Solutions ManualINTRODUCTION TO MECHATRONICS AND MEASUREMENT SYSTEMS 2nd edition 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 edu/~dga/mechatronics. However. if you notice any errors or have suggestions or advice concerning the textbook's content or approach." Only a few of the open-ended problems that do not have a unique answer are left for your creative solutions. More information.engr.Solutions Manual This manual contains solutions to the end-of-chapter problems in the second edition of "Introduction to Mechatronics and Measurement Systems. and it should be relatively free from errors.edu. a suggested laboratory syllabus. We hope it helps you provide your students with an enjoyable and fruitful learning experience in the cross-disciplinary subject of mechatronics. and other supplemental material are provided on the Internet at: http://www. including an example course outline.colostate. Thank you for choosing our book. We will post corrections for reported errors on our Web site. please feel free to contact us via e-mail at dga@engr. 2 Introduction to Mechatronics and Meaurement Systems .colostate.html We have class-tested the textbook for several years. MathCAD files for examples in the book. $ π' --------- × ρ = 1. as long as you are consistent in your application.1 5 S0$. 5 0$. the result will be negative. 5 5 5 -----------------5 5 × NΩ NΩ ( × ) ( × ) -------------------------------------------------- × × × a = 2 = red. ± L = 1000 m 5 ρ/ ------$ Ω (a) 5 × ± so NΩ ≤ 5 ≤ NΩ (b) 5 × ± so NΩ ≤ 5 ≤ NΩ (c) 5V 5 5 (d) 5S 5 5 -----------------5 5 2. Furthermore. When in parallel. --------------------------------5 0$. Introduction to Mechatronics and Measurement Systems 3 .1 5 0.6 Place two 100Ω resistors in parallel and you immediately have a 50Ω resistance. If you assume the wrong current direction. b = 0 = black.1 5 0. 5 0$. 2.2 2.1 ------------------------------5 0.5 No.001628 m. the trim pot could be 0Ω perhaps causing a short.4 In series. the trim value will not be additive with the fixed resistor. 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. 2. × . 5 0$.3 5 5 NΩ ± so NΩ ≤ 5 V ≤ NΩ 5 S0. you will obtain correct answers.1 D = 0.Solutions Manual 2. d = gold 2.7 x 10-8 Ωm.1 5 0.06408 in = 0. c = 1 = brown. so 5 HT 5 5 5 5 G9 & ---------GW From KVL.9 5 5 OLP -----------------5 → ∞ 5 5 2.8 and for one resistor. 4 . & G9 ------GW GW Introduction to Mechatronics and Meaurement Systems G9 ------.V ------5 HT From Ohm’s Law and Question 2. ---------GW GW G9 ------GW Therefore.11 9 9 . V 2. . -------& HT 2. 5 5 ------------5 G9 & ---------GW & HT G9 ------GW 5 5 5 --------------------------------------------------5 5 5 5 5 5 .7 . 9 . 5 5 5 5 5 5 V Therefore.V ---------------------------------------------------5 5 5 5 5 5 ---------------------------------------------------5 5 5 .Solutions Manual From KCL.10 . . 9 2. & G9 ------. -------5 HT . -----. . . 9 9 9 ------V ------V ------V 5 5 5 9V so from Ohm’s Law -------5 HT Therefore.---- ---- . 5 5 5 --------------------------------------------------- . 