Table Of Content

Solutions Manual INTRODUCTION TO MECHATRONICS AND MEASUREMENT SYSTEMS 4th edition 2012(c) SOLUTIONS MANUAL David G. Alciatore and Michael B. Histand Department of Mechanical Engineering Colorado State University Fort Collins, CO 80523 Introduction to Mechatronics and Measurement Systems 1 Solutions Manual This manual contains solutions to the end-of-chapter problems in the third edition of "Introduction to Mechatronics and Measurement Systems." Only a few of the open-ended problems that do not have a unique answer are left for your creative solutions. More information, including an example course outline, a suggested laboratory syllabus, MathCAD files for examples in the book, and other supplemental material are provided on the Internet at: mechatronics.colostate.edu We have class-tested the textbook for several years, and it should be relatively free from errors. However, if you notice any errors or have suggestions or advice concerning the textbook's content or approach, please feel free to contact us via e-mail at [email protected]. We will post corrections for reported errors on our Web site. Thank you for choosing our book. We hope it helps you provide your students with an enjoyable and fruitful learning experience in the cross-disciplinary subject of mechatronics. 2 Introduction to Mechatronics and Measurement Systems Solutions Manual 2.1 D = 0.06408 in = 0.001628 m. 2 D –6 A = ---------- = 2.082  10 4  = 1.7 x 10-8 m, L = 1000 m L R = ------- = 8.2 A 2.2 4 (a) R = 2110  20% so 168k R  252k 1 1 3 (b) R = 0710  20% so 5.6k R  8.4k 2 2 (c) R = R + R = 217k 20% so 174k R 260k s 1 2 s R R (d) R = --------1-------2---- p R + R 1 2 R R 1 2 R = ----------M---I-N---------M---I--N---- = 5.43k pMIN R +R 1 2 MIN MIN R R 1 2 R = ----------M---A---X---------M---A--X---- = 8.14k pMAX R + R 1 2 MAX MAX 2 1 2.3 R = 1010 , R = 2510 1 2 R = ----R----1--R-----2---- = ---1---0---------1---0---2--------2---5---------1---0---1---- = 20 101 R + R 2 1 1 2 10 10 +25 10 a = 2 = red, b = 0 = black, c = 1 = brown, d = gold 2.4 In series, the trim pot will add an adjustable value ranging from 0 to its maximum value to the original resistor value depending on the trim setting. When in parallel, the trim pot could be 0 perhaps causing a short. Furthermore, the trim value will not be additive with the fixed resistor. 2.5 When the last connection is made, a spark occurs at the point of connection as the completed circuit is formed. This spark could ignite gases produced in the battery. The negative terminal of the battery is connected to the frame of the car, which serves as a ground reference throughout the vehicle. Introduction to Mechatronics and Measurement Systems 3 Solutions Manual 2.6 No, as long as you are consistent in your application, you will obtain correct answers. If you assume the wrong current direction, the result will be negative. 2.7 Place two 100 resistors in parallel and you immediately have a 50 resistance. 