Introduction to Mechatronics and Measurement Systems 3rd Ed Solutions Manual Mmzzhh

Solutions Manual

Introduction to Mechatronics and Measurement Systems 1

INTRODUCTION TO MECHATRONICS AND

MEASUREMENTSYSTEMS

3rd edition

SOLUTIONS MANUAL

David G. Alciatoreand

Michael B. Histand

Department of Mechanical EngineeringColorado State UniversityFort Collins, CO 80523

Solutions Manual

2 Introduction to Mechatronics and Measurement Systems

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-endedproblems 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 examplesin the book, and other supplemental material are provided on the Internet at:

www.engr.colostate.edu/mechatronics

We have class-tested the textbook for several years, and it should be relatively free fromerrors. However, if you notice any errors or have suggestions or advice concerning the textbook'scontent or approach, please feel free to contact us via e-mail at [email protected] 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 anenjoyable and fruitful learning experience in the cross-disciplinary subject of mechatronics.

Solutions Manual


2.1 D = 0.06408 in = 0.001628 m.

ρ = 1.7 x 10-8 Ωm, L = 1000 m

2.2

(a) so

(b) so

(c) so

(d)

2.3 ,

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 tothe original resistor value depending on the trim setting. When in parallel, the trim potcould be 0Ω perhaps causing a short. Furthermore, the trim value will not be additive withthe fixed resistor.

2.5 No, as long as you are consistent in your application, you will obtain correct answers. Ifyou 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.

A πD2

4---------- 2.082 10 6–×= =

R ρLA------- 8.2Ω= =

R1 21 104× 20%±= 168kΩ R1 252kΩ≤ ≤

R2 07 103× 20%±= 5.6kΩ R2 8.4kΩ≤ ≤

Rs R1 R2+ 217kΩ 20%±= = 174kΩ Rs 260kΩ≤ ≤

RpR1R2

R1 R2+-------------------=

RpMIN

R1MINR2MIN

R1MINR2MIN

+-------------------------------- 5.43kΩ= =

RpMAX

R1MAXR2MAX

R1MAXR2MAX

+---------------------------------- 8.14kΩ= =

R1 10 102×= R2 25 101×=

RR1R2

R1 R2+------------------- 10 102×( ) 25 101×( )

10 102× 25 101×+--------------------------------------------------- 20 101×= = =

Solutions Manual


2.7 From KCL,

so from Ohm’s Law

Therefore, so

2.8 From Ohm’s Law and Question 2.7,

and for one resistor,

Therefore,

2.9

2.10

From KVL,

so

Therefore,

so or

2.11

and

From KCL,

Is I1 I2 I3+ +=

VsReq--------

VsR1------

VsR2------

VsR3------+ +=

1Req-------- 1

R1------ 1

R2------ 1

R3------+ += Req

R1R2R3R2R3 R1R3 R1R2+ +----------------------------------------------------=

VIs

Req--------

IsR2R3 R1R3 R1R2+ +

R1R2R3--------------------------------------------------------------------------------------------------------= =

V I1R1=

I1R2R3

R2R3 R1R3 R1R2+ +---------------------------------------------------- Is=

R1R2R1 R2+-------------------

R1 ∞→lim

R1R2R1

------------- R2= =

I CeqdVdt------- C1

dV1dt

---------- C2dV2dt

----------= = =

V V1 V2+=

dVdt-------

dV1dt

----------dV2dt

----------+=

ICeq-------- I

C1------ I

C2------+= 1

Ceq-------- 1

C1------ 1

C2------+= Ceq

C1C2C1 C2+-------------------=

V V1 V2= =

I1 C1dV1dt

---------- C1dVdt-------= = I2 C2

dV2dt

---------- C2dVdt-------= =

I I1 I2+ C1dVdt------- C2

dVdt-------+

dVdt------- C1 C2+( )= = =

Solutions Manual


Since

2.12

From KVL,

Since

2.13

From KCL, so

Therefore, so or

2.14 , regardless of the resistance value.

2.15 From Voltage Division,

2.16 Combining R2 and R3 in parallel,

and combining this with R1 in series,

(a) Using Ohm’s Law,

I CeqdVdt-------=

Ceq C1 C2+=

I I1 I2= =

V V1 V2+ L1dIdt----- L2

dIdt-----+

dIdt----- L1 L2+( )= = =

V LeqdIdt-----=

Leq L1 L2+=

V LdIdt----- L1

dI1dt------- L2

dI2dt-------= = =

I I1 I2+= dIdt-----

dI1dt-------

dI2dt-------+=

VL---- V

L1------ V

L2------+= 1

L--- 1

L1------ 1

L2------+= L

L1L2L1 L2+------------------=

Vo 1V=

Vo40

10 40+------------------ 5 15–( ) 8V–= =

R23R2R3

R2 R3+------------------- 2 3( )

2 3+------------ 1.2k= = =

R123 R1 R23+ 2.2k= =

I1VinR123---------- 5V

2.2k---------- 2.27mA= = =

Solutions Manual


(b) Using current division,

(c) Since R2 and R3 are in parallel, and since Vin divides between R1 and R23,

2.17(a) From Ohm’s Law,

(b)

2.18

(a)

(b)

(c)

2.19 Using superposition,

I3R2

R2 R3+-------------------I1

25---2.27mA 0.909mA= = =

V3 V23R23

R1 R23+---------------------Vin

1.22.2-------5V 2.73V= = = =

I4Vout V1–

R24----------------------- 14.2V 10V–

6k------------------------------- 0.7mA= = =

V5 V6 V56 Vout V2– 14.2V 20V– 5.8V–= = = = =

R45 R4 R5+ 5kΩ= =

R345R3R45

R3 R45+--------------------- 1.875kΩ= =

R2345 R2 R345+ 3.875kΩ= =

ReqR1R2345

R1 R2345+-------------------------- 0.795kΩ= =

VAR345

R2 R345+-----------------------Vs 4.84V= =

I345VA

R345---------- 2.59mA= =

I5R3

R3 R45+---------------------I345 0.97mA= =

VR21

R2R1 R2+-------------------V1 0.909V= =

Solutions Manual


2.20

2.21

2.22 It will depend on your instrumentation, but the oscilloscope typically has an inputimpedance of 1 MΩ.

2.23 Since the input impedance of the oscilloscope is 1 MΩ, the impedance of the source willbe in parallel, and the oscilloscope impedance will affect the measured voltage. Draw asketch of the equivalent circuit to convince yourself.

2.24

(a) ,

(b) ,

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

VR22

R1R1 R2+-------------------i1 9.09V= =

VR2 VR21VR22

+ 10.0V= =

R45R4R5

R4 R5+------------------- 0.5kΩ= =

IV1 V2–

R1 R2+------------------- 0.5mA–= =

VAR45

R3 R45+--------------------- V1 V2–( ) 0.238– V= =

R45 R4 R5+ 9kΩ= =

R345R3R45

R3 R45+--------------------- 2.25kΩ= =

R2345 R2 R345+ 4.25kΩ= =

ReqR1R2345

R1 R2345+-------------------------- 0.81kΩ= =

R23R2R3

R2 R3+-------------------=

VoutR23

R1 R23+---------------------Vin=

R23 9.90kΩ= Vout 0.995Vin=

R23 333kΩ= Vout 1.00Vin=

Solutions Manual


2.25

(a)

(b)

For a larger load impedance, the output impedance of the source less error.

2.26 It will depend on the supply; check the specifications before answering.

2.27

Combining R2 and L in series and the result in parallel with C gives:

Using voltage division,

where

so

Therefore,

2.28 With steady state dc Vs, C is open circuit. So

so and

2.29(a) In steady state dc, C is open circuit and L is short circuit. So

VoutR2

R1 R2+-------------------Vin=

Vout10

10.05-------------Vin 0.995Vin= =

Vout500

500.05----------------Vin 0.9999Vin= =

Vin 5 45°⟨ ⟩=

ZR2LCR2 ZL+( )ZC

R2 ZL+( ) ZC+------------------------------------- 1860.52 60.25°–⟨ ⟩ 923.22 1615.30j–= = =

VCZR2LC

R1 ZR2LC+---------------------------Vin=

R1 ZR2LC+ 1000 923.22 1615.30j–+ 2511.57 40.02°–⟨ ⟩= =

VC1860.52 60.25°–⟨ ⟩2511.57 40.02°–⟨ ⟩--------------------------------------------5 45°⟨ ⟩ 3.70 24.8°⟨ ⟩ 3.70 0.433rad⟨ ⟩= = =

VC t( ) 3.70 3000t 0.433+( )Vcos=

VC Vs 10V= = VR10V= VR2

Vs 10V= =

Solutions Manual


(b)

So

2.30

(a) ,

,

(b) ,

,

(c) ,

,

IVs

R1 R2+------------------- 0.025mA= =

ω π=

ZCj–

ωC-------- 106

π--------j 106

π-------- 90°–∠ Ω= = =

ZLR2ZL R2+ jωL R2+ 105 20πj+( )Ω 105 0.036°Ω∠= = = =

ZCLR2

ZCZLR2

ZC ZLR2+

------------------------ 91040 28550j–( )Ω 95410 17.4°–∠ Ω= = =

Zeq R1 ZCLR2+ 191040 28550j–( )Ω 193200 8.50°–∠ Ω= = =

IsVsZeq-------- 0.0259 8.50°∠ mA= =

IZC

ZC ZLR2+

------------------------Is 0.954 17.44°–∠( )Is 0.0247 8.94°–∠ mA= = =

I t( ) 24.7 πt 0.156–( )cos µA=

ω πradsec--------= f ω

2π------ 0.5Hz= =

App 2A 4.0= = dcoffset 0=

ω 2πradsec--------= f ω

2π------ 1Hz= =

App 2A 2= = dcoffset 10.0=

ω 2πradsec--------= f ω

2π------ 1Hz= =

App 2A 6.0= = dcoffset 0=

Solutions Manual


(d) ,

,

2.31

2.32

2.33

2.34 For , , and f = 60 Hz,

2.35 From Ohm’s Law,

Since ,

giving

or

For a resistor, , so the smallest allowable resistance would need a power rating of

at least:

so a 1/2 W resistor should be specified.The largest allowable resistance would need a power rating of at least:

