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Lecture 7Lecture 7
Poles and ZerosPoles and ZerosStabilityStability
Poles and ZerosPoles and Zeros
Transfer Function ModelsTransfer Function Models
General RepresentationGeneral Representation
whwhere zi are the zeros pi are the poles n ≥ m to have a physically realizable system
The dynamic behavior of a transfer function model can be characterized by the numerical value of its poles and zeros
)())(()())(()(
21
21
0
0
mn
mmn
i
ii
m
i
ii
pspspsazszszsb
sa
sbsG
Transfer Function ModelsTransfer Function Models
Transfer Function can be expressed in Transfer Function can be expressed in gaingain//time time constant formconstant form, i.e. factoring out , i.e. factoring out bbmm and and aann
Pole–Zero cancellation happens when the Pole–Zero cancellation happens when the numerator and denominator terms cancels each numerator and denominator terms cancels each otherother
)1)(1()1)(1(
)())(()())(()(
21
21
21
ssssK
pspspsazszszsbsG
ba
mn
mm
PolePole
The factor/s of the denominator of the transfer function
It is the value wherein the transfer function approaches infinity as the value of s approaches the pole
pole of order n where as
sgsf x
i
ni
i
1
)(
)()(
Example 1Example 1
Pole of order 1 or simple pole at Pole of order 1 or simple pole at ss = 0 = 0
Pole of order 2 at Pole of order 2 at ss = 5 and pole of order 3 = 5 and pole of order 3 at at ss = – 7 = – 7
32 )7()5(2)(
ssssf
ssf 3)(
ZeroZero
The factor/s of the numerator of the transfer function
It is the value wherein the transfer function approaches 0 as the value of s approaches the zero
pole of order m where sg
assf
y
i
mi
i
)(
)()( 1
Complex Plane PlotComplex Plane Plot
Graphical representation of a rational transfer Graphical representation of a rational transfer function in the complex plane which helps to function in the complex plane which helps to convey certain properties of the systemconvey certain properties of the system
x–axis is real part y–axis is imaginary part
)12)(1()(
222
21
ssssKsG
Pole–Zero PlotPole–Zero Plot
ZeroZeross = – 2 = – 2o in ploto in plot
PolePoless = ±0.5 = ±0.5iix in plotx in plot
412
2)(
sssf
Effects of Pole LocationEffects of Pole Location
Left Half Plane (LHP) results in a stable system, i.e. stable response
Right Half Plane (RHP) results in unstable system, i.e. unstable step response
x
x
x
Real axis
Imaginary axis
x
x
x
Real axis
Imaginary axis
Effects of Pole LocationEffects of Pole Location
Faster response when pole is farther from imaginary axis
Complex pole results to oscillatory response
p = a + bj where j = √– 1
Real axis
Imaginary axis
x
xx → complex poles
Effects of Pole LocationEffects of Pole Location
More oscillatory transient response when pole is farther from real axis
Pole at the origin, i.e. 1/s term in Transfer Function Model, results in an integrating element/process
Effects of Zero LocationEffects of Zero Location
Zeros have no effect on system stability Zero in Right Half Plane (RHP) results in
inverse response to a step change in the input
x y
t
inverse response
Real
axis
Imaginary axis
Effects of Zero LocationEffects of Zero Location
Zero in Left Half Plane (LHP) results in overshoot during a step response
StabilityStability
StabilityStability
Most industrial processes are stable without Most industrial processes are stable without feedback control and are said to be feedback control and are said to be open–loop open–loop stablestable or or self–regulatingself–regulating
An openAn open––loop stable process will return to the loop stable process will return to the original steady state after a transient disturbance, original steady state after a transient disturbance, i.e. one that is not sustained, occursi.e. one that is not sustained, occurs
By contrast there are a few processes, such as By contrast there are a few processes, such as exothermic chemical reactors, that can be exothermic chemical reactors, that can be open-open-loop unstableloop unstable
StabilityStability
DefinitionDefinitionAn unconstrained linear system is said to be stable if the An unconstrained linear system is said to be stable if the output response is bounded for all bounded inputs or output response is bounded for all bounded inputs or BIBO stability. Otherwise it is said to be unstable.BIBO stability. Otherwise it is said to be unstable.
