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Solutions Manual
INTRODUCTION 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
Solutions Manual
This manual contains solutions to the end-of-chapter problems in the second 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: http://www.engr.colostate.edu/~dga/mechatronics.html 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 Meaurement Systems
Solutions Manual 2.1
D = 0.06408 in = 0.001628 m. $
π' ---------
×
ρ = 1.7 x 10-8 Ωm,
±
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
2.2
2.3
5 5
NΩ ± so NΩ ≤ 5 V ≤ NΩ
5 S0,1
5 0,1 5 0,1 ------------------------------5 0,1 5 0,1
5 S0$;
5 0$; 5 0$; --------------------------------5 0$; 5 0$;
× , 5 5 5 -----------------5 5
×
NΩ NΩ
( × ) ( × ) -------------------------------------------------- × ×
×
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 0Ω perhaps causing a short. Furthermore, the trim value will not be additive with the fixed resistor.
2.5
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.
2.6
Place two 100Ω resistors in parallel and you immediately have a 50Ω resistance.
Introduction to Mechatronics and Measurement Systems
3
Solutions Manual From KCL, , V
2.7
, , , 9 9 9 ------V ------V ------V 5 5 5
9V so from Ohm’s Law -------5 HT Therefore, -------5 HT
,V ------5 HT
From Ohm’s Law and Question 2.7, 9
2.8
and for one resistor, 9
,V ---------------------------------------------------5 5 5 5 5 5 ---------------------------------------------------5 5 5
, 5
5 5 --------------------------------------------------- , 5 5 5 5 5 5 V
Therefore, ,
2.9
5 5 OLP -----------------5 → ∞ 5 5
2.10
,
5 5 ------------5
G9 & ---------GW
& HT G9 ------GW
5 5 5 --------------------------------------------------5 5 5 5 5 5
- ---- ---- - ------ so 5 HT 5 5 5
5 G9 & ---------GW
From KVL, 9
9 9
so G9 G9 ---------- ---------GW GW
G9 ------GW Therefore, , -------& HT 2.11
9
9
,
G9 & ---------GW
, , ------ ------ so -------& & & HT
------ ------ or & HT & &
& & -----------------& &
9 & G9 ------- and , GW
G9 & ---------GW
& G9 ------GW
From KCL, ,
4
, ,
& G9 ------- & G9 ------GW GW
Introduction to Mechatronics and Meaurement Systems
G9 ------- ( & & ) GW
Solutions Manual Since ,
& HT G9 ------GW & HT
2.12
,
,
& &
,
From KVL, 9 Since 9
/ G, ----- / G, ----GW GW
9 9
/ HT G, ----GW / /
/ HT
2.13
9
/ G, ----GW
G, ----- ( / / ) GW
G, / ------- GW
G, / ------- GW
From KCL, ,
, ,
9 Therefore, ---/
9- ----9 ---- / /
G, G, ------- ------GW GW
G, ----GW
so
so --/
2.14
9R
2.15
From Voltage Division, 9 R
2.16
Combining R2 and R3 in parallel,
- ---- ---- - or / / /
/ / ----------------/ /
9 , regardless of the resistance value.
5
------------------ ( ± )
5 5 -----------------5 5
± 9
( )----------
N
and combining this with R1 in series, 5 (a)
5 5
N
Using Ohm’s Law, ,
9 LQ ---------5
9 ---------N
P$
Introduction to Mechatronics and Measurement Systems
5
Solutions Manual (b)
Using current division, 5 ------------------, 5 5
, (c)
--- P$
P$
Since R2 and R3 are in parallel, and since Vin divides between R1 and R23, 5 --------------------9 5 5 LQ
9
9
------- 9
9
2.17 (a)
From Ohm’s Law, 9 RXW ± 9 ----------------------5
, (b)
9
(a)
5
9
9
9 RXW ± 9
9 ± 9 ------------------------------N
P$
9 ± 9
2.18
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)
,
9$ ---------5
, 2.19
5 5
P$
5 --------------------, 5 5
P$
Using superposition, 9 5
6
5 ------------------9 5 5
Introduction to Mechatronics and Meaurement Systems
9
± 9
Solutions Manual
9 5
9 5 9 5
9 5
2.20
5 , 9$
5 5 -----------------5 5 9 ± 9 -----------------5 5
5 ------------------L 5 5
9 9
NΩ ± P$
5 -------------------- ( 9 ± 9 ) 5 5
± 9
2.21
It will depend on your instrumentation, but the oscilloscope typically has an input impedance of 1 MΩ.
2.22
Since the input impedance of the oscilloscope is 1 MΩ, 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.
2.23
5 9 RXW
5 5 -----------------5 5 5 --------------------9 5 5 LQ
(a)
5
NΩ , 9 RXW
9 LQ
(b)
5
NΩ , 9 RXW
9 LQ
When the impedance of the load is lower (10k vs. 500k), the accuracy is not as good. 2.24
It will depend on the supply; check the specifications before answering.
