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This is the solutinal manual for Mechatronics 3rd edition
Typology: Study notes
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David G. Alciatore and Michael B. Histand
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:
www.engr.colostate.edu/mechatronics
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 David.Alciatore@colostate.edu. We will post corrections for reported errors on our Web site.
Thank you for choosing our book. We hope it helps you provide your students with an enjoyable and fruitful learning experience in the cross-disciplinary subject of mechatronics.
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,
From KVL,
so
Therefore,
so or
and
From KCL,
I (^) s = I 1 + I 2 +I (^3)
V (^) s R (^) eq
V (^) s R (^1)
V (^) s R (^2)
V (^) s R (^3)
R (^) eq
= + +------ R (^) eq
I (^) s R (^) eq
I (^) s R 2 R 3 + R 1 R 3 +R 1 R (^2) R 1 R 2 R (^3)
= ----------------------------------------------------^ I (^) s
R 1 →∞
lim
I C (^) eqdV dt
dV (^1) dt
dV (^2) dt
dV dt
dV (^1) dt
dV (^2) dt
C (^) eq
C (^) eq
= +------ C (^) eq
dV (^1) dt
---------- (^) C 1 dV dt
dV (^2) dt
---------- (^) C 2 dV dt
I I 1 + I 2 C 1 dV dt
------- (^) C 2 dV dt
Since
From KVL,
Since
From KCL, so
Therefore, so or
2.14 , regardless of the resistance value.
2.15 From Voltage Division,
2.16 Combining R 2 and R 3 in parallel,
and combining this with R 1 in series,
(a) Using Ohm’s Law,
I C (^) eqdV dt
C (^) eq =C 1 +C (^2)
V V 1 + V 2 L 1 dI dt
----- (^) L 2 dI dt
V L (^) eqdI dt
L (^) eq = L 1 +L (^2)
V LdI dt
dI (^1) dt
dI (^2) dt
I = I 1 +I 2 dI dt
dI (^1) dt
dI (^2) dt
V (^) o =1V
V (^) o^40 10 + 40
= = ------------ = 1.2k
R 123 = R 1 + R 23 = 2.2k
V (^) in R (^123)
2.2k
= = ---------- = 2.27mA
2.22 It will depend on your instrumentation, but the oscilloscope typically has an input impedance of 1 MΩ.
2.23 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.
(a) ,
(b) ,
When the impedance of the load is lower (10k vs. 500k), the accuracy is not as good.
= -------------------i 1 = 9.09V
= ------------------- = 0.5kΩ
= ------------------- = –0.5mA
R 45 = R 4 + R 5 = 9kΩ
= --------------------- = 2.25kΩ
R 2345 = R 2 + R 345 = 4.25kΩ
R (^) eq
= -------------------------- = 0.81kΩ
V (^) out
= ---------------------V (^) in
R 23 = 9.90kΩ V (^) out = 0.995V (^) in
R 23 = 333kΩ V (^) out = 1.00V (^) in
(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.
Combining R 2 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
(a) In steady state dc, C is open circuit and L is short circuit. So
V (^) out
=^ -------------------V (^) in
V (^) out^10
= -------------V (^) in = 0.995V (^) in
V (^) out^500
= ----------------V (^) in = 0.9999V (^) in
V (^) in = 5 〈 45 °〉
= ------------------------------------- = 1860.52 〈 – 60.25°〉 = 923.22 – 1615.30j
=^ ---------------------------V (^) in
R 1 + Z (^) R 2 LC = 1000 + 923.22 – 1615.30j =2511.57 〈 – 40.02°〉
= --------------------------------------------5 〈 45 °〉 = 3.70 〈 24.8°〉 =3.70 〈 0.433rad〉
V (^) C ( )t = 3.70 cos(3000t +0.433)V
V (^) C = V (^) s = 10V V (^) R 1 = 0V V (^) R 2 = V (^) s = 10V
(d) ,
,
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 0 rad sec
= -------- f ω 2 π
= ------ = 0Hz
A (^) pp = 2A = 0 dc (^) offset = sin ( π)+ cos( π) = – 1
V (^) rms^2 R
V (^) rms
V (^) pp 2
V (^) rms^2 R
V (^) m = 2V (^) rms =169.7V
V (^) rms = 120V V (^) m = 2V (^) rms = 169.7V
V t( ) = V (^) m sin ( 2 πf +φ) =169.7 sin( 120 πt +φ)
10mA ≤ I ≤100mA
10mA 3V R
≤ -------^ ≤100mA
100mA
10mA
2 R
2 30 Ω
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:
2 300 Ω
= ------ =10mA
10 = ( I 1 – 10m)2k +( I 1 – 10m–I 2 )3k
10 – 5 = ( I 1 – 10m–I 2 )3k – I 2 4k
( 5k)I 1 – ( 3k)I 2 = 60
( 3k)I 1 – ( 7k)I 2 = 35
I 1 = 12.12mA I 2 =0.1923mA
V (^) out = I 2 R 4 – V 2 =–4.23V
P 1 = I 1 V 1 = 121mW P 2 = I 2 V 2 = 0.962mW P 3 = – I 2 V 3 = –1.92mW
This is a sin wave with half the amplitude of the input with a period of 1s.
