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HW5Solution

# HW5Solution - ARO 322 Aerospace Feedback Control Systems...

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Unformatted text preview: ARO 322 Aerospace Feedback Control Systems Homework #5 Solution 1. Comment on the stability of a system with the closed~loop transfer function given by each of the following: (a) a: WWW (s+1)(sz+s+1) = 5(ns +2) (b) TU) (s+1){sz+s+l) 5 (c) To) = W (a) To) == (e) To) a {f} To) 2 m Solution (a) The roots of the denominator or the eigenvlaues are: s = —1 s = «0.5 i 0.8661“ The roots were calculated using roots command in MATLAB. All the roots are in the left-half of the complex plane. Therefore, the system is stable. This can also be seen from the step plot shown below: in» mm“ (b) The roots of the denominator are: S = —1 S = ~05 \$0.866i Ali the roots are in the ieft—half of the complex plane. Therefore, the system is stable. However, since the zero is on the right hand side, this is non-minimum phase system. The response shows some faulty behavior at the beginning and the response is slow compared to the problem in (a). This is seen in the step plot shown below: (c) The roots of the denominator are: s m—l s = —~0.5 i 0.8661' All the roots are in the left-half of the Complex plane. Therefore, the system is stable. However, since there is no zero as compared to (a), the system response is slower and less stable. This will become evident when we cover the root locus. However, the root locus plot is shown for both the systems, which shows that the system for (a) is more stable and in the left half~plane than the system in (c) 1‘0 ,. ....in.in......z..€1\\\\1‘1L:2,_..!11|..1......,..25M1“w+1Ls,..,..._1111||I.,25.. muse) MW ‘-_..-._.......Wﬂmvauuu-.._.-.._.,1»mww,~m.‘u...._ 5 4 \$3 MWWWAMW -_--- -meﬁﬁuw, 2 1 00“ M 4 3 2 1- O 1 2 3 E (d) The roots of the denominator polynomiai are: S=m1.7321 s =1.7321 s=1 The system is unstable since two pokes are on the right hand side. The step plot shown beiow shows instability: supmmu (e) The roots of the denominator or the eigenvlaues are: s = Mi 5 = 0.5 \$0.866i The system is unstable since two poies are on the right hand side. The step piot shown below shows instability: 3:9th “1000 2 "'21 a a Tin-(alt) (i) The roots of the denominator or the eigenviaues are: s = —1 S = i1.7321i Since there are two poles 011 the imaginary axis, the system is marginally stable as shown in the following ﬁgure. gum-mm- 50 40 30 Mind 1O 2. Use the Routthurtwitz criterion to determine the number of roots in the right- half plane, in the left~half plane, and on the imaginary axis for the foliowing characteristic equations: (a) ss+234+533+4s2+65 z 0 (b) 34+33+s+05 :0 _ (c) s4+s3+532+53+2 a: {3 (t1) s4+253+332+23+5 m 0 (e) s4+s3+532+23+4 m 0 (f) 34+232+1=~ﬁ 0 (g) Verify all resuits using MATLAB to find-the mot locations. LE3} 5’5 "5?“ 26"} +ENSB'3”L¥}\$ 1411;; W) s L3 U“ -+Q\$3~+1-5*5 1+ q\$+g :20 “7% “4,, WE. hawk GHQ {DO | g gm. 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For the cruise control system of an automobile shown in ﬁgure below, ﬁnd the ranges (1pr and K; for which the system is stable. {Disturbance torque Engine and ioad Speed 40 kmlh ' 33+1 Compensator Actuator Carburetor Throttle position, em Sensor where Gc(s)=Kp+£{— S Neglect the disturbance torque. Ckmmevistih ngym U, Wm- 14px;» K1 2:. g; *5 WWW!) -+ (N) 353+ q51+4§+ 94P5+Kr 3:0 “’3 353 +952 .4. QM+K.p)§—~}dkl—ma WM“ Rabi/k qma ,3: g 3 3 1+ KP 5 2 H K: S ’ WK], S O K: L} ...
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HW5Solution - ARO 322 Aerospace Feedback Control Systems...

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