Automatic control systems 9th golnaraghi kuo

So lu tio ns M an ua l Ninth Edition Farid Golnaraghi • Benjamin C Kuo 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    Automatic Control Systems, 9th Edition       Chapter 2 Solutions  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     Golnaraghi, Kuo  Automatic Control Systems, 9th Edition         Chapter 2 Solutions  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     Golnaraghi, Kuo  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    Automatic Control Systems, 9th Edition       Chapter 2 Solutions  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    ;  Golnaraghi, Kuo  Automatic Control Systems, 9th Edition       Chapter 2 Solutions  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     Golnaraghi, Kuo  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 Real Axis   Part(c)  2‐7    1.5 2.5   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 Real Axis 2‐8    0.2 0.4   Automatic Control Systems, 9th Edition      2-5)  Chapter 2 Solutions  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) , , 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     Golnaraghi, Kuo  − jω (ω + 4) ; 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  ⎤ ⎡ x1 ⎤ ⎡ ⎤ ⎥⎥ ⎢⎢ x2 ⎥⎥ ⎢⎢1/ 3⎥⎥ + f ⎥ ⎢ x3 ⎥ ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎦ ⎣ x4 ⎦ ⎣1/ 3⎦ 0⎤ 0 ⎥⎥ 0 1⎥ ⎥ 0 0⎦ C = [1 0] D=0   10‐98     Golnaraghi, Kuo  Automatic Control Systems, 9th Edition     Chapter 10 Solutions  Use ACSYS State tool and follow the design process stated in Example 10-17-1: 10‐99     Golnaraghi, Kuo  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  10‐100     Golnaraghi, Kuo  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  10‐101     Golnaraghi, Kuo  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 10‐102     Golnaraghi, Kuo  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: Comparing equation (1) and (2) gives: 2 10‐103    (2) 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    Automatic Control Systems, 9th Edition     Chapter 10 Solutions  10‐105     Golnaraghi, Kuo  Automatic Control Systems, 9th Edition     Chapter 10 Solutions  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     Golnaraghi, Kuo  Automatic Control Systems, 9th Edition     Chapter 10 Solutions  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     Golnaraghi, Kuo  Automatic Control Systems, 9th Edition     Chapter 10 Solutions  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     Golnaraghi, Kuo  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    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    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 As 0, because , 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    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    ...   ‐1.0000 ‐ 0.0000i    Zeros:       1    2‐3    Golnaraghi, Kuo Automatic Control Systems, 9th Edition       Chapter 2 Solutions  Golnaraghi, Kuo 2-4) Mathematical representation: In all cases... Nyquist plots (polar plots):   Part(a)  2‐6    Golnaraghi, Kuo Automatic Control Systems, 9th Edition       Chapter 2 Solutions  Golnaraghi, Kuo Nyquist Diagram 15 10 Imaginary Axis -5 -10... Gd=1/(s*(s*T+1)) figure(4) nyquist(Gd) 2‐11    Golnaraghi, Kuo Automatic Control Systems, 9th Edition       Chapter 2 Solutions  Golnaraghi, Kuo %Part(e) T=3.5 L=0.5 Ge=pade(exp(-1*s*L),2)/(s*T+1)