Colorado School of Mines CHEN403
Stability Analysis of Feedback Control Systems
Stability Analysis of Feedback Control Systems ....................................................................................... 1
Introduction ............................................................................................................................................................ 1
Routh-Hurwitz Stability Criterion .................................................................................................................. 2
Example #1......................................................................................................................................................... 4
Direct Substitution ............................................................................................................................................... 6
Example #1 by Direct Substitution ........................................................................................................... 6
Example with P & PI Control........................................................................................................................ 7
Example with Dead Time ........................................................................................................................... 11
Root-Locus Analysis.......................................................................................................................................... 13
Introduction
Process
Final Control
Element (Actuator)
Controller
′
y sp
+
ε′
Gc ( s )
-
C′
Ga ( s )
ym′
L′
GL ( s )
M′
Gp ( s )
+
+
y′
Gm ( s )
Measuring Element
Remember that for the generalized closed-loop system shown above the overall transfer
function is:
y=
GpG f Gc
1 + GpG f Gc Gm
where: Gsp =
y sp +
GpG f Gc
1 + GpG f Gc Gm
Gd
d
1 + GpG f Gc Gm
and Gload =
Stability Feedback Control Systems
Gd
1 + GpG f Gc Gm
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Colorado School of Mines CHEN403
Review the section Poles & Zeros in the notes Transfer Functions for a discussion of
process stability. The definition of stability we will use for bounded input, bounded output
stability is:
A dynamic system is considered stable if for every bounded input it produces a
bounded output, regardless of its initial state.
The stability will be dictated by the characteristic poles of the transfer functions Gsp and
Gload . The characteristic equation to give these poles is the same (since the denominator is
the same):
1 + GpG f Gc Gm = 0 .
Note that if the control is cut right before the comparator, we get the open loop transfer
function:
ym
= GpG f Gc Gm = GOL
y sp
so the characteristic equation can be written as:
1 + GOL = 0 .
For example, a process with a transfer function Gp = 1/( s − 1) is unstable since it has a
positive pole s = +1 . However, let’s put in a P controller with Gm = G f = 1 . Then, the closed
loop characteristic equation will be:
1+
1
⋅1 ⋅ Kc ⋅1 = 0 ⇒ s − 1 + Kc = 0 .
s −1
Now, the pole is at s = 1 − K c . This will be a stable dynamic system if Re( s ) = s < 0 , or:
1 − Kc < 0 ⇒ Kc > 1 .
Routh-Hurwitz Stability Criterion
We do not necessarily need to know the poles to determine stability — just the knowledge
of which side of the complex plane the poles lie may be enough. We can set up a Routh
array to determine this. The steps are:
•
Express the characteristic equation as an expanded polynomial:
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Colorado School of Mines CHEN403
± (1 + GpG f Gc Gm ) = an s n + an−1 s n−1 + ⋯ + a1 s + a0 .
where a0 > 0 . This subscript notation is different from many other texts that
use:
± (1 + GpG f Gc Gm ) = a0 s n + a1 s n−1 + ⋯ + an−1 s + an .
•
If any of the coefficients are negative, then there is at least one root with a
positive real part and the system is unstable.
•
Set up a table that looks like the following:
Row
1
an
an-2
an-4
an-6
2
an-1
an-3
an-5
an-7
3
b1
b2
b3
4
c1
c2
c3
5
d1
d2
6
e1
e2
where the elements are found from equations like:
an−1an−2 − anan−3
aa
= an−2 − n n−3
an−1
an−1
a a −a a
aa
b2 = n−1 n− 4 n n−5 = an−4 − n n−5
an−1
an−1
b a −a b
ab
c1 = 1 n−3 1 2 = an−3 − 1 2
b1
b1
b a −a b
a b
c2 = 1 n−5 n−1 3 = an−5 − n−1 3
b1
b1
b1 =
The rule is to look at the square matrix above the element to be calculated. Use
the values from column 1 & the column just to the right of the element of
interest. Multiply the off diagonal terms, subtract the product of the diagonal
terms, and divide by the element just above. So, for example, term d2 will be:
d2 =
c1b3 − b1c3
bc
= b3 − 1 3 …
c1
c1
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Colorado School of Mines CHEN403
•
•
If all of the elements in the 1st column are positive, then the system is stable.
