Solution 5.8.4.6 Wewish to factor the polynomial s 3 ;3s 2 +4s +8 Using root locus techniques. Wedosoby nding the roots of the equation s 3 ;3s 2 +4s +8=0: WeknowfromtheRouth test that because of the negativecoecientofthe s 2 term there will be at least one root to the rightofthe origin. Wecanreformulate this as a root locus problem as follows. Divide both sides of the equation by s 3 ;3s 2 ;; to obtain s 3 ;3s 2 s 3 ;3s 2 + 4s +8 s 3 ;3s 2 =0;; or 1+ 4(s+2 s 3 ;3s 2 =0: Wecaneasily factor s 3 ;3s 2 ,toobtain 1+ 4(s+2) s 2 (s;3) =0: Nowconsider the root locus problem 1+ K(s +2) s 2 (s;3) =0: The root locus is shown in Figure 1. The root locus shows the solutions to the equation s 3 ;3s 2 + K(s +2)=0;; for all possible values of K. What wewantisthesolution for K =0:5. It is clear from the gure that wecould have three real solutions or one real root and twocomplexroots. If wearetosolve using root locus methods, then what wecando is searchalong the real axis between s =0and s = ;2andto nd the value of s for which K =4.Figure 2 shows howwewould makethatcalculation. The MATLAB program 1 Im(s) Re(s) -2 2 3 Figure 1: Root Locus Im(s) Re(s) 2 s 3 -2 s - 3 s + 2 Figure 2: Root Locus 2 -2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0 10 20 30 40 50 60 Figure 3: K versus s p=[1-348] roots(p) z=2 p1 = 0 p2 = 0 p3 = -3 s=linspace(0,-2,9);; K=(abs(s+p1).^2.*abs(s+p3))./abs(s+ z);; plot(s,K) makes the plot shown in Figure 3 The gain of four seems to be in the neighborhood of s = ;1. Acutally,ifafter running the program wetype: K' wegenerate: EDU>K' 3 ans = 0 0.11607142857143 0.58333333333333 1.68750000000000 4.00000000000000 8.85416666666667 20.25000000000000 58.18750000000000 EDU> We see that the fth entry is K =4whichisthe gain we are after. Wecan then type: s(5) to obtain EDU>s(5) ans = -1 EDU> Wenowknow one of the roots is s = ;1. If we divide out the term s+1,we will obtain a second order polynomial in s that wecansolveby the quadratic formula. Thus: s 3 ;3s 2 +4s +8 s +1 = s 2 ;4s +8=(s;2+j2)(s;2;j2): Thus, s 3 ;3s 2 +4s +8=(s +1)(s;2+j2)(s;2;j2): 4