Lecture8 Stability of control system (to be continued)

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ChE 170
Lecture 8
Stability of control system
Response of control system for a unit-step
change in set point
For a unit-step change in R, the transform of the
response is
 As K c increases, the system response
becomes more oscillatory
 Beyond a certain value of Kc , the
successive amplitudes of the response
grow rather than decay; this type of
response is called unstable.
Definition of stability (linear systems)
Stable system - one for which the output response is bounded for
all bounded inputs.
Unstable system - exhibiting an unbounded response to a bounded
input is unstable.
STABILITY CRITERION
*Let
open-loop TF
Basic single-loop control system.
Characteristic equation
of the control system
**Where
is the open-loop transfer function
because it relates the measured variable B to the Stability criterion: A linear control
system is unstable if any real parts of the
set point R if the feedback loop is disconnected
from the comparator (i.e., if the loop is opened). roots of its characteristic equation are
on, or to the right of, the imaginary axis.
Otherwise the system is stable.
At RHP:
Variation
exponentially grow.
The farther from the
origin, the faster is
the growth
At LHP:
Variation
exponentially
decay.
The farther from the
origin, the faster is
the decay
Limit of stability!!!
Is when the pole is purely
imaginary (no real part)
CRITERION OF STABILITY
Characteristic equation of the
control system: 1 + G = 0
Characteristic equation of the control system: 1 + G = 0
Equivalent to:
Where,
*The stability of a control system is determined
solely by its open-loop transfer function through
the roots of the characteristic equation.
*If the real part of the roots of the characteristic
equation is negative (-), the system is stable.
HW
Consider the liquid-level control system shown .The tanks are noninteracting. The following information is
known:
1. The resistances on the tanks are linear. These resistances were tested separately, and it was found that, if
the steady-state flow rate q cfm is plotted against steady-state tank level h ft, the slope of the line dq/dh
is 2 ft2/min.
2. The cross-sectional area of each tank is 2 ft2.
3. The control valve was tested separately, and it was found that a change of 1 psi in pressure to the valve
produced a change in flow of 0.1 cfm.
4. There is no dynamic lag in the valve or the measuring element.
(a) Draw a block diagram of this control system, and in each block give the transfer
function, with numerical values of the parameters.
ROUTH TEST FOR STABILITY
 a purely algebraic method for determining how many roots of the
characteristic equation have positive real part
 if there are no roots with positive real parts, the system is stable
 Is limited to systems that have polynomial characteristic equations,
which means it cannot be used to test the stability of a control
system containing a transportation lag
When and How to apply the ROUTH TEST FOR
STABILITY
1. Write the characteristic equation in the form:
Where ao is positive. (If ao is originally negative, both sides are multiplied by -1).
Coefficients of equation: ao , a1 , a2 , . . . , an-1 , an
*If any coefficient is negative, the system is definitely unstable, NO need to apply Routh test
*If all the coefficients are positive, the system may be stable or unstable, apply Routh test
ROUTH TEST FOR STABILITY
1. Arrange the coefficients of characteristic equation (in
standard form) into the first two rows of the Routh array
2. Determine the remaining rows from the following:
Coefficient of characteristic equation in
standard polynomial form
3. Apply Routh test criterion: STABLE if
 (Theorem 1) all elements of the first column of the Routh
array are positive and nonzero
 (Theorem 2) there is NO sign changes in the first column.
Note: the number of sign changes is equal to the number of
roots with positive real parts
Apply Routh test criterion: STABLE if
 (Theorem 3) If one pair of roots is on the imaginary axis, equidistant
from the origin, and all other roots are in the left half plane,
 The location of the pair of imaginary roots can be found by solving the
equation:
where the coefficients C and D are the elements of the array in the (n - 1)th
row as read from left to right, respectively.
 this last rule to be of value in the root-locus method
n=7
Row of (n-1)
Example 6
Given the characteristic equation: s4 + 3s3 + 5s2 + 4s + 2 = 0
Determine the stability by the Routh criterion.
 Since all the coefficients am positive, the system may be stable, therefor Routh test is
applicable
n=4
Coefficient of characteristic equation in standard form
Calculated by equation
Example 7
Given the characteristic equation:
a. Determine the values of Kc for which the control system is stable.
b. For the value of Kc for which the system is on the threshold of
instability, determine the roots of the characteristic equation with the
help of Theorem 3
The characteristic equation 1 + G ( s ) = 0 becomes:
Rearrangement of this equation for use in the Routh test gives
Routh array:
<0
> -1
Theorem 1: Stable if elements 1st column are positive and non-zero
>0
>0
Therefore system will
be stable: -1 < < 10
Example 8
Determine the stability of the system shown below for which a PI
controller is used.
Limit of stability of a control system
(Direct substitution method – Frequency response stability)
To get the limit of
stability let
into
the equation and solve
of ultimate
,
Limit of stability of a control system
(Direct substitution method – Frequency response stability)
• The imaginary axis divides the complex plane into stable and unstable
regions for the roots of characteristic equation, as indicated in Fig. 11.26.
• On the imaginary axis, the real part of s is zero, and thus we can write s=jw.
Substituting s=jw 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.
Example 9
Use the direct substitution method to determine Kc,u for the system
with the characteristic equation given as:
Solution: Let s  jω
Where
gives 10 jω3  17ω2  8 jω  1  K c,u  0
and
Group the real and imaginary
term of the equation:
is the sustained oscillation
,
ω
ω
If both the real and imaginary parts are identically zero:
Solving:
ω
ω
ω
,
**To get the location of these purely imaginary pole, apply theorem 3 of the Routh stability
test
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