Alciatore - Introduction to Mechatronics and Measurement Systems (4th Edition) Solutions PDF

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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.

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Introduction to Mechatronics and Measurement Systems

Solutions Manual 2.1

D = 0.06408 in = 0.001628 m. 2

–6 ---------- = 2.082  10 A = D 4

 = 1.7 x 10-8 m, L = 1000 m ------- = 8.2 R = L A 2.2 4

(a)

R 1 = 21  10  20% so 168k  R 1  252k

(b)

R 2 = 07  10  20% so 5.6k   R 2  8.4k 

(c)

R s = R 1 + R 2 = 217k  20% so 174k   R s  260k 

(d)

R1 R2 R p = -----------------R1 + R2

3

R 1MIN R 2MIN - = 5.43k R pMIN = ------------------------------R 1MIN + R 2MIN R 1MAX R 2MAX R pMAX = --------------------------------- = 8.14k R 1MAX + R 2MAX 2.3

2

R 1 = 10  10 , R 2 = 25  10

1

2 1 R1 R2 1  10  10   25  10 - - = -------------------------------------------------R = -----------------= 20  10 2 1 R1 + R2 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 s = I 1 + I 2 + I 3 V V V V so from Ohm’s Law -------s- = -----s- + ------s + ------s R1 R2 R3 R eq R1 R2 R3 1 1 1 1 Therefore, -------- = ------ + ------ + ------ so R eq = ---------------------------------------------------R eq R1 R2 R3 R2 R3 + R1 R3 + R1 R2 Is Is From Ohm’s Law and Question 2.8, V = ------- = ---------------------------------------------------R2 R3 + R1 R3 + R1 R2 R eq ---------------------------------------------------R1 R2 R3

2.9

and for one resistor, V = I 1 R 1 R2 R3 Therefore, I 1 =  ---------------------------------------------------- I s R2 R3 + R1 R3 + R1 R2

2.10

R1 R2 R1 R2 - = R2 lim  -------------------  = -----------R1   R 1 + R 2 R1

2.11

dV dV dV I = C eq------- = C 1 ---------1- = C 2 ---------2dt dt dt From KVL, V = V 1 + V2 so dV dV dV ------- = ---------1- + ---------2dt dt dt Therefore, C1 C2 I - = ----1 - = ----I - + ----1- + ----I- so ------1- or ------C eq = -----------------C eq C eq C1 C2 C1 C2 C1 + C2

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Introduction to Mechatronics and Measurement Systems

Solutions Manual 2.12

V = V 1 = V2 dV 1 dV 2 dV dV I 1 = C 1 ---------- = C 1 ------- and I 2 = C 2 ---------- = C 2 ------dt dt dt dt From KCL, dV dV dV I = I1 + I 2 = C1 ------- + C 2 ------- = -------  C 1 + C 2  dt dt dt dV Since I = C eq------dt C eq = C 1 + C 2

2.13

I = I 1 = I2 From KVL, dI dI dI V = V 1 + V 2 = L1 ----- + L 2 ----- = -----  L 1 + L 2  dt dt dt dI Since V = Leq----dt L eq = L 1 + L 2

2.14

dI 1 dI 2 dI V = L----- = L 1 ------- = L 2 ------dt dt dt From KCL, I = I1 + I 2 VV- + ----Therefore, V ---- = ----L1 L2 L

so

dI = dI 1 + dI ------- -------2 ----dt dt dt

L1 L2 11- + ----so 1--- = ----or L = ----------------L1 + L2 L L1 L2

2.15

V o = 1V , regardless of the resistance value.

2.16

40 From Voltage Division, V o = ------------------  5 – 15  = –8V 10 + 40

Introduction to Mechatronics and Measurement Systems

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Solutions Manual 2.17

Combining R2 and R3 in parallel, R2 R3 2  3 - = 1.2k R 23 = ------------------ = ----------R2 + R3 2+3 and combining this with R1 in series, R 123 = R 1 + R 23 = 2.2k (a)

Using Ohm’s Law, V in 5V = 2.27mA = ---------I 1 = ---------2.2k R 123

(b)

Using current division, R2 - I = 2--- 2.27mA = 0.909mA I 3 = -----------------5 R2 + R3 1

(c)

Since R2 and R3 are in parallel, and since Vin divides between R1 and R23, R 23 1.2 V 3 = V 23 = --------------------- V in = ------- 5V = 2.73V 2.2 R 1 + R 23

2.18 (a)

From Ohm’s Law, V out – V 1 14.2V – 10V - = ------------------------------- = 0.7mA I 4 = ---------------------6k R 24

(b)

