Σ G H R C + Solution 6.8.4.1 (a) This problem is simple enough that it can be worked without knowledge of the root lcous, although that knowledge would make things easier. The closed loop transfer function is T c (s) = C(s) R(s) = 0:1 (s;j10)(s+j10) 1+ 0:1Ks (s;j10)(s+j10) = 0:1 s 2 +0:1Ks+100 The roots are s = ;0:05K 0:5 p 0:01K 2 ; 400 The smallest gain that will yield real roots is K = p 40;;000 = 200;; yielding twopoles at s = ;10 Note: For the studentfamiliar with the root locus, the locus o the real axis is part of a circle centered at the zero at the origin, as shown in Figure 1 The gain that gives twopoles at s = ;10 is then K = js ;j10jjs+ j10j 0:1jsj j s=;10 = 200 1 -10 s - j10 s + j 10 Re(s) j Im(s) Figure 1: Root Locus (b) The second engineer's criticism of the design is that it will not tracka step input. Toevaluate this we must use the closed loop transfer function, because wedonot have unity feedback. Thus, e ss =lim t!1 c(t) = lim s!0 sR(s)[1;T c (s)] = lim s!0 s( 1 s )  1; 0:1 s 2 +0:1Ks+100  = lim s!0  1; 0:1 s 2 +0:1Ks+100  = 0:999 Thus the tracking error is almost one hunderd per cent, and the second engineer is right,the system will not trackastep. (c) As can be seen from the expression e ss = lim s!0  1; 0:1 s 2 +0:1Ks+ 100  ;; 2 Nothing is accomplished byadding gain in the feedbacklook. However, if a gain K f is added in the forward loop, then e ss = lim s!0  1; 0:1K f s 2 +0:1Ks+100  = 1; 0:1K f 100 : By making K f =1000, wecanmakethesteady state error to a step zero. Thus, if gain is to be added, it should be added in the forward loop. 3