + jω d )(s + ζω n − jω d )
2
Im
ωn 1 − ζ 2
n
jd
Re
n
ζ =
σ
= cos β
ωn
ωd = ωn 1 − ζ 2
ωn :
ωd :
ζ:
undamped natural frequency
damped natural frequency
damping ratio
B9
Unit step response:
R(s) = 1/s
ζωn
s + ζωn
1
−
C(s) = −
s (s + ζωn )2 + ωd 2 (s + ζωn )2 + ωd 2
c(t ) = 1 − e
−ζω n t


ζ
 cos ω t +
sin ωd t 
d


1−ζ 2


t ≥0
o
r
c (t ) = 1 −
β =
1 − ζω n t
e
sin (ω n β t + θ
β
1−ζ
θ
2
e(t) = r(t) - c(t) = e
−ζω n t
= tan
)
−1
t ≥0
β
ζ



 cosω t + ζ
sin ωd t
d
2


1− ζ


2.0
1.0
0
1
2
3
4
5
6
7
8
9
10
11 12
Unit step response curves of a second order system
t≥0
B10
Unit step response:
c(t ) = 1 − cos ω n t
t≥0
Unit step Response:
R(s) = 1/s
ωn
ωn
C ( s)
= 2
=
R ( s ) s + 2ζω n + ω n 2 ( s + ω n ) 2
2
2
1
C(s) = s (s + ω )2
n
c(t ) = 1 − e −ωn t (1 + ωnt )
t≥0
B11
Unit step Response:
C(s) =
R(s) = 1/s
ωn
2
1
s + ζωn + ωn ζ 2 −1 s + ζωn −ωn ζ 2 −1 s
(
)(
)
 e−s1t e−s2t 


−
c(t ) = 1 +
2
s
s
2 ζ −1  1
2 
ωn
with
(
s1 = ζ +
(
ζ
2
)
−1 ω
)
⋅
t ≥0
n
s2 = ζ − ζ 2 − 1 ω n
if |s2| << |s1|, the transfer function can be approximated by
s2
C (s)
=
R(s)
s + s2
and
for R(s) = 1/s
c (t ) = 1 − e − s 2 t
with
(
t≥0
)
s2 = ζ − ζ 2 −1 ωn
B12
Unit step response curves of a critically damped system.
B13
Transient Response Specifications
Unit step response of a 2nd order underdamped system:
Allowable tolerance
0.05 or
0.02
~5
0
0
td delay time: time to reach 50% of c(\ Ior the first time.
tr rise time :
time to rise from 0 to 100% of c(’
tp peak time : time required to reach the first peak.
Mp maximum overshoot :
ts settling time :
c (t p ) − c (∞ )
c (∞)
⋅ 100%
time to reach and stay within a 2% (or
5%) tolerance of the final value c(’
0.4 < ζ < 0.8
Gives a good step response for an underdamped system
B14
Rise time tr
time from 0 to 100% of c(’
ζ
c(tr ) =1 ⇒ 1− e−ζ ωdtr (cosωd tr +
cosω d t r +
1−ζ
ζ
1−ζ
2
2
sinωd tr ) =1
sinω d t r = 0
ωd
1−ζ 2
t
ω
=
−
=
−
tan d r
ζ
σ
tr =
Peak time tp:
dc ( t )
dt
t=t p
1
tan
ωd
−1
ωd 


 σ 
time to reach the first peak of c(t)
= 0
⇒ (sin ωd t p )
sin ω d t p = 0
tp =
π
ωd
ωn
1−ζ
2
e
− ζω n t p
=0
B15
Maximum overshoot Mp:
t = tp =
π
ωd
(
)
M p = c(t p ) = 1− e−ζωn π / ωd (cosπ +
−
=e
ζωnπ
ωd
−ζπ
= e
1−ζ 2
= e
ζ
1− ζ
2
sinπ )
−σπ
ωd
Settling time ts:
2

