Exam (elaborations) TEST BANK FOR Digital Control Engineering - Analysis and Design 2nd Edition By Antonio Visioli, M. Sami Fadali (Solution Manual)

Exam (elaborations) TEST BANK FOR Digital Control Engineering - Analysis and Design 2nd Edition By Antonio Visioli, M. Sami Fadali (Solution Manual) Chapter 1 Solutions 1.1 A fluid level control system includes a tank, a level sensor, a fluid source and an actuator to control fluid inflow. Consult any classical control text1 to obtain a block diagram of an analog fluid control system. Modify the block diagram to show how the fluid level could be digitally controlled. Water Level Reference Level DAC ADC Computer Tank Level Sensor Actuator & InflowValve Block diagram of water level digital control system. 1.2 If the temperature of the fluid of Problem 1.1 is to be regulated together with its level, modify the analog control system to achieve the additional control (Hint: an additional actuator and sensor are needed). Obtain a block diagram for the two-input-two-output control system with digital control. Reference Water Level Level DAC ADC Computer Tank Level Sensor Actuator & InflowValve Reference Temperature Temperature Sensor Heater Temperature Block diagram of water level and temperature digital control system. Note that the DAC and ADC can have more than one input and output channel. 1.3 Position control servos are discussed extensively in classical control texts. Draw a block diagram for a DC motor position control system after consulting your classical control text. Modify the block diagram to obtain a digital position control servo. For the angular position sensor we could use a potentiometer, which is often packaged with an ADC to give a digital output. 1See for example: J. Van deVegte, Feedback Control Systems, Prentice Hall, Englewood Cliffs, NJ, 1994. 2 Angular Position Reference Position DAC ADC Motor Computer & Load Angular Position Sensor Block diagram of DC motor digital position control system. 1.4 Repeat Problem 1.3 for a velocity control servo. For the angular velocity sensor we could use a tachometer, which is often combined with an ADC to give a digital output. Alternatively, we could use an optical encoder , which has a digital output. Angular Velocity Reference Velocity DAC ADC Motor Computer & Load Angular Velocity Sensor Block diagram of DC motor digital velocity control system. 1.5 A ballistic missile is required to follow a predetermined flight path by adjusting its angle of attack  (the angle between its axis and its velocity vector v). The angle of attack is controlled by adjusting the thrust angle  (angle between the thrust direction and the axis of the missile). Draw a block diagram for a digital control system for the angle of attack including a gyroscope to measure the angle  and a motor to adjust the thrust angle . Fig. P1.1 Missile angle of attack control. 3  Reference Angle DAC ADC Computer Missile Angle Sensor Angle Actuator Thruster Block diagram of digital missile control system. 