Mathematical Models

1
Mathematical Models

• SYSTEMS AND CONTROL I ECE 09.321
• 10/09/07 Lecture 9
• ROWAN UNIVERSITY
• College of Engineering
• Prof. John Colton
• DEPARTMENT OF ELECTRICAL COMPUTER ENGINEERING
• Fall 2007 - Semester One

2
Some Administrative Items

• Quiz 2 will be on Wednesday 10/17 950 AM to
  1040 AM.
• Feedback Control System Characteristics and
  Performance
• Through Lecture 8, through textbook Chapter 4
• Through Problem Set 5
• Problem Set 5 is due today, 10/10.
• Problem Set 6 is due Wednesday 10/17.

3
Feedback Control System Performance

• Second Order Systems
• Additional poles and zeros
• Transient Response and s plane

4
Second-order System Step Response
5
Second-order System Step Response
6
Second-order System Step Response
Y(s)/R(s) G(s)/1 G(s) ?n2/s 2 2 ? ?ns
?n2 For r(t) u(t) ? R(s) 1/s Y(s) ?n2/
ss 2 2 ? ?ns ?n2 poles s 0, s ? ?n
?n (? 2 1)1/2 For ? gt 1, there are three real
poles For 0 lt ? lt 1, there is one real pole (s
0) plus two complex conjugate poles For 0 lt ? lt
1, Y(s) ? y(t) 1 (1/ß)e- ? ?nt sin(?nt ?)
u(t) where ß (1- ? 2 ) ½ and ? cos -1 ?
7
Second-order System s plane
Ims
0 lt ? lt 1
j?n (1 ? 2) 1/2
Res
? ?n
j?n (1 ? 2) 1/2
Y(s) ?n2/ss 2 2 ? ?ns ?n2 poles s 0, s
? ?n j?n (1 ? 2)1/2
8
Second-order System Step Response
y(t) s(t) 1 (1/ß)e ? ?nt sin(?n ß t
?) u(t) 0 lt t lt 1
9
Second-order System Step ResponseStandard
Performance Measures
Tr Rise time 0 100 normally used for
underdamped systems 10 90 normally used
for overdamped systems Tp Peak time Time to
reach the first peak in an underdamped
response Overshoot PO 100(Mpt fv)/fv
where Mpt and fv are peak and final values Ts
Settling time Time required to settle within a
specified percentage of final value
10
Second-order System Step ResponseStandard
Performance Measures
Tr Rise time 0 100 normally used for
underdamped systems 10 90 normally used
for overdamped systems Tp Peak time Time to
reach the first peak in an underdamped
response Overshoot PO 100(Mpt fv)/fv
where Mpt and fv are peak and final values Ts
Settling time Time required to settle within a
specified percentage of final value
11
Second-order System Step ResponseStandard
Performance Measures
12
Second-order System Step ResponseStandard
Performance Measures
13
Second-order System Step ResponseStandard
Performance Measures
Tr Rise time 0 100 normally used for
underdamped systems 10 90 normally used
for overdamped systems Tp Peak time Time to
reach the first peak in an underdamped
response Overshoot PO 100(Mpt fv)/fv
where Mpt and fv are peak and final values Ts
Settling time Time required to settle within a
specified percentage of final value
14
Second-order System Impulse Response
15
Second-order System Impulse Response
h(t) d/dt s (t) ? H(s) sYstep (s) H(s)
?n2/ s 2 2 ? ?ns ?n2 h(t) (?n / ß)e- ?
?nt sin(?n ß t ) u(t)
16
Second-order System Settling Time
y(t) s(t) 1 (1/ß)e ? ?nt sin(?n ß t
?) u(t) 0 lt t lt 1
17
Second-order System Settling Time
y(t) s(t) 1 (1/ß)e ? ?nt sin(?n ß t
?) u(t) 0 lt t lt 1
Settling time For d .02 (2 band around final
value of 1.0)
e ? ?nTs lt .02 ? ? ?nTs 4
Ts 4t 4/ ?
?n i.e. The settling time is about 4 time
constants of the dominant roots of the
characteristic equation
18
Second-order System Peak Time
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Second-order System Peak Time
y(t) s(t) 1 (1/ß)e ? ?nt sin(?n ß t
?) u(t) 0 lt t lt 1
20
Second-order System Peak Time
y(t) s(t) 1 (1/ß)e ? ?nt sin(?n ß t
?) u(t) 0 lt t lt 1
Peak time For the impulse response function
h(t) d/dt s(t) ? H(s) H(s) sYstep (s)
h(t) (?n / ß) ) e- ? ?nt sin(?n ß t ) u(t)
The first
zero in h(t) occurs at ?n ßt p
Tp p / ?n (1- ? 2)½
21
Second-order System Maximum Value
22
Second-order System Maximum Value
y(t) s(t) 1 (1/ß)e ? ?nt sin(?n ß t
?) u(t) 0 lt t lt 1 Mpt y(Tp) 1 1/(1-
? 2)½ e ? p/(1- ? 2)½ sin(cos-1 ? ) Mpt 1
e ? p/(1- ? 2)½
23
Second-order System Overshoot
24
Second-order System Overshoot
PO 100(Mpt fv)/fv 100 1 e ? p/(1- ?
2)½ 1 / 1 PO 100 e ? p/(1- ? 2)½
25
Transient Response and s plane
26
Transient Response and s plane
27
Transient Response and s plane
T(s) Y(s)/R(s) ?Pk(s)?k(s)/ ? (s) For a unit
step function input R(s) 1/s
M N Y(s)
T(s)/s A0 /s ? Ai /(s si) ? (Bks Ck
) /( s 2 2aks ak2 ?k2)
i 1 k 1 poles
of the system must be real s si or complex
conjugate pairs s ak j ?k
M N y(t) 1 ? Ai e sit ? Dk e
akt sin (?kt ?k ) u(t) i 1
k 1 For the system to be stable, the
real part of the poles must be in the left half
of the s plane.
28
Impulse Responses and Pole Locations
29
Effects of other poles and zeros
Ims
0 lt ? lt 1
j?n (1 ? 2) 1/2
Res
? ?n
j?n (1 ? 2) 1/2

• A necessary condition for stability is that the
  poles must be in the left hand s plane,
• with negative real parts.
• Moving a zero closer to a pole will reduce the
  contribution of that pole
• to the overall response
• Moving a pole closer to the Ims axis (Re s
  0 makes it more
• influential in the response
• Dominant poles are usually those closest to the
  Ims axis, subject to the
• relative values of the residues at all the
  poles.