ω 0radsec--------= f ω

2π------ 0Hz= =

App 2A 0= = dcoffset π( )sin π( )cos+ 1–= =

PVrms

2

R----------- 100W= =

VrmsVpp

2--------- 2( )⁄ 35.36V= =

PVrms

2

R----------- 12.5W= =

Vm 2Vrms 169.7V= =

Vrms 120V= Vm 2Vrms 169.7V= =

V t( ) Vm 2πf φ+( )sin 169.7 120πt φ+( )sin= =

I 5V 2V–R

--------------------- 3VR

-------= =

10mA I 100mA≤ ≤

10mA 3VR

------- 100mA≤ ≤

3V100mA------------------ R 3V

10mA---------------≤ ≤ 30Ω R 300Ω≤ ≤

P V2

R------=

P 3V( )2

30Ω--------------- 0.3W= =

Solutions Manual


so a 1/4 W resistor would provide more than enough capacity.

2.36 Using KVL and KCL gives:

The first loop equation gives:

Using this in the other two loop equations gives:

or

Solving these equations gives:

and

(a)

(b) , ,

2.37 Using KVL and KCL gives:

The first loop equation gives:

P 3V( )2

300Ω--------------- 0.03W= =

V1 IR1R1=

V1 I1 IR1–( )R2 I1 IR1

I2––( )R3+=

V3 V2– I1 IR1I2––( )R3 I2R4–=

IR1

V1R1------ 10mA= =

10 I1 10m–( )2k I1 10m– I2–( )3k+=

10 5– I1 10m– I2–( )3k I24k–=

5k( )I1 3k( )I2– 60=

3k( )I1 7k( )I2– 35=

I1 12.12mA= I2 0.1923mA=

Vout I2R4 V2– 4.23V–= =

P1 I1V1 121mW= = P2 I2V2 0.962mW= = P3 I– 2V3 1.92mW–= =

V1 IR1R1=

V1 I1 IR1–( )R2 I1 IR1

I2––( )R3+=

V3 V2– I1 IR1I2––( )R3 I2R4–=

Solutions Manual


Using this in the other two loop equations gives:

or

Solving these equations gives:

and

(a)

(b) , ,

2.38

Using the product formula trigonometric identity,

Therefore,

2.39

Using the double angle trigonometric identity,

Therefore,

IR1

V1R1------ 10mA= =

10 I1 10m–( )2k I1 10m– I2–( )2k+=

10 5– I1 10m– I2–( )2k I21k–=

4k( )I1 2k( )I2– 50=

2k( )I1 3k( )I2– 25=

I1 12.5mA= I2 0mA=

Vout I2R4 V2– 5V–= =

P1 I1V1 125mW= = P2 I2V2 0mW= = P3 I– 2V3 0mW= =

Pavg1T--- V t( )I t( ) td

0

T

∫VmIm

T-------------- ωt φV+( )sin ωt φI+( )sin td

0

T

∫= =

PavgVmIm

2T-------------- φV φI–( )cos 2ωt φ+ V φI+( )cos–( ) td

0

T

∫=

PavgVmIm

2-------------- φV φI–( )cos

VmIm2

-------------- θ( )cos= =

Irms1T--- Im

2 ωt φI+( )2sin td0

T

∫=

IrmsIm2

T----- 1

2--- 2 ωt φI+( )[ ]cos– td

0

T

∫=

Solutions Manual


2.40

This is a sin wave with half the amplitude of the input with a period of 1s.

2.41

2.42 RL = Ri = 8Ω for maximum power

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

IrmsIm2

T----- T

2--- Im

2-----= =

R23R2R3

R2 R3+------------------- 5kΩ= =

VoR23

R1 R23+---------------------Vi

12--- 2πt( )sin= =

NpNs------

VpVs------ 120V

24V------------- 5= = =

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 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).

3.4(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 halves of thenegative "humps" (below 0.5 V) clipped off.

3.5 Using superposition, Ohm’s Law, and current division,

I1left

1V

2R R2----+

----------------- 25R-------= =

I2left

12---I1left

– 15R-------–= =

Solutions Manual


,

3.6 For Vin > 0,

For Vin < 0,

The positive "bumps" of the resulting waveform are half the amplitude (5 vs. 10) of theoriginal, and the lower bumps are the same.

3.7 In steady state dc, the capacitor is equivalent to an open circuit. Therefore, the steady statecurrent through the capacitor is 0 and the steady state voltage across the capacitor is Vout.

(a) For Vs=10V, the diode is forward biased and is equivalent to a short circuit.Therefore, the equivalent resistance of the two horizontal resistors is R/2 and fromvoltage division,

(b) For Vs=−10V, the diode is reverse biased and is equivalent to an open circuit.Therefore, the circuit simplifies to two series resistors and

I4left

12---I1left

15R-------= = I4right

1V

R 2R3

-------+----------------- 3

5R-------= =

I2right

2RR 2R+-----------------I4right

25R-------= =

I1rightI4right

I2right– 1

5R-------= =

I1 I1leftI1right

+3

5R------- 0>= =

I2 I2leftI2right

+1

5R------- 0>= =

I3 0=

I4 I4leftI4right

+4

5R------- 0>= =

Vdiode 1V I4R–15---V 0>= =

Vout12---Vin 5 πt( )sin= =

Vout Vin 5 πt( )sin= =

Vcapacitor VoutR

R2---- R+--------------Vs

23---Vs 6.66V= = = =

Vcapacitor VoutR

R R+--------------Vs

12---Vs 5– V= = = =

Solutions Manual


3.8 There are three possible states of the diodes. When only the left diode is forward biased,. When only the right diode is forward biased, . When both diodes

are reverse biased, . In this case, the circuit is a voltage divider and

The upper limit of this state is when corresponding to .Vout remains at 5V when Vin increases above 10V. The lower limit of the double reversebiased state is when corresponding to . Vout remains at−5V when Vin decreases below −10V. It is not possible for both diodes to be reverse biasedat the same time in this circuit.The resulting output signal Vout is a sin wave scaled by 1/2 with the peaks clipped off at

.

3.9(a) output passes first (positive) hump only

(b) output is 5.1V over the whole input cycle

3.10 Use a resistor in series with the LED where:

The required resistance value is

(a)

(b)

3.11

Vout VH= Vout VL=

VL Vout VH< <

VoutRL

Ri RL+------------------Vin

12---Vin= =

Vout VH 5V= = Vin 10V=

Vout VL 5– V= = Vin 10– V=

10V±

I5V VLED–

R---------------------------=

R5V VLED–( )

Imax--------------------------------≥

R 5V50mA---------------≥ 100Ω=

R 5V 2V–( )50mA

--------------------------≥ 60Ω=

VE 5V VLED– VCE– 2.8V= =

Vin saturation( ) VE VBE+ 2.8V 0.7V+ 3.5V= = =

Solutions Manual


3.12(a) For the LED to be ON, the transistor must be in saturation and

When the LED is off, IB = 0 and

So for the LED to be ON,

(b) When the transistor is fully saturated,

and

and

Assume

If I1 is the current through the horizontal 1k resistor and I2 is the current through theright 1k resistor, then

and

3.13 When the transistor is in full saturation,

and

and

In the collector-to-emitter circuit,

giving

VB VLED VBE+ 1V 0.7V+ 1.7V= = =

VB12---Vin=

Vin 2VB> 3.4V=

VB VLED 0.7V+ 1.7V= = VC VLED 0.2V+ 1.2V= =

IC5V VC–

330Ω--------------------- 3.8V

330Ω------------- 11.5mA= = =

IB1

100---------IC 0.115mA= =

I1 I2 IB+VB1k------- 0.115mA+ 1.815mA= = =

Vin VB 1k( )I1+ 3.52V= =

VCE 0.2V= VBE 0.7V=

Iout IB IC+IC

100--------- IC+ 1.01IC= = =

Vout Vs ICRC– VCE– IoutRout= =

5V IC 1k( )– 0.2V– 1.01IC 1k( )=

Solutions Manual


Now we can solve for the collector and emitter currents:

and

Therefore,

and the minimum required input voltage is:

3.14 1: a resistor (e.g., 1k) to limit the base current while ensuring the transistor is in fullsaturation2: 24 Vdc capable of at least 1A of current

3: power diode capable of carrying at least 1A for flyback protection4: ground

3.15 A voltage source (e.g., 5V) and current limiting series resistor (e.g., 330 Ω) is required onthe LED side. On the phototransistor side, a pull-up resistor (e.g, 1k) and a voltage source(e.g., 5V) is required on the collector lead and ground is required on the emitter lead.

3.16 1: nothing required2: 24 Vdc capable of at least 1A of current

3: power diode capable of carrying at least 1A for flyback protection4: ground

3.17 The type of BJT required is an npn and an additional resistor must be added in series withthe open collector output to pull up the voltage enough to bias the BE junction of the BJT.

