Factorization of the Third-Degree Characteristic Equation
Abstract Generated abstract
This paper presents a simplified computational method for factoring a third degree characteristic equation with positive coefficients, under the usual Routh Hurwitz stability condition and assuming no multiple roots. The equation is represented as a product of a first order and a second order factor, leading to relations among three positive time constants, after which existence intervals for the constants and their product are derived. Two iterative cases are distinguished according to the ratio of the limiting quantities, and formulas are given for successive approximations with checks against the original coefficient. Worked examples indicate that preliminary determination of admissible intervals can reduce the number of iterations compared with an existing iteration method and make the calculation more directed.
Full Text
FACTORIZATION OF THE THIRD-DEGREE CHARACTERISTIC EQUATION
Yu. M. Trifonov
(Presented by the Department of ATPP)
A whole series of works [1, 2, 3, 4] has been devoted to the question of factorization or of determining the roots of a characteristic equation of the 3rd degree. The methods set forth in these works make it possible to solve the problem, but require the expenditure of a considerable amount of labor. The proposed method reduces the amount of computational work if the equation contains no multiple roots. This condition is practically always satisfied, since in the overwhelming majority of cases the characteristic equations of automatic-control systems do not contain multiple roots [3].
The method of factorization considered below can be applied to a characteristic equation of the form
\[ a_3p^3+a_2p^2+a_1p+1=0, \tag{1} \]
where
\[ a_1>0,\quad a_2>0,\quad a_3>0. \]
If the Routh–Hurwitz stability condition is satisfied, then
\[ a_1\cdot a_2>a_3. \]
Represent expression (1) in the form
\[ (T_1p+1)(T_2p^2+T_3p+1)=0 \tag{2} \]
where \(T_1>0,\ T_2>0,\ T_3>0\).
After transformation we obtain
\[ T_1T_2p^3+(T_1T_3+T_2)p^2+(T_1+T_3)p+1=0. \tag{3} \]
Comparing expressions (1) and (3), we see that
\[ \begin{aligned} T_1T_2&=a_3, &&(4-a)\\ T_1T_3+T_2&=a_2, &&(4-b)\\ T_1+T_3&=a_1. &&(4-c) \end{aligned} \tag{4} \]
Let us determine the limits that bound the interval of existence of \(T_1\). From expression (4-b) it follows that
\[ T_1=\frac{a_2-T_2}{T_3}. \]
For \(T_1\) to be positive, it is necessary that the inequality \(T_2 < a_2\) hold. From equation \((4-a)\) we have
\[ T_2=\frac{a_3}{T_1}, \quad \text{whence} \quad T_1>\frac{a_3}{a_2}. \]
The second limit of existence of \(T_1\) is determined from equation \((4-в)\)
\[ T_3=a_1-T_1. \]
For \(T_3>0\), it is necessary that the condition \(T_1<a_1\) hold. Thus, \(T_1\) and \(T_3\) exist in the intervals
\[ \frac{a_3}{a_2}<T_1<a_1; \quad 0<T_3<a_1-\frac{a_3}{a_2}. \]
To solve system \((4)\), it is necessary to find the interval of existence of \(T_1\cdot T_3\).
A. To this end, first find
\[ \lim_{T_3\to 0}T_1=a_1, \quad \text{then} \quad \lim_{T_3\to 0}T_2=\frac{a_3}{a_1}. \]
From equation \((4-б)\) we have
\[ T_1T_3 \leq a_2-\frac{a_3}{a_1}=c' \tag{5} \]
For
\[ \lim_{T_3\to a_1-\frac{a_3}{a_2}}T_1=\frac{a_3}{a_2}, \quad \lim_{T_3\to a_1-\frac{a_3}{a_2}}T_2=a_2, \quad T_1T_3=0. \]
\(T_1T_3\) exists in the interval
\[ 0<T_1T_3\leq c'. \]