−1 1 − ζ

sin ωd t + tan
c(t ) = 1 −
ζ
1 − ζ 2 
e−ζωnt
approximate ts using envelope curves:




env ( t ) = 1 ±
e − ζω n t
1− ζ
Pair of envelope curves for the unit-step response curve
2% band:
ts =
4
4
=
σ ζωn
3
3
5% band t s = σ = ζω
n
2
B16
Settling time ts versus ζ curves {T = 1/(ζωn) }
6T
5T
4T
3T
2T
T
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
B17
Impulse response of second-order systems
C ( s) =
ωn 2
s 2 + 2ζω n s + ω n 2
R(s) = 1
underdamped case (0< ζ < 1):
c(t ) =
ωn
1−ζ
2
e−ζω n t sin ωn 1 − ζ 2 t
the first peak occurs at
t = t0
tan
t0 =
t≥0
ω
−1
n
1− ζ
ζ
1− ζ
2
2
and the maximum peak is
2 

−
ζ
ζ
1
−
1

c(t0 ) = ωn exp  −
tan
 1− ζ 2
ζ 

B18
critically damped case ( ζ = 1):
c (t ) = ω n te − ω n t
2
t≥0
overdamped case ( ζ >1):
c (t ) =
ωn
2 ζ
2
−1
e − s1t −
ωn
2 ζ
2
−1
e − s2t
where
(
= (ζ +
s1 = ζ − ζ
2
ζ
2
s2
)
− 1 )ω
−1 ωn
n
Unit-impulse response for 2nd order systems
t≥0
B19
Remark:
Impulse Response = d/dt (Step Response)
Unit-impulse response
Relationship between tp, Mp and the unit-impulse response curve of a system
Unit ramp response of a second order system
C (s) =
ωn
2
s + 2ζω n + ω n
2
2
⋅
1
s2
R(s) = 1/ s2
for an underdamped system (0 < ζ < 1)


2ζ
2ζ 2 − 1
−ζω nt  2ζ

c (t ) = t −
+e
cos ω d t +
sin ω d t
2
 ωn

ωn
ωn 1 − ζ


and the error:
e(t) = r(t) – c(t) = t – c(t)
at steady-state:
e (∞ ) = lim e (t ) =
t→∞
2ζ
ωn
t≥0
B20
Examples:
a. Proportional Control
R(s)
+
C(s)
1
s (Js + F )
E(s)
K
-
ωn
C ( s)
= 2
R ( s ) s + 2ζω n + ω n 2
2
with
K
J
n²
F
J
= n ζ =
F
2 JK
Choose K to obtain ‘good’ performance for the closed-loop system
For good transient response:
0.4 < ζ <0.8
n
sufficiently large
Æ
Æ
acceptable overshoot
good settling time
For small stead- state error in ramp response:
e (∞ ) = lim e (t ) =
t →∞
2ζ
2F
=
⋅
ωn
2 Kζ
ζ
F
=
K
K
Æ
large K
Large K reduces e( ∞ ) but also leads to small ζ and large Mp
Æcompromise necessary
B21
b. Proportional plus derivative control:
C(s)
E(s) Kp+K/s
R(s)
1
s (Js + F )
+
-
K p + Kd s
C(s )
= 2
R(s ) Js + (F + Kd )s + K p
with
ς =
F + K
2
K
p
d
ω
J
K
=
n
p
J
The error for a ramp response is:
s 2 J + sF
⋅ R (s )
E (s ) = 2
s J + s (F + K d ) + K p
and at steady-state:
e (∞ ) = lim sE (s ) =
s→0
using
z =
K
p
K
d
F
Kp
C (s ) ω n
s+ z
=
⋅ 2
2
R (s )
z
s + 2ζω n s + ω n
2
Choose Kp, Kd to obtain ‘good’ performance of the closed-loop system
For small steady-state error in ramp response
For good transient response
ÆK
d
Æ
Kp large
so that 0.4 < ζ <0.8
B22
c. Servo mechanism with velocity feedback
s Θ(s)
R(s) +
K
Js + F
+
-
-
Θ(s)
1
s
Kh
Transfer function
Θ(s)
K
= 2
R ( s ) Js + (F + KK h )s + K
where
ς =
F + KK
KJ
2
K
J
ωn =
e(∞ ) =
h
F
K
(not affected by velocity feedback)
for a ramp
Choose K, Kh to obtain ‘good’ performance for the closed-loop system
For small steady-state error in ramp response
For good transient response
Remark:
ÆK
h
Æ
K large
so that 0.4 < ζ <0.8
The damping ratio ζ can be increased without
affecting the natural frequency ωn in this case.
B23
Effect of a zero in the step response of a 2nd order system
C (s ) ω n
s+z
=
⋅ 2
R (s )
z s + 2ζω n s + ω n 2
2
ζ = 0 .5
Unit-step response curves of 2nd order systems
B24
Unit step Response of 3rd order systems
C(s )
ωn p
= 2
R(s ) s + 2ζωn s + ωn2 (s + p)
2
(
)
0<ζ<1
e− pt
e−ξωnt
−
•
c(t ) = 1 − 2
βζ (β − 2) +1 βζ 2 (β − 2) +1
[
]
(
R(s)=1/s
)


