control chap5
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CONTROL SYSTEMS THEORY
Transient response stability
CHAPTER 5STB 35103
Objective To determine the stability of a system
represented as a transfer function.
Introduction In chapter 1, we learnt about 3
requirements needed when designing a control system Transient response Stability Steady-state errors
Introduction What is stability?
Most important system specification. We cannot use a control system if the system is
unstable Stability is subjective
From Chapter 1, we have learned that we can control the output of a system if the steady-state response consists of only the forced response. But the total response, c(t)
forced naturalc t c t c t
Introduction Using this concept we can summarize the
definitions for linear, time-invariant systems.
Using natural response; A system is stable if the natural response
approaches zero as time approaches infinity. A system is unstable if the natural response
approaches infinity as time approaches infinity. A system is marginally stable if the natural
response neither decays nor grows but remains constant or oscillates.
Introduction A system is stable if every bounded input
yields a bounded output. (bounded = terkawal). We call this statement the bounded-input, bounded-output (BIBO).
Using the total response (BIBO) A system is stable if every bounded input
yields a bounded output. A system is unstable if any bounded input
yields an unbounded output.
Introduction We can also determine the stability of a
system based on the system poles. Stable systems have closed-loop transfer
functions with poles only in the left half-plane. Unstable systems have closed-loop transfer
functions with at least one pole in the right half-plane and/or poles of multiplicity greater than 1 on the imaginary axis.
Marginally stable systems have closed-loop transfer functions with only imaginary axis poles of multiplicity 1 and poles in the left half-plane.
Introduction Figure 5.1 a indicates closed-loop poles for
a stable system.
Figure 5.1 a - Closed-loop poles and response for stable system
Introduction Figure 5.1 a indicates closed-loop poles for
an unstable system.
Figure 5.1 b - Closed-loop poles and response for unstable system
Stability summary
Introduction In order for us to know the stability of our
system we need to draw the system poles. To find the poles we need to calculate the roots of the system polynomials.
Try to get the system poles for the systems in Figure 5.1 a and Figure 5.1 b.
Introduction What about this system? Can you find the
root locus for this polynomial?
A method to find the stability without solving for the roots of the system is called Routh-Hurwitz Criterion.
Figure 5.2 – Close loop system with complex polynomial.
Routh-Hurwitz Criterion We can use Routh-Hurwitz criterion method
to find how many closed-loop system poles are in the LHP, RHP and on the jω-axis
Disadvantage : We cannot find their coordinates
The method requires two steps: Generate a data table called a Routh table Interpret the Routh table to tell how many close-
loop system poles are in the left half-plane, the right half-plane, and on the jω-axis
Routh-Hurwitz Criterion Example
Figure 5.3 displays an equivalent closed loop transfer function. In order to use Routh table we are only going to focus on the denominator.
Figure 5.3 – Equivalent closed-loop transfer function
Routh-Hurwitz Criterion First step (1)
Based on the denominator in Figure 5.3, the highest power for s is 4, so we can draw initial table based on this information. We label the row starting with the highest power to s0.
s4
s3
s2
s1
s0
Routh-Hurwitz Criterion Input the coefficient values for each s
horizontally starting with the coefficient of the highest power of s in the first row, alternating the coefficients.
s4 a4 a2 a0
s3 a3 a1 0
s2
s1
s0
Remaining entries are filled as follows. Each entry is a negative determinant of entries in the previous two rows divided by the entry in the first column directly above the calculated row.
Routh-Hurwitz Criterion
Routh-Hurwitz Criterion
Routh-Hurwitz Criterion
Routh-Hurwitz Criterion
Routh-Hurwitz Criterion
Routh-Hurwitz Criterion
Routh-Hurwitz Criterion
Routh-Hurwitz Criterion
Routh-Hurwitz Criterion
Routh-Hurwitz Criterion Example 6.1
Make a Routh table for the system below
Answer: get the closed-loop transfer function
Routh-Hurwitz Criterion We can multiply any row in Routh table by
a positive constant without changing the rows below.
Routh-Hurwitz Criterion Interpreting the basic Routh table
In this case, the Routh table applies to the systems with poles in the left and right half-planes.
Routh-Hurwitz criterion declares that the number of roots of the polynomial that are in the right half-plane is equal to the number of sign changes in the first column.
Routh-Hurwitz Criterion
Routh-Hurwitz Criterion If the closed-loop transfer function has all poles in the
left half of the s-plane, the system is stable. The system is stable if there are no sign changes in
the first column of the Routh table. Example:
Routh-Hurwitz Criterion
Based on the table, there are two sign changes in the first column. So there are two poles exist in the right half plane. Which means the system is unstable.