ApplicabilityApplicability– Any linear control system comprised of linear Any linear control system comprised of linear
elementselements– Nonlinear systems operating near the point of Nonlinear systems operating near the point of
linearizationlinearization
StabilityStability
Asymptotically stability is when the variables of the stable control system always decrease from their initial value and do not show permanent oscillations– Permanent oscillations occur when a pole has a real part exactly
equal to zero (in the continuous time case) or a modulus equal to one (in the discrete time case)
Marginally stability is when a simply stable system response neither decays nor grows over time, and has no oscillations– System transfer function has non–repeated poles at complex
plane origin, i.e. their real and complex component is zero in the continuous time case
Closed Loop SystemClosed Loop System
Transfer FunctionTransfer Function
Km
Gm
PYsp
Ym
Y’sp E–+
G*d
GpGvGcU ++
D
YV
pvcmOL
spOL
pvcm
OL
pd
sppvcm
pvcm
pvcm
pd
GGGGG where
YG
GGGKD
GGG
YGGGG
GGGKD
GGGGGG
Y
11
11*
*
Closed Loop System
Characteristic Equation of Closed Loop System– Is the denominator of the transfer function– Can be simplified to 1 + GOL where GOL is the
open loop transfer function– Used to solve for poles by equation to zero,
i.e. 1 + GOL = 0
Stability for Closed Loop Stability for Closed Loop SystemSystemGeneral Stability CriterionGeneral Stability Criterion
The feedback control system is stable if and only if The feedback control system is stable if and only if allall roots of the characteristic equation are negative or have roots of the characteristic equation are negative or have negative real parts. Otherwise, the system is unstable.negative real parts. Otherwise, the system is unstable.
AssumptionsAssumptions– Set–point changes rather than disturbance changes Set–point changes rather than disturbance changes
were consideredwere considered– Closed–loop transfer function was a ratio of Closed–loop transfer function was a ratio of
polynomialspolynomials– Poles are all distinctPoles are all distinct
Stability Region in Complex Plane
A system is stable if the poles of the transfer function lie strictly in the closed left half of the complex plane, i.e. the real part of all the poles is less than zero
Closed Loop Response
Stable System
Closed Loop Response
Unstable System
Example 1Example 1
Determine whether system is stable or not givenDetermine whether system is stable or not given
Solution:Solution:Solving for 1 + Solving for 1 + GGOLOL
Equating 1 + Equating 1 + GGOLOL = 0 = 0
)12()15.0(10
ss
sGOL
)12(53
)12()15.0(1011
2
ss
ssss
sGOL
0532 ss
Example 1Example 1
Solving for Solving for ss
Inference on systemInference on system– System is stable since real part of pole/s is negativeSystem is stable since real part of pole/s is negative– Behavior is oscillatory due to the presence of Behavior is oscillatory due to the presence of
imaginary termimaginary term
2113
2113
2113
22093
21
s js
or
js
Example 2Example 2
Determine whether system is stable or not givenDetermine whether system is stable or not given1 + 1 + GGOLOL = = ss + 0.2 + 0.2KKcc – 1 – 1
Solution:Solution:Equating 1 + Equating 1 + GGOLOL = 0 = 0
ss + 0.2 + 0.2KKcc – 1 = 0 – 1 = 0ss = 1 – 0.2 = 1 – 0.2KKcc
For system to be stable For system to be stable ss should be less than 0, should be less than 0, i.e. i.e. KKcc > > 5 5
Example 3Example 3
Determine whether system is stable or not givenDetermine whether system is stable or not given
Solution:Solution:Solving for 1 + Solving for 1 + GGOLOL
Equating 1 + Equating 1 + GGOLOL = 0 = 0
)15)(12(
ssKG c
OL
)15)(12(1710
)15)(12(11
2
ssKss
ssKG cc
OL
01710 2 cKss
Example 3Example 3
Solving for Solving for ss
For system to be stable, For system to be stable, ss should be less than 0, should be less than 0, i.e. 40(i.e. 40(KKcc + 1) > 0 or + 1) > 0 or KKcc > – 1 > – 1
