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So lu tio ns M an ua l Ninth Edition Farid Golnaraghi • Benjamin C Kuo CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition A Chapter 2 Solution ns Golnarraghi, Kuo C Chapter 2 2‐1 (a) Poless: s = 0, 0, −1, − −10; (b) Poles: s = −2,, −2; Zeross: s = −2, ∞, ∞, ∞. Zeros: s = 0 The pole and zero at s = −1 ccancel each otther. ( Poles: s = 0, −1 + j, −1 − j; (c) (d) Poles: s = 0, −1, −2, ∞. Zeross: s = −2. 2-2) a) b) c) 2-3) M MATLAB code e: 2‐1 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 2 Solutions Golnaraghi, Kuo clear all; s = tf('s') 'Generated transfer function:' Ga=10*(s+2)/(s^2*(s+1)*(s+10)) 'Poles:' pole(Ga) 'Zeros:' zero(Ga) 'Generated transfer function:' Gb=10*s*(s+1)/((s+2)*(s^2+3*s+2)) 'Poles:'; pole(Gb) 'Zeros:' zero(Gb) 'Generated transfer function:' Gc=10*(s+2)/(s*(s^2+2*s+2)) 'Poles:'; pole(Gc) 'Zeros:' zero(Gc) 'Generated transfer function:' Gd=pade(exp(-2*s),1)/(10*s*(s+1)*(s+2)) 'Poles:'; pole(Gd) 'Zeros:' zero(Gd) 2‐2 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 2 Solutions Golnaraghi, Kuo Poles and zeros of the above functions: (a) Poles: 0 0 ‐10 ‐1 Zeros: ‐2 (b) Poles: ‐2.0000 ‐2.0000 ‐1.0000 Zeros: 0 ‐1 (c) Poles: 0 ‐1.0000 + 1.0000i ‐1.0000 ‐ 1.0000i Zeros: ‐2 Generated transfer function: (d) using first order Pade approximation for exponential term Poles: 0 ‐2.0000 ‐1.0000 + 0.0000i ‐1.0000 ‐ 0.0000i Zeros: 1 2‐3 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 2 Solutions Golnaraghi, Kuo 2-4) Mathematical representation: In all cases substitute and simplify The use MATLAB to verify R= 10 22 + ω + ω 102 + ω ; −ω (ω + 1)(ω + 100) ω 10( jω + 2) 2 φ1 = tan −1 + ω −ω ( jω + 1)( jω + 10) 10( jω + 2) ( − jω + 1)(− jω + 10) 22 + ω = × −ω ( jω + 1)( jω + 10) (− jω + 1)(− jω + 10) −ω 10( jω + 2)( − jω + 1)(− jω + 10) a) = φ2 = tan −1 + ω 2 −ω (ω + 1)(ω + 100) 1+ ω2 jω + − jω + − jω + 10 =R −ω 22 + ω + ω 102 + ω 2 = R(e jφ1 e jφ2 e jφ3 ) φ3 = tan −1 10 + ω 10 102 + ω φ = φ1 + φ2 + φ3 R= 10 + ω + ω ; (ω + 1) (ω + 9) −ω 10 −1 + ω φ = tan ( jω + 1) ( jω + 3) 10 (− jω + 1)(− jω + 1)(− jω + 3) 1+ ω2 = × ( jω + 1)( jω + 1)( jω + 3) ( − jω + 1)(− jω + 1)(− jω + 3) −ω 10(− jω + 1)(− jω + 1)(− jω + 3) −1 + ω b) = = φ tan (ω + 1) (ω + 9) 1+ ω2 − jω + − jω + − jω + R = −ω 1+ ω2 1+ ω2 + ω2 = R (e jφ1 e jφ2 e jφ3 ) φ3 = tan −1 + ω + ω2 φ = φ1 + φ2 + φ3 2‐4 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 2 Solutions Golnaraghi, Kuo 10 jω ( j 2ω + − ω ) −10 j (2 − ω − j 2ω ) = × ω ( j 2ω + − ω ) (2 − ω − j 2ω ) 10(−2ω − (2 − ω ) j ) c) = ω (4ω + (2 − ω ) ) =R −2ω − (2 − ω ) j = R (e 4ω + (2 − ω ) jφ ) 10 4ω + (2 − ω ) 10 R= = ; 2 2 ω (4ω + (2 − ω ) ) ω 4ω + (2 − ω ) −2 − ω φ = tan −1 4ω + (2 − ω ) −2ω 4ω + (2 − ω ) 2 R= 10ω + ω + ω −ω e −2 jω 10 jω ( jω + 1)( jω + 2) −1 22 + ω = φ tan − j (− jω + 1)(− jω + 2) −2 jω = e 10ω (ω + 1)(ω + 2) d) − jω + − jω + −2 jω − jπ / −ω =R e 2 2 + ω 1+ ω φ2 = tan −1 + ω = R (e jφ1 e jφ2 e jφ3 ) ; 2 + ω2 1+ ω2 φ = φ1 + φ2 + φ3 MATLAB code: clear all; s = tf('s') 'Generated transfer function:' Ga=10*(s+2)/(s^2*(s+1)*(s+10)) figure(1) 