B. Suppose that the product \(T_1\cdot T_3=c''\).
Then consider the system of equations
\[ \begin{cases} T_1\cdot T_3=c'', & (6)\\ T_1+T_3=a_1. & (4-в) \end{cases} \]
Express \(T_3\) from equation \((4-в)\) and substitute it into equation \((6)\); then we obtain
\[ T_1^2-a_1T_1+c''=0, \]
\[ T_1=\frac{a_1}{2}\pm \sqrt{\frac{a_1^2}{4}-c''}, \]
since \(T_1>0\) and is a real number, then
\[ \frac{a_1^2}{4}-c''\geq 0. \]
Hence
\[ c''\leq \frac{a_1^2}{4}. \tag{7} \]
In order to choose the first approximation \(T_1\cdot T_3\), it is necessary to compare the following inequalities:
\[ 0<T_1T_3\leq c', \tag{8-а} \]
\[ 0<T_1T_3\leq c''. \tag{8-б} \]
From consideration of expressions (8–a) and (8–b) it follows that
\[ 0<T_1\cdot T_3 \leq c_1\,(8-\mathrm{b}), \]
where \(c_1\) is equal to the smaller of the numbers \(c'\) and \(c''\).
When the intervals of existence of \(T_1\) and \(T_1\cdot T_3\) have been found, the given characteristic equation of the 3rd degree can be factored. The following cases are possible:
Case 1
\[ \frac{c'}{c''}<1, \]
then
\[ c_1=a_2-\frac{a_3}{a_1}. \]
Let us represent graphically expressions (4–b) and (8–b) by solid lines in Fig. 1. It is evident from the figure that \(c\) can only decrease or be equal to the number \(c_1\). A decrease of the number \(c\) means that the hyperbola is shifted toward the origin (see the dashed line in Fig. 1). Therefore, if we now determine the point of intersection of the hyperbola \(c_1\) with the straight line, then by this the interval of existence of \(T_1\) is substantially reduced. The system of equations (4–b) and (8–b) is equivalent to the equation
Fig. 1. Graphical representation of equations (4–b) and (8–b).
\[ T^2-a_1T+c_1=0, \]
from which we have
\[ T_{1,3}=\frac{a_1}{2}\pm \sqrt{\frac{a_1^2}{4}-c_1}. \tag{9} \]
In the first approximation the values of \(T_1\) and \(T_3\) are equal to
\[ T_1'=\frac{a_1}{2}+\sqrt{\frac{a_1^2}{4}-c_1}, \]
\[ T_3'=\frac{a_1}{2}-\sqrt{\frac{a_1^2}{4}-c_1}. \]
Check. We compute the value \(T_1'T_3' : T_2'=a_2'\) and compare it with \(a_2\); if \(a_2'\) differs from \(a_2\), then we determine \(c_2\) by the formula
\[ c_2=a_2-\frac{a_3}{T_1'}. \tag{10} \]
Then we again determine \(T_1'\cdot T_3' : T_2'=a_2\) and carry out the check. The calculation is continued until the required accuracy is obtained (Example No. 1).
Case 2.
\[ \frac{c'}{c''}>1. \]
In this case, after determining the intervals of existence of \(T_1\) and \(T_1\cdot T_3\), one should determine \(T_1'\) from the expression
\[ T_1'=\frac{a_3}{a_2-c_1}, \tag{11} \]
then from (4—a) and (4—в) we obtain:
\[ T_3'=a_1-T_1', \]
\[ T_2'=\frac{a_3}{T_1'}. \]
Check. We compute the value \(T_1'T_3' : T_2'=a_2'\) and compare it with \(a_2\). If \(a_2'\) differs from \(a_2\), then we determine \(c_2\), and from it find \(T_1''\). The formula for determining \(c_2\) is obtained after substituting the value \(T_1'\) into expression (9):
\[ T_1'=\frac{a_1}{2}+\sqrt{\frac{a_1^2}{4}-c_2}, \]
whence
\[ c_2=T_1'(a_1-T_1'). \tag{12} \]
We carry out the computations until \(c_{n-1}-c_n\simeq 0\) with the required accuracy.
Example 1.
Consider the equation
\[ 7p^3+3p^2+5p+1=0. \]
According to (1) we have
\[ a_3=7,\quad a_2=3,\quad a_1=5. \]
Step 1. Let us check whether the Routh–Hurwitz condition is satisfied:
\[ 3\cdot 5>7. \]
Step 2. Determine the interval of existence of the time constant \(T_1\)
\[ \frac{7}{3}=2.33<T_1<5. \]
Step 3. From expression (8-c), determine the interval of existence of \(T_1\cdot T_3\)
\[ c'=a_2-\frac{a_3}{a_1},\qquad c'=3-\frac{7}{5}=1.6; \]
\[ c''=\frac{a_1^2}{4},\qquad c''=\frac{5^2}{4}=6.25. \]
Consequently, \(0<T_1\cdot T_3<1.6\).