V 5 = V 6 = V 56 = V out – V 2 = 14.2V – 20V = –5.8V

(a)

R 45 = R 4 + R 5 = 5k

2.19

R 3 R 45 - = 1.875k R 345 = -------------------R 3 + R 45 R 2345 = R 2 + R 345 = 3.875k R 1 R 2345 R eq = ------------------------- = 0.795k R 1 + R 2345

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Introduction to Mechatronics and Measurement Systems

Solutions Manual

(b)

R 345 V = 4.84V V A = ----------------------R 2 + R 345 s

(c)

VA I 345 = ---------= 2.59mA R 345 R3 I 5 = --------------------- I 345 = 0.97mA R 3 + R 45

2.20

This circuit is identical to the circuit in Question 2.19. Only the resistance values are different: (a)

R 45 = R 4 + R 5 = 4k R 3 R 45 - = 2.222k R 345 = -------------------R 3 + R 45 R 2345 = R 2 + R 345 = 6.222k R 1 R 2345 - = 1.514k R eq = ------------------------R 1 + R 2345

(b)

R 345 V A = ----------------------- V s = 3.57V R 2 + R 345

(c)

VA = 1.61mA I 345 = ---------R 345 R3 I 5 = --------------------- I 345 = 0.89mA R 3 + R 45

2.21

Using superposition, R2 V R2 1 = ------------------- V 1 = 0.909V R1 + R2 R1 V R2 2 = ------------------- i 1 = 9.09V R1 + R2 V R2 = V R2 1 + V R22 = 10.0V

Introduction to Mechatronics and Measurement Systems

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Solutions Manual

2.22

R4 R5 - = 0.5k R 45 = -----------------R4 + R5 V1 – V2 I = ------------------ = – 0.5mA R1 + R2 R 45 V A = --------------------- V 1 – V 2 = – 0.238 V R 3 + R 45

2.23

R 45 = R 4 + R 5 = 9k R 3 R 45 - = 2.25k R 345 = -------------------R 3 + R 45 R 2345 = R 2 + R 345 = 4.25k R 1 R 2345 R eq = ------------------------- = 0.81k R 1 + R 2345

2.24

Using loop currents, the KVL equations for each loop are: V 1 – I out R 1 = 0 V 2 – I5 R5 – I3 R3 – V 1 = 0 – I6 R6 + I 5 R5 = 0 I 3 R 3 – I 24 R 4 – I 24 R 2 = 0 and using selected KCL node equations, the unknown currents are related according to: I out = I 2 + I 3 + I V1 I V1 = I out –  I 5 + I 6  I 3 = I 5 + I 6 – I 24 This is now 7 equations in 7 unknowns, which can be solved for Iout and I6. The output voltage is then given by: V out = V 2 – I 6 R 6

2.25

8

It will depend on your instrumentation, but the oscilloscope typically has an input impedance of 1 M.

Introduction to Mechatronics and Measurement Systems

Solutions Manual 2.26

2.27

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. R2 R3 R 23 = -----------------R2 + R3 R 23 V out = --------------------- V in R 1 + R 23 (a)

R 23 = 9.90k , V out = 0.995V in

(b)

R 23 = 333k , V out = 1.00V in

When the impedance of the load is lower (10k vs. 500k), the accuracy is not as good.

2.28

R2 V out = ------------------- V in R1 + R2 (a)

10 V out = ------------- V in = 0.995Vin 10.05

(b)

500 V out = ---------------- V in = 0.9999Vin 500.05

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: R1 R2 R3 R TH = ---------------------------------------------------R2 R3 + R1 R3 + R1 R2

2.31

V in = 5  45 Combining R2 and L in series and the result in parallel with C gives: R 2 + Z L Z C ZR2 LC = ------------------------------------ = 1860.52  – 60.25 = 923.22 – 1615.30j  R 2 + ZL  + ZC

Introduction to Mechatronics and Measurement Systems

9

Solutions Manual Using voltage division, Z R2 LC -V V C = -------------------------R 1 + Z R2 LC in where R 1 + Z R2 LC = 1000 + 923.22 – 1615.30j = 2511.57  –40.02 so 1860.52  –60.25 V C = -------------------------------------------- 5  45 = 3.70  24.8 = 3.70  0.433rad 2511.57  –40.02 Therefore, V C  t  = 3.70 cos  3000t + 0.433 V 2.32

With steady state dc Vs, C is open circuit. So V C = V s = 10V so V R1 = 0V and V R2 = V s = 10V

2.33 (a)

In steady state dc, C is open circuit and L is short circuit. So Vs I = ------------------ = 0.025mA R1 + R2

(b)