20)1(40497
cKs
Stability Test
Direct Substitution Method Routh Stability Criterion Root Locus Diagram Bode Stability Criterion Nyquist Stability Criterion
Direct Substitution MethodDirect Substitution Method
Imaginary axis divides the complex plane into stable and unstable regions for the roots of characteristic equation
On the imaginary axis, the real part of s is zero, and thus we can write s = j. Substituting s = j into the characteristic equation allows us to find a stability limit such as the maximum value of Kc
As the gain Kc is increased, the roots of the characteristic equation cross the imaginary axis when Kc = Kcm
Direct Substitution MethodDirect Substitution Method
Methodology– s = j is substituted in the characteristic
equation– Kc is equated to Kcm
– Both the real part and imaginary part is equated to 0 and value of and Kcm is computed
– Stability region is determined for Kc
Example 4Example 4
Determine whether system is stable or not givenDetermine whether system is stable or not given1010ss33 + 17 + 17ss22 + 8 + 8s + s + 11 + + KKcc = 0 = 0
Solution:Solution:Substituting Substituting s = j and Kc = Kcm in the characteristic equation
– – 1010jj33 – 17 – 1722 + 8 + 8jj + + 11 + + KKcmcm = 0 = 0oror
(1(1 + + KKcm cm – 17– 1722) + ) + jj (8 (8 – 10 – 1033)) = 0= 0
Example 4Example 4
Equating both real and imaginary part to zeroEquating both real and imaginary part to zero11 + + KKcm cm – 17– 1722 = 0 = 0
88 – 10 – 1033 = 0= 0
Solving Solving from 88 – 10 – 1033 = 0= 0 results to = 0 → KKcmcm = – 1 = – 1
= ±0.894 → KKcmcm = 12.6 = 12.6
Region of stability would beRegion of stability would be – – 1 < 1 < KKcc < 12.6 < 12.6
Example 5Example 5
Determine whether system is stable or not givenDetermine whether system is stable or not given
Solution:Solution:Solving for 1 + Solving for 1 + GGOL OL = 0= 0
1
31
21 sss
KG cOL
0
13
12
111
sss
KG cOL
Example 5Example 5
ss33 + 6 + 6ss22 + 11 + 11s s + 6(1+ 6(1 + + KKcc) = 0) = 0 Substituting Substituting s = j and Kc = Kcm in the
characteristic equation– – jj33 – 6 – 622 + 11 + 11jj + + 6(1 6(1 + + KKcmcm) = 0) = 0
oror(6(6 + 6+ 6KKcm cm – 6– 622) + ) + jj (11 (11 – – 33)) = 0= 0
0
13
12
1
)1(6116 23
sss
Ksss c
Example 5Example 5
Equating both real and imaginary part to zeroEquating both real and imaginary part to zero66 + 6+ 6KKcm cm – 6– 622 = 0 = 0
1111 – – 33 = 0= 0
Solving Solving from 11 – – 33 = 0= 0 results to = 0 → KKcmcm = – 1 = – 1
= ±3.32 → KKcmcm = 10 = 10
Region of stability would beRegion of stability would be – – 1 < 1 < KKcc < 10 < 10
Routh Stability Criterion
Developed by E J Routh in 1905 a.k.a.
– Routh–Hurwitz Stability Criterion– Routh Test
Purely algebraic method Used to establish stability in single input single
output (SISO) linear time invariant control system
Routh Stability Criterion
Applied to systems with characteristic equation that has a polynomial form.
Hence, it can not be used to systems with time delays or transport lag, i.e. e –s term
For system with e –s term, Padé approximation is done on the time delay term/s
positive is a where asa sasa
n
nn
nn 001
11
Routh Stability Criterion
It is necessary (but not sufficient) that all the coefficients of the characteristic equation, i.e. an, an – 1, . . . , a1 and a0, be positive else the system is unstable. Hence, no need to perform the Routh Test
Flow Process for Performing Flow Process for Performing Stability AnalysisStability Analysis
Example 6Example 6
Determine the stability of system that has Determine the stability of system that has characteristic equationcharacteristic equation
ss44 + 5 + 5ss33 + 3 + 3ss22 + 1 + 1 = 0= 0
Solution:Solution:Since the Since the s term is missing, its coefficient is zero
Thus, the system is unstable. Recall that a necessary condition for stability is that all of the coefficients in the characteristic equation must be positive.