3.18 The upper FET is a p-channel enhancement mode MOSFET and the lower is an n-channelenhancement mode MOSFET. When Vin = 5V, the upper MOSFET doesn’t conduct butthe bottom one does, so Vout = 0V. When Vin = 0V, the upper MOSFET conducts but thebottom one doesn’t, so Vout = Vcc.

3.19 The requirements are that

and

IRF530N is a good choice

IC4.8V2.01k------------- 2.39mA= =

Iout 1.01IC 2.41mA= =

Vout IoutRout 2.41mA( ) 1k( ) 2.41V= = =

Vin Vout VBE IBRB+ + 2.41V 0.7V 0.239mA( ) 1k( )+ + 3.13V= = =

Id cont( ) 10A> Pd max( ) Ion2 Ron> 100Ron=

Solutions Manual


3.20(a) cutoff

(b) ohmic

(c) saturation

(d) cutoff

Solutions Manual


4.1(a) linear

(b) nonlinear

(c) linear

(d) linear

(e) nonlinear

(f) nonlinear

(g) linear

4.2 The Fourier Series is:

and the fundamental frequency is

4.3 Since Vrms = 120V and f = 60Hz, the Fourier Series is:

and the fundamental frequency is 60Hz.

4.4

So

But , so

But sine of any multiple of π is 0, so An = 0

F t( ) 5 2πt( )sin=

f0ω02π------ 1Hz= =

F t( ) 2Vrms 2πft( )sin 169.7 120πt( )sin= =

An2T--- F t( ) nω0t( )cos td

0

T

∫2T--- nω0t( )cos td

0

T2---

∫ nω0t( )cos tdT2---

T

∫–= =

An2

nω0T------------- nω0t( )sin[ ]0

T2---

nω0t( )sin[ ]T2---

T–

=

ω02πT------=

An1

nπ------ nπ( )sin 2nπ( )sin– nπ( )sin+( )=

Solutions Manual


4.5 Using MathCAD:

4.6

(a)

So the bandwidth is: 0 Hz to 7.17 Hz

t 0 0.01, 3.0.. Va t( ) 1π

sin 2 π. t.( )2

n 2 4, 10.. Vb t( ) 1π

sin 2 π. t.( )2

2π

n

cos 2 n. π. t.( )n 1( ) n 1( ).

.

m 2 4, 50.. Vc t( ) 1π

sin 2 π. t.( )2

2π

m

cos 2 m. π. t.( )m 1( ) m 1( ).

.

0 1 2 30.5

0

0.5

1

Va t( )

t0 1 2 3

0.5

0

0.5

1

1.5

Vb t( )

t

0 1 2 30.5

0

0.5

1

1.5

Vc t( )

t

fH 61 1

2-------–

1--------------------- 10 6–( )+ 7.17Hz= =

Solutions Manual


(b) T = 1s

,

(c) Using MathCAD:

4.7

(a)

n ω (rad/sec) f (Hz) Ain Aout/Ain Aout = (Aout/Ain)Ain

1 ω0 1 4/π 1 4/π2 3ω0 3 4/3π 1 4/3π3 5ω0 5 4/5π 1 4/5π4 7ω0 7 4/7π 3/4 3/7π5 9ω0 9 4/9π 1/4 1/9π

>5 (2n-1)ω0 (2n-1) 4/fπ 0 0

fo1T--- 1Hz= = ω0 2πf0 2πrad

sec--------= =

Vout4π--- 2πt( )sin 4

3π------ 6πt( )sin 4

5π------ 10πt( )sin 3

7π------ 14πt( )sin 1

9π------ 18πt( )sin+ + + +=

t 0 0.01, 2..

Vout t( ) 4π

sin 2 π. t.( ). 43 π.

sin 6 π. t.( ). 45 π.

sin 10 π. t.( ). 37 π.

sin 14 π. t.( ). 19 π.

sin 18 π. t.( ).

0 1 22

0

2

Vout t( )

2.0

t

ωc1

RC-------- 1000rad

sec--------= =

Solutions Manual


(b) T = 1s, f = 1Hz, ω0 = 2π

where

(c) Using MathCAD:

4.8

F t( ) Ain n( )AoutAin---------- n( ) ωnt( )sin

n 1=

∞

∑=

Ain n( ) 42n 1–( )π

-----------------------=

AoutAin---------- n( ) 1

1ωnωc------

2+

----------------------------=

ωn 2n 1–( )2π=

n 1 100.. ω c 1000 ωn 2 n. 1( ) 2. π.

t 0 0.01, 2..

F t( )

n

42 n. 1( ) π.

1

1ωn

ω c

2

. sin ωn t..

0 1 21

0

1

F t( )

t

ωL12

-------radsec-------- 0.707rad

sec-------- 0.113Hz= = =

ωH 3 1.5 1 12

-------– +

radsec-------- 3.44rad

sec-------- 0.547Hz= = =

ωL bandwidth ωH≤ ≤

Solutions Manual


4.9

(a)

(b) and (c)

4.10

To find the cut-off frequency, set the amplitude ratio magnitude to :

Solving for the frequency gives

Using this expression gives:

n Ain Aout

1 1 12 2 23 3 34 4 1.335 5 0

1radsec-------- ω 5rad

sec--------≤ ≤

VoR

R 1jωC----------+

---------------------Vi=

VoVi------ jωRC

jωRC 1+------------------------=

12

-------

VoVi------ ωRC

ωRC( )2 1+--------------------------------- 1

2-------= =

ωc1

RC--------=

VoVi------

ωωc------

ωωc------ 2

1+

----------------------------=

Solutions Manual


and now we can plot the frequency response in terms of the dimensionless frequency ratio

:

4.11

Using and ,

ωrωωc------=

ω r 0 0.01, 5.0.. A r ω rω r

ω r2 1

0 2 4 60

0.5

1

A r ω r

ω r

φVoVi------ arg 1( )∠ 1 ωRCj+( )∠– 0 ωRC

1------------ atan– ωRC( )atan–= = = =

ωc1

RC--------= ωr

ωωc------ ωRC= =

φ ωr( )atan–=

0 0.5 1 1.5 2 2.580

60

40

20

00

68.199

φ ω r

2.50 ω r

Solutions Manual


4.12 Using MathCAD:

4.13 When the mass is in static equilibrium,

so

n 1 20.. t 0 0.01, 2..

ωn 2 n. 1( ) 2. π. F t( )

n

42 n. 1( ) π.

sin ωn t..

ATTn 1 ATT1 .25 ATT2 .25 ATT3 .25

F low t( )

n

42 n. 1( ) π.

ATTn. sin ωn t..

ATTn 1 ATT17 .25 ATT18 .25 ATT20 .25

F high t( )

n

42 n. 1( ) π.

ATTn. sin ωn t..

0 1 22

1

0

1

2

F t( )

t0 1 21

0

1

F low t( )

t

0 1 22

1

0

1

2

F high t( )

t

Ming kXout=

Xoutgk--- Min=

Solutions Manual


and the static sensitivity is

4.14 We generally assume that the displayed voltage is the gain times the input voltage. Thisassumption will be in error if the oscilloscope is dc coupled and some of the frequencies inthe signal exceed the bandwidth of the oscilloscope.

4.15 KVL gives:

where q is the charge on the capacitor. Putting this in standard form gives:

where q is the dependent variable, the time constant (τ) is RC, and the sensitivity (K) is C.Using the general solution for a 1st order system,

Therefore, the step response output voltage (which is the voltage across the capacitor) is

4.16 The rate of change of internal energy is equal to the rate of heat transfer:

so

Defining (thermal capacitance) and (thermal resistance) and

converting into standard form gives:

where the time constant is and the sensitivity is K=1.

K gk---=

IR 1C----q+ Vin=

RC( )q· q+ CVin=

q t( ) CAi 1 etτ--–

–

=

Vout t( ) 1C----q t( ) Ai 1 e

tτ--–

–

= =

ddt----- Ein( ) Q· in=

mc( )dTout

dt------------- hA( ) Tin Tout–( )=

Ct mc= Rt1

hA-------=

CtRt( )dTout

dt------------- Tout+ 1( )Tin=

τ RtCtmchA-------= =

Solutions Manual


4.17 Plotting the data Xout(t) shows a steady state asymptote of approximately 5 indicating that:

Plotting shows a near linear relation indicating that the system

can be modeled as 1st order. The slope of the line is approximately −1/3 indicating a timeconstant of

τ = 3 sec

4.18 The damped natural frequency is always smaller if there is damping in the system.

4.19(a) mechanical rotary (applied torque, torsion spring, rotary damper, and rotary inertia):

(b) electrical (voltage source and series resistor, inductor, and capacitor):

or

(c) hydraulic (pump with inlet in reservoir, long pipe with friction loss and fluid inertia,and tank):

V = P

where V =

4.20 Given the differential equation:

Applying the procedure results in:

KAin 5=

Z t( ) 1Xout t( )KAin

-----------------– ln=

Jθ·· Bθ· kθ+ + τext=

Lq·· Rq· 1C----q+ + Vs= LI· RI 1

C---- I td∫+ + Vs=

IdQdt------- RQ 1

C----+ +

Q td∫

τx· out xout+ Kxin=

τs 1+( )Xout s( ) KXin s( )=

G s( )Xout s( )Xin s( )------------------ K

τs 1+( )-------------------= =

G jω( ) K1 τωj+------------------=

XoutXin

------------- G jω( ) K

1 τω( )2+----------------------------= =

φ G jω( )( )arg 0 τω( )atan– τω( )atan–= = =

Solutions Manual


4.21

, ,

Therefore, the steady state response is:

4.22 The equation of motion is

Taking the Laplace transform of both sides gives the transfer function:

Now

So the steady state response is

F0 20N= ω 0.75radSec--------=

ωnkm---- 1.095= = ωr

ωωn------ 0.685= = ς b

2 km--------------- 0.456= =

X0F0 k⁄------------ 1

1 ωr2

–[ ]2

4ς2ωr2

+

------------------------------------------------- 1.219= =

X0X0

F0 k⁄------------ F0

k----- 2.032m= =

φ 2ς1ωr----- ωr–-----------------

1–

tan– 49.6°– 0.866rad–= = =

x t( ) X0 ωt φ+( )sin 2.032 0.75t 0.866–( )msin= =

mx··out bx· out kxout+ + kxin=

G s( )Xout s( )Xin s( )------------------ k

ms2 bs k+ +-------------------------------= =

G jω( ) kk mω2–( ) bωj+

-----------------------------------------=

XoutXin

------------- G jω( ) kk mω2–( )2 bω( )2+

-----------------------------------------------------= =

φ bωk mω2–-------------------- 1–

tan–=

xout t( ) XinXoutXin

------------- ωt φ+( )sin=

Solutions Manual


Using MathCAD,

Only the first input results in an acceptable output.