 =  6

ZC

6

10 – j- = 10 -------- j = -------  –90  = ------  C 5

5

ZLR2 = Z L + R 2 = jL + R 2 =  10 + 20  j  = 10 0.036  Z C ZLR 2 ZCLR 2 = -----------------------=  91040 – 28550j  = 95410 – 17.4  ZC + Z LR 2 Zeq = R 1 + ZCLR 2 =  191040 – 28550j  = 193200 – 8.50  V I s = -------s- = 0.0259 8.50 mA Zeq ZC I = -----------------------I =  0.954 – 17.44 I s = 0.0247 – 8.94 mA ZC + Z LR 2 s

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Introduction to Mechatronics and Measurement Systems

Solutions Manual

So I  t  = 24.7 cos  t – 0.156  A 2.34 (a)

- = 0.5Hz  =  rad -------- , f = ----sec 2 A pp = 2A = 4.0 , dcoffset = 0

(b)

rad - = 1Hz  = 2 -------- , f = ----2 sec A pp = 2A = 2 , dcoffset = 10.0

(c)

rad   = 2 -------- , f = ----- = 1Hz sec 2 A pp = 2A = 6.0 , dcoffset = 0 rad   = 0 -------- , f = ------ = 0Hz sec 2

(d)

A pp = 2A = 0 , dcoffset = sin    + cos    = – 1 2

2.35

V rms P = ----------= 100W R

2.36

V pp V rms =  ---------   2  = 35.36V 2 2

V rms P = ----------= 12.5W R 2.37

Vm =

2V rms = 169.7V

2.38

For V rms = 120V , V m =

2V rms = 169.7V , and f = 60 Hz,

V  t  = V m sin  2f +   = 169.7 sin  120t +  

Introduction to Mechatronics and Measurement Systems

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Solutions Manual 2.39

From Ohm’s Law, – 2V3VI = 5V -------------------= -----R R Since 10mA  I  100mA , 3V 10mA  -------  100mA R giving 3V 3V ---------------- R  --------------- or 30  R  300 100mA 10mA 2

------ , so the smallest allowable resistance would need a power rating of For a resistor, P = V R at least: 2

 3V  P = --------------- = 0.3W 30 so a 1/2 W resistor should be specified. The largest allowable resistance would need a power rating of at least: 2

3V  - = 0.03W P = -------------300 so a 1/4 W resistor would provide more than enough capacity. 2.40

Using KVL and KCL gives: V 1 = I R1 R 1 V 1 = I 1 – I R1 R 2 +  I 1 – I R1 – I 2 R 3 V 3 – V 2 =  I 1 – I R1 – I 2 R 3 – I 2 R 4 The first loop equation gives: V I R1 = ------1 = 10mA R1 Using this in the other two loop equations gives: 10 =  I1 – 10m 2k + I1 – 10m – I 23k 10 – 5 = I 1 – 10m – I 23k – I 2 4k

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Introduction to Mechatronics and Measurement Systems

Solutions Manual or  5k I 1 –  3k I 2 = 60  3k I 1 –  7k I 2 = 35 Solving these equations gives: I 1 = 12.12mA and I 2 = 0.1923mA

2.41

(a)

V out = I 2 R 4 – V 2 = –4.23V

(b)

P 1 = I 1 V 1 = 121mW , P 2 = I 2 V 2 = 0.962mW , P 3 = – I 2 V 3 = – 1.92mW

Using KVL and KCL gives: V 1 = I R1 R 1 V 1 = I 1 – I R1 R 2 +  I 1 – I R1 – I 2 R 3 V 3 – V 2 =  I 1 – I R1 – I 2 R 3 – I 2 R 4 The first loop equation gives: V I R1 = ------1 = 10mA R1 Using this in the other two loop equations gives: 10 =  I1 – 10m 2k + I1 – 10m – I 22k 10 – 5 = I 1 – 10m – I 22k – I 2 1k or  4k I 1 –  2k I 2 = 50  2k I 1 –  3k I 2 = 25 Solving these equations gives: I 1 = 12.5mA and I 2 = 0mA (a)

V out = I 2 R 4 – V 2 = –5V

(b)

P 1 = I 1 V 1 = 125mW , P 2 = I 2 V 2 = 0mW , P 3 = – I 2 V 3 = 0mW

Introduction to Mechatronics and Measurement Systems

13

Solutions Manual

2.42

P avg

T

T

0

0

Vm Im 1 - sin  t +  V  sin t +  I  dt = ---  V  t I  t  dt = ------------T  T

Using the product formula trigonometric identity, T

P avg

Vm Im = --------------   cos   V –  I  – cos  2t +  V +  I   dt 2T 0

Therefore, Vm Im Vm Im P avg = -------------- cos  V –  I  = -------------- cos    2 2 T

2.43

I rms =

1--- 2 2 I sin  t +  I dt T m 0

Using the double angle trigonometric identity, 2T

I rms =

Im  1 -----  --- – cos 2 t +  I   dt  T 2 0

Therefore, 2

I rms =

2.44

I m  T I ----- --- = ----mT  2 2

R2 R3 R 23 = ------------------ = 5k R2 + R3 R 23 - V = 1--- sin  2t  V o = -------------------R 1 + R 23 i 2 This is a sin wave with half the amplitude of the input with a period of 1s.