If some of the elements in the 1st column are negative, the number of roots with
a positive real part will be equal to the number of sign changes in the 1st column.
We can use the criteria to pick valid controller parameters that keep the system stable.
Example #1
In the following system, determine the value of the gain that makes the system unstable if
(a) τD = 0.25 min and (b) τD = 0.5 min . Use the following data:
mɺ = 250 lb/min
ρ = 62.5 lb/ft³
V1 = 4 ft³
V2 = 5 ft³
V3 = 6 ft³
Cˆ = 1 Btu/lb°F
p
The energy balance around each tank will be:
(
d ρVi Hˆ i
dt
) = mH
ɺ ˆ
i −1
dT
ɺ ˆ i + Qi ⇒ ρVi Cˆ p i = mC
ɺ ˆ p (Ti −1 − Ti ) + Qi
− mH
dt
So, it terms of deviation variables & taking the Laplace transform we get 1st order systems
of the form:
Ti′ =
Kq
ρV
1
1
Ti′−1 +
Qi′ where τi = i & K q =
mɺ
ɺ ˆp
τi s + 1
τi s + 1
mC
and where Q2′ = Q3′ = 0 since there are no heaters in these tanks. The following information
block diagram shows the feedback control system.
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Colorado School of Mines CHEN403
1
τ1 s + 1
T0′
Tsp′
+
-
Gc ( s )
Q′
Kq
τ1 s + 1
+
+
T1′
1
τ2 s + 1
T2′
1
τ3 s + 1
T3′
The system parameters will be:
K1 =
1
1
=
= 0.004 °F/(Btu/min)
ɺ ˆ p 250 ⋅ 1
mC
ρV1 62.5 ⋅ 4
=
= 1.00 min
mɺ
250
ρV 62.5 ⋅ 5
τ2 = 2 =
= 1.25 min
mɺ
250
ρV 62.5 ⋅ 6
τ3 = 3 =
= 1.50 min
mɺ
250
τ1 =
The characteristic equation will be:
1 + Gc G f G1G2G3 = 1 + K c (1 + τD s )
K1
1
1
τ1 s + 1 τ2 s + 1 τ3 s + 1
0.004
1
1
s + 1 1.25s + 1 1.50s + 1
( s + 1)(1.25s + 1)(1.50s + 1) + 0.004K c (1 + τD s ) = 0
= 1 + K c (1 + τD s )
1.875s3 + 4.625s 2 + 3.75s + 1 + 0.004τD K c s + 0.004K c = 0
The Routh array will be:
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Colorado School of Mines CHEN403
Row
1
2
1
1.875
3.75 + 0.004K c τD
2
4.625
1 + 0.004K c
3
4.625(3.75 + 0.004K c τD ) − 1.875(1 + 0.004K c )
4.625
1 + 0.004K c
4
The important element is:
4.625(3.75 + 0.004K c τD ) − 1.875(1 + 0.004K c )
4.625
= 3.345 + 0.004K c τ D − 0.00162K c
So, the system will be stable as long as:
3.345 + 0.004K c τD − 0.00162K c > 0
(0.00162 − 0.004τD ) K c < 3.345
So: for τD = 0.25 , 0.00062K c < 3.345 ⇒ K c < 5380 — this case has conditional stability.
For τD = 0.50 , −0.00038K c < 3.345 ⇒ K c > −8800 — this case has unconditional stability.
Direct Substitution
The direct substitution method can give both the ultimate values of the controller settings
and the period of oscillation at the ultimate settings. It can also be applied to systems with
dead time without having to make any approximations to the e −θs term.
The procedure is to substitute s = jω (where j ≡ −1 ) and set the real & imaginary parts
to zero. These expressions will give the ultimate values for ω (the frequency of oscillation
at this stability limit) & the associated controller settings. A time delay term poses no
problems since:
e − jωθ = cos ( ωθ ) − j sin ( ωθ ) .