βζ ζ 2 (β − 2) +1
2
2
2
−
βζ
(
β
)
ζ
ω
ζ
ω
−
−
+
t
t
sin
1
2
cos
1

n
n 


1− ζ 2
p
β =
where
ζω n
Unit-step response curves of the third-order system, =
The effect of the pole at s = -p is:
• Reducing the maximum overshoot
• Increasing settling time
0.5
B25
Transient response of higher-order systems
C ( s ) b0 s m + ... + bm −1 s + bm K (s + z1 )...(s + z m )
=
= n
(s + p1 )... (s + p n )
R (s)
s + ... + d n −1 s + a n
n>m
Unit step response
m
C (s) =
K ∑ (s + zi )
i=1
∑ (s + p )∑ (s
q
r
j
j =1
0 < ζk <1
2
+ 2ζ kωk s + ωk
k =1
k=1,…,r
and
2
)
⋅
1
s
q + 2r = n
2
r
bk (s +ζ kωk ) + ckωk 1−ζ k
a q aj
C(s) = + ∑
+∑
2
s j=1 s + pj k=1
s2 + 2ζ kωk +ωk
q
c(t ) = a + ∑ a j e
j =1
r
− p jt
r
2
+ ∑ bk e −ζ nω k t cos ωk 1 − ζ k t 