2.45

No. A transformer requires a time varying flux to induce a voltage in the secondary coil.

2.46

Np V 120V ------ = ------p = ------------- = 5 24V Vs Ns

2.47

RL = Ri = 8 for maximum power

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Introduction to Mechatronics and Measurement Systems

Solutions Manual 2.48

The BNC cable is far more effective in shielding the input signals from electromagnetic interference since no loops are formed.

Introduction to Mechatronics and Measurement Systems

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Solutions Manual 3.1

For Vi > 0, Vo = 0. For Vi < 0, Vo = Vi. The resulting waveform consists only of the negative "humps" of the original cosine wave. Each hump has a duration of 0.5s and there is a 0.5s gap between each hump.

3.2 (a)

output passes first (positive) hump only

(b)

output passes second (negative) hump only

(c)

output passes first (positive) hump only

(d)

output passes second (negative) hump only

(e)

output passes full first (positive) hump and 1/2-scale second hump

(f)

output passes full wave

3.3

When the diode is forward biased, the output voltage is -0.7V, so the output signal is chopped off at -0.7V instead of 0V.

3.4

When the switch is closed and the circuit is in steady state, the current through the load is constant, and the diode is reverse biased (i.e., there is no diode current). When the switch is opened, the inductor generates a forward voltage to oppose a decrease in current. Now the diode forms a circuit with the load, allowing the current to dissipate through the resistor. If there were no diode, and the switched were opened, because the current would attempt to decrease instantaneously, the inductor would generate a very large voltage which would create an arc (current through air) across the switch contacts.

3.5

Forward bias (Vin > Vout + 0.7V) is required for charging. "Leaking" causes voltage decay (i.e., Vout decreases slowly).

3.6

For Vi > 0, Vo = Vi. For Vi < 0, Vo = -Vi. The resulting waveform is a full wave rectified sin wave where there are two positive "humps" for each period of Vi (0.5 sec).

3.7

16

See the data sheet for a LM7815C voltage regulator.

Introduction to Mechatronics and Measurement Systems

Solutions Manual 3.8 (a)

For Vi > 0.5V, Vo = 0.5V. For Vi < 0.5V, Vo = Vi. The resulting waveform is the original sine wave with the top halves of the positive "humps" (above 0.5 V) clipped off.

(b)

For Vi < 0.5V, Vo = 0.5V. For Vi > 0.5V, Vo = Vi. The resulting waveform is the original sine wave with the bottom of the negative "humps" (below 0.5 V) clipped off.

3.9

Using superposition, Ohm’s Law, and current division, 1V = -----2I 1left = ----------------5R R 2R + ---2 1 1I 2left = – --- I 1left = – -----2 5R 1 3 1 1V I 4left = --- I 1left = ------- , I 4right = ----------------- = ------2 5R 5R 2R R + ------3 2R 2 I 2right = ---------------= ------I R + 2R 4right 5R 1I 1right = I 4right – I 2right = -----5R 3 I 1 = I 1left + I 1right = -------  0 5R 1 I 2 = I 2left + I 2right = -------  0 5R I3 = 0 4 I 4 = I 4left + I 4right = -------  0 5R 1 V diode = 1V – I4 R = --- V  0 5

Introduction to Mechatronics and Measurement Systems

17

Solutions Manual 3.10

With I2=I3=0, I1 and I4 are equal. The current (I=I1=I4) is: 2 + 1V- = ------V -------------------I = 1V 3R 2R + R and the voltage of node A relative to node B is: 1 V AB = 1V – I  2R  = – 1V + I  R  = – --- V 3 Therefore, the voltage polarity on the left diode is incorrect.

3.11

When the left diode is forward biased and the right diode is reverse biased, V out = V H and when the right diode is forward biased and the left diode is reverse biased, V out = V L When both diodes are reverse biased, RL V out = -----------------V Ri + RL i Therefore, the output is a scaled version of the input chopped off below VL and above VH.

3.12

1 For Vin > 0, V out...