Unformatted text preview: ME 575 Handouts Analysis of Feedback Systems
• Typical Classical Feedback Controller
Structure
• Nominal Sensitivity Functions
• Stability of Nominal Feedback System
– Internal Stability
– Routh’s Stability Test
– Root Locus Method
– Nyquist Stability Criterion ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 1 TwoDegreeofFreedom Closed Loop
Di (s)
Reference
Value R(s) H(s) R(s) E(s) C(s) U (s ) − Ym(s) U D (s ) x0 G(s) Do (s) Y (s) Plant
N (s )
Noise Controller ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Sensor Slide 2 1 ME 575 Handouts ClosedLoop Transfer Functions ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 3 Sensitivity Functions NG (s)NC (s)
G(s)C(s)
Complementary
=
:
sensitivity function
1 + G(s)C(s) DG (s)DC (s) + NG (s)NC (s)
DG (s)DC (s)
1
S(s) =
=
: Sensitivity function
1+ G(s)C(s) DG (s)DC (s) + NG (s)NC (s)
NG (s)DC (s)
G(s)
Input disturbance
Si (s) =
=
:
sensitivity function
1+ G(s)C(s) DG (s)DC (s) + NG (s)NC (s)
DG (s)NC (s)
C(s)
Control
Su (s) =
=
:
sensitivity function
1 + G(s)C(s) DG (s)DC (s) + NG (s)NC (s)
T (s) = ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 4 2 ME 575 Handouts Sensitivity Function
• The sensitivity function S describes how error
signal E is related to the reference input R.
• The sensitivity function S also describes how
much a change in the plant G (perhaps due to
parameter errors) affects the closedloop transfer
function T (a.k.a. complementary sensitivity).
• A controller C with high gain will make CG large,
CG
therefore making S small, exactly as desired. ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 5 Sensitivity Function
• Relationship Between Sensitivities S ( s) + T ( s) = 1
Si (s) = S (s)G(s)
Su (s) = S (s)C (s)
• Practical Implications ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 6 3 ME 575 Handouts Internal Stability
⎡
⎢
⎡ E ( s) ⎤
⎢
⎥
1⎢
⎢
⎢U D ( s ) ⎥ =
⎢
⎥ 1 + CG ⎢
⎢ Ym ( s ) ⎥
⎢
⎣
⎦
⎢
⎣ ⎤ ⎡ R (s) ⎤
⎥
⎥⎢
⎢ Di ( s ) ⎥
⎥
⎥
⎥⎢
⎥ ⎢ Do ( s ) ⎥
⎥
⎥⎢
⎥ ⎢ N (s) ⎥
⎦⎣
⎦ This system is internally stable if the internal signals
decay to zero when any of the inputs are impulses. ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 7 Internal Stability
Theorem:
The feedback system is internally stable if and only if
• the sensitivity function S has no poles in the RHP
(s ≥ 0)
• there is no cancellation of unstable poles between
the controller C and the plant G. ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 8 4 ME 575 Handouts RouthHurwitz Stability Criterion
Allows checking for stability without having to
solve for the poles.
QUICK TEST: Hurwitz Necessary Condition
n
Given the char. eq.
p( s) = a s k = 0 ∑
k =0 k a stable system must have: (i)
(ii) ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 9 Hurwitz Necessary Condition
Ex: p(s) = s2 − 3s + 2 = 0 stable? Ex: p(s) = s3 + s2 + 2s + 8 = 0 stable? We need a sufficient condition.
ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 10 5 ME 575 Handouts Routh Array
n For p( s ) = ∑ ak s k = 0
k =0 n s
s n−1
s n−2
s n −3 an
an−1
b1
c1 an−2
an−3
b2
c2 an−4
an−5
b3
c3 L
L 0
0 b1 = b2 = an
an −1 an − 2
an − 3 −an −1
an
an −1 an − 4
an −5 −an −1 , , M M
s0
c1 = an −1
b1 an −3
b2
−b1 M
ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 11 Routh Sufficient Condition
The number of roots of p(s) having positive real
parts (in the RHP) equals the number of sign
changes in the 1st column of the Routh array.
Ex: p(s) = s3 + s2 + 2s + 8 ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 12 6 ME 575 Handouts Root Locus
• Definition
Root Locus is the plot of the roots of the following equation on the
complex s plane when the parameter K varies from 0 → ∞:
varies 1+ K ⋅ N (s)
= 0 or D ( s ) + K ⋅ N ( s ) = 0
D( s) where N(s) and D(s) are known polynomials in factorized form: N ( s ) = ( s − z1 )( s − z2 )L ( s − z N z )
D( s ) = ( s − p1 )( s − p2 )L ( s − pN P ) z1 , z2 , …, zNz , are called the finite openloop zeros.
are
openp1 , p2 , …, pNp , are called the finite openloop poles.
are
open ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 13 Root Locus
Methods of obtaining root locus:
• Given a value of K, numerically solve the characteristic equation for a
K, numerically
set of roots. Repeat this for a set of K values and plot the
corresponding roots on the complex plane.
