Question

Figure 6 shows a feedback control system for which G(s) = 6 (s + 1)3 J and K(s) is the transfer function of a compensator. (6] When K= 4/3 the loop gein in one at a frequency of w = is c] TE KIS) = (s+1), GOS) KO) = _6 (stil? and the DC loop gain is

Please explain part b and C in detail.

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sol given ka will Characterstic en I + C7 () K (3) + KX6 (S+02 Gbook 1+ 4 x 6 3 (5+1)3 $+1) 3 +8 $+1)2 (5+1) +8 2 (5+25+1)(5+© introdi oducing one zero zero at (-1) K (5) 2 (5+1) GS).Ksia 6 S+U2 gerin lim Gojok (8) Sao lim 6 sto (4+1)2 Steady stepe gLGOHB) at Wzo rad /sec LG (SIH() z-ztant w af waoo bad I see LG 60-H (8) 2 -180 Im pím Re Wzorad/see W202

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