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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 closedloop 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 steadystate 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 straightline segments
marked) shown below, determine the transfer function. "ﬁnalﬂittiEH""'"""'
Iliiiiiiii‘"nuIllﬂiiilﬁﬂiii
III‘ IllllllIIIIIIII nia;=!llIIIIIIII
IIIIIlllIlllllllIIIIIIIIILIIIII 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
closedloop system at K=1. 2). For what range of K is the closedloop 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. (959) 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 zerofrequency 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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This note was uploaded on 01/14/2012 for the course EEL 3657 taught by Professor Staff during the Spring '08 term at University of Central Florida.
 Spring '08
 Staff

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