CH-03 (523171), страница 3
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It indicatesthat the mass reaches 50% of the unit-step disturbance in approximately 1.1 seconds.© 2001 by CRC Press LLCAn associated problem of the mechanical vibration problem is to find the magnitude and time of overshoot when the mass reaches the farthest point as illustratedin Figure 1. Instead of Equation 13, for calculation of overshoot we examine theequation:X(t ) = 1 − e −.2 t sin(t + 1.37)(14)To determine the maximum of X(t), we differentiate Equation 14 with respectto t to derive the expression for the first derivative of X(t).
That is:dX (t ) dt = e −.2 t [.2 sin(t + 1.37) − cos(t + 1.37)](15)The magnitude and time of maximum X(t) can then be determined by settingEquation 15 equal to zero. In so doing, the fourth option of the program FindRootis again applied using the bounds tr = 1 and tr = 2 to find that Xmax is equal to 1.523or overshoot is equal to 53% and occurs at t = 3.145 seconds. See Figure 2 fordefinitions of these design specifications.EXTENDED NEWTON-RAPHSON METHODThe iterative method of Newton-Raphson for solving a either polynomial ortranscendental equation of one variable can be extended into solution of multipleequations of multiple variables. Consider the case of two equations of two variables,u(x,y) = 0 and v(x,y) = 0.
Let (xg,yg) be a guessing solution of these two equations.In that neighborhood, the Taylor’s series for f(x,y) and g(x,y) are:()()()(16)()()()(17)u(x r , y r ) = u x g , y g + u,x x g , y g ∆x + u,y x g , y g ∆y + …andv(x r , y r ) = v x g , y g + v,x x g , y g ∆x + v,y x g , y g ∆y + …where u,∂u/∂x and v,y∂v/∂y, and the root location (xr,yr) is predicted using theadjustments x and y. That is,x r = x g + ∆xy r = y g + ∆yand(18,19)Since it is hoped that u(xr,yr) and v(xr,yr) would both be equal to zero,Equations 16 and 17 therefore can be expressed, after dropping the higher orderterms of x and y, in the forms of:()()()(20)()()()(21)u,x x g , y g ∆x + u,y x g , y g ∆y = − u x g , y gandv,x x g , y g ∆x + v,y x g , y g ∆y = − v x g , y g© 2001 by CRC Press LLCCramer’s rule can then be applied to obtain x and y as() (u v() (u∆x = − uv,y + vu,y− u , y v, x)(22)v − u , y v, x)(23),x , yand∆y = + uv,x − vu,x,x , ywhere u, v, and their derivatives are to be evaluated at (xg,yg).
Equations 22 and 23are to be continuously applied to adjust the guessing values of (xr,yr) until bothu(xr,yr) and v(xr,yr) are negligibly small.Program NewRaphG has been developed by use of the iterative equations 22and 23. Both QuickBASIC and FORTRAN versions of this program are listedbelow along with a sample application of solving the intercepts of two ellipses,f(x,y) = (x/3)2 + (y/4)2–1 and g(x,y) = (x/4)2 + (y/3)2–1.QUICKBASIC VERSIONFORTRAN VERSION© 2001 by CRC Press LLCSample ApplicationMATLAB APPLICATIONSHere, we provide a m file called NewRaphG.m as a companion of the FORTRAN and QuickBASIC versions:© 2001 by CRC Press LLCFor using this function, the problem to be solved needs to be defined by creatingtwo m files, in which the equations involved and the expressions for their firstderivatives should be spelled out.
In case of solving the sample problem used inFORTRAN and QuickBASIC versions, first we may define the equation as:Next, the expressions for their first derivatives may then be written as:© 2001 by CRC Press LLCTo solve this problem, the interactive application of MATLAB proceeds asfollows (some displays have been rearranged for saving spaces):Notice that the initial values are taken as X(1) = –2 and X(2) = 2, a toleranceof 10–5 and the iteration is limited to 20 trials.
The solutions are found after fourNewton-Raphson trials when the sum of the absolute values of the two equations isequal to 3.7361x10–8.MATHEMATICA APPLICATIONSMathematica applies the Newton’s method in its function FindRoot which canbe applied for solving a polynomial, or, transcendental equation, and also for multipleequations. We illustrate its applications by using the examples discussed earlier.First, the root near X = 1.75 for a third-order polynomial is found:In[1]: = FindRoot[{X^3–6*X^2 + 11*X–6 = = 0}, {X,1.75}]Out[1] = {X -> 2.}The solution is X = 2.