Routh Array Generation
Given a polynomial of the form:ansn + an–1sn–1 + ● ● ● + a1s + a0 = 0
an–1
anan–3
an–2
z1
b1
Row1
● ● ● an–5
an–4 ● ● ●
b3b2 ● ● ● c1 c2 ● ● ●
● ●
●
3
2
4
n + 1
● ●
●
a
aaaab
aaaaab
n
nnnn
n
nnnn
1
5412
1
3211
b
baabc
bbaabc
nn
nn
1
33512
1
21311
Theorems on Routh TestTheorems on Routh Test
Theorem 1Theorem 1A necessary and sufficient condition for all the roots A necessary and sufficient condition for all the roots of the characteristic equation to have a negative real of the characteristic equation to have a negative real parts (or stable system) is that all the elements of the parts (or stable system) is that all the elements of the first column in the Routh array be positive and first column in the Routh array be positive and nonzerononzero
Theorem 2Theorem 2If some of the elements in the first column are If some of the elements in the first column are negative, the number of roots with a positive real negative, the number of roots with a positive real part, i.e. in the right hand plane, is equal to the part, i.e. in the right hand plane, is equal to the number of sign changes in the first columnnumber of sign changes in the first column
Theorems on Routh TestTheorems on Routh Test
Theorem 3Theorem 3If one pair of the roots is on the imaginary axis, If one pair of the roots is on the imaginary axis, equidistant from the origin, and all other roots are in equidistant from the origin, and all other roots are in the left half plane, all the elements of the nthe left half plane, all the elements of the nthth row will row will vanish and none of the elements preceding row will vanish and none of the elements preceding row will vanish. The location of the pair of imaginary roots vanish. The location of the pair of imaginary roots can be found by solving the equationcan be found by solving the equation
CCss22 + D = 0 + D = 0where C and D are the elements of the where C and D are the elements of the ((nn–1)–1)thth row row read left to right, respectivelyread left to right, respectively
Example 7Example 7
Determine the stability of system that has Determine the stability of system that has characteristic equationcharacteristic equation
ss44 + 3 + 3ss33 + 5 + 5ss22 + 4 + 4ss + 2 + 2 = 0= 0Solution:Solution:
31
45
2
11/3
Row1 2
226/11
3
2
4
5System is stable since all terms in the 1st column is positive (Theorem 1)
Example 8Example 8
Determine the stability of system that has Determine the stability of system that has characteristic equationcharacteristic equation
ss66 + + s s55 + + 44ss44 + 3 + 3ss33 + 2 + 2ss22 + 4 + 4ss + 2 + 2 = 0= 0Solution:Solution:
11
34
–12/5
1
Row1 2
–25
32
45
42
222
–74/126
27
System is unstable since not all terms in the 1st column is positive
2 roots are in the right half plane due to 2 sign changes
Example 9Example 9
Determine value of Determine value of KKcc to have a stable system to have a stable system
Solution:Solution:Solving for 1 + Solving for 1 + GGOLOL
1
31
2)1( sss
KG cOL
1
31
2)1(
)1(6116
13
12
)1(11
23
sss
Kssssss
KG ccOL
Example 9Example 9
Characteristic equationCharacteristic equationss33 + 6 + 6ss22 + 11 + 11ss + 6(1 + + 6(1 + KKcc)) = 0= 0
61
6(1 + Kc)11
10 – Kc
Row1
6(1 + Kc)
3
2
4
For system to be stable, all terms in the 1st column should be greater than zero (Theorem 1)
10 – Kc > 0 → Kc < 106 (1 + Kc) > 0 → Kc > –1
Region of stability –1 < Kc < 10
Example 9Example 9
When When KKcc = 10, system is on the verge of = 10, system is on the verge of instability. The Routh array becomesinstability. The Routh array becomes
61
6611
0
Row1
66
3
2
4
According to Theorem 3, the (n–1)th row is the coefficient C and D used in solving the imaginary roots.
