3. Introduction
In chapter 1, we learnt about 3
requirements needed when designing a
control system
Transient response
Stability
Steady-state errors
4. 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 ) = cforced ( t ) + cnatural ( t )
c(t)
5. 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.
6. 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.
7. 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 halfplane.
8. 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
9. 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
11. 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.
12. Introduction
What about this system? Can you find the
root locus for this polynomial?
Figure 5.2 – Close loop system with complex polynomial.
A method to find the stability without
solving for the roots of the system is
called Routh-Hurwitz Criterion.
13. 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 halfplane, the right half-plane, and on the jω-axis
14. Routh-Hurwitz Criterion
Example
Figure 5.3 – Equivalent closed-loop transfer function
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.
15. 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
16. 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
17. Routh-Hurwitz Criterion
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.
28. 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 halfplanes.
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.
30. 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:
31. 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.
32. 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.
P ( s ) = 3s + 9s + 6s + 4s + 7 s + 8s + 2s + 6
7
6
5
4
3
2
40. Routh-Hurwitz Criterion: Special cases
Two special cases can occur:
Routh table has zero only in the first column of
a row
s3
3
0
s2
3
4
0
s1
1
0
1
2
Routh table has an entire row that consists of
zeros.
s3
1
3
0
s2
3
4
0
s1
0
0
0
41. 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
42. Routh-Hurwitz Criterion: Special cases
Zero only in the first column
Stability via epsilon method
Example 6.2
Determine the stability of the closed-loop
transfer function
10
T ( s) = 5
s + 2 s 4 + 3s 3 + 6 s 2 + 5s + 3
43. 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
Zero in first
column
s0
44. 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
47. 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 ‘-’
48. 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.
49. Routh-Hurwitz Criterion: Special cases
Zero only in the first column
Stability via reverse coefficients
Example 6.3
Determine the stability of the closed-loop
transfer function
10
T ( s) = 5
s + 2 s 4 + 3s 3 + 6 s 2 + 5s + 3
50. Routh-Hurwitz Criterion: Special cases
Solution:
First step is to write the denominator in
reverse order (123653 to 356321)
D ( s ) = 3s + 5s + 6 s + 3s + 2s + 1
5
4
3
2
We can form the Routh table using D(s)
values.
51. Routh-Hurwitz Criterion: Special cases
The Routh table indicates two signal
changes. Thus the system is unstable and
has two right-half plane poles.
52. 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.
53. Routh-Hurwitz Criterion: Special cases
Example 6.4
Determine the number of right-half-lane poles
in the closed-loop transfer function
10
T ( s) = 5
s + 7 s 4 + 6s 3 + 42s 2 + 8s + 56
56. 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.
57. Routh-Hurwitz Criterion: Special cases
Return to the row immediately above the
row of zeros and form the polynomial.
The polynomial formed is
P ( s ) = s + 6s + 8
4
2
58. Routh-Hurwitz Criterion: Special cases
Next we differentiate the polynomial with
respect to s and obtain
dP ( s )
ds
= 4 s 3 + 12 s + 0
We use the coefficient above to replace
the row of zeros. The remainder of the
table is formed in a straightforward
manner.
60. 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.