2‐5 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 2 Solutions Golnaraghi, Kuo Nyquist(Ga) 'Generated transfer function:' Gb=10*s*(s+1)/((s+2)*(s^2+3*s+2)) figure(2) Nyquist(Gb) 'Generated transfer function:' Gc=10*(s+2)/(s*(s^2+2*s+2)) figure(3) Nyquist(Gc) 'Generated transfer function:' Gd=pade(exp(-2*s),1)/(10*s*(s+1)*(s+2)) figure(4) Nyquist(Gd) Nyquist plots (polar plots): Part(a) 2‐6 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 2 Solutions Golnaraghi, Kuo Nyquist Diagram 15 10 Imaginary Axis -5 -10 -15 -300 -250 -200 -150 -100 -50 Real Axis Part(b) Nyquist Diagram 1.5 Imaginary Axis 0.5 -0.5 -1 -1.5 -1 -0.5 0.5 1.5 2.5 Real Axis Part(c) 2‐7 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 2 Solutions Golnaraghi, Kuo Nyquist Diagram 80 60 40 Imaginary Axis 20 -20 -40 -60 -80 -7 -6 -5 -4 -3 -2 -1 Real Axis Part(d) Nyquist Diagram 2.5 1.5 Imaginary Axis 0.5 -0.5 -1 -1.5 -2 -2.5 -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 Real Axis 2‐8 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition 2-5) Chapter 2 Solutions Golnaraghi, Kuo In all cases find the real and imaginary axis intersections G ( jω ) = 10 10(− jω + 2) 10 = = ( jω − 2) (ω + 4) (ω + 4) Re {G ( jω )} = cos φ = a) (ω + 4) −ω Im {G ( jω )} = sin φ = (ω + 4) − jω (ω + 4) ; , , φ = tan −1 (ω + 4) −ω (ω + 4) R= 10 (ω + 4) lim ω →0 G ( jω ) = 5; φ = tan −1 = −90o −0 lim ω →∞ G ( jω ) = 0; φ = tan −1 = −180o −1 Real axis intersection @ jω = Imaginary axis int er sec tion does not exist b&c) ∞ =1 0o =0 -180o Therefore: Re{ G(jω) } = Im {G(jω)} = 2‐9 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition ⎡ x&2 ⎤ ⎡ 1/3*f+49/6x1 ⎤ ⎢ x& ⎥ = ⎢1/3*f-49/30x ⎥ 1⎦ ⎣ 4⎦ ⎣ ⎡ x&1 ⎤ ⎡ ⎢ x& ⎥ ⎢ 49/6 0 ⎢ 2⎥ = ⎢ ⎢ x&3 ⎥ ⎢ 0 ⎢ ⎥ ⎢ ⎣ x&4 ⎦ ⎣-49/30 0 ⎡ ⎢ 49/6 A=⎢ ⎢ ⎢ ⎣-49/30 ⎡ ⎤ ⎢1/ 3⎥ B=⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣1/ 3⎦ Chapter 10 Solutions Golnaraghi, Kuo ⎤ ⎡ x1 ⎤ ⎡ ⎤ ⎥⎥ ⎢⎢ x2 ⎥⎥ ⎢⎢1/ 3⎥⎥ + f ⎥ ⎢ x3 ⎥ ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎦ ⎣ x4 ⎦ ⎣1/ 3⎦ 0⎤ 0 ⎥⎥ 0 1⎥ ⎥ 0 0⎦ C = [1 0] D=0 10‐98 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions Golnaraghi, Kuo Use ACSYS State tool and follow the design process stated in Example 10-17-1: 10‐99 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions The A matrix is: Amat = 0 1.0000 0 0 8.1667 0 0 0 0 0 0 1.0000 ‐1.6333 0 0 0 Characteristic Polynomial: ans = s^4‐49/6*s^2 Eigenvalues of A = Diagonal Canonical Form of A is: Abar = 0 0 0 0 0 0 0 0 0 0 2.8577 0 0 0 0 ‐2.8577 Eigen Vectors are T = 0 0 0.3239 ‐0.3239 0 0 0.9256 0.9256 1.0000 ‐1.0000 ‐0.0648 0.0648 0 0.0000 ‐0.1851 ‐0.1851 State‐Space Model is: a = x1 x2 x3 x4 x1 0 1 0 0 x2 8.167 0 0 0 x3 0 0 0 1 x4 ‐1.633 0 0 0 b = u1 x1 0 x2 0.3333 Golnaraghi, Kuo 10‐100 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions x3 0 x4 0.3333 c = x1 x2 x3 x4 y1 1 0 1 0 d = u1 y1 0 Continuous‐time model. Characteristic Polynomial: ans = s^4‐49/6*s^2 Equivalent Transfer Function Model is: Transfer function: 4.441e‐016 s^3 + 0.6667 s^2 ‐ 2.22e‐016 s ‐ 3.267 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ s^4 ‐ 8.167 s^2 