4.23

gives C = 0

x(0) = 0 gives A = -C

gives B = 0

Therefore,

4.24 The volume in the tank is

V =

The pressure at the bottom of the tank is

Solving the volume expression for h and substituting into the pressure expression gives

V = V

m 0.10 k 100000 b 10 X in 0.05

X out ω( )k X in.

k m ω2.2

b ω.( )2φ ω( ) angle k m ω2. b ω.,

X out 10( ) 0.05= φ 10( ) 0.057 deg=

X out 1000( ) 0.5= φ 1000( ) 90 deg=

X out 10000( ) 5.05 10 4.= φ 10000( ) 179.421 deg=

xh t( ) eςωnt–

A ωdt( )cos B ωdt( )sin+[ ]=

xp t( ) C=

x t( ) xh t( ) xp t( )+=

x ∞( )F0k-----=

x· 0( ) 0=

x t( )F0k----- 1 e

ςωnt–ωdt( )cos–( )=

πD2

4----------h

P ρgh γh= =

P 4γπD2----------=

1C----

Solutions Manual


so

4.25 Since , and ,

4.26 Element [flow] analogies:

Analogous free body diagram equations [KVL]:

The resulting analogous electrical circuit follows:


C πD2

4γ----------=

F ma= a x··= x· QA----=

PA ρLA( ) Q·

A---- =

P ρLA------- Q· IQ·= =

Fin vin[ ] Vin Iin[ ]→

k1 vin vm–[ ] C1 Iin Im–[ ]→

m vm[ ] L Im[ ]→

b1 vm[ ] R1 Im[ ]→

k2 vm[ ] C2 Im[ ]→

Vin VC1=

VC1VR1

VC2VL+ +=

+Vin

L

C1 R1

C2

F1 v1[ ] V1 I1[ ]→

m v1[ ] L I1[ ]→

k1 v1[ ] C1 I1[ ]→

Solutions Manual




4.28 Hydraulic elements are direct analogies to electrical elements. The capacitors are replacedby tanks, and the resistor and inductor are replaced by a long pipe with flow resistance andinertance.



b1 v1 v2–[ ] R1 I1 I2–[ ]→

k2 v1 v2–[ ] C2 I1 I2–[ ]→

F2 v– 2[ ] V2 I– 2[ ]→

V1 VC2– VR1

– VC1– VL=

VR1V+ C2

V2– 0=

+

+V2

V1

L

C1

R1 C2

Fin v1[ ] Vin I1[ ]→

b1 v1 v2–[ ] R1 I1 I2–[ ]→

k1 v1 v2–[ ] C1 I1 I2–[ ]→

m v2[ ] L I2[ ]→

k2 v2 v3–[ ] C2 I2 I3–[ ]→

b2 v3[ ] R2 I3[ ]→

Vin VR1V+ C1

=

VR1V+ C1

VL VC2+=

Solutions Manual



4.30

where

, ,

VC2VR2

=

+Vin

I1

I2

(I -I )1 2

(I -I )2 3

I3

R1

C1

C2 R2

L

k z x–( ) b z· x·–( ) µm1g x·( )sgn–+ m1x··=

k z x–( ) b z· x·–( ) F1–+ m– 2z··=

rF1– I2θ··=

z y rθ–= z· y· rθ·–= z·· y·· rθ··–=

Solutions Manual


5.1 The power dissipated by each resistor is

and

To be able to use 1/4 W resistors, the following must be true:

or

or

(a) for GAIN=1, RF > 100Ω

(b) for GAIN=10, RF > 10kΩ

5.2

(a)

(b)

so

so

5.3 With RF replaced by a short, the op amp circuit becomes a buffer so the gain is 1.

Vin2

R-------- 25

R------=

Vout2

RF----------

GAIN( )Vin[ ]2

RF------------------------------------ 25GAIN2

RF------------------------= =

25R------ 0.25< R 100Ω>

25GAIN2

RF------------------------ 0.25< RF GAIN2( )100Ω>

V+ IR1=

Vo 1R3R4------+

V+ 1R3R4------+

R2I= =

V+ V- Vo= =

I1 I2V+R2-------

VoR2------= = =

V+ I1R1+ Vo I3R3+=

I3R1R3------I1

R1R2R3-------------Vo= =

I I1 I3+ Vo1

R2------

R1R2R3-------------+

= =

VoR2R3

R1 R2+-------------------I=

Solutions Manual


5.4

(a)

but I3 = 0, so

(b) same as (a)

5.5

5.6 If VA denotes the voltage at the output of the first op amp,

and from Ohm’s Law, the current from voltage source V1 is

If VB denotes the voltage at the inverting input of the second op amp,

and from Ohm’s Law,

and

where I4 is the current through the vertical resistor and I2 is the current through the feedbackresistor of the second op amp. From this,

Now applying Ohm’s Law to the resistor between the op amps gives

where I3 is the current through the resistor.

V- V+R2

R1 R2+-------------------V1 5V= = =

Vout V- V2 I3R3–+=

Vout V- V2+ 10V= =

V+ V- Vi= =

V4 1R3R2------+

V+=

I4V4R4------

R2 R3+

R2R4-------------------Vi= =

VA V- V+ 0V= = =

I1V1 VA–

R--------------------

V1R------= =

VB V- V+ V2= = =

I4VBR

-------V2R------= = I2

VB Vout–

R------------------------

V2 Vout–

R-----------------------= =

Vout V2 I2R–=

I3VA VB–

R---------------------

V2R------–= =

Solutions Manual


From KCL, the current out of the first op amp is

The negative sign indicates that the current is actually into the op amp.From KCL,

Therefore,

5.7 because of positive feedback.

5.8 Applying Ohm’s Law to both resistors gives

and

From KCL,

Since ,

For R1 = R2 = RF = R,

5.9 Applying Ohm’s Law to both resistors gives

and

From KCL,

Since ,

Iout1I3 I1–

V2R------–

V1R------–

1R---- V1 V2+( )–= = =

I2 I3 I4–V2R------–

V2R------– 2

R----–

V2= = =

Vout V22V2

R----------–

R– 3V2= =

Vo Vi≠

I1V1R1------= I2

V2R2------=

IF I1 I2+=

Vo IFRF+ 0=

Vo RFV1R1------

V2R2------+

–=

Vo V1 V2+( )–=

I1V1 V3–

R1-------------------= I2

V2 V3–

R2-------------------=

IF I1 I2+=

Vo IFRF+ V3=

Vo V3 RFV1 V3–

R1-------------------

V2 V3–

R2-------------------+

–=

Solutions Manual


For R1 = R2 = RF = R,

5.10 From voltage division,

From Ohm’s Law,

The output voltage can be found with:

Simplifying gives

For R1=R2=R,

5.11 and

5.12 Using superposition,

Vo V3 V1 V3– V2 V3–+( )– 3V3 V1 V2+( )–= =

V- V+RF

RF R2+------------------- V2= =

I1V1 V- –( )

R1------------------------=

Vo V- I1RF–RF

RF R2+------------------- V2

V1RF

RF R2+------------------- V2–

R1--------------------------------------------------RF–= =

VoR1V2 V1 RF R2+( ) RFV2–( )–

R1 RF R2+( ) RF⁄----------------------------------------------------------------------------=

VoV2 RF R1+( ) V1 RF R2+( )–

R1 RF R2+( ) RF⁄---------------------------------------------------------------------=

VoRFR------ V2 V1–( )=

Voutin1

RFR------+

Vin= Voutref

RFR------ Vref–=

Vout VoutinVoutref

+ 1RFR------+

VinRFR------ Vref–= =

Vo1

R4R3------V3–=

Vo21

R4R3------+

R5R3 R5+-------------------V4=

Solutions Manual


5.13

so

but IR = IL, so

5.14

(a)

(b)

(c)

(d)

From Ohm’s Law,

and

From KCL,

but from Ohm’s Law,

so

Vo Vo1Vo2

+R4R3------V3– 1

R4R3------+

R5R3 R5+-------------------V4+= =

V+ V- 0= =

Vi LdILdt

--------= IL1L--- Vi td∫=

Vo V- IRR+=

VoRL---- Vi td∫=

VoRFR------ Vi– 2Vi–= =

Vo1

RC-------- Vi td∫– Vi td∫–= =

VoRFR------ V1 V2+( )– 4Vi–= =

V- V+ 0V= =

I1V15k------

Vi5k------= = I2

V210k---------

Vi10k---------= =

IF I1 I2+ Vi1

5k------ 1

10k---------+

= =

IF0 Vo–

5k---------------=

Vo 5kIF– Vi 1 12---+

–32---Vi–= = =

Solutions Manual


5.15

5.16

5.17 The limit on the feedback resistor current is:

Therefore,

5.18

so the fall-off frequency is 105Hz.