+
-
-
+
Routh-Hurwitz Criterion Exercise 1
Make a Routh table and tell how many roots of the following polynomial are in the right half-plane and in the left half-plane.
7 6 5 4 3 23 9 6 4 7 8 2 6P s s s s s s s s
Routh-Hurwitz CriterionAnswer
Routh-Hurwitz CriterionAnswer
Routh Hurwitz Exercise 2
Routh Hurwitz Solution
Routh Hurwitz Exercise 3
Solution
Routh Hurwitz Exercise 4
Routh Hurwitz Solution
Routh-Hurwitz Criterion: Special cases Two special cases can occur:
Routh table has zero only in the first column of a row
Routh table has an entire row that consists of zeros.
s3 1 3 0
s2 3 4 0
s1 0 1 2
s3 1 3 0
s2 3 4 0
s1 0 0 0
Routh-Hurwitz Criterion: Special cases Zero only in the first column
There are two methods that can be used to solve a Routh table that has zero only in the first column.
1. Stability via epsilon method
2. Stability via reverse coefficients
Routh-Hurwitz Criterion: Special cases Zero only in the first column
Stability via epsilon method
Example 6.2Determine the stability of the closed-loop
transfer function
5 4 3 2
10
2 3 6 5 3T s
s s s s s
Routh-Hurwitz Criterion: Special cases Solution: We will begin forming the Routh table
using the denominator. When we reach s3 a zero appears only in the first column.
s5 1 3 5
s4 2 6 3
s3 0 7/2 0
s2
s1
s0Zero in first column
Routh-Hurwitz Criterion: Special cases If there is zero in the first column we
cannot check the sign changes in the first column because zero does not have ‘+’ or ‘-’.
A solution to this problem is to change 0 into epsilon (ε).
s5 1 3 5
s4 2 6 3
s3 0 7/2 0
s2
s1
s0
ε
Routh-Hurwitz Criterion: Special cases We will then calculate the determinant for
the next s values using the epsilon.
Routh-Hurwitz Criterion: Special cases If we all the columns and rows in the
Routh table we will get
Routh-Hurwitz Criterion: Special cases We can find the number of poles on the
right half plane based on the sign changes in the first column. We can assume ε as ‘+’ or ‘-’
Routh-Hurwitz Criterion: Special cases There are two sign changes so there are
two poles on the right half plane. Thus the system is unstable.
Routh-Hurwitz Criterion: Special cases Zero only in the first column
Stability via reverse coefficients
Example 6.3Determine the stability of the closed-loop
transfer function
5 4 3 2
10
2 3 6 5 3T s
s s s s s
Routh-Hurwitz Criterion: Special cases Solution: First step is to write the denominator in
reverse order (123653 to 356321)
We can form the Routh table using D(s) values.
5 4 3 23 5 6 3 2 1D s s s s s s
Routh-Hurwitz Criterion: Special cases The Routh table indicates two signal
changes. Thus the system is unstable and has two right-half plane poles.
Routh-Hurwitz Criterion: Special cases Entire row is zero
the method to solve a Routh table with zeros in entire row is different than only zero in first column.
When a Routh table has entire row of zeros, the poles could be in the right half plane, or the left half plane or on the jω axis.
Routh-Hurwitz Criterion: Special cases Example 6.4
Determine the number of right-half-lane poles in the closed-loop transfer function
5 4 3 2
10
7 6 42 8 56T s
s s s s s
Routh-Hurwitz Criterion: Special cases Solution: Start with forming the initial Routh table
Routh-Hurwitz Criterion: Special cases We can reduce the number in each row
Routh-Hurwitz Criterion: Special cases We stop at the third row since the entire
row consists of zeros.
When this happens, we need to do the following procedure.
Routh-Hurwitz Criterion: Special cases Return to the row immediately above the
row of zeros and form the polynomial.
The polynomial formed is
4 26 8P s s s
Routh-Hurwitz Criterion: Special cases Next we differentiate the polynomial with
respect to s and obtain
We use the coefficient above to replace the row of zeros. The remainder of the table is formed in a straightforward manner.
34 12 0dP s
s sds
Routh-Hurwitz Criterion: Special cases The Routh table when we change zeros
with new values
Routh-Hurwitz Criterion: Special cases Solve for the remainder of the Routh table
There are no sign changes, so there are no poles on the right half plane. The system is stable.
Example 1- normal
Example 2 - special case
Example 3 – special case
Example 4 – special case
Exercise Given that G(s) is the open loop transfer
function for a unity feedback system, find the range of K to yield a stable system
Solution
Exercise Find the range of K to yield a stable
system given the closed loop transfer function below
Solution
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