Pole, Zero Form: Zero/pole/gain: 4.4409e‐016 (s+1.501e015) (s+2.214) (s‐2.214) ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ s^2 (s‐2.858) (s+2.858) The Controllability Matrix [B AB A^2B ] is = Smat = 0 0.3333 0 2.7222 0.3333 0 2.7222 0 0 0.3333 0 ‐0.5444 0.3333 0 ‐0.5444 0 The system is therefore Not Controllable, rank of S Matrix is = rankS = 4 Golnaraghi, Kuo 10‐101 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions Mmat = 0 ‐8.1667 0 1.0000 ‐8.1667 0 1.0000 0 0 1.0000 0 0 1.0000 0 0 0 The Controllability Canonical Form (CCF) Transformation matrix is: Ptran = 0 0 0.3333 0 0 0 0 0.3333 ‐3.2667 0 0.3333 0 0 ‐3.2667 0 0.3333 The transformed matrices using CCF are: Abar = 0 1.0000 0 0 0 0 1.0000 0 0 0 0 1.0000 0 0 8.1667 0 Bbar = 0 0 0 1 Cbar = ‐3.2667 0 0.6667 0 Dbar = 0 Golnaraghi, Kuo 10‐102 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions Golnaraghi, Kuo Note incorporating –K in Abar: Abar K= 0 1.0000 0 0 0 0 1.0000 0 0 0 0 1.0000 ‐k1 ‐k2 8.1667‐k3 ‐k4 System Characteristic equation is: ‐k4*s^4+(8.1667‐k3 )*s^3‐k2*s‐k1=0 From desired poles we have: >> collect((s-210)*(s-210)*(s+20)*(s-12)) ans = -10584000+s^4-412*s^3+40500*s^2+453600*s Hence: k1=10584000, k2=40500, k3=412+8.1667and k4=1 10-65) If system is: and 0.707, then 1.414 The 2nd order desired characteristic equation of the 2 On the other hand: where the characteristic equation would be: (2) Comparing equation (1) and (2) gives: 2 10‐103 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition which means and Chapter 10 Solutions Golnaraghi, Kuo 10-66) Using ACSYS we can convert the system into transfer function form USE ACSYS as illustrated in section 10-19-1 1) 2) 3) 4) 5) 6) 7) 8) 9) Activate MATLAB Go to the folder containing ACSYS Type in Acsys Click the “State Space” pushbutton Enter the A,B,C, and D values Note C must be entered here and must have the same number of columns as A We us [1,1] arbitrarily as it will not affect the eigenvalues Use the “Calculate/Display” menu and find the eigenvalues Next use the “Calculate/Display” menu and conduct State space calculations Next verify Controlability and find the A matrix Follow the design procedures in section 10-17 (pole placement) 10‐104 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions Golnaraghi, Kuo 10‐105 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions Golnaraghi, Kuo The A matrix is: Amat = -1 -2 -2 -1 1 -1 Characteristic Polynomial: ans = s^3+3*s^2+5*s+5 Eigenvalues of A = Diagonal Canonical Form of A is: Abar = -0.6145 + 1.5639i 0 -0.6145 - 1.5639i 0 -1.7709 Eigen Vectors are T= -0.8074 -0.8074 -0.4259 0.2756 + 0.1446i 0.2756 - 0.1446i -0.7166 -0.1200 + 0.4867i -0.1200 - 0.4867i 0.5524 State-Space Model is: a= x1 x2 x3 x1 -1 -2 -2 x2 -1 x3 -1 10‐106 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions Golnaraghi, Kuo b= u1 x1 x2 x3 c= x1 x2 x3 y1 1 d= u1 y1 Continuous-time model Characteristic Polynomial: ans = s^3+3*s^2+5*s+5 Equivalent Transfer Function Model is: Transfer function: s^2 + s + s^3 + s^2 + s + Pole, Zero Form: Zero/pole/gain: (s+1.333) (s+1) (s+1.771) (s^2 + 1.229s + 2.823) 10‐107 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions Golnaraghi, Kuo The Controllability Matrix [B AB A^2B ] is = Smat = -4 0 1 -5 The system is therefore Controllable, rank of S Matrix is = rankS = Mmat = 3 1 0 The Controllability Canonical Form (CCF) Transformation matrix is: Ptran = -2 2 3 The transformed matrices using CCF are: Abar = 1.0000 0 1.0000 -5.0000 -5.0000 -3.0000 10‐108 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions Golnaraghi, Kuo Bbar = 0 Cbar = Dbar = Using Equation (10-324) we get: sI − ( A − BK ) = s + (3 + k3 ) s + (5 + k2 ) s + (5 + k1 ) = Using a 2nd order prototype system, for 5, then For overshoot of 4.33%, nd Then the desired order system will have a characteristic equation: 0.707 s + 2ζωn s + ωn = s + s + = The above system poles are: s1,2 = −1 ± j One approach is to pick K=[k1 k2 k3] values so that two poles of the system are close to the desired second order poles and the third pole reduces the effect of the two system zeros that are at z=-1.333 and z=-1 Let’s set the third pole at s=-1.333 Hence (s+1.333)*(s^2+2*s+2)= s^3+3.33*s^2+4.67*s+2.67 and K=[-2.37 -0.37 0.33] Y 3( s + 1) = R s + 2s + Use ACSYS control tool to find the time response First convert the transfer function to a unity feedback system to make compatible to the format used in the Control toolbox G= 3( s + 1) s2 − s −1 10‐109 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions Golnaraghi, Kuo Overshoot is about 2% You can adjust K values to obtain alternative results by repeating this process 10-67) a) According to the circuit: If , , then 10‐110 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions Golnaraghi, Kuo or 1 Therefore: 1 0 b) Uncontrollability condition is: According to the state-space of the system, C is calculated as: 1 0, because As , then the system is controllable c) Unobservability condition is: According to the state-space of the system, C is calculated as: Since det H 0, because R or L ∞, then the system is observable 10‐111 CuuDuongThanCong.com https://fb.com/tailieudientucntt Automatic Control Systems, 9th Edition Chapter 10 Solutions Golnaraghi, Kuo d) The same as part (a) 1 1 1 For controllability, we define G as: 1 1 ,and then If 0, which means the system is not controllable For observability, we define H as: 1 1 If , then 0, which means the system is not observable 10‐112 CuuDuongThanCong.com https://fb.com/tailieudientucntt ... 60 40 Imaginary Axis 20 -2 0 -4 0 -6 0 -8 0 -7 -6 -5 -4 -3 -2 -1 Real Axis Part(d) Nyquist Diagram 2.5 1.5 Imaginary Axis 0.5 -0 .5 -1 -1 .5 -2 -2 .5 -1 -0 .8 -0 .6 -0 .4 -0 .2 0.2 0.4 Real Axis ... Magnitude (dB) -1 0 -2 0 -3 0 -4 0 -5 0 Phase (deg) -4 5 -9 0 -1 35 -1 80 -1 10 10 10 10 Frequency (rad/sec) Part(e) Bode Diagram Magnitude (dB) -2 0 -4 0 -6 0 -8 0 -1 00 Phase (deg) -1 20 -9 0 -1 80 -2 70 -1 10 10... 40 Imaginary Axis 20 -2 0 -4 0 -6 0 -3 .5 -3 -2 .5 -2 -1 .5 -1 -0 .5 Real Axis Part(e) Nyquist Diagram 0.8 0.6 0.4 Imaginary Axis 0.2 -0 .2 -0 .4 -0 .6 -0 .8 -1 -0 .8 -0 .6 -0 .4 -0 .2 0.2 0.4 0.6 0.8
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