2. -----.so -------& & & HT -----. .7. G9 & ---------GW . -----.and .or & HT & & & & -----------------& & 9 & G9 ------. . GW G9 & ---------GW & G9 ------GW From KCL. -----. . 9 9 9 so G9 G9 ---------.( & & ) GW . . / G. 5 (a) 5 5 N Using Ohm’s Law.( / / ) GW G. . 9 LQ ---------5 9 ---------N P$ Introduction to Mechatronics and Measurement Systems 5 . ------GW GW G.( ± ) 5 5 -----------------5 5 ± 9 ( )---------- N and combining this with R1 in series. . . ----. ----. ----GW so so --/ 2.13 9 / G. / ------- GW G. / ------- GW From KCL.16 Combining R2 and R3 in parallel. ---/ 9. regardless of the resistance value.---- ---- . 9 Therefore.or / / / / / ----------------/ / 9 .Solutions Manual Since .14 9R 2. . & & .----9 ---- / / G. & HT G9 ------GW & HT 2. ------.15 From Voltage Division. ----GW / / / HT 2. 5 -----------------. ----GW G.12 . ----GW GW 9 9 / HT G. G. From KVL. 9 Since 9 / G. . 9 R 2. . 9 RXW ± 9 ----------------------5 . (b) 9 (a) 5 9 9 9 RXW ± 9 9 ± 9 ------------------------------N P$ 9 ± 9 2. 9$ ---------5 . 9 5 6 5 ------------------9 5 5 Introduction to Mechatronics and Meaurement Systems 9 ± 9 . 5 5 P$ Using superposition.Solutions Manual (b) Using current division. 5 5 . (c) --.9 9 2. 2. 5 ------------------. 5 --------------------9 5 5 LQ 9 9 ------.18 5 5 NΩ 5 5 -------------------5 5 NΩ 5 5 NΩ 5 HT 5 5 ------------------------5 5 NΩ (b) 9$ 5 -----------------------9 5 5 V 9 (c) . and since Vin divides between R1 and R23.P$ P$ Since R2 and R3 are in parallel.19 5 5 P$ 5 --------------------.17 (a) From Ohm’s Law. 2. 500k).23 5 9 RXW 5 5 -----------------5 5 5 --------------------9 5 5 LQ (a) 5 NΩ .24 It will depend on the supply. but the oscilloscope typically has an input impedance of 1 MΩ. 2. 2. check the specifications before answering. the impedance of the source will be in parallel. 9$ 5 5 -----------------5 5 9 ± 9 -----------------5 5 5 ------------------L 5 5 9 9 NΩ ± P$ 5 -------------------. 9 RXW 9 LQ (b) 5 NΩ . Draw a sketch of the equivalent circuit to convince yourself.20 5 . 2. 9 RXW 9 LQ When the impedance of the load is lower (10k vs. the accuracy is not as good. and the oscilloscope impedance will affect the measured voltage.Solutions Manual 9 5 9 5 9 5 9 5 2.22 Since the input impedance of the oscilloscope is 1 MΩ.21 It will depend on your instrumentation.25 9 LQ 〈 °〉 Combining R2 and L in series and the result in parallel with C gives: = 5 /& ( 5 = / )= & -----------------------------------( 5 =/ ) =& 〈 ± °〉 ± M Introduction to Mechatronics and Measurement Systems 7 .( 9 ± 9 ) 5 5 ± 9 2. V . V Introduction to Mechatronics and Meaurement Systems ∠± ° P$ . 9& = 5 /& --------------------------9 5 = 5 /& LQ where 5 = 5 /& ± M 〈 ± °〉 so 〈 ± °〉 〈 °〉 -------------------------------------------- 〈 ± °〉 9& 〈 °〉 〈 UDG〉 Therefore.26 FRV ( W )9 With steady state dc Vs. So 9& 9V 9 so 9 5 9 and 9 5 9V 9 2.27 (a) In steady state dc.∠± ° Ω π Mω/ 5 ( πM )Ω ( ± M )Ω ( ± M )Ω ∠°Ω ∠± ° Ω ∠± ° Ω ∠° P$ =& ------------------------. C is open circuit and L is short circuit.Solutions Manual Using voltage division. C is open circuit. 