1.6 A system is proposed to remotely control a missile from an earth station. Due to cost and technical constraints, the missile coordinates would be measured every 20 seconds for a missile speed of up to 500 m/s. Is such a control scheme feasible? What would the designers need to do to eliminate potential problems? If the missile is only observed every 20 seconds with speeds of up to 500 m/s, the missile position could change drastically between measurements. This makes the control scheme unrealistic. The missile coordinates need to be measured at a much higher rate. 1.7 The control of the recording head of a dual actuator hard disk drive (HDD) requires two types of actuators to achieve the required high areal density. The first is a coarse voice coil motor (VCM) with large stroke but slow dynamics and the second is a fine piezo-electric transducer (PZT) with a small stroke and fast dynamics. A sensor measures the head position and the position error is fed to a separate controller for each actuator. Draw a block diagram for a dual actuator digital control system for the HDD2. Control Computer Reference Position Position Sensor + Fine Controller DAC ADC DAC Coarse Controller VCM PZT Recording Head 2 J. Ding, F. Marcassa, S.-C. Wu, and M. Tomizuka, “Multirate control for Computational Saving”, IEEE Trans. Control Systems Tech., Vol. 14, No. 1, January 2006, pp. 165-169. 1 Chapter 2 Solutions 2.1 Derive the discrete-time model of Example 2.1 from the solution of the system differential equation with initial time kT and final time(k+1)T. The volumetric fluid balance gives the analog mathematical model C h q dt dh   i  where  = R C is the fluid time constant for the tank. The solution of this equation is        t t i t t e t q d C h t e h t 0 ( ) 0 ( ) 1 ( ) / ( ) 0 ( ) /     Let qi be constant over each sampling period T, i.e. qi(t) = qi(k) = constant, for t in the interval [kT, (k+1)T). Then (i) Let t0 = kT, t = (k + 1)T (ii) Simplify the integral as follows with : (k 1)T      1  ( ) 1 ( ) ( ) 1 ( ) 1 ( ) / 0 / ( 1) [( 1) ] / ( 1) [( 1) ] / e q kT C e d q kT C e d q kT C e q kT d C i T T i i k T kT k T k T kT i k T                                       k T T kT d d 0, ( 1) , :      We thus reduce the differential equation to the difference equation h(k 1) e / h(k) R1 e / qi (k)   T    T  2.2 For each of the following equation, determine the order of the equation then test it for (i) Linearity. (ii) Time-invariance. (iii) Homogeneousness. (a) y(k+2) = y(k+1) y(k) + u(k) (b) y(k+3) + 2 y(k) = 0 (c) y(k+4) + y(k-1) = u(k) (d) y(k+5) = y(k+4) + u(k+1)  u(k) (e) y(k+2) = y(k) u(k) The results are summarized below Problem Order Linear Time-invariant Homogeneous (a) 2 No Yes No (b) 3 Yes Yes Yes (c) 5 Yes Yes No (d) 5 Yes Yes No (e) 2 No Yes No 2 2.3 Find the transforms of the following sequences using Definition 2.1 (a) {0, 1, 2, 4, 0, 0,...} (b) {0, 0, 0, 1, 1, 1, 0, 0, 0,...} (c) {0, 20.5 , 1, 20.5 , 0, 0, 0, ... } From Definition 2.1, {u0, u1 , u2 , ... , uk , ... } transforms to U z ukz k k ( )      0 . Hence: (a) Z 0,1,2,4,0,0,... z 1  2z 2  4z 3 (b) Z 0,0,0,1,1,1,0,0,... z 3  z 4  z 5 (c) Z 0,20.5 ,1,20.5 ,0,0,... 20.5 z 1  z 2  20.5 z 3 2.4 Obtain closed forms of the transforms of Problem 2.3 using the table of z-transforms and the time delay property. Each sequence can be written in terms of transforms of standard functions (a) {0, 1, 2, 4,0,0,...} = {0, 1, 2, 4, 8, 16,...}  {0, 0, 0, 0, 8, 16,...}={f(k)}{g(k)} where        0, 0 f ( ) 2 , 0 1 k k k k         0, 4 g( ) 8 2 , 4 4 k k k k   ( 2) 8 2 8 2 0,1,2,4,0,0,... 3 3 1 4          z z z z z z z Z z z (b) {0, 0, 0, 1, 1, 1, 0, 0,...} = {0, 0, 0, 1, 1, 1, 1, 1,...}  {0, 0, 0, 0, 0, 0, 1, 1, 1, 1,...} = {f(k)} {g(k)} where       0, 3 1, 3 f ( ) k k k       0, 6 1, 6 g( ) k k k   ( 1) 1 1 1 0,0,0,1,1,1,0,0,... 5 3 3 6          z z z z z z z Z z z (c) {0,2-0.5,1,2-0.5,0,0,...