Modeling and Analysis of Dynamic Systems 2nd Edition

1 Review Problems 1. Write a user-defined function with function call C=temp_conv(F) that converts the temperature from Farenheit Fto Celsius C. Execute the function for the case of F = 86. Solution function C = temp_conv(F) C = (F-32)*100/180; >> C = temp_conv(86)C =30 >> C = temp_conv(86)C =30 >> C = temp_conv(86) C = 30 2. Write a user-defined function

439 00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0 0.2 0.4Open-loop response 0.4Open-loop response 0.4 Open-loop response 0.6 0.8 11.2 11.2 1 1.2 1.4 Step Response Time (sec)Figure PS10-4 No5 Time (sec)Figure PS10-4 No5 Time (sec) Figure PS10-4 No5 Amplitude Closed-loop response 6. Consider the feedback control system shown in Figure 10.43. a. Find the values for kpand TIsuch that the maximum overshoot in the response to a unit-step reference input is

419 Problem Set 10.1 1. Draw a block diagram for the feedback control of a liquid-level system, which consists of a valve with a control knob (0100%) and a liquid-level sensor. Clearly label essential components and signals. Solution Figure PS10-1 No1 Figure PS10-1 No1 Figure PS10-1 No1 2. Draw a block diagram for the feedback control of

459 c. The frequency response function is 2 jM 255 22 jM 12 12 25 2 0.01 1 j M (j j M 144 2 0.01 1 G The basic terms include: 1. One constant term 25 144 ( 25 10 144 20log 15 dB) 2. Two second-order terms: one in the numerator with n rad/s and and one in the denominator with n

479 479 479 8. Consider Problem 7 in Problem Set 10.7. Using the full-state feedback controller obtained in Part (b), build a Simulink block diagram to simulate the resulting feedback control system. Find the closed-loop response if the initial conditions are x1(0) = 0.1 and x2(0) = 0. Solutionyx' = Ax+Bu y =

23 In Problems 1924, (a) Find the inverse Laplace transform using the partial-fraction expansion method. (b) Repeat in MATLAB. 19. 34 (1) s ss Solution (a) Expand as 34 34 ( ) 41 (1) 1 (1) AB A sABABsA AB ss s s ss Therefore,134 4 1 4(1) 1tsess s s L(b)>> syms s>> ilaplace((3*s+4)/s/(s+1))ans =4 - 1/exp(t) Therefore,134 4 1 4(1) 1tsess s s L(b)>>

11 Problem Set 2.1 In Problems 14, (a) Perform 12 /zz and express the result in rectangular form, (b) Verify that 12 1 2 //zz z z , (c) Repeat Part (a)in MATLAB. 1.3 2 j j Solution (a) 31313 2222 jj j j jj (b)22 2 2 4 2j ,22j(c)>> z1 = -3-j; z2 = 2*j; z1/z2ans =-0.5000 + 1.5000i (b)22 2 2 4 2j ,22j(c)>> z1

35 Problem Set 3.1 In Problems 18 perform the indicated operations, if defined, for the vectors and matrices below. 0 30 2 1 0 1 , , , 1 12 15 4 2 3 AB vw 1. T B+vw Solution 22317 10T B+vw 22317 10T B+vw 223 17 10 T B+vw 2. T vB w Solution Operation is undefined! Operation is undefined! Operation is undefined! 3. TT ww vv Solution15TT ww

50 5. 110 220 001 A Solution (a) Block diagonal matrix; () 0,1,3 O A . For 10 O we solve >@ 111 110 0 1 2 2 0 0 1 001 0 0 Av 0 v v For 21O we solve >@222010 0 0 2 1 0 0 0000 0 1 AIv 0 v vFor 33O we

120 2 12 12 (2 1) 3 (3 2) 0 ss X X F XsX Solve for 1 X : 1322303 2 (3 2)65213132FssFXssssss Since the output is x, the transfer function is 132(3 2)65sX ssFsss 1322303 2 (3 2)65213132FssFXssssss Since the output is x, the transfer function is 132(3 2)65sX ssFsss 132 2 3 03 2 (3 2) 65 213 132 F ssF Xsss ss s Since

65 Problem Set 4.1 In Problems 110 express the system model, assuming general initial conditions, in (a) Configuration form, (b) Standard, second-order matrix form. 1. 11 12 1 22 12 2 2( ) 10 2( ) 0 t xx xx e xx xx Solution (a) The generalized coordinates are 11 qx and 22 qx . Then 11 12 1121212 2 1 2 2121222( )10

