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Solution Manual for Control Systems Engineering, 8th Edition By Norman S. Nise All Chapters | Complete Guide | Academic Year | Graded A+
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All Chapters | Complete Guide | Academic Year | Graded A+
Chapter Title 1 Introduction to Control Systems 2 Modeling in the Frequency Domain 3 Modeling in the Time Domain 4 Time Response 5 Reduction of Multiple Subsystems 6 Stability 7 Steady-State Errors 8 Root Locus Techniques 9 Design via Root Locus
Chapter Title 10 Frequency Response Techniques 11 Design via Frequency Response 12 Design via State Space 13 Digital Control Systems
Control Systems Engineering, 8th Edition by Norman S. Nise is a comprehensive textbook that provides a balanced presentation of control systems theory and applications. This solution manual contains verified solutions to all end-of-chapter problems and design problems. Key Features: Complete solutions for all chapters Step-by-step problem-solving approach MATLAB verification where applicable Detailed rationales and explanations Perfect for students and instructors
PROBLEM 1. A system is described by the differential
Using block diagram reduction techniques:
PROBLEM 2. Find the Laplace transform of f(t)=e−2tcos(3t) f( t)= e−2 tcos(3 t). SOLUTION: Using the Laplace transform property: L[e−atcos(ωt)]=s+a(s+a)2+ω2L[ e− atcos( ωt)]=( s+ a) 2 + ω 2 s+ a For a=2 a=2, ω=3 ω=3: F(s)=s+2(s+2)2+9=s+2s2+4s+13 F( s)=( s+2) 2 +9 s+ = s 2 +4 s+13 s+ ANSWER: F(s)=s+2s2+4s+13 F( s)= s 2 +4 s+13 s+ PROBLEM 2. Find the inverse Laplace transform of F(s)=2s+5s2+5s+6 F( s)= s 2 +5 s+62 s+5. SOLUTION: Factor the denominator: s2+5s+6=(s+2)(s+3) s 2 +5 s+6=( s+2)( s+3) Partial fraction expansion: F(s)=As+2+Bs+3 F( s)= s+2 A+ s+3 B 2s+5=A(s+3)+B(s+2) 2 s+5= A( s+3)+ B( s+2) Solving for A and B: Let s=−2 s=−2: 2(−2)+5=1=A(1)⇒A=12(−2)+5=1= A(1)⇒A=
mx¨+bx˙+kx=f(t) mx¨+ bx˙+ kx= f( t) Taking the Laplace transform with zero initial conditions: ms2X(s)+bsX(s)+kX(s)=F(s) ms 2 X( s)+ bsX( s)
PROBLEM 3. Write the state equations for the system described by: d2ydt2+3dydt+2y=u(t) dt 2 d 2 y+3 dtdy+2 y= u( t) SOLUTION: Choose state variables: x1=y x 1 = yx2=y˙ x 2 = y˙ Then: x˙1=x2 x˙ 1 = x 2 x˙2=y¨=−3y˙−2y+u=−3x2−2x1+u x˙ 2 = y¨ =−3 y˙−2 y+ u=−3 x 2 −2 x 1 + u In matrix form: x˙=[01−2−3]x+[01]u x ˙=[0−21−3] x +[ 01 ] uy=[10]x y=[ 10 ] x ANSWER: State equations as shown above
PROBLEM 3. Find the transfer function Y(s)/U(s) Y( s)/ U( s) from the state-space representation: x˙=[01−6−5]x+[01]u x ˙=[0−61−5] x +[ 01 ] uy=[10]x y=[ 10 ] x SOLUTION: The transfer function is given by: T(s)=C(sI−A)−1B T( s)= C ( s I − A )−1 B Compute sI−A=[s−16s+5] s I − A =[ s 6 −1 s+5] The determinant is: det(sI−A)=s(s+5)+6=s2+5s+6det( s I − A )= s( s+5)+6= s 2 +5 s+6( sI−A)−1=1s2+5s+6[s+51−6s]( s I − A )−1= s 2 +5 s+61[ s+5−6 1 s] Then: C(sI−A)−1B=1s2+5s+6[10][s+51−6s][01] C ( s I − A ) −1 B = s 2 +5 s+61[ 10 ][ s+5−6 1 s][ 01 ]=1s2+5s+6[10] [1s]=1s2+5s+6= s 2 +5 s+61[ 10 ][ 1 s]= s 2 +5 s+ ANSWER: T(s)=1s2+5s+6 T( s)= s 2 +5 s+
