Solution 5.8.4.7 Wewish to factor the polynomial s 4 +13s 3 +76s 2 +192s + 128 Using root locus techniques. Wedosoby nding the roots of the equation s 4 +13s 3 +76s 2 +192s +128=0: Wecan reformulate this as a root locus problem as follows. Divide both sides of the equation by s 4 +13s 3 +76s 2 ;; to obtain s 4 +13s 3 +76s 2 s 4 +13s 3 +76s 2 + 192s +128 s 4 +13s 3 +76s 2 =0;; or 1+ 192(s+128=192) s 4 +13s 3 +76s 2 =0: Wecaneasily factor s 4 +13s 3 +76s 2 ,toobtain 1+ 192(s+128=192) s 2 (s +6:5; j5:8095)(s+6:5+j5:8095) =0: Nowconsider the root locus problem 1+ K(s +128=192) s 2 (s +6:5; j5:8095)(s+6:5+j5:8095) =0: The root locus is shown in Figure 1. From the root locus weseethat for K =152 there will either be four complex roots, or tworealroots and twocomplexroots. Further, weknow that the real roots lie between s = ;128=192 and s = ;1.Sowebegin by searching along the real axis to the left of s = ;128=192 for the point where the gain is 192. Figure 2 shows howwecalculate K for s = ;2. For s = ;2we get K = [ jsjjsjjs+6:5;j5:8095jjs+6:5+j5:8095j js +128=192j ] j s=;2 = 162 1 -8 -6 -4 -2 0 2 -10 -8 -6 -4 -2 0 2 4 6 8 10 Real Axis I mag A x i s Figure 1: Root Locus Im(s) Re(s) -(128/192) 2 s + 152/320 s s +6.5 -j5.8095 s +6.5 +j5.8095 Figure 2: Gain Calculation 2 s -1 -2 -3 -4 -5 K 192 162 177.4 192 207.7 Table 1: Searchforthevalue K =192 The value of K wehavefound is too small. This tells us that the root we are looking for is to the left of s = ;1. So wenextguess s = ;3, for whichwe obtain: K = jsjjsjjs+6:5;j5:8095jjs+6:5+j5:8095j js +128=192j s=;1 = 177:4 This is closer, and Table 1 summarizes the searchofthe value 192. Wehave now found the tworeal rools. To nd the twocomplex roots, we divide out the term (s + 1)(s+4)=s 2 +5s +4: s 4 +13s 3 +76s 2 +192s +128 s 2 +5s +4 = s 2 +8s 2 +32 = (s +4; j4)(s+4+j4) Thus s 4 +13s 3 +76s 2 + 192s +128 = (s +1)(s+4)(s +4; j4)(s+4+j4) 3