85 SolutionWith the assumption of zero initial conditions, Laplace transformation of the governing equation yields Transfer function21()1( ) ( ) 1() 1Is CsLs R I s V sCs V s LCs RCsLs R Cs SolutionWith the assumption of zero initial conditions, Laplace transformation of the governing equation yields Transfer function21()1( )

105 105 105 20. The block diagram representation of a system model is presented in Figure 4.33, where U, 1 X ,2 X , and Y denote the Laplace transforms of the input, the two state variables, and the output. (a) Derive the state-space form directly from the block diagram. (b) Find the transfer function directly from the block diagram. Figure

208 eq ()mx k x f t eq ()mx k x f t eq ()mx k x f t Figure Review5 No3 b. Taking the Laplace transform of both sides of the preceding equation with zero initial conditions results in 2eq() 1()XsFs ms k c. The state, the input, and the output are12,,xxuf

194 b. Using the differential equation obtained in Part (a), determine the state-VSDFHUHSUHVHQWDWLRQ8VHa,1a, DQG1as tKHVWDWHYDULDEOHVDQGXVH2DQG2as the output variables. Figure 5.99 Problem 4. Solution a. Assume that a1 !! . The free-body diagram is shown below. Applying the moment equation to each gear, +: OO MI Gear 1: a1a1 a KI W Gear 2: 1a 1 1 11 KrFI Gear 3: 22222rF

154 Rearranging the equations, we have 11 11 1 2 1 2 2 1 ()mx bx k k x k x f 22 21 2 3 2 2 ()mx kx k k x f Figure PS5-2 No14 Figure PS5-2 No14 Figure PS5-2 No14 b. The differential equations can be expressed in second-order matrix form as 12 211

174 174 174 15. Consider the system shown in Figure 5.73, where a uniform sphere of mass mand radius rrolls along an inclined plane of 30. A translational spring of stiffness kis attached to the sphere. Assuming that there is no slipping between the sphere and the surface, draw the necessary free-body diagram

134 Problem Set 5.1 1. If the 50-kg block in Figure 5.12 is released from rest at A, determine its kinetic energy and velocity after it slides 5 m down the plane. Assume that the plane is smooth. Figure 5.12 Problem 1. Solution Choosing point Bas the datum and applying the principle of conservation of

262 Figure 6.74 Problem 6. Solution The transfer function Vo(s)/Vi(s) is oi212(1)( 1()))(RCsRC sVsVs s RC oi212(1)( 1()))(RCsRC sVsVs s RC o i 2 12 (1)( 1 ( )) ) ( RCs RC s Vs Vs s RC MATLAB Figures Problem 1 i(t) vC(t)V+-Voltage SensorSwitchf(x)=0SolverConfigurationSimulink-PSConverterScopeRe si st o rPS-SimulinkConverterElectrical ReferenceDC Voltage SourceCurrent Sensor+-Capacitor vC(t)V+-Voltage SensorSwitchf(x)=0SolverConfigurationSimulink-PSConverterScopeRe si st o rPS-SimulinkConverterElectrical ReferenceDC Voltage SourceCurrent Sensor+-Capacitor vC(t) V + - Voltage Sensor Switch f(x)=0 Solver Configuration Simulink-PS Converter Scope Re si st o r PS-Simulink Converter Electrical Reference DC Voltage

222 Problem Set 6.1 1. Determine the equivalent resistance Req for the circuit shown in Figure 6.8. Figure 6.8 Problem 1. Solution The equivalent resistance for the parallel-connected resistors R1and R3is eq1 1 3 111 RRR or 13 eq1 13 RR RRR They are connected with the resistor R2in series. The equivalent resistance for the circuit is13 12 23 13eq 2 eq1

242 Figure 6.38 Problem 2. Solution Because the current drawn by the op-amp is very small, i.e., 0ii || , we have at node 1, 12 ii o 1 12 vv vv RR or 21 1o 1 2 ()Rv Rv R R v and at node 2,34ii 212vv vRR or 22 1 2()Rv R R v where vv|. Eliminating v_and v+gives21 1o 22Rv