PROBLEM 4. A second-order system has a natural frequency ωn=10 ωn=10 rad/s and a damping ratio ζ=0.5 ζ=0.5. Find the peak time, percent overshoot, and settling time. SOLUTION:
A=55=1 A=55=15s(s2+2s+5)=1s+−s−2s2+2s+5 s( s 2 +2 s+5) = s 1 + s 2 +2 s+5− s− Inverse Laplace: y(t)=1−e−tcos(2t)−12e−tsin(2t) y( t)=1− e− tcos(2 t)− e− tsin(2 t) ANSWER: y(t)=1−e−t(cos2t+12sin2t) y( t)=1− e− t(cos2 t+ 21 sin2 t)
PROBLEM 5. Reduce the block diagram to a single transfer function. SOLUTION: Step-by-step reduction:
PROBLEM 5. Use Mason's gain formula to find the transfer function of the signal-flow graph. SOLUTION: Mason's gain formula: T(s)=∑kPkΔkΔ T( s)=Δ∑ kPkΔ k Where: Pk Pk = forward path gain ΔΔ = 1 - (sum of all loop gains) + (sum of products of non-touching loops)
PROBLEM 6. Use the Routh-Hurwitz criterion to determine the stability of the system with characteristic equation: s4+2s3+3s2+4s+5=0 s 4 +2 s 3 +3 s 2 +4 s+5= SOLUTION: Construct the Routh array:
PROBLEM 7. Find the steady-state error for a unit step input for the system: G(s)=10s(s+2) G( s)= s( s+2) SOLUTION: For a unity feedback system, the error is: E(s)=11+G(s)R(s) E( s)=1+ G( s)1 R( s) For a unit step input R(s)=1/s R( s)=1/ s: E(s)=11+10s(s+2)⋅1s=s(s+2)s(s+2)+10⋅1s=s+2s(s2+2s+10) E ( s)=1+ s(s+2)10 1 ⋅ s 1 = s( s+2)+10 s( s+2)⋅ s 1 = s( s 2 +2 s+10) s+ Using the final value theorem: ess=lims→0sE(s)=lims→0s+2s2+2s+10=210=0.2 ess= s→0lim sE( s)= s→0lim s 2 +2 s+10 s+2=102=0. ANSWER: ess=0.2 ess=0. PROBLEM 7. For the system in Problem 7.1, find the position error constant Kp Kp, velocity error constant Kv Kv, and acceleration error constant Ka Ka. SOLUTION: The open-loop transfer function is:
G(s)=10s(s+2) G( s)= s( s+2) Position error constant: Kp=lims→0G(s)=lims→010s(s+2)=∞ Kp= s→0lim G( s)= s→0lim s( s+2)10=∞ Velocity error constant: Kv=lims→0sG(s)=lims→010s+2=5 Kv= s→0lim sG( s)= s→0lim s+ = Acceleration error constant: Ka=lims→0s2G(s)=lims→010ss+2=0 Ka= s→0lim s 2 G( s)= s→0lim s+210 s= ANSWER: Kp=∞,Kv=5,Ka=0 Kp=∞, Kv=5, Ka=
PROBLEM 8. Sketch the root locus for the system: G(s)=Ks(s+2)(s+4) G( s)= s( s+2)( s+4) K SOLUTION: Steps for sketching root locus:
%OS=e−πζ/1−ζ2×100%=20%% OS= e− πζ/1− ζ 2 ×100%=20%ζ=−ln(0.2)π2+ln2(0.2)=0.456 ζ= π 2 +ln 2 (0.2) −ln(0.2)=0. Settling time: Ts=4ζωn=2⇒ωn=4ζ×2=40.456×2=4.386 rad/s Ts = ζωn 4 =2⇒ωn= ζ×24=0.456×24=4.386rad/s Desired poles: s=−ζωn±jωn1−ζ2=−2±j3.9 s=− ζωn± jωn1− ζ 2 =−2± j3.
PROBLEM 10. Sketch the Bode plot for: G(s)=100s(s+10) G( s)= s( s+10) SOLUTION:
G(s)=Ks(s+5) G( s)= s( s+5) K to meet the specifications: Kv=20 Kv=20, PM=45° PM=45° SOLUTION:
PROBLEM 12. Design a state feedback controller for the system: x˙=[0100]x+[01]u x ˙=[ 0010 ] x +[ 01 ] u to place the closed-loop poles at s=−3±j3 s=−3± j 3. SOLUTION:
