EECE 460 Homework, Spring 2008

B-8-19. Consider the case where G(s) has one pole in the right-half s plane. From the Nyquist plot of G(j#) shown on the next page, the -1 + jO point is encircled by the G(j@) locus once clockwise and once counterclockwise. Hence N= 0. Since G(s) has one pole in the right-half s plane; we have P = 1. Since Z=NtP=otl =| the system is unstable. Next, consider the case where G(s) has no pole in the right-half s plane, but has one zero in the right-half s plane. The -1 + jO point is encircled by the G(j@) locus once clockwise and once counterclockwise. Hence, N = 0. Since G(s) has no poles in the right-half s plane, we have P= 0. Therefore, Z=oNtP=0+d=0 The system is stable. (Note that the presence of a zero of G(s) in the right- . half s plane does not affect the stability of the system.) BS 28. Gis) = 20(s+/) T= “S(s*+ 25+ 10)(s45) The phase margin, gain margin, phase crossover frequency, and gain crossover frequency are obtained by use of the command {Gm, pm, wep, weg] = margin(sys) A MATLAB program to solve this problem is given below. The Bode diagram shown below verifies the phase margin, gain margin, phase crossover frequency, and gain crossover frequency obtained with MATLAB. % tees Bode Diagram sees num=[0 0 0 20 20]; den=conv([1 2 10 O},1 5); ae coy we i : bode(sys,w) title(Bode Diagram of G(s) = 20(s+1)/[s(s*2+2s+10\s+: [Gm,pm,wep,weg] = margin(sys); — GmdB = 20*1og10(Gm); [GmdB pm wep wog] 9.9293 103.6573 4.0131 0.4426 Bode Diagram of G(s) = 20(s+1)/js(s*+2s+10Xe+5)) i ==aull es et anit a Can i Ti rt 10" a #826 40° 40131 10" 0 Frequency (red/sec) 8-35: G(s) = 10 / (s*(0.17 s + 1))