Step 4. Determine the ratio
\[ \frac{c'}{c''}=\frac{1.6}{6.25}=0.256. \]
Since \(\dfrac{c'}{c''}<1\), we compute \(T_1\) by formula (9)
\[ T_1'=\frac{5}{2}+\sqrt{6.25-1.6}=4.66; \]
\[ T_3'=5-4.66=0.34;\qquad T_2'=\frac{7}{4.66}=1.5. \]
Check. \(a_2'=4.66\cdot 0.33+1.5=3.04\).
Step 5. Carry out the second approximation
\[ C_2=3-\frac{7}{4.66}=1.5, \]
then
\[ T_1=4.7;\qquad T_3=0.3;\qquad T_2=\frac{7}{4.7}=1.49. \]
Check. \(a_2''=4.7\cdot 0.3+1.49=3.0\)
We have obtained complete agreement of \(a_2''\) with \(a_2\).
Example 2.
Solve the following equation:
\[ 0.025p^3+0.225p^2+0.75p+1=0. \]
Similarly, we carry out steps 1, 2, 3.
-
\(0.225\cdot 0.75>0.025.\)
-
\[ \frac{0.025}{0.225}=0.111<T_1<0.75. \]
3.
\[
c'=0.225-\frac{0.025}{0.75}=0.22;\qquad
c''=\frac{0.75^2}{4}=0.14;\qquad
\frac{c'}{c''}=1.56.
\]
Consequently, \(0<T_1\cdot T_3<0.14\).
- Use formulas (11) and (12)
\[ T_1'=\frac{0.025}{0.225-0.140}=0.294;\qquad c_2=0.294(0.75-0.294)=0.134; \]
\[ T_1''=\frac{0{,}025}{0{,}225-0{,}134}=0{,}275;\quad c_3=0{,}275(0{,}75-0{,}275)=0{,}130; \]
\[ T_1'''=\frac{0{,}025}{0{,}225-0{,}130}=0{,}263;\quad c_4=0{,}263(0{,}75-0{,}263)=0{,}128. \]
\[ T_1^{\mathrm{IV}}=\frac{0{,}025}{0{,}225-0{,}128}=0{,}254;\quad c_5=0{,}254(0{,}75-0{,}254)=0{,}126. \]
\[ T_1^{\mathrm{V}}=\frac{0{,}025}{0{,}225-0{,}126}=0{,}253;\quad c_6=0{,}253(0{,}75-0{,}253)=0{,}125. \]
\[ T_1^{\mathrm{VI}}=\frac{0{,}025}{0{,}225-0{,}125}=0{,}25;\quad c_7=0{,}25(0{,}75-0{,}25)=0{,}125. \]
Check.
\[ T_3=0{,}75-0{,}25=0{,}5;\quad T_2=\frac{0{,}025}{0{,}25}=0{,}1, \]
\[ a_2=0{,}25\cdot 0{,}5+0{,}1=0{,}225. \]
A complete agreement was obtained.
Let us compare the proposed method with the iteration method [2] for third-order equations. In example 2, computed by the iteration method [1], the solution was obtained after 10 iterations. When this same example is solved by the present method, the result is obtained after 6 approximations with the same accuracy.
Conclusion
Determining the interval in which the time constants exist before carrying out the calculation reduces the number of iterations and makes the calculation more purposeful.
The convergence of the computations is the greater, the stronger the inequality
\[ \frac{c'}{c''}\gtrless 1. \]
REFERENCES
-
G. N. Balansanov. Fundamentals of Automation of Technological Processes in the Hydrometallurgy of Rare and Radioactive Metals. Atomizdat, 1960.
-
O. M. Kryzhanovskii. On an iteration method for determining approximate values of the roots of equations. Avtomatika i telemekhanika, XI, No. 3, 1950.
-
B. N. Naumov. Transient Processes in Linear Automatic-Control Systems. Gosenergoizdat, 1960.
-
K. Lanczos. Practical Methods of Applied Analysis. Fizmatgiz, 1961.