2.42 RL = Ri = 8Ω for maximum power
2.43 The BNC cable is far more effective in shielding the input signals from electromagnetic interference since no loops are formed.
I (^) rms
I (^) m^2 T
I (^) m 2
= ------------------- = 5kΩ
V (^) o
---------------------V (^) i^1 2
= = ---^ sin( 2 πt)
N (^) p N (^) s
V (^) p V (^) s
3.1 For V (^) i > 0, Vo = 0.
For V (^) i < 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.
(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 V (^) i > 0, Vo = Vi.
For V (^) i < 0, Vo = -V (^) i. The resulting waveform is a full wave rectified sin wave where there are two positive "humps" for each period of V (^) i (0.5 sec).
(a) For V (^) i > 0.5V, Vo = 0.5V. For V (^) i < 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 V (^) i < 0.5V, Vo = 0.5V. For V (^) i > 0.5V, Vo = Vi. The resulting waveform is the original sine wave with the bottom halves of the negative "humps" (below 0.5 V) clipped off.
3.5 Using superposition, Ohm’s Law, and current division,
I (^1) left 1V 2R R 2
I (^2) left^1 2
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 reverse biased state is when corresponding to. V (^) out remains at −5V when Vin decreases below −10V. It is not possible for both diodes to be reverse biased at 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 .
(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)
V (^) out = V (^) H V (^) out =V (^) L V (^) L < V (^) out <V (^) H
V (^) out
R (^) i +R (^) L
------------------V (^) in^1 2
= = ---V (^) in
V (^) out = V (^) H = 5V V (^) in = 10V
V (^) out = V (^) L = – 5 V V (^) in = – 10 V
I (^) max
50mA
50mA
V (^) in ( saturation) = V (^) E + V (^) BE = 2.8V + 0.7V = 3.5V
(a) For the LED to be ON, the transistor must be in saturation and
When the LED is off, I (^) B = 0 and
So for the LED to be ON,
(b) When the transistor is fully saturated, and and
Assume
If I 1 is the current through the horizontal 1k resistor and I 2 is the current through the right 1k resistor, then
and
3.13 When the transistor is in full saturation,
and
and
In the collector-to-emitter circuit,
giving
= ---V (^) in
V (^) in > 2V (^) B = 3.4V
= = ------------- =11.5mA
= ---------I (^) C = 0.115mA
1k
= = ------- + 0.115mA =1.815mA
V (^) in = V (^) B + (1k )I 1 =3.52V
I (^) out I (^) B + I (^) C
V (^) out = V (^) s – I (^) C R (^) C– V (^) CE =I (^) out R (^) out
5V – I (^) C ( 1k)– 0.2V = 1.01I (^) C ( 1k)
(a) cutoff
(b) ohmic
(c) saturation
(d) cutoff
(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.
So
But , so
But sine of any multiple of π is 0, so A (^) n = 0
F t( ) = 5 sin( 2 πt)
f (^0)
ω 0 2 π
= ------ = 1Hz
F t( ) = 2V (^) rms sin ( 2 πft) = 169.7 sin( 120 πt)
A (^) n^2 T
--- F t( ) cos (n ω 0 t)dt 0
T
∫
--- cos ( nω 0 t)dt 0
T -- 2 -
∫ cos^ (n^ ω^0 t)dt T 2 ---
T = = – ∫
A (^) n^2 nω 0 T
------------- [ sin( nω 0 t)] 0
T 2 --- nω [ sin( 0 t)]T -- 2 -
ω 0 2 π T
A (^) n^1 nπ
=^ ------^ ( sin ( nπ)– sin( 2nπ)+sin( nπ))