9& ( W ) 2. = & = /5 V ( ∠± ° ).M π =/ 5 = & = /5 -----------------------= & = /5 = &/5 = HT P$ 5 = &/5 9 -------V= HT -------. (b) ω π ± M------ω& =& = /5 . 8 -------. So 9V -----------------5 5 . I VHF ω $ +] .Solutions Manual So . GF RIIVHW π UDG -------.. 9 P VLQ ( πI φ ) VLQ ( πW φ ) Introduction to Mechatronics and Measurement Systems 9 .(W) FRV ( πW ± ) µ$ 2.30 9 UPV 9 UPV ----------5 : SS 9 -------. GF RIIVHW ω----π +] .. I VHF ω $ SS (c) ω -----π UDG π -------..31 9P 2.⁄ ( ) 9 3 9 UPV ----------5 2. and f = 60 Hz. I VHF ω $ SS (d) $ π UDG -------.32 For 9 UPV : 9 UPV 9 9 . 9 P 9(W) 9 UPV 9 . I VHF ω +] .29 3 2. GF RIIVHW ω----π UDG -------.28 (a) $ SS (b) ω----π $ $ $ SS +] . GF RIIVHW VLQ ( π ) FRV ( π ) ± 2.. 9 ± 9 --------------------5 . 5 5 ( . )N ± . 5 )5 ( . ± P ± .5 9 -----5 P$ Using this in the other two loop equations gives: ( . ± . ± .. 5 ± . 5 The first loop equation gives: . ± .or Ω ≤ 5 ≤ Ω ----------------P$ P$ For a resistor.34 Using KVL and KCL gives: 9 9 .Solutions Manual 2. )5 ± . 5 ± . 9 ------5 Since P$ ≤ .≤ P$ 5 giving 9 . ± P ± . N Introduction to Mechatronics and Meaurement Systems . so the smallest allowable resistance would need a power rating of 5 at least: ( 9 ) --------------Ω 3 : so a 1/2 W resistor should be specified. )5 9 ± 9 ( . ≤ P$ . The largest allowable resistance would need a power rating of at least: (-------------9 ) Ω 3 : so a 1/4 W resistor would provide more than enough capacity. )N ± 10 ( .33 From Ohm’s Law.≤ 5 ≤ -------------9 . ± P )N ( . 2. P$ ≤ 9 ------. 3 9 -----. 5 ± .35 (a) 9 RXW (b) 3 P$ and . 3 ±.5 9 -----5 P$ Using this in the other two loop equations gives: ( . 2. 5 ± . Solving these equations gives: . 5 ± 9 . ± ( N ). ± . 9 P: . 9 ± P: Using KVL and KCL gives: 9 9 . 9 P$ ± 9 P: .36 3 DYJ P$ ± 9 P: . ± . 5 ± 9 . )5 ± .∫ 9 ( W ). ± P ± . . (a) 9 RXW (b) 3 . )N ± ( . 5 5 ( . . 9 P: 7 9P . N or ( N ). 3 .P -------------. ± . ( W ) GW 7 P$ and .∫ VLQ ( ωW φ 9 ) VLQ ( ωW φ . 5 )5 ( . Solving these equations gives: .Solutions Manual or ( N ). )N ± . ) GW 7 Introduction to Mechatronics and Measurement Systems 11 . ± ( N ). 9 P: . ± P )N ( . ± ( N ). 3 --. ± ( N ). ( N ). 3 ±. )5 9 ± 9 ( . ± P ± . 5 The first loop equation gives: . ( N ). 9 7 2. P -------------.38 5 9R 5 5 -----------------5 5 5 --------------------9 5 5 L .FRV ( θ ) 7 2.40 RL = Ri = 100Ω for maximum power 2. P 7 ----. 7 . ) ) GW 7 ∫ 3 DYJ Therefore.∫ --.Solutions Manual Using the product formula trigonometric identity. UPV --.P -------------. UPV .VLQ ( πW ) This is a sin wave with half the amplitude of the input with a period of 1s. UPV 2. ) ± FRV ( ωW φ 9 φ . 2. 12 9S -----9V 9 ------------9 Introduction to Mechatronics and Meaurement Systems .( FRV ( φ 9 ± φ .± FRV [ ( ωW φ .--7 . ) 3 DYJ 9P .FRV ( φ 9 ± φ .41 The BNC cable is far more effective in shielding the input signals from electromagnetic interference since no loops are formed.37 . ) GW 7 Using the double angle trigonometric identity. . ) ] GW 7 Therefore. 9P .P ----.∫ . 7 9P .39 1 ------S 1V 2.P ------------. P VLQ ( ωW φ .P ---- NΩ --.
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