+−+

+5VVin

LED

330Ω

comparator

+−+

+5VVin

LED

330Ω

open collectorcomparator

IFVoutRF

---------- 10VRF

---------- 10mA<= =

RF10V

10mA---------------> 1kΩ=

closed loop gainRFR------ 10= =

Solutions Manual


6.1 11111111111111112 = 216 − 1 = 65,535

6.2

(a) 128 = 100000002 since 27 = 128

(b) 127 = 11111112 since (27 − 1) = 127

6.3(a) 128 = 8016 since 8(16) = 128

(b) 127 = 7F16 since 7(16) + 15 = 127

6.4(a) 1 1

1101 13

+ 1001 + 9

10110 22

(b) 1101 13

- 1001 - 9

0100 4

(c) 1101 13

x 1001 x 9

1101 27

0000 9

0000 117

1101

1110101

(d) 111

111 7

+ 111 + 7

1110 14

Solutions Manual


(e) 111 7

x 111 x 7

11

1111

111 49

111

111

110001

6.5(a)

(b)

6.6

6.7(a) The output (C) is high (5V) iff both inputs are low (0V).

A B C0 0 10 1 01 0 01 1 0

AB

C

A

BC

AA

AA A=

C AB A B+= =

Solutions Manual


(b)

(c)

(d)

6.8

6.9

A B C0 0 00 1 01 0 01 1 1

A B C0 0 00 1 11 0 11 1 0

A B C0 0 00 1 11 0 11 1 1

C AB( ) AB( ) AB= =

C AABBAB AAB BAB+ A A B+( ) B A B+( )+= = =

C AB BA+ A B⊕= =

C AB A B+= =

A

B

C

A

B

C

Solutions Manual


6.10

6.11

(a)

1

(b)

A

6.12

6.13 Equation 6.19:

A B C D E F

0 0 0 0 1 0

0 0 1 0 0 1

0 1 0 0 1 0

0 1 1 0 0 1

1 0 0 0 1 0

1 0 1 0 0 1

1 1 0 1 1 0

1 1 1 1 1 1

A

B

C

D

E

F

1 0⋅ 1 0 1+( )⋅ 0 1 0+( )⋅+ +

1 1⋅ 1 1⋅ 0 1⋅+ +

1 1 0+ +

A B⋅ A A B+( )⋅+

A B A B+ +( )

A A 1+( )

A 1( )

X AB BC+ BC C+ + ABBC BC C+ + BC A 1+( ) C+ C B 1+( ) C= = = = =

Solutions Manual


Multiplying (ANDing) both sides by A gives:

Simplifying, gives:

Thus, the identity is valid.

Equation 6.20:

6.14 Equation 6.21:

A B C

0 0 0 0 0 0 0 00 0 1 0 1 0 0 00 1 0 1 0 0 0 00 1 1 1 1 1 1 11 0 0 1 1 0 1 11 0 1 1 1 0 1 11 1 0 1 1 0 1 11 1 1 1 1 1 1 1

A A B⋅( )+ A B+=

A A⋅ A A B⋅ ⋅+ A A⋅ A B⋅+=

A 0+ A A B⋅+=

A A 1 B+( )⋅=

A A=

A B+( ) A B+( ) AA AB BA BB+ + + A A B B+( )+ A A+ A= = = =

A B+( ) A C+( )⋅

AA AC BA BC+++

A 1 C+( ) B A C+( )+

A 1( ) B A C+( )+

A BA BC+ +

A 1 B+( ) BC+

A 1( ) BC+

A BC+

A B+ A C+ BC A B+( ) A C+( )⋅ A BC+

Solutions Manual


6.15 Equation 6.22:

6.16 Equation 6.23:

6.17

0 0 0 0 01 0 1 1 10 1 1 0 11 1 1 1 1

A B C AB BC

0 0 0 0 0 0 0 00 0 1 0 0 1 1 10 1 0 0 0 0 0 00 1 1 0 1 0 1 11 0 0 0 0 0 0 01 0 1 0 0 1 1 11 1 0 1 1 0 1 11 1 1 1 1 0 1 1

A B C AB AC

0 0 0 0 0 0 0 00 0 1 0 0 1 1 10 1 0 0 0 0 0 00 1 1 0 0 0 0 01 0 0 0 0 0 0 01 0 1 0 1 1 1 11 1 0 1 0 0 1 11 1 1 1 1 0 1 1

A B A B+ AB A B+( ) AB+

BC AB BC BC+ + AB C+

BC AB AC BC+ + AB BC+

Solutions Manual


6.18(a)

(b)

(c)

A B C AB BC

0 0 0 0 0 0 0 10 0 1 0 0 1 1 00 1 0 0 0 0 0 10 1 1 0 1 0 1 01 0 0 0 0 0 0 11 0 1 0 0 1 1 01 1 0 1 0 0 1 11 1 1 1 1 0 1 1

A B C ABC

0 0 0 0 10 0 1 0 00 1 0 0 00 1 1 0 01 0 0 0 01 0 1 0 01 1 0 0 01 1 1 1 0

A B C AB BC

0 0 0 0 0 0 0 00 0 1 0 0 1 1 10 1 0 0 0 0 0 00 1 1 0 1 0 1 11 0 0 0 0 0 0 01 0 1 0 0 1 1 11 1 0 1 0 0 1 11 1 1 1 1 0 1 1


A B C+ +


Solutions Manual


6.19

6.20

The circuit is called a multiplexer because P allows one of two (multiple) inputs to passthrough to the output.

6.21

For the unallowed code CD=11, the output (Y) would be:

In this state, the alarm would go off when windows or doors are disturbed or when motionis detected. This state is the same as state 2 (CD = 10).

6.22 The simplified Boolean expression is:

Using DeMorgan’s Laws, the all-AND representation is:

and the all-OR representation is:

The original expression contains 1 AND operations, 1 OR operation, and one inversion,requiring 3 ICs. The all-AND version contains 2 ANDs and 2 inversions, requiring 2 ICs.The all-OR version contains 2 ORs and 4 inversions, also requiring 2 ICs.

A B

0 0 1 1 1 10 1 1 0 0 01 0 0 1 0 01 1 0 0 0 0

P A B X0 0 0 00 0 1 00 1 0 10 1 1 11 0 0 01 0 1 11 1 0 01 1 1 1

A B A B+ A B⋅

X AP BP+=

Y AD A B+( )C+=

Y A A B+( )+ A B+= =

X B C A+( )⋅=

B C A⋅( )⋅ B C A⋅( )⋅=

B C A+( )+

Solutions Manual


6.23

6.24 Segment c is OFF only for the digit 2, so the output of the logic circuit must be:

or more simply

6.25

6.26

6.27 The required Boolean expression is:

which can be implemented with the following logic circuit:

The complete truth table (including the sub expression ) is:

A B C X

0 0 0 1 00 0 1 1 10 1 0 1 00 1 1 1 11 0 0 1 01 0 1 1 11 1 0 0 01 1 1 0 0

Y A D⋅( ) A B+( ) C⋅+=

Y A D+( ) A B+( ) C++=

A B C D

X DCBA= X CBA=

X ABC A B+( )C+ ABC A B+( )C ABCABC= = =

X AA C C+( ) BC+ A1 BC+ A BC+ A B C++ A B C++= = = = =

X C A B+( )⋅=

A

B

C

X

A B+( )

A B+( )

Solutions Manual


6.28

6.29

6.30

Ci-1 Ai Bi Si Ci

0 0 0 0 00 0 1 1 00 1 0 1 00 1 1 0 11 0 0 1 01 0 1 0 11 1 0 0 11 1 1 1 1

X PAB PAB PAB PAB+ + +=

X PA B B+( ) PB A A+( )+=

X PA PB+=

Si Ci 1– AiBi Ci 1– AiBi Ci 1– AiBi Ci 1– AiBi+ + +=

Ci Ci 1– AiBi Ci 1– AiBi Ci 1– AiBi Ci 1– AiBi+ + +=

preset

clear

J

K

CK

Q

toggle set

reset

set

reset

toggle

Solutions Manual


6.31

6.32

T Q

↑ 1 1

↓ 1 1

0 1 1

1 1 1

0 0 1 1 01 1 0 0 1

CK D Q

↑ x 1 1

↓ 0 1 1 0 1

↓ 1 1 1 1 0

0 x 1 1

1 x 1 1

x x 0 1 1 0x x 1 0 0 1

Preset Clear Q

Q0 Q0

Q0 Q0

Q0 Q0

Q0 Q0

Preset Clear Q

Q0 Q0

Q0 Q0

Q0 Q0

Solutions Manual


6.33

6.34

Solutions Manual


6.35

6.36

6.37

D Q

Q

CK

D Q

Q

CK

D Q

Q

CK

D Q

Q

CK

S1 S2 S3 S4

E

J

K

Q

Q

CK

1

1

Solutions Manual


6.38

6.39 see 6.49

6.40

6.41 The CMOS LOW can sink only 1mA per gate which is enough to drive only two LS TTLinputs (which require 0.36 mA per gate).