2.8 From KCL, I = I +I + I s 1 2 3 V V V V so from Ohm’s Law ------s-- = ------s + ------s + ------s R R R R eq 1 2 3 1 1 1 1 R R R Therefore, -------- = ------ + ------ + ------ so R = ---------------------1-------2-------3----------------- R R R R eq R R + R R + R R eq 1 2 3 2 3 1 3 1 2 I I 2.9 From Ohm’s Law and Question 2.8, V = -----s--- = ---------------------------s------------------------- R R R + R R + R R eq -----2-------3------------1-------3-------------1-------2- R R R 1 2 3 and for one resistor, V = I R 1 1 R R Therefore, I = ------------------------2-------3---------------------I 1 R R +R R +R R  s 2 3 1 3 1 2 R R R R 2.10 lim --------1-------2---- = -----1-------2- = R R1R1+ R2 R1 2 dV dV dV 2.11 I = C ------- = C --------1-- = C --------2-- eq dt 1 dt 2 dt From KVL, V = V +V 1 2 so dV dV dV ------- = --------1-- + --------2-- dt dt dt Therefore, I I I 1 1 1 C C -------- = ------ + ------ so -------- = ------ + ------ or C = --------1-------2---- C C C C C C eq C + C eq 1 2 eq 1 2 1 2 4 Introduction to Mechatronics and Measurement Systems Solutions Manual 2.12 V = V = V 1 2 dV dV dV dV I = C --------1-- = C ------- and I = C --------2-- = C ------- 1 1 dt 1 dt 2 2 dt 2 dt From KCL, dV dV dV I = I + I = C ------- + C ------- = -------C + C  1 2 1 dt 2 dt dt 1 2 dV Since I = C ------- eq dt C = C + C eq 1 2 2.13 I = I = I 1 2 From KVL, dI dI dI V = V + V = L ----- + L ----- = -----L +L  1 2 1dt 2dt dt 1 2 dI Since V = L ----- eqdt L = L +L eq 1 2 dI dI dI 2.14 V = L----- = L ------1- = L ------2- dt 1 dt 2 dt dI dI dI From KCL, I = I +I so ----- = ------1- + ------2- 1 2 dt dt dt V V V 1 1 1 L L Therefore, ---- = ------ + ------ so --- = ------ + ------ or L = -------1-------2---- L L L L L L L +L 1 2 1 2 1 2 2.15 V = 1V, regardless of the resistance value. o 40 2.16 From Voltage Division, V = ------------------5–15 = –8V o 10+ 40 Introduction to Mechatronics and Measurement Systems 5 Solutions Manual 2.17 Combining R and R in parallel, 2 3 R R 23 R = --------2-------3---- = ------------ = 1.2k 23 R + R 2+3 2 3 and combining this with R in series, 1 R = R + R = 2.2k 123 1 23 (a) Using Ohm’s Law, V 5V I = ------i--n-- = ---------- = 2.27mA 1 R 2.2k 123 (b) Using current division, R 2 I = -----------2--------I = ---2.27mA = 0.909mA 3 R + R 1 5 2 3 (c) Since R and R are in parallel, and since V divides between R and R , 2 3 in 1 23 R 1.2 V = V = -----------2--3--------V = -------5V = 2.73V 3 23 R +R in 2.2 1 23 2.18 (a) From Ohm’s Law, V –V 14.2V –10V I = -----o---u--t------------1- = ------------------------------- = 0.7mA 4 R 6k 24 (b) V = V = V = V –V = 14.2V–20V = –5.8V 5 6 56 out 2 2.19 (a) R = R +R = 5k 45 4 5 R R R = --------3-------4--5---- = 1.875k 345 R +R 3 45 R = R +R = 3.875k 2345 2 345 R R R = --------1-------2--3--4--5----- = 0.795k eq R + R 1 2345 6 Introduction to Mechatronics and Measurement Systems Solutions Manual R (b) V = -----------3--4---5-------V = 4.84V A R + R s 2 345 V (c) I = -------A--- = 2.59mA 345 R 345 R I = ------------3---------I = 0.97mA 5 R + R 345 3 45 2.20 This circuit is identical to the circuit in Question 2.19. Only the resistance values are different: (a) R = R +R = 4k 45 4 5 R R R = --------3-------4--5---- = 2.222k 345 R +R 3 45 R = R +R = 6.222k 2345 2 345 R R R = --------1-------2--3--4--5----- = 1.514k eq R + R 1 2345 R (b) V = -----------3--4---5-------V = 3.57V A R + R s 2 345 V (c) I = -------A--- = 1.61mA 345 R 345 R I = ------------3---------I = 0.89mA 5 R + R 345 3 45 2.21 Using superposition, R V = -----------2--------V = 0.909V R21 R +R 1 1 2 R V = -----------1--------i = 9.09V R22 R + R 1 1 2 