from my lecturer :3 control engineering (time response)

1. 3/28/2014
1
KCEC 2117
Control Engineering
Analyses of Transient and Steady State
Time response
Dr. Yap Hwa Jen
What is next
 After modeling, we look at the response of the system
based on the input
 Some fundamental properties of systems
 1st order
 2nd order
2

2. 3/28/2014
2
Transient and steady-state response
y(t) = ytr (t)
transient
+ yss (t)
steady state
3
Steady-state response
 The final value of the system, should the system is stable
 Can also be found by FinalValue Theorem
 When is the FVT applicable?
 F(s) should have no poles in the right half of the complex plane (Real part should not be +v).
 F(s) should have no poles on the imaginary axis, except at most one pole at s=0.
 E.g.
lim
t®¥
f (t) = lim
s®0
sF(s);
only if lim
t®¥
f (t) is finite
4
no)(
yes)(
2


s
A
sF
s
A
sF

3. 3/28/2014
3
Steady-state response
 Another use of FVT is to calculate DC gain
 DC gain is the ratio of output to input after all transients
have decayed
 Assume input is unit step
)(lim;
1
)(lim=gainDC
00
sF
s
ssF
ss 

5
1st order systems
 Example:
 RC circuit
 Thermal systems
 Assume zic
C(s)
R(s)
=
1
Ts+1
6

4. 3/28/2014
4
1st order systems: unit step response
Tt
etc
sTs
sC
TssR
sC
/
1)(
1
1
1
)(
1
1
)(
)(








7
T
e
Tdt
dc
t
Tt
t
11
:0)(tSlope
0
/
0 




1st order systems: unit ramp response
Tt
TeTttc
sTs
sC
/
2
)()(
1
1
1
)(





8
 
     
 
  Te
eTte
tctrte
te
Tt




and
)1(
thenis,signal,errorThe
/

5. 3/28/2014
5
1st order systems: unit impulse response
C(s) =
1
Ts +1
1
c(t) =
1
T
e-t/T
9
Specifications
 Rise time, tr (10%-90%)
 Settling time, ts (2% or 5%)
 Time constant,T or t
10

6. 3/28/2014
6
Example
11
Characteristics of First Order Systems
Example: Obtain the
transfer function of the
system shown in Figure
(s)/T(s) and find its
time constant and the
final value under unit
step input, J=0.01 Kgm2
and B=0.04
Characteristics of First Order Systems
Solution
1250
25
4
100
040010
1
1
ss
s.
BJssT
s
.
.
)(
)(
The time constant =0.25sec
The final value= 25rad/sec
0 0.5 1 1.5
0
5
10
15
20
25
StepResponse
Time(sec)
Amplitude
11
Example
11
Characteristics of First Order Systems
Example: Obtain the
transfer function of the
system shown in Figure
(s)/T(s) and find its
time constant and the
final value under unit
step input, J=0.01 Kgm2
and B=0.04
Characteristics of First Order Systems
Solution
1250
25
4
100
040010
1
1
ss
s.
BJssT
s
.
.
)(
)(
The time constant =0.25sec
The final value= 25rad/sec
0 0.5 1 1.5
0
5
10
15
20
25
StepResponse
Time(sec)
Amplitude
12

7. 3/28/2014
7
2nd order system
b
u(t)
s
m
k
s
m
b
s
m
skbsms
sX
s
sU
kbsmssU
sX
1
1
11
)(
;
1
)(let
1
)(
)(
2
2
2













13
2nd order system
m =1,b = 2,k = 5
X(s)
U(s)
=
1
s2
+ 2s+ 5
Case 1:
14

8. 3/28/2014
8
2nd order system
m =1,b = 2,k =1
X(s)
U(s)
=
1
s2
+ 2s+1
Case 2:
15
2nd order system
m =1,b = 4,k = 3
X(s)
U(s)
=
1
s2
+ 4s+3
Case 3:
16

9. 3/28/2014
9
2nd order system
n
nn
nn
n
ss
sssR
sC





frequency,natural
ratio,damping
02:equationsticCharacteri
2)(
)(
22
22
2



17
2nd order system
18
 







 














2
1
2
2
1
tanshift,phase
1frequency,naturaldamped
where
sin
1
1
1)(
nd
d
t
tety n
 In general, the output response with a unit step input &
zic:

10. 3/28/2014
10
Various damping ratios
19
2nd order systems
Case ζ Roots of
characteristic
equation
Example
Case 1:
Underdamped
Pair of complex
poles
Case 2:
Critically damped
Two equal poles
Case 3:
Overdamped
Two distinct,
negative real poles
0 <z <1 s = -1± j2
z >1
z =1
s = -1,-2
s = -1,-1
20

11. 3/28/2014
11
2nd order system: characteristics
21
2nd order system: specifications
 Rise time, tr (10%-90%)
 Settling time, ts (2% or 5%)
 Delay time, td (0%-50%)
 Peak time, tp
 Maximum (percent) overshoot, Mp
 These specifications can be determined from the plot of
the response. Additionally these specifications also apply
for systems of higher orders.
22

12. 3/28/2014
12
Transient response specifications
(for 2nd order systems)
 Rise time,
 Peak time,
 Max overshoot,
 Setting time,
2
1, 




 nd
d
rt
d
pt

















2
1
eM p
criterion)(2%
4
n
st


23
Tutorials
24
1. Case-4: m=1, b=0, k=1
2. For all Cases (1, 2, 3 and 4), find the
 time response equation, y(t)
 damping ratio (ζ) and natural frequency (ωn)
 damped natural frequency (ωd)
 Rise time (tr), Peak time (tp) & Setting time (ts)
 Max overshoot (Mp)

13. 3/28/2014
13
Poles and zeros
 Say we have a transfer function of a system:
 The zero(s) of the system are the roots of the numerator
 The pole(s) of the system are the roots of the
denumerator/characteristic eqn.
 What are the zeros of the system above? What are the
poles?
 What is the order of the system?
)3)(2(
1
)(



ss
s
sG
25
Pole zero map
 It is convenient to draw the pole(s) and zero(s) of the
system in a graphical manner
)3)(2(
1
)(



ss
s
sG
26

14. 3/28/2014
14
Example
 Draw the pole-zero map (pzmap) of this system
)52)(5(
)2)(5.0(
)(



sss
ss
sG
27
Example
28

15. 3/28/2014
15
Effects of pole(s) position
 Let’s start with a simple, 1st order system
 What is the pole of the system?
 Plot the pzmap and the response of the system to a unit
step input
1
1
)(1


s
sG
29
30

16. 3/28/2014
16
Effects of pole(s) position
 Another system
2
2
)(2


s
sG
31
Effects of pole(s) position
 What about G3? What is the expected response?
 What is your conclusion from this?
10
10
)(3


s
sG
32

17. 3/28/2014
17
Effects of pole(s) position
 What if we have a pole on the right half plane (RHP) of
the pzmap? Eg.:
1
1
)(4


s
sG
33
What can be concluded from the info about the position
of the pole on the pzmap?
34

18. 3/28/2014
18
Finding:
A system is stable only if all the poles of
the system are located in the LHP of the
pzmap!
35
Effect of pole position
 Do the same to 2nd order systems
 Remember we have 3 cases for 2nd order systems:
underdamped, critically damped and overdamped
36

19. 3/28/2014
19
Effect of pole position
 An example 2nd order system:
 What are the poles of this system? Write the complex
pole in terms of
 Compare the denumerator with the standard
characteristic eqn. for a 2nd order system:
 How are these related to each other? :
)1)(1(
1
22
1
)( 21
jsjsss
sG




djs  
02 22
 nnss 
dn  ,,,
37
Effect of pole position
z = cosb
wn = wd
2
+s 2
wd =wn 1-z2
s =zwn
38

20. 3/28/2014
20
Time function vs pole location
39
Effects of zero locations
 The zero affects the transient response:
 Example: G2 has a zero near a pole:
tt
tt
eetg
ssss
s
sG
eetg
ssss
sG
2
2
2
2
1
1
64.118.0)(
2
64.1
1
18.0
)2)(1(1.1
)1.1(2
)(
22)(
2
2
1
2
)2)(1(
2
)(

















40

21. 3/28/2014
21
Effects of zero locations
G1 G2
41
Effects of zero locations
 Time response plot:
Generally, a zero near
a pole reduces the
amount of that term
in the total response
42

22. 3/28/2014
22
Effect of zero location
 Observe these two systems, one with zero in RHP







)(
)42.11)(42.11(
)1(1
)(
)(
)42.11)(42.11(
1
)(
2
2
1
1
th
jsjs
s
sH
th
jsjs
sH
43
Effect of zero location
 Observe these two systems, one with zero in RHP
A zero in the RHP results
in an initial opposite
response  non-
minimum phase system
44