6.42

6.43 See info in TTL Data Book.

6.44 The input is the same: some sort of clock signal. Three of the four outputs look the same(the 3 least significant bits), but in the decade counter the most significant bit resets all thebits on the count of 10. The binary counter will continue to increment bits until 16 isreached. In the binary counter the output code provides 16 combinations, but in the decimalcounter the output code provides 10 different output combinations.

6.45 The output goes high when the signal increases through 4V and low when it decreasesthrough 1V.

D Q

Q

CKsensor

NCbutton

CL

5V

5V

1k

74LS00 4011B

5V

1k

c QDQCQBQA=

e QD QC QB QA+ + +( ) QD QC QB QA+ + +( ) QD QC QB QA+ + +( ) QD QC QB QA+ + +( )=

Solutions Manual


6.46 where

But when t = ∆T, VCAPACITOR = 2/3 Vcc, so

so

and

6.47 See a Linear Circuits data and/or applications book.

6.48 Assume that the digital event is a digital pulse that can be applied to the input of the counter.Cascade 3 74LS90's and connect the output of the third to an LED. Refer to the IC specsheet for the appropriate wiring.

6.49

VCAPACITOR Vcc 1 etτ--–

–

= τ RC=

23--- 1 e

∆TRC--------–

–

= e∆TRC--------– 1

3---=

∆T RC 3( )ln 1.1RC≈=

D Q

CK

NO NC

5V

5V

7474

74124

7409 7490

CQ

O

B

S

one-shot

counter

Solutions Manual


6.50 See "Case Study 2" in Chapter 11.

6.51 See "Case Study 2" in Chapter 11.

B

S

Q

O

C

Solutions Manual


7.1

list p=16f84include <p16F84.inc>

; Define counter variable locationsc1 equ 0x0cc2 equ 0x0dc3 equ 0x0e

; Initializes PORTA to output all zerosbcf STATUS, RP0 ; select bank 0clrf PORTA ; initialize all pin values to zerobsf STATUS, RP0 ; select bank 1clrf TRISA ; designate all PORTA pins as outputsbcf STATUS, RP0 ; select bank 0

; Main program loopstart

bsf PORTA, 0 ; turn on the LED connected to RA0call pause ; pause for 1 secondbcf PORTA, 0 ; turn off the LED connected to RA0call pause ; pause for 1 second

goto start

PIC16F84

RA2

RA3

RA4

MCLR

Vss

RB0

RB1

RB2

RB3

RA1

RA0

OSC1

OSC2

Vdd

RB7

RB6

RB5

RB4

1

2

3

4

5

6

7

8

9 10

11

12

13

14

15

16

17

18

5V 22 pF

22 pF

4 MHz

1 k

5V

0.1 Fµ

330 LED

Solutions Manual


; Subroutine to pause for approximately 1 secondpause

; Initialize counter variablesmovlw 0x00movwf c1movlw 0x00movwf c2movlw 0xFAmovwf c3

loopincfsz c1, Fgoto loopincfsz c2, Fgoto loopincfsz c3, Fgoto loopreturn ; end of subroutine

end ; end of instructions

7.2PIC16F84

RA2

RA3

RA4

MCLR

Vss

RB0

RB1

RB2

RB3

RA1

RA0

OSC1

OSC2

Vdd

RB7

RB6

RB5

RB4

1

2

3

4

5

6

7

8

9 10

11

12

13

14

15

16

17

18

5V 22 pF

22 pF

4 MHz

1 k

5V

0.1 Fµ

330LED

NO

5V

Solutions Manual


’ Program to turn an LED on and off at 1 Hz while a pushbutton switch’ is being held down

’ Define variable names for the I/O pinsmy_button Var PORTB.0my_led Var PORTA.0

begin:While (my_button == 1) ’ while the switch is held down

’ Turn on the LEDHigh my_led

’ Wait for 1/2 secPause 500

’ Turn off the LEDLow my_led

’ Wait for 1/2 secPause 500

Wend

Goto begin ’ continue

End ’ end of instructions

7.3 ’ Program to perform the functionality of the Pot statement

’ assumed variables:’ pin: I/O pin identifier’ scale: byte variable containing maximum time constant’ var: byte variable containing measured time constant

’ Charge the capacitorHigh pinPause scale

’ Discharge the capacitor and measure the discharge timevar = 0Low pinInput pinWhile (pin == 1) ’ while the capacitor is not discharged

Pause 1var = var + 1

Wend

Solutions Manual


7.4 ’ Subroutine to perform a software debounce on pin RB0debounce:

’ Define pin RB0 as an inputpin Var PORTB.0Input pin

’ Wait for the pushbutton switch to be pressed (1st bounce)loop:

If (pin == 0) Then loop

’ Wait 10 milliseconds for the switch bounce to settlePause 10

’ Wait for the pushbutton switch to be released (1st bounce)loop2:

If (pin == 1) Then loop2

’ Wait 10 milliseconds for the switch bounce to settlePause 10

’ End of subroutineReturn

7.5 ’ Subroutine to perform a simulated D/A conversion, holding a voltage for approximately’ 1 secondD_to_A:

’ Define variables:’ digital_value: predefined byte variable indicating the relative voltage valuepin Var PORTA.0 ’ output pini Var BYTE ’ counter variable used in For loop

’ Maintain filtered PWM signal for approximately 1 secondFor i = 1 To 1000

’ Hold the pin high for the portion of a second based on the digital valueHigh pin’ Pause for the appropriate number of 1/255 (approximately) incrementsPause (4*digital_value)

’ Hold the pin low for the remainder of the secondLow pin’ Pause for the appropriate number of 1/255 (approximately) incrementsPause (4 * (255 − digital_value))

Next i

’ End of subroutineReturn

Solutions Manual


7.6 The polling loop continues to run after the alarm as been activated, and the if the door andwindows are closed the alarm will go off. An improved design would branch off to anothersection of code when the alarm is activated that would wait for some sort of alarm resetsignal before deactivating the alarm.

7.7 ’ PicBasic Pro program to perform the control functions of the security system example’ using interrupts

’ Define variables for I/O port pinsdoor_or_window Var PORTB.0 ’ signal Amotion Var PORTB.1 ’ signal Bc Var PORTB.2 ’ signal Cd Var PORTB.3 ’ signal Dalarm Var PORTA.0 ’ signal Y

’ Define constants for use in IF comparisonsOPEN Con 1 ’ to indicate that a door OR window is openDETECTED Con 1 ’ to indicate that motion is detected

’ Initialize interruptsOPTION_REG = $7F ’ enable PORTB pull-upsOn Interrupt Goto myintINTCON = $88 ’ enable interrupts on RB4 through RB7

’ Main loop waiting for sensors to change value (i.e., wait for interrupts)always:

Low alarm ’ keep the alarm low until a sensor changes valueGoto always ’ continue

’ Interrupt service routine that runs until sensors return to inactive statesDisable ’ disable interrupts during the interrupt service routinemyint:

While ((door_or_window == OPEN) Or (motion == DETECTED))If ((c == 0) And (d == 1)) Then ’ operating state 1 (occupants sleeping)

If (door_or_window == OPEN) ThenHigh alarm

ElseLow alarm

EndifElse

If ((c == 1) And (d == 0)) Then ’ operating state 2 (occupants away)If ((door_or_window == OPEN) Or (motion == DETECTED)) Then

High alarmElse

Low alarmEndif

Solutions Manual


Else ’ operating state 3 or NA (alarm disabled)Low alarm

EndifEndif

WendINTCON.1 = 0 ’ clear the interrupt flagResume ’ end of interrupt service routine

Enable ’ allow interrupts again

End

7.8

PIC16F84

RA2

RA3

RA4

MCLR

Vss

RB0

RB1

RB2

RB3

RA1

RA0

OSC1

OSC2

Vdd

RB7

RB6

RB5

RB4

1

2

3

4

5

6

7

8

9 10

11

12

13

14

15

16

17

18

5V 22 pF

22 pF

4 MHz1 k

5V

0.1 Fµ

20x2 LCD character display

Vss Vcc Vee RS R/W E DB4 DB5 DB6 DB7

1 2 3 4 5 6 11 12 13 145V

5V20kpot

5V1 k

5k potΩ

0.1 Fµ

Solutions Manual


’ Displays the scaled resistance value of a potentiometer on an LCD.

’ Define variables, pin assignments, and constantsvalue Var BYTE ’ scaled potentiometer valuepercentage Var WORD ’ displayed potentiometer percentage valuepot_pin Var PORTB.0 ’ pin to which the potentiometer and series capacitor are

’ attached (RB0)SCALE Con 200 ’ value for Pot statement scale factor

loop:Pot pot_pin, SCALE, value ’ read the potentiometer valuepercentage = (value * 100) / 255 ’ convert to percentage value’ Display the percentage value on LCD displayLcdout $FE, 1, "pot value = ", DEC percentage, " %"

Goto loop ’ continue to sample and display the potentiometer value

End

7.9 Use a combination of Figures 7.11 and 7.13 along with the associated code. Use byte arrayvariables called "digits_prev" and "digits" to store the digits of the entered numbers, where"digits_prev" contains the digits of the previous number entered and "digits" contains thedigits of the current number being entered. Add appropriate processing statements to thekeypad code to keep track of and store the digits of the current number. Here are excerptsof code needed in the implementation:

’ Initialize arrays of the two 5-digit numbersdigits_prev Var BYTE[5]digits Var BYTE[5]

’Initialize other variablesi Var BYTE ’ For loop counterdisplay Var BYTE[5] ’ number being displayed

’ When enter key is pressed, initialize the digits for the next number to be enteredFor i = 0 To 4

digits_prev[i] = digits[i]digits[i] = 10 ’ to indicate a missing digit (for numbers with < 5 digits)

Next i

’ Display the numbers on the LCD displayLcdout $FE, 1 ’ clear the displayFor i = 0 to 4

display[i] = digits_prev[i]Next iGosub display_digitsLcdout $FE, $C0 ’ go to the next line of the displayFor i = 0 to 4

display[i] = digits[i]Next i

Solutions Manual


Gosub display_digits

’ Subroutine to display the digits of a number stored in an array of five elementsdisplay_digits:For i = 0 To 4

If (display[i] != 10) ’ skip missing digitsLcdout DEC display[i]

EndifNext iReturn

7.10 Pin RA4 is an open-collector output. The two possible states are open-circuit and ground.The pull-up resistor results in a logic high signal (5V) at Vee when RA4 is in the open-circuit state. Vee is at logic low (0V) when RA4 is grounded.