V = V + V = 10.0V R2 R2 R2 1 2 Introduction to Mechatronics and Measurement Systems 7 Solutions Manual R R 2.22 R = --------4-------5---- = 0.5k 45 R +R 4 5 V –V I = -----1-------------2- = –0.5mA R + R 1 2 R V = -----------4--5--------V –V  = –0.238V A R + R 1 2 3 45 2.23 R = R +R = 9k 45 4 5 R R R = --------3-------4--5---- = 2.25k 345 R +R 3 45 R = R + R = 4.25k 2345 2 345 R R R = --------1-------2--3--4--5----- = 0.81k eq R + R 1 2345 2.24 Using loop currents, the KVL equations for each loop are: V –I R = 0 1 out 1 V – I R – I R –V = 0 2 5 5 3 3 1 –I R + I R = 0 6 6 5 5 I R –I R –I R = 0 3 3 24 4 24 2 and using selected KCL node equations, the unknown currents are related according to: I = I + I +I out 2 3 V 1 I = I –I + I  V out 5 6 1 I = I + I –I 3 5 6 24 This is now 7 equations in 7 unknowns, which can be solved for I and I . The output out 6 voltage is then given by: V = V –I R out 2 6 6 2.25 It will depend on your instrumentation, but the oscilloscope typically has an input impedance of 1 M. 8 Introduction to Mechatronics and Measurement Systems Solutions Manual 2.26 Since the input impedance of the oscilloscope is 1 M, the impedance of the source will be in parallel, and the oscilloscope impedance will affect the measured voltage. Draw a sketch of the equivalent circuit to convince yourself. R R 2.27 R = --------2-------3---- 23 R +R 2 3 R V = -----------2--3--------V out R + R in 1 23 (a) R = 9.90k, V = 0.995V 23 out in (b) R = 333k , V = 1.00V 23 out in When the impedance of the load is lower (10k vs. 500k), the accuracy is not as good. R 2.28 V = -----------2--------V out R + R in 1 2 10 (a) V = -------------V = 0.995V out 10.05 in in 500 (b) V = ----------------V = 0.9999V out 500.05 in in For a larger load impedance, the output impedance of the source less error. 2.29 It will depend on the supply; check the specifications before answering. 2.30 With the voltage source shorted, all three resistors are in parallel, so, from Question 2.8: R R R R = ---------------------1-------2-------3----------------- TH R R + R R + R R 2 3 1 3 1 2 2.31 V = 545 in Combining R and L in series and the result in parallel with C gives: 2 R +Z Z Z = ----------2-------------L---------C----- = 1860.52–60.25 = 923.22–1615.30j R2LC R + Z + Z 2 L C Introduction to Mechatronics and Measurement Systems 9 Solutions Manual Using voltage division, Z R LC V = --------------2-------------V C R +Z in 1 R LC 2 where R + Z = 1000+923.22–1615.30j = 2511.57–40.02 1 R LC 2 so 1860.52–60.25 V = --------------------------------------------545 = 3.7024.8 = 3.700.433rad C 2511.57–40.02 Therefore, V t = 3.70cos3000t +0.433V C 2.32 With steady state dc V , C is open circuit. So s V = V = 10V so V = 0V and V = V = 10V C s R R s 1 2 2.33 (a) In steady state dc, C is open circuit and L is short circuit. So V I = ------------s------- = 0.025mA R + R 1 2 (b)  =  6 6 –j 10 10 Z = -------- = --------j = --------–90 C C   5 5 Z = Z + R = jL+ R = 10 + 20j = 10 0.036 LR L 2 2 2 Z Z C LR Z = ---------------------2--- = 91040– 28550j = 95410–17.4 CLR2 Z + Z C LR 2 Z = R + Z = 191040–28550j = 193200–8.50 eq 1 CLR 2 V I = ------s-- = 0.02598.50mA s Z eq Z I = -------------C-----------I = 0.954–17.44I = 0.0247–8.94mA Z + Z s s C LR 2 10 Introduction to Mechatronics and Measurement Systems