The second example is for finding a root near T = 0.5 fora transcendental equation described inside the first pair of braces:In[2]: = FindRoot[{Exp[-.2*T]*Sin[T + 1.37] = = 0.5},{T,0.5}]Out[2] = {T -> 1.09911}For solving two simultaneous transcendental equations, two examples are presented below. The first is to find one of the intercepts of two ellipses and the secondis to find one of the intercepts of a circle of radius equal to 2 and a sine curve.In[3]: = (FindRoot[{(X/3)^2 + (Y/4)^2 = = 1,(X/4)^2 + (Y/3)^2 = = 1},{X,–2}, {Y,2}]Out[3] = {X -> –2.4, Y -> 2.4}In[4]: = FindRoot[{x = = Sqrt[4y^2],y = = Sin[2*x]},{x,1.95},{y,–0.6}]Out[4] = {x -> 1.90272, y -> –0.616155}© 2001 by CRC Press LLC3.4 PROGRAM BAIRSTOW — BAIRSTOW’S METHOD FORFINDING POLYNOMIAL ROOTSProgram Bairstow is developed for finding the roots of polynomials based on theNewton-Raphson’s iterative method for two variables (see program NewRaphG).Let a nth-order polynomial be denoted as:P(x) = x N + a1x N −1 + a 2 x N −2 + … + a N −2 x 2 + a N −1x + a N(1)Notice that the highest term xN has a coefficient equal to 1; otherwise the entireequation must be normalized by dividing by that coefficient.
The Bairstow’s methodconsists of first selecting a trial divider D(x) = x2 + d1x + d2, and to obtain thequotient Q(x) = xN–2 + q1xN–3 + q2xN- 4 + ••• + qN–4x2 + qN–3x + qN–2 and a remainderR(x) = r1x + r2. The objective is to continuously adjust the values of d1 and d2 untilboth values of r1 and r2 are sufficiently small. It is apparent that both r1 and r2 aredependent of d1 and d2.
Taylor’s series expansions of r1 and r2 can be written as:r1 (d1 + ∆d1, d 2 + ∆d 2 ) = r1 (d1, d 2 ) + r1,d1 (d1, d 2 )∆d1+ r1,d 2 (d1, d 2 )∆d 2 + …(2)andr2 (d1 + ∆d1, d 2 + ∆d 2 ) = r2 (d1, d 2 ) + r2,d1 (d1, d 2 )∆d1+ r2,d 2 (d1, d 2 )∆d 2 + …(3)wherer1,d1 ≡ ∂r1 ∂d1, r2,d 2 ≡ ∂r2 ∂d 2and so on.The adjustments d1 and d2 are to be calculated so as to make the left-handside of Equations 2 and 3 both equal to zero and these adjustments are expected tobe small enough (if the guessed values of d1 and d2 values are sufficiently close tothose which make both r1 and r2 equal to zero) so that the second and higher derivativeterms in Equations 2 and 3 can be dropped. This leads to:r1,d1 (d1, d 2 )∆d1 + r1,d 2 (d1, d 2 )∆d 2 = − r1 (d1, d 2 )(4)r2,d1 (d1, d 2 )∆d1 + r2,d 2 (d1, d 2 )∆d 2 = − r2 (d1, d 2 )(5)and© 2001 by CRC Press LLCCramer’s rule can then be applied to obtain d1 and d2 as:() (rr− r1,d 2 r2,d1)(6)() (rr− r1,d 2 r2,d1)(7)∆d1 = − r1r2,d 2 + r2 r1,d 21,d1 2 ,d 2and∆d 2 = − r1r2,d1 + r2 r1,d11,d1 2 ,d 2where r1, r2, and their partial derivatives are to be evaluated at (d1,d2).
Equations 6and 7 are to be continuously applied to adjust the guessing values of (d1,d2) untilboth r1(d1,d2) and r2(d1,d2) are negligibly small.To calculate the adjustments d1 and d2 based on Equations 6 and 7, we needto find the partial derivatives ∂r1/∂d1, ∂r1/∂d2, ∂r2/∂d1, and ∂r2/∂d2. These derivativesare, however, depend on the d1 and d2, and the coefficients q’s in the quotient Q(x).This can be shown by actually carried out the division of P(x) by D(x).