274 which can be rearranged and gives the second equation 122 22 111 0vvv RRL or 1222 0Lv Lv R v In second-order matrix form, we have1121112 12 a222000vRRRvRRC RRCvvRvLL Assume that all the initial conditions are zero. Taking Laplace transform gives 12 1 2 1 1 12 a22 0RRCs R R R V s RRCsV sLs

285 Problem Set 7.1 1. A car has an internal volume of 2.8 m3. If the sun heats the car from a temperature of 20C to a temperature of 40C, what will the pressure inside the car be? Assume the pressure is initially 1 atm. Solution This is a constant-volume process, where 1 20 C

298 Solution a. The characteristic length of the junction is 3 4 body 3 c2 surface 1 0.001 4 Vr Lr Ar The Biot number of the junction is 4c70 0.001 2.92 10 0.140 6hLBi k u 4c70 0.001 2.92 10 0.140 6hLBi k u 4 c 70 0.001 2.92 10 0.1 40 6 hL Bi k u Thus, the junction can be treated as a lump-temperature system, and its

365 >> xRK4 = RK4System(f,t,x0);>> theta2 = xRK4(1,:); plot(t,theta2) >> xRK4 = RK4System(f,t,x0);>> theta2 = xRK4(1,:); plot(t,theta2) >> xRK4 = RK4System(f,t,x0); >> theta2 = xRK4(1,:); plot(t,theta2) Figure PS8-5No11 12. A nonlinear dynamic system model is derived as 3 2 ( ) , (0) 0 , 0 3xxutx t dd where ()ut is the unit-step function. (a) Build the Simulink model and

351 >> simple(expm(A*t))ans =[ 2 - 1/exp(t), 1/exp(t) - exp(5*t), 2*exp(5*t) - 2][ 2 - 2/exp(t), 2/exp(t) - exp(5*t), 2*exp(5*t) - 2][ 1 - 1/exp(t), 1/exp(t) - exp(5*t), 2*exp(5*t) - 1](b)>> A=[1 -6 10;2 -7 10;1 -6 10]; syms s>> simple(ilaplace(inv(s*eye(3)-A)))ans =[ 2 - 1/exp(t), 1/exp(t) - exp(5*t), 2*exp(5*t) - 2][

311 Problem Set 8.1 1. Consider a first-order system with time constant W and zero initial condition. Find the systems unit-impulse response for 1 4 W and 1 2 W , plot the two curves versus 02tdd in the same graph, and comment. Solution When 1 4 W , the unit-impulse response is 4 1() 4 t xt e When 12W , it is22() 2

331 Figure PS8-2No21-1 Figure PS8-2No21-2 22. The equations of motion of a mechanical system are given below, where ()t G denotes the unit-impulse. Assuming zero initial conditions, plot the response 1()xt by (a) Simulating the Simulink model of the system. (b)Using the impulse command. 11 12 2122 22 () 20 xx xx t xxxx G 050 100 150 -0.3 -0.2 -0.2 -0.2 -0.1 -0.1 -0.1 00 00 0 0 0.1 0.2 0.3 0.4 0.5 0.6 Contribution of f1 to x3 050 100

379 Problem Set 9.1 1. A lightly damped single-degree-of-freedom system is subjected to free vibration. The response of the system is shown in Figure (VWLPDWHWKHYDOXHRIWKHYLVFRXVGDPSLQJUDWLR. Figure 9.5 Problem 1. Solution Note that the first and fifth peak displacements are about 0.62 m and 0.09 m, respectively. Thus, the logarithmic decrement is 1 5 1 1 0.6 OQ OQ

399 % modal response to external excitations f0 = [0.15 0 0]'; omega = 0.15; figure(1); t = 0:0.1:100;q = zeros(length(t),3);for i = 1:3subplot(3,1,i);q(:,i) = U(:,i)'*f0/(omegar(i)^2-omega^2)*sin(omega*t);plot(t,q(:,i),'b-');end% system response to external excitationsx = zeros(3,length(t));for i = 1:length(t)x(:,i) = U*q(i,:)';endfigure(2);for i = 1:3subplot(3,1,i);plot(t,x(i,:),'b-');end t = 0:0.1:100;q = zeros(length(t),3);for i = 1:3subplot(3,1,i);q(:,i) = U(:,i)'*f0/(omegar(i)^2-omega^2)*sin(omega*t);plot(t,q(:,i),'b-');end% system response to