Example #1 by Direct Substitution
For the previous example the characteristic equation was:
1.875s3 + 4.625s 2 + 3.75s + 1 + 0.004τD K c s + 0.004K c = 0
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December 21, 2008
Colorado School of Mines CHEN403
Substituting s = jω :
1.875( jω) + 4.625( jω) + 3.75( jω) + 1 + 0.004τD K c ( jω) + 0.004K c = 0
3
2
−1.875 jω3 − 4.625ω2 + 3.75 jω + 1 + 0.004τD K c jω + 0.004K c = 0
(1 + 0.004K
c
)
(
)
− 4.625ω2 + jω 3.75 + 0.004τD K c − 1.875ω2 = 0
From the imaginary part:
3.75 + 0.004τD K cu − 1.875ω2 = 0 ⇒ ω2 =
3.75 + 0.004τD K cu
1.875
and from the real part:
4.625ω2 − 1
1 + 0.004K cu − 4.625ω = 0 ⇒ K cu =
.
0.004
2
Or we can use the result from the imaginary part first:
 3.75 + 0.004τD K cu 
1 + 0.004K cu − 4.625
=0
1.875


4.625 ⋅ 3.75 4.625 ⋅ 0.004τ D K c
1 + 0.004 K c −
−
=0
1.875
1.875
4.625 ⋅ 0.004τD 
4.625 ⋅ 3.75

K cu =
−1
 0.004 −

1.875
1.875


4.625 ⋅ 3.75 − 1.875
3,867
K cu =
=
.
0.004 ⋅ 1.875 − 4.625 ⋅ 0.004τD 1.875 − 4.625τD
So, for τD = 0.25 , K cu = 5380 which is what we found previously. Next, for τD = 0.50 ,
K cu = −8839 — it is not immediately obvious how to interpret that, but we know from the
Routh array analysis that this is a lower limit on K c , so this case has unconditional stability.
Example with P & PI Control
Let’s consider the following example where the actuator & measurement devices have nonnegligible dynamics.
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Colorado School of Mines CHEN403
5
10s + 1
L′
R′
+
1
s +1
Gc
-
10
10s + 1
+
+
Y′
1
s +1
For this problem the characteristic equation will be:
1
1
10
⋅
⋅
=0
s + 1 s + 1 10s + 1
2
( s + 1) (10s + 1) + 10Gc = 0
1 + Gc
10s3 + 21s 2 + 12s + 1 + 10Gc = 0
P Control – Routh Array Analysis. With P control, the characteristic equation is:
10 ⋅ s3 + 21 ⋅ s 2 + 12 ⋅ s + (1 + 10K c ) = 0
and the Routh array is as follows:
Row
1
2
1
10
12
2
21
1 + 10K c
3
4
12 −
10 (1 + 10K c )
21
1 + 10K c
Notice that the restriction on K c from the 4th row is exactly the same as the restriction on
K c from the polynomial’s coefficients, namely
1 + 10K c > 0 ⇒ K c > −
1
.
10
This is not really a restriction since we are considering only positive values for the
controller gain.
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Colorado School of Mines CHEN403
However, the 3rd row does give a restriction on K c :
12 −
10 (1 + 10K c )
21
> 0 ⇒ Kc <
12 × 21 − 10
= 2.42 .
100
There are two important features to the result, (1) the value of the controller gain that will
be on the edge of stability and (2) whether this value is a maximum controller gain to be
used or a minimum. In this case we see that the ultimate controller gain is a maximum –
this is typical of P control – the gain can be small & the larger that it is made, the more
likely the system will start going unstable.
P Control – Direct Substitution Analysis. With P control, the characteristic equation with the
substitution of s = ωj is:
−10 ⋅ ω3 j − 21 ⋅ ω2 + 12 ⋅ ωj + (1 + 10K c ) = 0
The imaginary part gives:
(
)
−10 ⋅ ω3u + 12 ⋅ ωu = 0 ⇒ ωu 12 − 10 ⋅ ωu2 = 0 ⇒ ωu2 =
12
3
⇒ ωu = 2
10
10
and the real part gives:
−21 ⋅ ω2u + (1 + 10K cu ) = 0 ⇒ K c =
21 ⋅ ω − 1
=
10
2
u
21 ⋅
12
−1
10
= 2.42
10
which is the same ultimate value as found by the Routh array analysis. The only difference
between the two techniques is that we are not told whether this K cu value is an upper or
lower limit on K c .