} = {0,2-0.5,1,2-0.5,0,-2-0.5,-1,-2-0.5,0,...}+ {0,0,0,0,2-0.5,1,2-0.5,0,-2-0.5,-1,-2-0.5,0,...} = {f(k)} + {g(k)} where       0, 0 sin( 4) , 0 f ( ) k k k k        0, 4 sin( 4) , 4 g( ) k k k k       2 1 2 1 2 cos( 4) 1 sin( 4) 2 cos( 4) 1 0,2 ,1,2 ,0,0,0,... sin( 4) 3 2 0.5 0.5 4 2 4 2 0.5 0.5               z z z z z z z z z z z    Z  2.5 Prove the linearity and time delay properties of the z-transform from basic principles. To prove linearity, we must prove homogeneity and additivity using Definition 2.1, (i) Homogeneity: Z f (k) Z f (k) 3                 0 f (0), f (1), f (2),..., f ( ),... f (0) f (1) 1 f (2) 2 ... f ( ) ... f ( ) i Z i z z i z i i z i                 0 f (0), f (1), f (2),..., f ( ),... f (0) f (1) 1 f (2) 2 ... f ( ) ... f ( ) i Z     i   z  z  i z i  i z i (ii) Additivity Z f (k)  g(k) Z f (k) Z g(k)             f ( ) g( ) f ( ) g( ) f (0) g(0) f (1) g(1) f (2) g(2) ... f ( ) g( ) ... f ( ) g( ) f (0) g(0), f (1) g(1), f (2) g(2),..., f ( ) g( ),... 0 0 1 2 i z i z k k z z i i z k k i i i i i i i Z Z Z Z                                To prove the time delay property, we write the transform of the delayed sequence   f ( ) f ( ) 0, f (0), f (1), f (2),..., f ( ),... f (0) f (1) f (2) ... f ( ) ... 1 0 1 1 2 3 1 z i z z k i z z z i z i i i Z Z                    2.6 Use the linearity of the z-transform and the transform of the exponential function to obtain the transforms of the discrete-time functions. (a) sin(kT) (b) cos(kT) (a) sin(k T) e e j jk T jk T       2            2cos( ) 1 sin( ) 2 1 1 2 1 2 sin( ) 1 2 2                            z T z T z z e e z e e z j z e z z e z j e e j k T j T j T j T j T j T j T jk T jk T          Z Z  Z  (b) 2 cos( ) e jk T e jk T k T                  2cos( ) 1 cos( ) 1 2 2 1 2 1 2 cos( ) 1 2 2 2 2                              z T z z T z z e e z z e e z z e z z e z k T e e j T j T j T j T j T j T jk T jk T          Z Z  Z  2.7 Use the multiplication by exponential property to obtain the transforms of the discrete-time functions. (a) ekTsin(kT) (b) ekTcos(kT) 4 The multiplication by exponential property with ak  e T k  e kT gives Z e kT f (k) F(e T z) (a)         T T T T T T kT z T e z e T e z e z T e z e k T T e z             2 2 2cos( ) 2 sin( ) 2cos( ) 1 sin( ) sin( )          Z  (b)           T T T T T T T kT z T e z e z T e z e z T e z e k T e z T e z              2 2 2 2 2 2cos( ) cos( ) 2cos( ) 1 cos( ) cos( )            Z  2.8 Find the inverse transforms of the following functions using Definition 2.1 and, if necessary, long division (a) F(z)  1 3z 1  4z 2 (b) F(z)  5z 1  4z 5 (c) 0.3 0.02 ( ) 2    z z F z z (d) 0.04 0.25 ( ) 0.1 2     z z F z z Use Definition 2.1 to obtain (a) Z 1 3z 1  4z 2  {1,3,4,0,0,0,..} (b) Z 5z 1  4z 5  {0,5,0,0,0,4,0,0,..} (c) 1 2 1 2 3 -1 2 -1 -1 2 0.07 0.006 0.3 0.07 ...... 0.3 0.09 0.006 0.3 0.02 0.3 0.02 0.3 0.02                    z z z z z z z z z z z z z 0.3 0.07 ..... 0.3 0.02 ( ) 1 2 3 2        z  z  z  z z F z z {f (k)}  {0,1,0.3,0.07,....} (d) 1 2 1 2 3 -1 2 -1 -1 2 0.244 0.035 0.14 0.244 ...... 0.14 0.0056 0.035 0.14 0.25 0.04 0.25 0.04 0.25 0.1                      z z z z z z z z z z z z z 0.14 0.244 .... 0.04 0.25 ( ) 1 2 3 2        z  z  z  z z F z z {f (k)}  {0,1,0.14,0.244,....} 2.9 For Problems 2.8.(c), (d), find the inverse transforms of the functions using partial fraction expansion and table look-up. 5 (c)                    0.2 1 0.1 10 1 0.1 0.2 1 0.3 0.02 ( ) 1 z z 2 z z z z z F z           0.1 0.2 ( ) 10 z z z F z z {f (k)}  10 0.1k   0.2k  (d)   25 0.04 0. 