6s2 + 66 = 0
s = ±√11
Example 10Example 10
Determine value of Determine value of KKcc to have a stable system to have a stable system
GGOLOL = 5s +2 = 5s +2KKccee–s–s
Solution:Solution:Solving for 1 + Solving for 1 + GGOLOL
1 + 1 + GGOLOL = 1 + 5s +2 = 1 + 5s +2KKccee–s–s
Using 1/1 PadUsing 1/1 Padéé approximation approximation
sse s
5.015.01
Example 10Example 10
Characteristic equationCharacteristic equation
2.52.5ss22 + (5.5 – + (5.5 – KKcc))ss + (1 + 2 + (1 + 2KKcc)) = 0= 0
5.5 – Kc
2.5 1 + 2Kc
1 + 2Kc
Row1
3
2
For system to be stable, all terms in the 1st column should be greater than zero (Theorem 1)
5.5 – Kc > 0 → Kc < 5.5 1 + 2Kc > 0 → Kc > –0.5
Region of stability –1 < Kc < 10
05.015.01251
ssKs c
Root Locus
The locus of the roots of the characteristic equation of the closed loop transfer function as the loop gain, Kc, of the feedback system is increased from zero to infinty
It is a useful tool for analyzing the transient response, as well as the stability of a single input single output dynamic systems
A system is stable if all of its poles are in the left hand side of the s–plane
Graphical procedure for finding roots of the characteristic equation, 1 + GOL = 0
Root LocusRoot Locus
MethodologyMethodology– Obtain characteristic equationObtain characteristic equation– Vary value of Vary value of KKcc
– Solve for the roots of the equationSolve for the roots of the equation– Plot the roots based on specific Plot the roots based on specific KKcc values values
– Connect the points based on increasing Connect the points based on increasing KKcc valuesvalues
Example 11Example 11
Plot the Root Locus Diagram of the characteristic Plot the Root Locus Diagram of the characteristic equationequation
ss33 + 6 + 6ss22 + 11 + 11ss + 6(1 + + 6(1 + KKcc)) = 0= 0
Solution:Solution:Rearranging equation toRearranging equation to
((ss + 1) ( + 1) (ss + 2) ( + 2) (ss + 3) + + 3) + KK = 0 = 0where where KK = 6 = 6KKcc
Example 11Example 11KK = 6 = 6KKcc Root 1Root 1 Root 2Root 2 Root 3Root 3––3030 1.21 1.21– 3.61 – 2.60– 3.61 – 2.60jj – 3.61 + 2.60– 3.61 + 2.60jj––66 0 0 – 3.00 – 1.41– 3.00 – 1.41jj – 3.00 + 1.41– 3.00 + 1.41jj 00 – 3– 3 – 2– 2 – 1– 10.230.23 – 3.1– 3.1 – 1.75– 1.75 – 1.15– 1.150.390.39 – 3.16– 3.16– 1.42– 1.42– 1.42– 1.421.581.58 – 3.45– 3.45– 1.28 – 0.75– 1.28 – 0.75jj – 1.28 + 0.75– 1.28 + 0.75jj 6.66.6 – 4.11– 4.11– 0.95 – 1.50– 0.95 – 1.50jj – 0.95 + 1.50– 0.95 + 1.50jj26.526.5 – 5.10– 5.10– 0.45 – 2.50– 0.45 – 2.50jj – 0.45 + 2.50– 0.45 + 2.50jj 6060 – 6.00– 6.00 – 3.32 – 3.32jj + 3.32+ 3.32jj100100 – 6.72– 6.72 0.35 – 4.00 0.35 – 4.00jj 0.35 + 4.000.35 + 4.00jj
-5
-4
-3
-2
-1
0
1
2
3
4
5
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2
Example 11Example 11
-5
-4
-3
-2
-1
0
1
2
3
4
5
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2X XX
X
X
Kc = 0
Kc = 10
Kc = 10
Kc = 10
X
Kc = 0.07
Kc = –5X
Kc = –5
Kc = –5
Kc = –1
Kc = –1
Kc = –1
X
X
Example 12Example 12
Consider a feedback control system that has the Consider a feedback control system that has the open loop transfer functionopen loop transfer function
Plot the root locus for 0 Plot the root locus for 0 ≤≤ KKcc ≤≤ 20 20
Solution:Solution:The characteristic equation 1 + The characteristic equation 1 + GGOL OL = 0 or= 0 or
((ss + 1)( + 1)(ss + 2)( + 2)(ss + 3) + 4 + 3) + 4KKcc = 0 = 0
)3)(2)(1(4
sssKG c
OL
Example 12Example 12
When Kc = 0, the roots are merely the poles of the open loop transfer function, i.e. – 1, – 2 and – 3
X denotes an open loop pole
Dots denote locations of the closed loop poles for different values of Kc
ReferencesReferences
1.1. Coughanowr, Donald R. Coughanowr, Donald R. Process Systems Process Systems Analysis and Control. Analysis and Control. 22ndnd ed. New York: ed. New York: McGraw–Hill, Inc, 1991.McGraw–Hill, Inc, 1991.
2.2. Seborg, Dale E. Seborg, Dale E. et al. et al. Process Dynamics and Process Dynamics and Control. Control. 22ndnd ed. New York: John Wiley & ed. New York: John Wiley & Sons, Inc, 2004.Sons, Inc, 2004.
3.3. http://www.wikipedia.orghttp://www.wikipedia.org