7.11 See the figure in the solution of Questions 6.49 for the "hardware solution." A "softwaresolution" would look something like:

B Var PORTB.0 ’ pin attached to the bounce-free, NO buttonS Var PORTB.1 ’ pin attached to the SPDT switchstate Var BYTE ’ state of the switch when the button is pressedcount Var BYTE ’ variable used to track the

’ Reset the counter variablecount = 0

’ Main looploop:

’ Wait for the button to be pressed, and store the state of the switchWhile (B == 0) : Wendstate = S

’ Wait for the button to be released, and increment the count if appropriateWhile (B == 1) : WendIf (state = 1) Then

count = count + 1Endif

goto loop

7.12 See Design Example 7.1 for driving the display and see Question 7.4 for how to debouncethe inputs (or use the Button statement). Here are some code excerpts that might be usefulin the implementation:

my_count Var BYTE ’ current count (0 to 99)first Var BYTE ’ first digit (tens place)second Var BYTE ’ second digit (ones place)

Solutions Manual


inc Var PORTB.0 ’ pin attached to "increment" buttondec Var PORTB.1 ’ pin attached to "decrement" buttonreset Var PORTB.2 ’ pin attached to "reset" button

UP Con 0DOWN Con 1

’ Process the inputIf (inc == DOWN) Then

my_count = my_count + 1If (my_count > 99) Then

my_count = 0Endif

Endif

If (dec == DOWN) ThenIf (my_count > 0) Then

my_count = my_count − 1Endif

Endif

If (reset == DOWN) Thenmy_count = 0Endif

Endif

’ Determine the digitsfirst = my_count / 10second = my_count − (10*first)

7.13 Go to www.microchip.com, click on "PICmicro Microcontrollers," click on "FLASH"under "Memory Type," then select the appropriate model from the table.

7.14 See the comments and program flow in the code of TDE A.4 on pages 297-299.

7.15 See the comments and program flow in the main code of TDE B.2 on pages 300-301.

7.16 See the comments and program flow in the "move," "move_steps," and "step_motor"subroutines on page 302.

7.17 See the comments and program flow in the "speed" and "get_AD_value" subroutines onpages 302-304.

7.18 See the comments and program flow in the "position" and "get_encoder" subroutine codeson pages 308-310.

Solutions Manual


7.19 See the comments and program flow in the "slave" code on pages 313-314.

Solutions Manual


8.1 A digital computer or microprocessor uses digital or discrete data, that is, data that aresimply strings of 1's and 0's that have no time correspondence. We have to add some typeof time coding to make sense of the data. Therefore we have to design interfaces that willchange (convert) analog information into a discretized form that will be compatible with acomputer. Again, additional code must be included to provide the time references.

8.2 > 2*(15 kHz) = 30 kHz

8.3(a) 1 sample per minute would probably suffice, so fs = 1/60 Hz

(b)

(c)

fs 2 120MHz( )≥ 240MHz=

fs 2 20kHz( )≥ 40MHz=

Solutions Manual


8.4 Using MathCAD:

ω 0 1 ω max 2

f maxω max

2 π.( )T 2 π.

ω 0f s 2 f max. ∆ t 1

f s

f max 0.318= T 6.283= f s 0.637= ∆ t 1.571=

V t( ) sin t( ) sin 2 t.( )

t 0.33 0 ∆ t0.33

, 2 T... t 0.67 0 ∆ t0.67

, 2 T... t 2 0 ∆ t2

, 2 T... t 10 0 ∆ t10

, 2 T...

0 2 4 6 8 10 122

1.5

1

0.5

0

0.5

1

1.5

2

V t 0.33

V t 0.67

V t 2

V t 10

t 0.33 t 0.67, t 2, t 10,

Solutions Manual


8.5 Using MathCAD:

8.6

a 2 π⋅:= b 0.9 a⋅:=

t.5 0 0.5, 40..:= t1 0 1, 40..:= t10 0 10, 40..:=

F t( ) 2cosa b−

2t⋅

sina b+

2t⋅

⋅:=

0 10 20 30 401

0

1

F t.5( )

t.5

0 10 20 30 401

0

1

F t1( )

t1

0 10 20 30 401

0

1

F t10( )

t10

Q 5V 5V–( )–4096

----------------------------- 2.44mV= =

Solutions Manual


8.7

An 11 bit A/D converter would suffice since 211 = 2048. A 12-bit converter would be theminimal acceptable standard size available.

8.8 N = 28 = 256Q = 10V / N = 0.0391 VThe digital state number for a given voltage V between 0 and 10 is the truncated value ofV/Q. The code is the binary equivalent of the state number.(a) 0/Q=0 corresponding to state 0: 00000000

(b) 1/Q=25.6 corresponding to state 25: 00011001

(c) 5/Q=127.9 corresponding to state 127: 01111111

(d) 7.5/Q=191.8 corresponding to state 191: 10111111

8.9

8.10

but it is clear in the truth table that , , and , so

Also,

8.11

digital output = 10111

8.12 more memory will be required

bit scale fraction bit value cumulative voltage5 1/2 1 -5V + 1/2(10 V) = 0 V4 1/4 0 0 V3 1/8 1 1.25 V2 1/16 1 1.875 V1 1/32 1 2.1875 V

N 5V 5V–( )–0.005V

----------------------------- 2000= =

10sec 5000samplessec

------------------- 12 bits

sample------------------

1byte8bits------------- 75000bytes=

B0 G2G1G0 G2G1G0+=

G1G0 G1= G2G1 G2= G2G1 G1=

B0 G1G0 G2+=

B1 G2G1G0 G2G1G0+ G1G0 G1= = =

Solutions Manual


8.13 See www.microchip.comresolution: 12 bitsarchitecture: successive approximation

8.14 See www.national.com

8.15

8.16

8.17

Vout1

12---V1– Vo–

18---Vs–= = =

Vout2

12---V2– 2Vo–

14---Vs–= = =

Vout3

12---V3– 4Vo–

12---Vs–= = =

Solutions Manual


9.1

9.2

9.3

9.4

9.5 During the transition from 3 (0011) to 4 (0100), any of the following 8 codes could result:0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111.

9.6

NC5VNO

NO SPSTSPDT NO DPST

NC5VNO

digitalsignal

NO5V

reset

V1circuit 1

V2circuit 2

360°rev

----------- 1rev

1000lines------------------------- 1line

2pulses-------------------- 0.18°

pulse--------------=

Solutions Manual


9.7

but , so

9.8

9.9

A πD2

4----------=

dA 2πD4----dD=

dDD

------- νdLL

-------–=

dAA

-------2πD

4---- νDdL

L-------–

πD2

4----------

------------------------------------ 2νdLL

-------–= =

A πD2

4---------- 0.0491in2

= =

σ PA---- 10 190psi,= =

Esteel 30 106× psi=

ε σE--- 3.40 10 4–× 340 10 6–× 340µε= = = =

F ∆R R⁄ε

---------------- 0.01 120⁄

3.40 10 4–×--------------------------- 0.245= = =

E 200 109× Pa= D 0.010m= P 50 103N×=

FG 2.115= R 120Ω=

A πD2

4---------- 7.854 10 5– m2×= =

σ PA---- 0.637 109× Pa= = ε σ

E--- 0.00318= =

∆R εFGR 0.808Ω= = R1 R ∆R+ 120.808Ω= = R2 R3 R4 120Ω= = =

Vo VexR1

R1 R4+-------------------

R2R2 R3+-------------------–

0.00168Vex= =

Solutions Manual


9.10 Combination of the two figures in the chapter.

9.11

9.12 so

but , so

9.13 Using MathCAD:

The sensitivity is approximately 0.055 mV / deg C.

IVex2R---------≈ 10V

2 350Ω( )---------------------- 14.2mA= =

VVex

2---------≈ 5V=

P IV 71.4mW= =

2R' 0.001RG< R' 0.0012

-------------RG<

R' L 0.050Ωm----

=

L 0.0012 0.050( )---------------------120m< 1.2m=

Type J Coefficents:

c0 0.0488683 c3 1.15692107.

c1 19873.1 c4 2.64918108.

c2 218615 c5 2.01844109.

Temperature/Voltage Relationship:

T V( )

0

5

i

ci Vi.

=

Approximate Sensitivity:

T 0( ) 0.049= T 0.03( ) 546.224=

DVDT 0.03 0T 0.03( ) T 0( )

DVDT 5.492 10 5.=

Solutions Manual


9.14 Using MathCAD:

The sensitivity is approximately 0.050 mV / deg C.

