INTRODUCTION TO MECHATRONICS AND MEASUREMENT SYSTEMS SOLUTIONS MANUAL

Transcription

1 NTODUCTON TO MECHATONCS AND MEASUEMENT SYSTEMS 4th edition 0(c) SOLUTONS MANUAL David G. Alciatore and Michael B. Histand Department of Mechanical Engineering Colorado State University Fort Collins, CO 8053 ntroduction to Mechatronics and Measurement Systems

2 This manual contains solutions to the end-of-chapter problems in the third edition of "ntroduction 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 nternet 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 at David.Alciatore@colostate.edu. 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. ntroduction to Mechatronics and Measurement Systems

3 . D in m. A D x 0-8 m, L 000 m L A % so 68k 5k (b) % so 5.6k 8.4k (c) s + 7k 0% so 74k s 60k (d) p pmn MN MN k + MN MN pmax MAX MAX k + MAX MAX.3 0 0, a red, b 0 black, c brown, d gold.4 n 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..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. ntroduction to Mechatronics and Measurement Systems 3

4 .6 No, as long as you are consistent in your application, you will obtain correct answers. f you assume the wrong current direction, the result will be negative..7 Place two 00 resistors in parallel and you immediately have a 50 resistance..8 From KCL, s so from Ohm s Law V s eq V s V s V s 3 Therefore, eq so eq From Ohm s Law and Question.8, V and for one resistor, V s eq s Therefore, s.0 lim dv dv C eq dv C C From KVL, V V + V so dv dv dv Therefore, C eq C C so C eq C or C C C C eq C + C 4 ntroduction to Mechatronics and Measurement Systems

5 . V V V dv C dv dv C and C dv C From KCL, Since dv C eq dv dv + C C C eq C + C dv C + C.3 From KVL, Since V d L eq ---- d V V + V L ---- d + L ---- L eq L + L d ---- L + L.4 V L d d ---- d L L From KCL, + so d ---- d d V Therefore, V V so or L L L L L L L L L L + L.5 V, regardless of the resistance value. V o 40.6 From Voltage Division, V o V ntroduction to Mechatronics and Measurement Systems 5

6 .7 Combining and 3 in parallel, and combining this with in series, k k Using Ohm s Law, V in V.7mA.k 3 (b) Using current division, mA mA (c) Since and 3 are in parallel, and since V in divides between and 3, 3 V 3 V V + in V.73V 3..8 (b) From Ohm s Law, V out V V V 0.7 m A 6k 4 V 5 V 6 V 56 V out V 4.V 0V 5.8V k k k 345 eq k ntroduction to Mechatronics and Measurement Systems

7 345 (b) V A V + s 4.84V 345 (c) 345 V A mA mA 45.0 This circuit is identical to the circuit in Question.9. Only the resistance values are different: k k k 345 eq k (b) 345 V A V + s 3.57V 345 (c) 345 V A mA mA 45. Using superposition, V V V V i V V V + V 0.0V ntroduction to Mechatronics and Measurement Systems 7

8 k V V m A + 45 V A V 3 + V 0.38V k k k 345 eq k Using loop currents, the KVL equations for each loop are: V out 0 V V and using selected KCL node equations, the unknown currents are related according to: out V V out This is now 7 equations in 7 unknowns, which can be solved for out and 6. The output voltage is then given by: V out V t will depend on your instrumentation, but the oscilloscope typically has an input impedance of M. 8 ntroduction to Mechatronics and Measurement Systems

9 .6 Since the input impedance of the oscilloscope is 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 V out V + in k, V out 0.995V in (b) 3 333k, V out.00v in When the impedance of the load is lower (0k vs. 500k), the accuracy is not as good..8 V out V + in 0 V out V 0.05 in 0.995V in (b) 500 V out V in V in For a larger load impedance, the output impedance of the source less error..9 t will depend on the supply; check the specifications before answering..30 With the voltage source shorted, all three resistors are in parallel, so, from Question.8: 3 TH V in 545 Combining and L in series and the result in parallel with C gives: Z LC + Z L Z C j + Z L + Z C ntroduction to Mechatronics and Measurement Systems 9

10 Using voltage division, V C Z LC V + in Z LC where j Z LC so V C rad Therefore, V C t 3.70cos3000t V.3 With steady state dc V s, C is open circuit. So V C V s 0V so V 0V and V V s 0V.33 n steady state dc, C is open circuit and L is short circuit. So V s mA + (b) Z C j j C Z L Z L + jl j Z CL Z C Z L j Z C Z L Z eq + Z CL j s V s mA Z eq Z C Z C Z L s s mA 0 ntroduction to Mechatronics and Measurement Systems

11 So t 4.7cost 0.56A.34 rad , f sec H z A pp A 4.0, dc offset 0 (b) rad , f sec A pp A, dc offset 0.0 (c) rad , f sec A pp A 6.0, dc offset 0 (d) 0 rad , f sec H z H z H z A pp A 0, dc offset sin + cos.35 P V rms W.36 V rms V pp V P V rms W.37 V m V rms 69.7V.38 For 0V, V m V rms 69.7V, and f 60 Hz, V rms Vt V m sinf sin0t + ntroduction to Mechatronics and Measurement Systems

12 .39 From Ohm s Law, Since 0mA 00mA, 5V V 3V mA 3V mA giving V 3V or mA 0mA For a resistor, P at least: V , so the smallest allowable resistance would need a power rating of P 3V W 30 so a / W resistor should be specified. The largest allowable resistance would need a power rating of at least: P 3V W 300 so a /4 W resistor would provide more than enough capacity..40 Using KVL and KCL gives: V V + 3 V 3 V 3 4 The first loop equation gives: V m A Using this in the other two loop equations gives: 0 0mk + 0m 3k 0 5 0m 3k 4k ntroduction to Mechatronics and Measurement Systems

13 or 5k 3k 60 3k 7k 35 Solving these equations gives:.ma and 0.93mA V out 4 V 4.3V (b) P V mw, P V 0.96mW, P 3 V 3.9mW.4 Using KVL and KCL gives: V V + 3 V 3 V 3 4 The first loop equation gives: V m A Using this in the other two loop equations gives: 0 0mk + 0m k 0 5 0m k k or 4k k 50 k 3k 5 Solving these equations gives:.5ma and 0mA V out 4 V 5V (b) P V 5mW, P V 0mW, P 3 V 3 0mW ntroduction to Mechatronics and Measurement Systems 3

14 .4 T P avg -- Vt t T 0 V m m t T sin + V sint + Using the product formula trigonometric identity, T 0 Therefore, P avg T V m m T cos V cost + V + 0 P avg V m m cos V V m m cos.43 T rms T -- sin m 0 t + Using the double angle trigonometric identity, Therefore, m rms ---- T T 0 -- cost + m rms ---- T T -- m k + 3 V o V + i -- sint 3 This is a sin wave with half the amplitude of the input with a period of s..45 No. A transformer requires a time varying flux to induce a voltage in the secondary coil..46 N p N s V p V V V s.47 L i 8 for maximum power 4 ntroduction to Mechatronics and Measurement Systems

15 .48 The BNC cable is far more effective in shielding the input signals from electromagnetic interference since no loops are formed. ntroduction to Mechatronics and Measurement Systems 5