• Use MATLAB. In MATLAB use the commands rlocus and
rlocus
rlocfind. You can use online help to find the usage for these
rlocfind
oncommands. 1+ K ⋅ 800
=0
s ( s + 50) >> G = tf(800,[1 50 0]);
>> rlocus(G);
>> [K, poles]=rlocfind(G); • Apply the following root locus sketching rules to obtain an approximate
approximate
root locus plot.
ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 14 7 ME 575 Handouts Root Locus Magnitude and Phase Conditions
Magnitude
The necessary and sufficient conditions for a point on the splane to be on
the root locus (or a value of s that satisfies the characteristic eq.) are that s
satisfies the following phase and magnitude conditions: Phase Condition (or Angle Criterion): ( ∠( s − z1 ) + L + ∠ s − zNz ) ( ) − ∠( s − p1 ) −L− ∠ s − pN p = 180o Sum of phase angles to all openloop zeros
− Sum of phase angles to all openloop poles = 180o Magnitude Condition:
K=
K= D(s)
N ( s) = s − p1 s − p2 L s − pN p
s − z1 s − z2 L s − z N z Multiplication of lengths to all openloop poles
Multiplication of lengths to all openloop zeros ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 15 Root Locus Sketching Rules
Rule 1: There is a branch of the root locus for each root of the
characteristic equation. The number of branches is equal to the
number of openloop poles or openloop zeros, whichever is
openopengreater.
Rule 2: Root locus starts at openloop poles (when K = 0) and ends at
0)
openopenloop zeros (when K → ∞). If the number of poles is greater
openthan the number of zeros, roots start at the excess poles and
terminate at zeros at infinity. If the reverse is true, branches will
branches
start at poles at infinity and terminate at the excess zeros.
Rule 3: Root locus is symmetric about the real axis, i.e., closedloop
closedpoles appear in complex conjugate pairs.
Rule 4: Along the real axis, the root locus includes all points to the left of
left
an odd number of real poles and zeros.
odd
ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 16 8 ME 575 Handouts Root Locus Sketching Rules
Rule 5: If number of poles NP exceeds the number of zeros NZ , then as
K → ∞, (NP − NZ) branches will become asymptotic to straight
branches
lines. These straight lines intersect the real axis with angles θk at
σ0 .
σ0 =
θk = ∑ p −∑z
i NP − NZ i = Sum of openloop poles − Sum of openloop zeros
# of openloop poles − # of openloop zeros (2k + 1)π
(2k + 1)180°
[rad] =
[deg] , k = 0, 1, 2, L
NP − NZ
NP − NZ If NZ exceeds NP , then as K → 0, (NZ − NP) branches behave as
branches
above.
Rule 6: Breakaway and/or breakin (arrival) points can be obtained by
breaksolving s in the following equations:
d
( K (s) ) = 0
ds
ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 17 Root Locus Sketching Rules
Rule 7: The departure (arrival) angle for a pole pi (zero zi) can be
calculated by slightly modifying the following equation:
∠( s − z1 ) + ∠( s − z2 ) + L + ∠( s − z N Z ) − ∠( s − p1 ) − ∠( s − p2 ) − L − ∠( s − pN p ) = 180° The departure angle θn from the pole pn can be calculated by replacing
the term ∠( s − pn )
with θn and replacing all the s’s with pn in the
other terms. Rule 8: If the root locus passes through the imaginary axis (the stability
stability
boundary), the crossing point jω and the corresponding gain K can be
found as follows:
– Replace s in the left side of the closedloop characteristic equation
closedwith jω to obtain the real and imaginary parts of the resulting
complex number.
– Set the real and imaginary parts to zero, and solve for ω and K.
This will tell you at what values of K and at what points on the jω
axis the roots will cross.
ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 18 9 ME 575 Handouts Root Locus Example
A feedback control system is proposed. The corresponding
block diagram is:
R(s) +
− Controller C (s ) Plant G (s) U(s) 1
s(s + 2) Y(s) Let C(s) = K. Sketch the root locus of the closedloop poles
closedas the controller gain K varies from 0 to ∞. Find closedloop characteristic equation:
closed ME575 Session 8 – Sensitivity Functions
and Stability School of Mechanical Engineering
Purdue University Slide 19 Root Locus Example
C (s ) = K C(s ) = K (s + 3)
Im s
j2 j1
−6 −4 −2 Im s
j2 j1
Re s −6 −4 −2 Re s −j1
−j2
ME575 Session 8 – Sensitivity Functions
and Stability −j1
−j2 School of Mechanical Engineering
Purdue University Slide 20 10 ...
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This note was uploaded on 12/09/2011 for the course ME 575 taught by Professor Meckl during the Fall '10 term at Purdue.
 Fall '10
 Meckl

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