9.15 Using MathCAD:

Type J Coefficents:

c0 0.100861 c3 7.80256107. c6 2.661921013.

c1 25727.9 c4 9.24749109. c7 3.940781014.

c2 767346 c5 6.976881011.


T V( )

0

7

i

ci Vi.

=

Approximate Sensitivity:

T 0( ) 0.101= T 0.015( ) 302.478= T 0.010( ) 213.286=

DVDT 0.015 0T 0.015( ) T 0( )

DVDT 4.961 10 5.=

Type J Coefficents:

c0 0.0488683 c3 1.15692107.

c1 19873.1 c4 2.64918108.

c2 218615 c5 2.01844109.


T V( )

0

5

i

ci Vi.

=

V 0

V 200 root T V( ) 200 V,( ) V 200 0.011= T V 200 200=

Solutions Manual


So

9.16

9.17

V200 0⁄ 11mV=

VT 100⁄ 30mV=

100 a0 a1V100 0⁄ a2V100 0⁄2 a3V100 0⁄

3 a4V100 0⁄4 a5V100 0⁄

5+ + + + +=

V100 0⁄ 5.26mV=

VT 0⁄ VT 100⁄ V100 0⁄+ 35.26mV= =

T VT 0⁄( ) 636.6°C=

VT 11⁄ 30mV=

11 a0 a1V11 0⁄ a2V11 0⁄2 a3V11 0⁄

3 a4V11 0⁄4 a5V11 0⁄

5+ + + + +=

V11 0⁄ 0.559mV=

VT 0⁄ VT 11⁄ V11 0⁄+ 30.559mV= =

T VT 0⁄( ) 556°C=

Solutions Manual


9.18 Using MathCAD:

So the change in voltage would be 5.85 mV.

9.19 From Equation 9.62:

so the transfer function is:

Substituting s=jω gives:

Type J Coefficents:

c0 0.0488683 c3 1.15692107.

c1 19873.1 c4 2.64918108.

c2 218615 c5 2.01844109.


T V( )

0

5

i

ci Vi.

=

V 0

V 120 root T V( ) 120 V,( ) V 120 6.356 10 3.= T V 120 120=

V 10 root T V( ) 10 V,( ) V 10 5.084 10 4.= T V 10 10=

∆ V V120 V 10 ∆ V 5.848 10 3.=

1ωn

2------s2 2ζ

ωn------s 1+ +

Xr s( ) 1ωn

2------s2

–

Xi s( )=

G s( )Xr s( )Xi s( )-------------

1ωn

2------s2

–

1ωn

2------s2 2ζ

ωn------s 1+ +

-------------------------------------= =

G jω( )

ωωn------ 2

ωωn------ –

22ζ ω

ωn------j 1+ +

-------------------------------------------------=

Solutions Manual


Therefore, the amplitude magnitude is:

9.20

(a)

(b)

(c) Since is assumed to be 1 for the accelerometer device for all ω’s,

XrXi------ G jω( )

ωωn------ 2

1 ωωn------ –

2

2

2ζ ωωn------

2+

---------------------------------------------------------------= =

ωnkm----

5000Nm----

0.05kg----------------- 316.2rad

sec-------= = =

ωωn------ 100

316.2------------- 0.316= =

ωωn------ 2

0.1=

ζ b2 km--------------- 30

2 5000 0.05( )----------------------------------- 0.949= = =

x··in actual Xinω2 5mm 100rad

sec-------

25 104× mm

sec2---------- 50 m

sec2----------= = = =

x··in actual 50 msec2----------

9.81m sec⁄g

---------------- ÷ 5.1g= =

Ha ω( ) 1

1 ωωn------ 2

–2

4ζ2 ωωn------ 2

+

------------------------------------------------------------------ 1.08= =

Xr1ωn

2------Ha ω( ) Xinω

2( ) 1

316.2radsec-------

2----------------------------- 1.08( ) 50 m

sec2----------

= =

Xr 5.4 10 4–× m 0.54mm= =

Ha ω( )

x··in measured ωn2Xr 1( ) 54 m

sec2----------= =

Solutions Manual


(d)

9.21

Therefore, the steady state output displacement response is:

φ tan 1–2ζ ω

ωn------

1 ωωn------ 2

–-----------------------

– tan 1– 2 0.949( ) 0.316( )1 0.1–

---------------------------------------- – 33.7°– 0.588rad–= = = =

xr t( ) Xr ωt φ+( )sin 0.54 100t 0.588–( )sin mm= =

m 1kg= k 2Nm----= b 2Ns

m-------= ω 1.25rad

s--------= Xi 0.010m=

ωnkm---- 1.414rad

s--------= = ωr

ωωn------ 0.884= = ς b

2 km--------------- 0.707= =

XrXi------

ωr2

1 ωr2

–( )2

4ς2ωr2

+

------------------------------------------------ 0.616= =

φ2ςωr

1 ωr2

–--------------- 1–

tan– 80.1– ° 1.398– rad= = =

Xr XiXrXi------ 0.00616m= =

xo t( ) xi t( ) xr t( )+ Xi ωt( )sin Xr ωt φ+( )sin+= =

xo t( ) 10 1.25t( )sin 6.16 1.25t 1.46–( )sin+( )mm=

Solutions Manual


10.1 Use a combination of a power transistor switch circuit and a diode clamp.

10.2 Electric motors and solenoids create changing magnetic fields which induce voltages innearby unshielded circuits.

10.3 At the maximum no-load speed, the motor torque is 0 so

10.4 See the documentation on the LMD18200.

10.5

10.6 Search the Internet.

ωmaxVinke

-------- 10V12V( ) 1000rpm( )⁄

----------------------------------------------- 833rpm= = =

IsVinR

-------- 6.67A= =

Ts ktVinR

-------- 0.8Nm= =

RESET

STEP

CW/CCW

B0

B1

φ1

φ2

φ3

φ4

Solutions Manual


10.7

Therefore, the minimum required number of steps per revolution is

To achieve the maximum speed,

Therefore, the required step rate is

10.8(a) servo motor

(b) ac induction motor

(c) series dc

(d) ac induction

(e) servo motor

(f) series dc

(g) stepper motor

(h) synchronous motor

(i) dc motor

(j) ac induction motor

(k) ac motor

(l) ac motor

10.9 The load speed is related to the motor speed with:

∆θd∆xr

------- 0.0010.05------------- 0.02rad= = =

∆θm 3∆θd 0.06rad 3.44°= = =

N 360°∆θm----------- 105= =

ωdvmax

r-----------

0.10cms

-------

0.05cm------------------ 2rad

s--------= = =

ωm 3ωd 6rads

--------= =

ωm∆θm---------- 100steps

s-------------=

Solutions Manual


where the gear box reduction ratio is:

(a)

The maximum acceleration occurs at start-up where:

(b) At steady state,

and

Combining these equations gives:

so

and

(c) Designating the torque on the motor (rotor) side of the gear box as Tgm and the torqueon the load side of the gear box as Tgl, the equations of motion for the motor rotor andload can be written as:

and

where the gear box torques are related by:

and the load and motor angular accelerations are related by:

Therefore, the motor equation of motion can be written as:

ωl rgωm=

rgMN-----=

J Jr Jlrg2

+=

αmaxTsJ

-----Ts

Jr Jlrg2

+---------------------= =

Tm rgTl rg kωl( )= = Tm Ts Tsωmωmmax

-------------–=

ωl rgωm=

krg2ωm Ts Ts

ωmωmmax

-------------–=

ωmTs

krg2 Ts

ωmmax

-------------+

------------------------------= ωl rgωmTs

krgTs

rgωmmax

------------------+

--------------------------------= =

Tm Tgm– Jrαm= Tgl Tl– Jlαl=

Tgm rgTgl=

αl rgαm=

Solutions Manual


This can be written as:

where Jeff is the total effective inertia seen by the motor, given by:

Designating the motor speed ωm as ω, the motor and load torques are given by:

and

and the motor equation can now be written as:

which can be written as:

where:

The particular (steady state) solution to the equation of motion is:

and the total general solution for the initial condition ω(0) = 0 is:

Therefore, when the motor is at 95% of its steady state speed,

so the time required for to reach this speed is:

Tm rg Tl Jlrgαm+( )– Jrαm=

Tm rgTl– Jeffαm=

Jeff rg2Jl Jr+=

Tm Ts 1 ωωmax------------–

= Tl kωl krgω= =

Ts 1 ωωmax------------–

rg krgω( )– Jeffdωdt-------=

Jeffdωdt------- bω+ Ts=

bTs

ωmax------------ krg

2+=

ωssTsb-----=

ω t( ) ωss 1 eb

Jeff-------– t

–

=

0.95ωss ωss 1 eb

Jeff-------– t

–

=

tJeffb

--------– 1 0.95–( )ln 2.996rg2Jl Jr+( )

Tsωmax------------ krg

2+

---------------------------= =

Solutions Manual


10.10

10.11

10.12 Search manufacturer catalogs or the Internet.

10.13 Required components: pressure regulator (e.g., 1500 psi), pneumatic cylinder (double-acting or spring return), valve. The required cylinder area is:

Therefore, the required cylinder diameter is:

A πd2

4--------- 0.785in2

= =

F PA 1000psi( ) 0.785in2( ) 785lb= = =

A πd2

4--------- π 10mm( )2

4--------------------------- 78.5mm2

= = =

P FA---- 2000N

78.5mm2----------------------- 25.5MPa= = =

A Fp--- 100

1500------------in2 0.0667in2

= = =

D 4Aπ

------- 0.29in= =

Introduction to Mechatronics and Measurement Systems 3rd Ed Solutions Manual Mmzzhh

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