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SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OFENGINEERING SYSTEMS 3RD EDITION KULAKOWSKISOLUTIONSSOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERINGSYSTEMS 3RD EDITION KULAKOWSKI2-CHAPTER 2SYSTEMS 3RD EDITION KULAKOWSKI2-SYSTEMS 3RD EDITION KULAKOWSKI2-SYSTEMS 3RD EDITION KULAKOWS...

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SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

SOLUTIONS

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

CHAPTER 2

2-1 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-2 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-3 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-4 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-5 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-6 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-7 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-8 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-9 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-10 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-11 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-12 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-13 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-14 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-15 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-16 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-17 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-18 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-19 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-20 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-21 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-22 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2-23 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

Problem 2.11

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SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

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SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

Problem 2.12

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SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

Problem 2.13 This is an algebraic equation and is hence easier to linearize. ∂ ∂ Tˆt ≅ (Tt ) Ω1 =Ω1 Ωˆ 1 + (T ) Ω =Ω ΔPˆ ∂Ω1 ∂ΔP t Δ 1P=Δ1P Δ P=Δ P

⎛ T ⎛ Ω ⎞⎛ 1 Tˆt ≅ ⎜ −2 0 ⎜1 − 1 ⎟⎜ − ⎜ 6 ⎝ Ω0 ⎠⎝ Ω0 ⎝

⎛ ⎞ ⎜ ⎟ ⎞⎞ 1 ˆ ⎜ ⎟ T Ω − Δ Pˆ ⎟ ⎟⎟ 1 0 P Δ ⎜ 1− ⎟ ⎠ ⎠ Ω1 =Ω1 ⎜ Δ P=Δ P P0 ⎟⎠ Ω1 =Ω1 ⎝ Δ P=Δ P

Which simplifies to:

T ⎛ Ω ⎞ˆ T0 Tˆt ≅ 0 ⎜ 1 − 1 ⎟ Ω ΔPˆ 1− ΔP 3 Ω0 ⎝ Ω0 ⎠ 1− P0

2-27 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

Problem 2.14 The loop equation for the circuit is: e i = Ri + L

di +eD dt

and the diode is modeled by this equation: i = I 0eαeD

which is clearly nonlinear. To solve the problem first we must solve the diode function for eD and substitute it back to the loop equations.

eD =

⎛ i ⎞ ln ⎜ ⎟ α ⎝I0⎠ 1

The loop equation is now:

e i = Ri + L

di 1 ⎛ i ⎞ + ln ⎜ ⎟ dt α ⎝ I 0 ⎠

Now follow the five steps: 1) Derive the nonlinear model Done (see above) 2) Determine the normal operation point: Generally, the NOP occurs at an equilibrium where the derivatives go to zero. Setting the di/dt term to zero, we find:

e i = Ri +

⎛i ⎞ ln ⎜ ⎟ α ⎝ I0 ⎠ 1

3) Introduce incremental variables: e i = e i + eˆi i = i + iˆ

2-28 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

(

)

e i + eˆi = R i + iˆ + L

(

1 ⎛ i + iˆ d ˆ i + i + ln ⎜ α ⎜ I0 dt ⎝

(

)

) ⎞⎟

⎟ ⎠

Before we linearize, let’s regroup:

e i + eˆi = Ri + Riˆ + L

(

di diˆ 1 ⎛ i + iˆ + L + ln ⎜ dt dt α ⎜ I0 ⎝

) ⎞⎟

⎟ ⎠

4) Linearize the nonlinear term:

(

) ⎟⎞ ≅ 1 ln ⎛ i

⎛ i +iˆ ln ⎜ α ⎜ I0 ⎝ 1

=

⎞ ∂ ⎡1 ⎛ i ⎞⎤ ⎜ ⎟ + ⎢ ln ⎜ ⎟ ⎥ ˆi ⎝ I0 ⎠ ∂ i ⎣α ⎝ I0 ⎠ ⎦ i =i

⎟ α ⎠

⎛ i ⎞ 1 ⎛I ⎞ ln ⎜ ⎟ + ⎜ 0 ⎟ iˆ α ⎝ I0 ⎠ α ⎝ i ⎠ 1

5) Substitute back in and re-group:

e i + eˆi = Ri + Riˆ + L

e i + eˆi = Ri +

di diˆ 1 ⎛ i ⎞ 1 ⎛ I ⎞ + L + ln ⎜ ⎟ + ⎜ 0 ⎟ iˆ dt dt α ⎝ I0 ⎠ α ⎝ i ⎠

⎛ i ⎞ diˆ 1 ⎛ I ⎞ ln ⎜ ⎟ + Riˆ + L + ⎜ 0 ⎟ iˆ α ⎝I0⎠ dt α ⎝ i ⎠

1

Notice that the definition of the NOP allows us to cancel the first term on the LHS with the first two terms on the RHS. Rearranging, we have the final answer: eˆi = L