PI Control – Routh Array Analysis. With PI control, the characteristic equation is:

1 
10 ⋅ s3 + 21 ⋅ s 2 + 12 ⋅ s + 1 + 10K c  1 +
=0
τI s 

10K c
10 ⋅ s 4 + 21 ⋅ s3 + 12 ⋅ s2 + (1 + 10K c ) s +
=0
τI
and the Routh array is as follows:
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Colorado School of Mines CHEN403
Row
1
2
3
1
10
12
10K c
τI
2
21
1 + 10K c
12 −
3
10 (1 + 10K c )
10K c
τI
21
4
10K c
τI
1 + 10K c −
10 (1 + 10K c )
12 −
21
5
10K c
τI
21
Here the 3rd row gives us the same restriction on K c as the P control situation. Now we
also have a restriction on the integral time from the 4th row:
10K c
4410K c
τI
.
> 0 ⇒ τI >
1 + 10K c −
10 (1 + 10K c )
1 + 10K c )( 242 − 100K c )
(
12 −
21
21
There are two things to note here: (1) the ultimate value of τI that leads to instability is a
minimum value of integral time & (2) the actual value will depend upon whatever value of
controller gain is used. That idea that the integral time can be reduced until instability
occurs makes intuitive sense, since very large values of τI “turns off” the integral action in
the PI controller.
One rule-of-thumb states that the K c value chosen for PI control should be ½ the ultimate
value from P control. Using this criteria, K c here should be 1.21; however, we’ll use K c = 1
to simplify the math. Then the values of τI that lead to a stable system are:
τI >
4410
= 2.82 .
11 × 142
PI Control – Direct Substitution Analysis. With PI control, the characteristic equation using
K c = 1 and with the substitution of s = ωj is:
10 ⋅ ω4 − 21 ⋅ ω3 j − 12 ⋅ ω2 + 11ωj +
Stability Feedback Control Systems
10
= 0.
τI
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December 21, 2008
Colorado School of Mines CHEN403
The imaginary part gives:
−21 ⋅ ω3u + 11ωu = 0 ⇒
(11 − 21 ⋅ ω ) ω
2
u
u
= 0 ⇒ ωu2 =
11
11
⇒ ωu =
21
21
and the real part gives:
10 ⋅ ωu4 − 12 ⋅ ωu2 +
10
10
= 0 ⇒ τIu =
=
2
12 ⋅ ωu − 10 ⋅ ωu4
τIu
10
11
 11 
12 ⋅ − 10  
21
 21 
2
= 2.82
which is the same ultimate value as found by the Routh array analysis. Again, the only
difference between the two techniques is that we are not told whether this τIu value is an
upper or lower limit on τI .
Example with Dead Time
Let’s examine the stability of the following system when using P control.
L′
R′
+
-
Kc
+
+
2e − s
5s + 1
C′
The characteristic equation for the closed loop is:
2e − s
1 + Kc
=0 ⇒
5s + 1
(5s + 1) + 2K c e − s = 0 .
Using a 1st order Padé approximation:
e
−θs
1
1 − θs
2 = 2 − θs
≈
1
1 + θs 2 + θs
2
the characteristic equation becomes:
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Colorado School of Mines CHEN403
2− s
≈0
2+ s
(5s + 1)(2 + s ) + 2K c ( 2 − s ) = 0
(5s + 1) + 2K c
(5s
2
+ 11s + 2) + 2K c ( 2 − s ) = 0
5s2 + (11 − 2K c ) s + ( 2 + 4K c ) = 0 .
The first limit to stability is the coefficient on the s term. For it to remain positive, then:
11 − 2K c > 0 ⇒ K c < 5.5 .
Next, we’ll do the full Routh array analysis. The Routh array will be:
Row
1
2
1
5
2 + 4K c
2
11 − 2K c
3
2 + 4K c
The stability limit is still controlled by the same coefficient.
Now, let’s use the direct substitution method. The characteristic equations becomes:
(5ωj + 1) + 2K c e −ωj = 0
(5ωj + 1) + 2K c ( cos ω − j sin ω) = 0
[1 + 2K c cos ω] + j [5ω − 2K c sin ω] = 0 .
The ultimate values can be determined by setting each part equal to zero & doing some
algebraic manipulation:
1 + 2K c cos ω = 0 ⇒ K cu = −
and:
1
2cos ω
sin ω
=0
cos ω
5ω + tan ω = 0
5ω − 2K cu sin ω = 0 ⇒ 5ω +
The value for ω must be solved numerically & then substituted in to get K cu . When doing
this, we find ω = 1.6887 and K cu = 4.251 .
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Colorado School of Mines CHEN403
Note that the use of 1st order overestimates the ultimate value. One can use higher order
approximations to determine what order approximations will give good estimates to K cu .
The following table shows estimates using various order Padé approximations (the
calculations were performed using Mathematica). Notice that the a values from the
approximate tranfer functions very quickly approach the exact value from the direct
substitution method.
Order Padé
Approximation
1st
2nd
3rd
Exact
K cu
5.5
4.289
4.252
4.251
Root-Locus Analysis
Can plot in the complex plane the values of the roots of the characteristic equation as the
controller parameter change. These root loci plots can be useful to determine
characteristics of the response of the system.
For example, for two tanks in series with P control, assuming G f = Gm = 1 , then:
Gp =
Kp
( τ1 s + 1)( τ2 s + 1)
and Gc = K c
and the characteristic equation is:
1 + Gc Gp = 1 +
KcK p
=0
( τ1 s + 1)( τ2 s + 1)
( τ1 s + 1)( τ2s + 1) + K c K p = 0
τ1 τ2 s 2 + ( τ1 + τ2 ) s + 1 + K c K p = 0
so the roots of this equation are given by:
p1 , p2 =
=
− ( τ1 + τ2 ) ±
( τ1 + τ2 )
2
− 4τ1 τ2 (1 + K c K p )
2τ1 τ2
− ( τ1 + τ2 ) ±
( τ1 − τ2 )
2
− 4τ1 τ2K c K p
2τ1 τ2
From this, you can make the following observations:
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December 21, 2008
Colorado School of Mines CHEN403
•
When K c = 0 , then:
p1 , p2 =
•
− ( τ1 + τ2 ) ±
( τ1 − τ2 )
2
=
− ( τ1 + τ2 ) ± ( τ1 − τ2 )
=−
1
1
,−
τ1 τ2
2τ1 τ2
2τ1 τ2
the “standard” roots for a non-interacting series of processes.
We can get a critically damped response when the roots are repeated, or when:
( τ1 − τ2 )
2
− 4τ1 τ2K c K p = 0 ⇒ K c
(τ − τ )
= 1 2
2
4τ1 τ2K p
So, this means that the system will be overdamped when:
(τ − τ )
< 1 2
2
( τ1 − τ2 )
2
•
− 4τ1 τ2K c K p > 0 ⇒ K c
4τ1 τ2K p
Furthermore, for these values of K c the poles will remain negative real.
We can get an underdamped response when:
(τ − τ )
> 1 2
2
( τ1 − τ2 )
2
− 4τ1 τ2K c K p < 0 ⇒ K c
4τ1 τ2K p
The poles are then given by:
4τ1 τ2K c K p − ( τ1 − τ2 )
τ +τ
p1 , p2 = − 1 2 ± i
2τ1 τ2
2τ1 τ2
The real portion of these roots remain negative, whereas the imaginary part gets
bigger and bigger as K c increases.
2
These types of plots are more instructive for more complicated systems — for example, see
the plot on page 265 of SEM. This plot shows that as K c increases, an underdamped
system results. However, the real portion of the complex conjugate roots also increases,
going positive at K c > 30 . So, the gain must be kept below this value.
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December 21, 2008