0.4 0.4 1.016 0.04 0.25 ( ) 0.1 2 2           z z z z z z z z z F z We obtain 0.04 0.25 ( ) 0.4 0.4 1.016 2 2       z z F z z z and use the identities          2 2 cos( ) 2 sin( ) sin( )        z e z e e z e k d d d Z k          2 2 cos( ) 2 [ cos( )] cos( )         z e z e z z e e k d d d Z k e  0.25  0.5 cos( ) .  . d d  0 04  1 611rad       0.04 0.25 0.4 0.02 2.018 0.4996 0.04 0.25 0.4 0.02 1.008 0.04 0.25 0.4 1.016 2 2 2 2 2 2              z z z z z z z z z z z z z z    0.4 ( ) 2.0570.5 sin(1.611 0.196) {f ( )} 0.4 ( ) 0.5 0.4cos(1.611 ) 2.018sin(1.611 )         k k k k k k k k      2 2 018 . 2 4 . 0 057 . 2          2.057 0.196 sin 1 0.4 sin(A+B) = sin(A) cos(B) + cos(A) sin(B) 2.10 Solve the following difference equations (a) y(k+1)  0.8 y(k) = 0, y(0) = 1 (b) y(k+1)  0.8 y(k) = 1(k), y(0) = 0 (c) y(k+1)  0.8 y(k) = 1(k), y(0) = 1 (d) y(k+2) + 0.7 y(k+1) + 0.06 y(k) = (k), y(0)=0, y(1)=2 (a) y(k+1)  0.8 y(k) = 0, y(0) = 1 z-transform 0.8 ( ) 0.8 ( ) 0 ( )       z zY z z Y z Y z z f (k)  0.8k ,k  0,1,2,... (b) y(k+1)  0.8 y(k) = 1(k), y(0) = 0 z-transform ( 0.8)( 1) ( ) 1 ( 0.8) ( )        z z Y z z z z Y z z            0.8 1 1 5 1 ( 0.8)( 1) ( ) 1 z z z z z Y z 6 f (k)  51 0.8k ,k  0,1,2,... (c) y(k+1)  0.8 y(k) = 1(k), y(0) = 1 The solution is the sum of the solutions from (a) and (b) f (k)  51 0.8k   0.8k ,k  0,1,2,... (d) y(k+2) + 0.7 y(k+1) + 0.06 y(k) = (k), y(0)=0, y(1)=2 z-transform ( 0.1)( 0.6) ( 2 0.7 0.06) ( ) 1 2 ( ) 2 1          z z z z Y z z Y z z 0.6 0.667 0.1 16.667 16 ( 0.1)( 0.6) ( ) 2 1          z z z z z z z z Y z 0.6 0.667 0.1 ( ) 16.667 16      z z z Y z z y(k)  16.667 (k) 16 0.1k  0667 0.6k 2.11 Find the transfer functions corresponding to the difference equations of Problem 2.2 with input u(k) and output y(k). If no transfer function is defined, explain why. (a) and (e) are nonlinear and (b) is homogeneous. They have no transfer functions. (c) y(k+4) + y(k1) = u(k) Z-transform (z 4  z 1 )Y(z) U(z) 1 ( ) 5   z G z z (d) y(k+5) = y(k+4) + u(k+1)  u(k) z-transform (z5  z 4 )Y(z)  (z 1)U(z) 5 4 4 ( ) 1 1 z z z G z z     2.12 Test the linearity with respect to the input of the systems for which you found transfer functions in 2.11. (c) y(k+4) + y(k1) = u(k) The transfer function of the system is 1 ( ) 5   z G z z For inputs u1(k) and u2(k), we have outputs ( ), 1,2 1 ( ) ( ) ( ) 5     U z i z Y z G z U z z i i i We now as input try the linear combination 7 ( ) ( ) ( ) 1 ( ) 1 ( ) ( ) ( ) ( ) ( ) ( ) 1 2 5 1 5 2 1 2 Y z Y z U z z U z z z Y z G z U z z u k u k u k                (d) y(k+5) = y(k+4) + u(k+1)  u(k) Repeat above steps using the transfer function of (d). 2.13 If the rational functions of Problems 2.8.(c), (d), are transfer functions of LTI systems, find the difference equation governing each system. (c) 0.3 0.02 ( ) 2    z z F z z y(k+2 + 0.3 y(k+1) + 0.02 y(k) = u(k+1) (d) 0.04 0.25 ( ) 0.1 2     z z F z z y(k+2 + 0.04 y(k+1) + 0.25 y(k) = u(k+1)  0.1 u(k) 2.14 We can use z-transforms to find the sum of integers raised to various powers. This is accomplished by first recognizing that the sum is the solution of the difference equation f(k) = f(k1) + a(k) where a(k) is the kth term in the summation. Evaluate the following summations using z-transforms (a)   n k k 1 (b)   n k k 1 2 (a) We consider the difference equation f(k) = f(k1) + k Z-transform             