1 diˆ ⎛ + ⎜R + α dt ⎝

⎛ I0 ⎜i ⎝

⎞⎞ ˆ ⎟⎟i ⎠⎠

2-29 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

Problem 2.15 To solve the problem, we need to derive the equation of motion for the ball. Combining the equation for the magnetic force with the information in the free body diagram, give us the following nonlinear equation of motion. mx = mg − K i

i2 x2

Part (a) asks to find the relationship between the coil current and the position of the ball for a steady state equilibrium. This is found by setting the derivatives (wrt time) equal to zero. 0 = mg − K i

i2 x2

which leads to:

i =

mg x Ki

Part (b) asks for the linearized equation describing small perturbations about the nominal equilibrium point. Using the Taylor series expansion, the following equation can be derived: mxˆ = −

2K i i ˆ 2K i i 2 i+ xˆ 2 x x3

2-30 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

Problem 2.16 Since the shaft connecting J1 and J2 is rigid, we can lump them together as a single rotational inertia. Free Body Diagrams of the two bodies (the combined inertias and the rack) lead to the following two equations of motion:  Ti − RFrack = ( J 1 + J 2 )Ω

− Frack − Fk − Fb = mx Where Frack is the internal force transmitting the motion between the rack and the pinion (J2), Fk is the force exerted by the spring to the rack (assumed positive in compression) and Fb is the damping force exerted on the pinion. Solve the second equation for Frack and substitute into the first, thus eliminating the force.  − Rmx Ti + RFk + RFb = ( J1 + J 2 ) Ω

To eliminate the rotational variable, we need to derive the kinematic constraint relating the pinion motion to that of the rack. The sign is important here as the figure defines positive rotation and positive translation to be inconsistent with each other.

RΩ = − x Taking the time derivative of the constraint equation and substituting into the previous equation gives us the answer:

Ti ⎡ ⎛ J1 + J 2 ⎞ ⎤ ⎟ ⎥x + bx + kx = − ⎢m + ⎜ 2 R ⎝ R ⎠⎦ ⎣

2-31 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

Problem 2.17 There are three energy storing elements in this system, hence a third order differential equation can be expected. Examining J1, the flexible drive and J2 separately, the following three equations can be written:

 Ti − BΩ1 − Tshaft = J 1Ω 1 Tshaft = K (θ 1 − θ 2 )

 Tshaft = J2 Ω 2 Combining the three equations to eliminate Tshaft and W1, leads to the following solution:

J 1J 2  J 2 B  Ω2+ Ω + (J1 + J2 )Ω 2 + BΩ2 = Ti K K 2 Alternatively, in d/dt form: J 1J 2 d 3 Ω 2 K

dt

3

+

J 2 B d 2 Ω2 K

dt

2

+ ( J1 + J 2 )

dΩ 2 + BΩ 2 = Ti dt

2-32 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

Problem 2.18 This is a great problem because it appears simple, but it’s quite difficult. In later chapters, it will be a great example of how a state space approach can help clarify thinking. The forces acting on the mass quickly lead to this equation: F − Fk 1 − Fk 2 = mx Assuming that the reference point for measuring x is the static equilibrium point. The spring force in the second spring is easily derived: Fk 2 = k 2 x Now it gets interesting. Let’s introduce a new displacement variable, x’, which is the displacement of the point linking the spring k1 with the damper. Fk1 = k1( x − x′) Fb = bx′ Now, notice that the force in the spring k1 must be the same as the force in the damper. We can combine the two equations and eliminate the variable x’.

 k1 + k1 F k1 = k1 x F b To make this a little more straightforward, it is advantageous to introduce the “D”operator to combine our first equation with the definitions of Fk1 and Fk2. The result is the following equation: mk1 2 k1 k2 ⎤ ⎡ 3 ⎢ mD + b D + (k 1 + k 2 ) D + b ⎥ x = ⎣ ⎦

k1 ⎤ ⎡ ⎢D + b ⎥ F ⎣ ⎦

Which, in traditional form becomes:

m

d 3x dt 3

+

mk 1 d 2 x d x k 1k 2 dF k1 x= + (k1 + k 2 ) + + F 2 b dt b dt b dt

2-33 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

Problem 2.19 From Problem 1.15, we already know the linearized version of the force: 2 ˆm = 2K i i ˆi − 2K i i ˆx F x2 x3

For a nominal value of 0.015 m for x, the corresponding current is 2.144 amps, and the nominal force is 0.6671 N. The following Matlab Script computes the nonlinear force for various values of displacement and a constant current. It also computes the linearized approximation of the force. % % Script to Solve P2.19 % m=0.068; Ki=3.2654E-5 xbar=0.015; % % Nominal current and force at xbar % ibar=(m*9.81/Ki)^0.5*xbar Fmbar=Ki*ibar^2/xbar^2; % % Compute linear coefficient for x-hat % Kx=2*Ki*ibar^2/xbar^3; % % set up the displacement matrix % x=[0.005:0.001:0.025]; % % compute nonlinear force and the linearized force % for i=1:length(x) Fm(i)=Ki*ibar^2./x(i)^2; Fmhat(i)=Fmbar-Kx*(x(i)-xbar); end % % Plot results % plot(x,Fm,x,Fmhat) grid title('Mag Lev Force vs. position') xlabel('equilibrium position (m)') ylabel('Force (N))')