The resultsare:q 1 = a 1 − d1andq 2 = a 2 − q1d1 − d 2(8,9)andq k = a k − q k −1d1 − q k −2 d 2 ,for k = 3, 4,…, N − 2(10)It can also be shown that the coefficients in the remainder R(x) are:r1 = a N −1 − q N −2 d1 − q N −3d 2andr2 = a N − q N −2 d 2(11,12)We notice that Equations 11 and 12 can be included in Equation 10 if k isextended to N and if the remainder is redefined as:R(x) = (x + d1 )q N −1 + q N(13)That is, r1 is renamed as qN–1 and r2 is equal to d1qN–1 + qN.
As a consequence,we need to replace r1 and r2 in Equations 6 and 7 by qN–1 and qN. For calculation ofthe adjustments d1 and d2, Equation 10 should be used for qN–1 and qN and toderive their partial derivatives respect to d1 and d2. Since all q’s are functions of d1and d2, to derive the partial derivatives of the last two q’s we must find the partialderivatives for all q’s starting with q1. From Equations 8 to 10, we can have:∂q1 ∂d1 = −1,∂q1 ∂d 2 = 0,∂q 2 ∂d1 = ( −∂ q1 ∂d1 ) d1 − q1 = d1 − q1,© 2001 by CRC Press LLC(14,15)(16)∂q 2 ∂d2 = −(∂q 1 ∂d2 )q 1 − 1 = −(17)∂q k ∂d1 = −(∂q k −1 ∂d1 )d1 − q k −1 − (∂q k −2 ∂d1 ) d 2(18)∂q k +1 ∂d 2 = −(∂q k ∂d 2 )d1 − q k −1 − (∂q k −1 ∂d 2 )d 2(19)and for k = 3,4,…,NIt can be concluded from the above results that:∂q k +1 ∂d 2 = ∂q k ∂d1for k = 1, 2,…, N − 1(20)Now, we can summarize the procedure of Bairstow’s method for factorizing aquadratic equation from an Nth-order polynomial as follows: (Some changes ofvariables are made in the computer programs to be presented next, such as q’s arechanged to b’s, d1 and d2 are changed to u and v, respectively, and c’s are introducedto represent the derivatives of q’s.)(1) Specify the values of N, a1 through aN, and a tolerance .(2) Assume an initial guessing values for d1 and d2 for the divider D(x).(3) Calculate the coefficients q1 through qN–2 for the quotient Q(x) usingEquations 8 to 10.(4) Also use Equation 10 to calculate the coefficients qN–1 and qN for theremainder R(x).(5) Test the absolute values of qN–1 and qN.
If they are both less than , tworoot of P(x) are to be calculated by use of the quadratic formulas. Theorder of P(x), N, is to be reduced by 2, and q1 through qN–2 are to becomea1 through aN–2, respectively, and return to Step 2. This looping continuesuntil the quotient Q(x) is of order two or one, for which the root(s) easilycan be calculated.(6) If the absolute value of either qN–1 or qN is greater than ε, calculate thepartial derivatives of qk with respect to d1, c’s using Equations 14, 16, and18 for k = 3,4,…,N. The derivatives of qk with respect to d2 are alreadyavailable due to Equation 20.(7) Use Equations 6 and 7 to calculate the adjustments d1 and d2, noticingthat r1 and r2 are to be replaced by qN–1 and qN, respectively.
The iterationis resumed by returning to Step 3.Both QuickBASIC and FORTRAN versions of the program Bairstow codedfollowing the steps described above are to be presented next.© 2001 by CRC Press LLCQUICKBASIC VERSION© 2001 by CRC Press LLCSample ApplicationAs an example, the polynomial P(x) = x4–5x3 + 13x2–19x + 10 = 0 is solved byapplication of the QuickBASIC version of the program Bairstow. The response onscreen is:The quotient in this case is a quadratic equation:Q(x) = [x − (1 + 2i)][x − (1 − 2i)] = (x − 1) − (2i) = x 2 − 2 x + 5 = 0.2FORTRAN VERSION© 2001 by CRC Press LLC2© 2001 by CRC Press LLCSample ApplicationConsider the polynomial P(x) = x3 + 2x2 + 3x + 4 = 0.
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