k =1
2
+ ∑ ck e−ζ k ω k t sin ωk 1 − ζ k t 


k =1
t≥0
Dominant poles: the poles closest to the imaginary axis.
B26
STABILITY ANALYSIS
m
G (s) =
B (s)
=
A(s)
∑bs
m −i
i
i=0
n
∑a
i
s n −i
i=0
Conditions for Stability:
A.
Necessary condition for stability:
All coefficients of A(s) have the same sign.
B.
Necessary and sufficient condition for stability:
A( s ) ≠ 0
for
Re[s] _ or, equivalently
All poles of G(s) in the left-half-plane (LHP)
Relative stability:
The system is stable and further, all the poles of the system
are located in a sub-area of the left-half-plane (LHP).
Im
Re
Re
B27
Necessary condition for stability:
A(s ) = a 0 s n + a1s n −1 + ... + a n −1 s + a n
= a 0 (s + p1 )(s + p 2 )...(s + p n )
= a 0 s n + a 0 ( p1 + p 2 + ... p n )s n −1
+ a 0 ( p1 p 2 + ... + p n −1 p n )s n − 2
+ a 0 ( p1 p 2 ... p n )
-p1 to -pn are the poles of the system.
If the system is stable
Æ all poles have negative real parts
Æ the coefficients of a stable polynomial have the same sign.
Examples:
A(s ) = s 3 + s 2 + s + 1
can be stable or unstable
A(s) = s3 −s2 +s +1
is unstable
Stability testing
Test whether all poles of G(s) (roots of A(s)) have
negative real parts.
Find all roots of A(s)
Easier Stability test?
Æ too many computations
B28
Routh-Hurwitz Stability Test
A(s) = α0sn + α1sn-1+… + αn-1 s +αn
sn
sn-1
sn-1
s2
s1
s0
α0
α1
b1
c1
……
e1
f1
g1
α2 α4
α3 α5
b2 b3
c2
…
…
e2
1 a0 a2 a1a2 − a0 a3
=
b1 =
− a1 a1 a3
a1
b2 =
1 a0 a4 a1a4 − a0 a5
=
a
a
− a1 1 5
a1
c1 =
1 a1 a3 a3b1 − a1b2
=
− b1 b1 b2
b1
etc
.
Properties of the Ruth-Hurwitz table:
1. Polynomial A(s) is stable (i.e. all roots of A(s) have
negative real parts) if there is no sign change in the first
column.
2. The number of sign changes in the first column is equal
to the number of roots of A(s) with positive real parts.
B29
Examples:
A(s) = a0s2 + α1 s + α2
s2
s1
a0
a1
s0
a2
α0 >0,
α0 <0,
a2
α1 > 0 ,
α1 < 0 ,
α2 > 0 or
α2 < 0
For 2nd order systems, the condition that all coefficients of
A(s) have the same sign is necessary and sufficient for
stability.
A(s) = α0 s3 + α1s2 + α2s +α3
s3
s2
s1
s0
a0
a1
a1 a 2 − a 0 a 3
a1
a3
α0 >0,
α1>0,
a2
a3
α3>0,
α1α2 − α0α3 > 0
(or all first column entries are negative)
B30
Special cases:
1. The properties of the table do not change when all the
coefficients of a row are multiplied by the same positive
number.
2. If the first-column term becomes zero, replace 0 by and
continue.
• If the signs above and below are the same, then
there is a pair of (complex) imaginary roots.
• If there is a sign change, then there are roots with
positive real parts.
Examples:
A(s) = s3 +2s2 + s +2
s3
s2
s1
s0
1
2
1
2
Æ pair of imaginary roots
0→ε
2
( s= ±j )
A(s) = s3- 3s +2 = (s-1)2(s+2)
s3
s2
s1
s0
1
0≈ε
2
−3−
ε
2
−3
2
Æ two roots with positive real parts
B31
3. If all coefficients in a line become 0, then A(s) has roots
of equal magnitude radially opposed on the real or
imaginary axis. Such roots can be obtained from the roots
of the auxiliary polynomial.
Example:
A(s)= s5 + 2s4 +24 s3 + 48s2 -25s -50
s5
s4
s3
1 24 − 25
2 48 − 50
0 0
Æ
auxiliary polynomial p(s)
p(s) = 2s4 + 48s2 - 50
dp( s)
= 8s 3 + 96 s
ds
8
96
s3
24 − 50
s2
0
s1 112.7
s 0 − 50
• A(s) has two radially opposed root pairs (+1,-1) and
(+5j,-5j) which can be obtained from the roots of p(s).
• One sign change indicates A(s) has one root with
positive real part.
Note:
A(s) = (s+1) (s-1)(s+5j)(s-5j)(s+2)
p(s) = 2(s2-1) (s2 +25)
B32
Relative stability
Question:
Have all the roots of A(s) a distance of at least
IURP WKH LPDJLQDU\ D[LV"
Im
Substitute s with
s = z - in A(s)
and apply the
Routh-Hurwitz
test to A(z)
Re
Closed-loop System Stability Analysis
C(s)
R(s)
K
G (s )
+
-
Question:
For what value of K is the closed-loop system
stable?
Apply the Routh-Hurwitz test to the denominator polynomial
of the closed-loop transfer function
KG ( s )
1 + KG ( s )
.
B33
Steady-State Error Analysis
R(s)
E(s)
C(s)
+
G(s)
-
H(s)
Evaluate the steady-state performance of the closed-loop
system using the steady-state error ess
ess = lim e(t ) = lim sE ( s )
t →∞
E (s) =
s →0
1
⋅ R ( s)
1 + G ( s) H (s )
for the following input signals:
Unit step input
Unit ramp input
Unit parabolic input
Assumption: the closed-loop system is stable
Question: How can we obtain the steady-state error ess of the
closed-loop system from the open-loop transfer
function G(s)H(s) ?
B34
Classification of systems:
For an open-loop transfer function
G (s) H (s) =
K (Ta s + 1)(Tb s + 1)
s N (T1s + 1)(T2 s + 1)
Type of system: Number of poles at the origin, i.e., N
Static Error Constants:
Kp, Kv, Ka
Open-loop transfer function:
G(s)H(s)
Closed-loop transfer function:
G tot ( s ) =
G (s)
1 + G (s)H (s)
Static Position Error Constant: Kp
Unit step input to the closed-loop system shown in fig, p.
B33.
1
1 + G ( 0) H ( 0)
R(s) = 1/s
ess = lim sE ( s ) =
Define:
K p = lim G ( s ) H ( s ) = G (0) H (0)
s→ 0
s→0
Type 0 system
Kp = K
Type 1 and higher
Kp= \
1
1+ Kp
ess = 0
ess =
B35
Static Velocity Error Constant: Kv
Unit ramp input to the closed-loop system shown if fig, p. B33.
R(s) = 1/s2
s
1
1
⋅ 2 = lim
s→0 1 + G ( s ) H ( s ) s
s → 0 sG ( s ) H ( s )
ess = lim
Define:
K v = lim sG ( s ) H ( s )
Type 0 system
Kv = 0
ess = \
Type 1 system
Kv = K
ess = 1/Kv
Type 2 and higher
Kv = \
ess = 0
s→0
Static Acceleration Error Constant:
Ka
Unit parabolic input to the closed-loop system shown in fig, p. B33
3
R(s) = 1/s
s
1
1
⋅ 3 = lim 2
s→0 1 + G ( s) H ( s) s
s→0 s G ( s) H ( s)
ess = lim
Define:
K a = lim s 2 H ( s ) G ( s )
Type 0 system
Ka = 0
Type 1 system
Ka = 0
ess = \
Type 2 system
Ka = K
ess = 1/ Ka
Type 3 and higher
Ka = \
ess = 0
s→ 0
ess = \
B36
Summary:
Consider a closed-loop system:
R(s)
E(s)
C(s)
+
G(s)
-
H(s)
with an open-loop transfer function:
G (s )H (s ) =
K (T a s + 1 ) ⋅ (T b s + 1 )...
s N (T1 s + 1 )⋅ (T 2 s + 1 )...
and static error constants defined as:
K p = lim G ( s ) H ( s ) = G (0) H (0)
s→0
K v = lim sG ( s ) H ( s )
s→0
K a = lim s 2 H ( s ) G ( s )
s→ 0
The steady-state error ess is given by:
Unit step
r(t) = 1
Type 0
ess =
1
1
(=
)
1+ K p
1+ K
Type 1
ess= 0
Type 2
ess= 0
Unit ramp
r(t) = t
ess= \
ess =
1
1
(= )
Kv
K
ess= 0
Unit
parabolic
r(t) = t2/2
ess= \
ess= \
ess =
1
1
(= )
Ka
K
B37
Correlation between the Integral of error in
step response and Steady-state error in
ramp response
R(s)
+
E(s)
-
E(s) = L[e(t)] =
∞
− st
substitute
lim
s→ 0
R(s)
1 + G(s)
∞
∫
0
1
=
KV
s→ 0 0
E (s) =
e(t )dt =
∞
=
∞
∫ e− st e(t)dt
0
lim E(s) = lim ∫ e
s→0
C(s)
G(s)
e(t)dt =
∞
∫ e(t)dt
0
R( s )
1 + G(s)
in the above eq.
∫ e(t) dt
step : R(s) =
0

1
1
1
⋅ =
=

s ⋅ G ( s)
1 + G ( s) s 

s →0
s →0
lim
lim
1
K
Steady-state error in unit-ramp input = essr
∞
essr = ∫
0
e(t)dt
v
1
s
B38
∞
c(t)
∫ e(t )dt
0
t
c(t)
r(t)
r(t)
essr
c(t)
0
t