As a result of running this script, the following plot is generated: 2-34 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

Mag Lev Force vs. position 7

6

5

Force (N))

4

3

2

1

0

-1 0.005

0.01

0.015 equilibrium position (m)

0.02

0.025

To get a handle on the linearization error, compute the difference between the two force vectors: >>err=Fm-Fmhat err = Columns 1 through 11 4.4472 2.7017 1.6845 1.0555 0.1084 0.0432 0.0098 -0.0000

0.6523

0.3891

0.2176

0.1529

0.2069

0.2656

Columns 12 through 21 0.0082 0.0302 0.0630 0.3282 0.3940 0.4625

0.1045

The result is easier to interpret when you plot it: >>plot(x,err) I manipulated this plot using MATLAB’s zoom tool to get this figure: 2-35 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

0.08

0.06

linerization error

0.04

0.02

0

-0.02

0.013

0.0135

0.014

0.0145

0.015 0.0155 0.016 displacement

0.0165

0.017

0.0175

0.018

We’re looking for the range of displacement that keeps the linearization error within 10%. For a nominal force of 0.67 N, we want the error to be under 0.067N. The plot indicates that the answer is: 0.0125 m < x < 0.018 m

2-36 © 2007 by John Gardner; duplication and distribution rights not granted

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

Problem 2.20 Free body diagrams of the two masses yield the following equations of motion:

m1x1 +b1 ( x1 − x2 ) + b2 x1 + k1 ( x1 − x2 ) + k2 ( x1 − x2 ) + k3 x1 = Fi m2  x 2 = b1 ( x1 − x 2 ) + k1 ( x1 − x2 ) + k2 ( x1 − x2 ) Re-arrange to group by terms:

m1x1 + ( b1 +b2 ) x1 + ( k1 + k2 + k3 ) x1 − b1 x2 −( k1 + k2 ) x2 = Fi m2  x2 + b1 x2 + (k1 + k2 ) x2 − b1 x1 − ( k1 + k2 ) x1 = 0 Now we need to combine these equations to eliminate the unwanted motion variable, x1. The most expedient approach is to use the “D” operator, solve the second equation for x1 and substitute it back into the first equation. ⎡⎣ m1 D 2 + ( b1 + b2 ) D + ( k1 + k2 + k3 )⎤⎦ x1 + ⎡⎣−b1 D − ( k1 + k2 ) ⎤⎦ x2 = Fi 2 ⎣⎡ m 2D + b1D + ( k 1 + k 2 )⎦⎤ x 2 + ⎡⎣−b1D − (k 1 + k 2 ) ⎤⎦ x 1 = 0

Solving for x1: x1 =

m2 D 2 + b1 D + ( k1 + k2 ) x2 ⎡⎣b1D + ( k1 + k 2 )⎤⎦

Substituting back: ⎡⎣ m1 D 2 + ( b1 + b2 ) D + ( k1 + k2 + k3 )⎤⎦

m2 D 2 + b1 D + ( k1 + k2 ) x2 + ⎡⎣ −b1 D − ( k1 + k2 )⎤⎦ x2 = Fi ⎡⎣ b1 D + ( k1 + k2 ) ⎤⎦

Now multiply through by the denominator and gather terms: [m 1m 2D 4 + (m1b1 + m 2 ( b1 + b 2 ) ) D 3 + (m1 ( k1 + k 2 ) + b1 ( b1 + b 2 ) + m 2 ( k 1 + k 2 + k 3 ) − b12 ) D 2

(

)

+ ( (b1 + b2 )( k1 + k2 ) + b1 (k1 + k2 + k3 )− 2b1 (k1 + k 2 )) D + (k1 + k 2 )(k 1 + k 2 + k 3 )− (k 1 + k 2 ) ]x 2 = [b1D + (k1 + k 2 )]Fi

Which can be transformed back to an input/output differential equation: 2-37 © 2007 by John Gardner; duplication and distribution rights not granted

2

SOLUTIONS MANUAL FOR DYNAMIC MODELING AND CONTROL OF ENGINEERING SYSTEMS 3RD EDITION KULAKOWSKI

2 d 4 x2 d 3 x2 2 d x2 + + + + + + + + + + − m b m b b m k k b b b m k k k b ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 1 2 1 1 2 1 1 2 2 1 2 3 1 dt4 dt3 dt2 d x2 2 + (( b1 + b2 )( k1 + k2 ) + b1 ( k1 + k2 + k3 ) − 2 b1 ( k1 + k 2 ) ) + ( k1 + k 2 )( k1 + k 2 + k 3 ) − ( k1 + k 2 ) x 2 dt dF = b1 i + ( k1 + k 2 ) Fi dt

m1m 2

(

2-38 © 2007 by John Gardner; duplication and distribution rights not granted

)...