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# final_so - Final Exam EEL 3657(Summer 2003 Name Student ID...

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Unformatted text preview: Final Exam EEL 3657 (Summer 2003) Name: Student ID#: Instructions 1. There are totally 4 problems in this exam. 2. Show all work for partial credit. 3. Assemble your work for each problem in logical order. 4. Justify your conclusion. I cannot read minds. Problem 1. (25 points) Consider the system shown below, + R a _ 3(32+4s+5) Y 1 5+1 1). Find the closed-loop characteristic equation. 2). Use Routh’s stability criterion, determine all values of K for which the system is stable. , , s+g Sci/dim} 0' GM“): H ’ ii:— _ Karler s CSL+LkS+kXS+O +Kts+3) : W S‘WWHH \$thsz Ce : shrﬁﬁﬂ (Hm +SI< =0 at : 4° ~K 5 I bl = 3K Ct : (S‘H'x) -‘ C1 : , Problem 2. (25 points) For the system shown below, what steady-state error can be expected for an input of 15u(t)? (u(t) is a unit step function.) Reduce the system [021711 eduhtglent unity feedback sy (em. _5_ - _ ﬂ -___5_— m;— G(S)" 5(s+3) — s(s+ll)+5(s+3) -sz+l6s +15 1+———— s(s+ll) Hence, e(oo) = lim G(s) = KP =§ . s-—>0 Finally, _15__=JL=A§ =11.25 1+Kp 1+1 4 6ss Problem 3. (20 points) For the Bode magnitude plot (with straight-line segments marked) shown below, determine the transfer function. "ﬁnal-ﬂitti-EH""'"""' Iliiiiiii-i‘"nu-Illﬂiii-lﬁﬂiii ----III-‘ -Illlll|l-IIIIIIII- nia;=!ll-IIIIIIII -IIIIIlll-Illlllll-IIIIIIIIILIIIII ll ‘ Frequency (rad/s) ‘ (W) {OK/‘4‘ka ; [Q3 meme} km 0x .4th “'20 (*3/ch / Wig“) {WM—b; (L 3" ‘ierrm I“ Tim team} Aseameh‘r 5% 0V 3(7):; "(POW/4a, W01) (30505853 0e ‘3 ‘KQJYM- H}: W‘nt M'ﬂme’wa Bride, Wﬂmﬁobv “5% , 6] :: 20 boa K ~ whatnot k :3 K 1 /8 f So G“): SCI‘V'Ei'): scans): “my Problem 4. (30 points) For a unity feedback system shown below, where G(s) = 3;“le ‘. k). K s(s + 2)(s +10) 9 1). Sketch the Nyquist plot for K=1, and use NyquiSt criterion to claim stability for the closed-loop system at K=1. 2). For what range of K is the closed-loop system stable? 3). What is the gain margin? 1. Substitute s = jw in Setting s = ja) 2. Substituting a) = 0 in the last equation, we get 50W = «>4- 3. Substituting w = 00 established. 51"”? 4. To ﬁnd the intersect(s) of the Nyquist plot with the real axis, if any, we rationalize by multiplying the numerator and the denominato gate of the denominator. Thus, Eq. (9-59) becom [—12w2 —jw( [—12w2 +jw(20 — w2)][— = [—1210 -j(20 — (02)] w[1440)2 + (20 - (02)] 5. To ﬁnd the possible intersects on the real axis, w zero. The result is ‘ Imlqtjwi . 0/.— 27b° ‘(20 — g) w[144w2 + (2 - (02)] jw(jw + in) + the zero-frequency property of L(jw), 90° , the property of the Nyquist plot at inﬁnite frequency is jw) r of the equation by the comple conju- s — 022)] 2w2 — jw(20 — a)2 )l e set the imaginary part of jw) to 2 0 The solutions of the last equation are: m = 00, which is known to be a solution at qtjw) = 0, and w==iV20 Since (0 is positive, the correct answer is w = V20 rad/s , we have the intersect on the real axis of the ij} 12 ' \ / 2 = ——— = a“ 0) 2880 rad/ sec ec. Substituting this frequency plane at -(D.OO4167 ...
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final_so - Final Exam EEL 3657(Summer 2003 Name Student ID...

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