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We have suggested a physical system by the components placed in parentheses. Find the characteristic equation and its roots, and determine whether the system is stable.
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The first step is to write the open-loop transfer function:
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+ lW.5s + 1) = o
42s + 1)
which is equivalent to
s2+3s+5=o
Solving by the quadratic formula gives
3 s=-* 2 JG-36 2
or fi Sl = $+jF s2 =
-3 fi
--j, 2
Since the real part of s1 and s2 is negative (+), the system is stable.
ROUTH TEST FOR STABILITY
The Routh test is a purely algebraic method for determining how many roots of the characteristic equation have positive real parts; from this it can also be determined whether the system is stable, for if there are no roots with positive real parts, the system is stable. The test is limited to systems that have polynomial characteristic equations. This means that it cannot be used to test the stability of a control system containing a transportation lag. The procedure for application of the Routh test is presented without proof. The proof is available elsewhere (Routh, 1905) and is mathematically beyond the scope of this text. The procedure for examining the roots is to write the characteristic equation in the form
a(# + qsn- + u*r2 + ... + a, = 0
(14.9)
where aa is positive. (If aa is originally negative, both sides are multiplied by -1.) In this form, it is necessary that all the coefficients uo, al, a2, . . . , a,-1, a,
he positive if all the roots are to lie in the left half plane. If any coefficient is
negative, the system is definitely unstable, and the Routh test is not needed to answer the question of stability. (However, in this case, the Routh test will tell
LINEAR
CLOSED-LOOP
SYSTEMS
us the number of roots in the right half plane.) If all the coefficients are positive, the system may be stable or unstable. It is then necessary to apply the following procedure to determine stability.
Routh Array
Arrange the coefficients of Eq. (14.9) into the first two rows of the Routh array, as follows:
1 a0 a2
2 3 4 5 6
a1 bl Cl dl
The array has been filled in for n = 7 in order to simplify the discussion. For any other value of n, the array is prepared in the same manner. In general, there are (n + 1) rows. For n even, the first row has one more element than the second row. The elements in the remaining rows are found from the formulas b1=
Cl =
ala2 - aoas a1
b2 = c2
ala4
- aoa5 a1
ha3 - albz bl
has - albs . . h
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The elements for the other rows are found from formulas that correspond to those just given. The elements in any row are always derived from the elements of the two preceding rows. During the computation of the Routh array, any row can be divided by a positive constant without changing the results of the test. (The application of this rule often simplifies the arithmetic.) Having obtained the Routh array, the following theorems are applied to determine stability.
Theorems of the Routh Test
1. The necessary and sufficient condition for all the roots of the characteristic equation [Eq. (14.9)] to have negative real parts (stable system) is that all elements of the first column of the Routh array (aa, a 1, b 1, c 1, etc.) be positive and nonzero.
STABILITY
2. If some of the elements in the first column are negative, the number of roots
with a positive real part (in the right half plane) is equal to the number of sign changes in the first column. 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, all the elements of the nth row will vanish and none of the elements of the preceding row will vanish. The location of the pair of imaginary roots can be found by solving the equation
(14.10) cs*+D=o 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. We shall find this last rule to be of value in the root-locus method presented in the next chapter.
The algebraic method for determining stability is limited in its usefulness in that all we can learn from it is whether a system is stable. It does not give us any idea of the degree of stability or the roots of the characteristic equation.
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