2.25
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
Solutions Manual Using voltage division, 9&
= 5 /& --------------------------9 5 = 5 /& LQ
where 5 = 5 /&
± M
〈 ± °〉
so 〈 ± °〉 〈 °〉 -------------------------------------------- 〈 ± °〉
9&
〈 °〉
〈 UDG〉
Therefore, 9& ( W ) 2.26
FRV ( W )9
With steady state dc Vs, C is open circuit. So 9&
9V
9 so 9 5
9 and 9 5
9V
9
2.27 (a)
In steady state dc, C is open circuit and L is short circuit. So 9V -----------------5 5
, (b)
ω
π ± M------ω&
=& = /5
,V ,
8
-------- M π
=/ 5 = & = /5 -----------------------= & = /5
= &/5 = HT
P$
5 = &/5 9 -------V= HT
-------- ∠± ° Ω π Mω/ 5
( πM )Ω
( ± M )Ω ( ± M )Ω
∠°Ω
∠± ° Ω ∠± ° Ω
∠° P$
=& ------------------------, = & = /5 V
( ∠± ° ), V
Introduction to Mechatronics and Meaurement Systems
∠± ° P$
Solutions Manual
So ,(W)
FRV ( πW ± ) µ$
2.28 (a)
$ SS (b)
ω----π
$
$
$ SS
+]
, GF RIIVHW ω----π
+]
, GF RIIVHW ω----π
UDG -------- , I VHF
ω
+]
, GF RIIVHW
π UDG -------- , I VHF
ω $ SS
(d)
$
π UDG -------- , I VHF
ω $ SS
(c)
ω -----π
UDG π -------- , I VHF
ω
$
+]
, GF RIIVHW
VLQ ( π ) FRV ( π )
±
2.29
3
2.30
9 UPV
9 UPV ----------5
:
SS 9 -------- ⁄ ( )
9
3
9 UPV ----------5
2.31
9P
2.32
For 9 UPV
:
9 UPV
9
9 , 9 P 9(W)
9 UPV
9 , and f = 60 Hz,
9 P VLQ ( πI φ )
VLQ ( πW φ )
Introduction to Mechatronics and Measurement Systems
9
Solutions Manual 2.33
From Ohm’s Law, 9 ± 9 --------------------5
,
9 ------5
Since P$ ≤ , ≤ P$ , P$ ≤ 9 ------- ≤ P$ 5 giving 9 - ≤ 5 ≤ -------------9 - or Ω ≤ 5 ≤ Ω ----------------P$ P$ For a resistor, 3
9 ------ , 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. 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. 2.34
Using KVL and KCL gives: 9 9
, 5 5
( , ± , 5 )5 ( , ± , 5 ± , )5
9 ± 9
( , ± , 5 ± , )5 ± , 5
The first loop equation gives: ,5
9 -----5
P$
Using this in the other two loop equations gives:
( , ± P )N ( , ± P ± , )N ±
10
( , ± P ± , )N ± , N
Introduction to Mechatronics and Meaurement Systems
Solutions Manual or ( N ), ± ( N ),
( N ), ± ( N ),
Solving these equations gives: ,
2.35
(a)
9 RXW
(b)
3
P$ and ,
, 5 ± 9 , 9
P$
± 9
P: , 3
, 9
P: , 3
±, 9
± P:
Using KVL and KCL gives: 9 9
, 5 5
( , ± , 5 )5 ( , ± , 5 ± , )5
9 ± 9
( , ± , 5 ± , )5 ± , 5
The first loop equation gives: ,5
9 -----5
P$
Using this in the other two loop equations gives:
( , ± P )N ( , ± P ± , )N ±
( , ± P ± , )N ± , N
or ( N ), ± ( N ),
( N ), ± ( N ),
Solving these equations gives: , (a)
9 RXW
(b)
3
, 5 ± 9 , 9
7
2.36
3 DYJ
P$
± 9
P: , 3
--- ∫ 9 ( W ), ( W ) GW 7
P$ and ,
, 9
P: , 3
±, 9
P:
7
9P ,P -------------- ∫ VLQ ( ωW φ 9 ) VLQ ( ωW φ , ) GW 7
Introduction to Mechatronics and Measurement Systems
11
Solutions Manual Using the product formula trigonometric identity, 7
9P ,P ------------- ( FRV ( φ 9 ± φ , ) ± FRV ( ωW φ 9 φ , ) ) GW 7 ∫
3 DYJ
Therefore, 9P ,P -------------- FRV ( φ 9 ± φ , )
3 DYJ
9P ,P -------------- FRV ( θ )
7
2.37
, UPV
--- ∫ , P VLQ ( ωW φ , ) GW 7
Using the double angle trigonometric identity, 7
, UPV
,P ----- ∫ --- ± FRV [ ( ωW φ , ) ] GW 7
Therefore,
, UPV
2.38
5 9R
5 5 -----------------5 5 5 --------------------9 5 5 L
, P 7 ----- --7
,P ----
NΩ --- VLQ ( πW )
This is a sin wave with half the amplitude of the input with a period of 1s.
2.39
1 ------S 1V
2.40
RL = Ri = 100Ω for maximum power
2.41
The BNC cable is far more effective in shielding the input signals from electromagnetic interference since no loops are formed.
12
9S -----9V
9 ------------9
Introduction to Mechatronics and Meaurement Systems
