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Electrical; Computers; Ethics
Refresher Notes
Spring 2013
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1FE Electrical General Notes
Electrical Circuits, Computers,
and Ethics
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2Introduction I. Electromagnetics
Electrostatics/Magnetostatics II. DC Circuits
KCL, KVL Series/Parallel Equivalent Circuits Node and Loop Analysis Thevenin/Norton Theorems Transient Response Sampling Theorem Instrumentation
III. AC Circuits Average and RMS Values of Waveforms Impedance and Admittance Complex Power Resonance Analog-to-Digital Conversion Transformers Rotating Machines
IV. Computers Terminology Spreadsheets Structured Programming
V. Ethics and Business Practices
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3I. Electromagnetics
Electrostatics/Magnetostatics
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4Introduction
Equivalent Units Amperes A = W/V = N/Tm Charge C = J/V = Nm/V Capacitance F = C/V = C/J = C/Nm = S/ = J/V Inductance H = Vs/A = Tm/A Energy J = Nm = VC = C/F = (Kgm)/s Force N = J/m = VC/m = (Kgm)/s Flux Density T = Ns/Cm = N/Am = Wb/m Voltage V = W/a = C/F = J/C Power W = J/s = VA = V/ Flux Wb = Vs = HA = Tm
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5Electrostatics
Charges
E F
+Q E = F/Q
Coulombs Force Law
+Q1 +Q2 F1 = -Q1E2 F2 = Q2E1
r
E1 E2
Page 193 from Supplied Reference Handbook
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6Electrostatics
Permittivity Note: On the FE exam, assume the permittivity is 0 = 8.85 x 10 F/m unless another value is provided.
ar
QE 24
Law sCoulomb'
+Q
E1 F1 F2 E2
Q1 +Q2 F1 = (-Q1)(-E2)
F1 = Q1E2
F2 = (+Q2)(-E1)
F2 = -Q2E1
Page 193 from Supplied Reference Handbook
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7Electrostatics
Electric Field Intensity
+Q Es
s
-Q Es
s
EL
L = Q/d = C/m
s = Q/A = C/m
(V/m or N/C)
Page 193 from Supplied Reference Handbook
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8Electrostatics
Example A point charge of 0.001 C is placed 10 m from a sheet charge of 0.001 C/m, and a 10 m diameter sphere of charge 0.001 C is placed half-way in between on a straight line, all in a vacuum. What is the Qp force on the point charge? F = Fsheet + Fsphere = Qpoint (Esheet + Esphere)
F
5m 10m
QP
QS
S
ESH Esph
(-)
+ +
S = - 0.001 C/m
Qs = 0.001 C
QP = 0.001 C
NmCm
C
mF
CF
rQQ
rQ
QF spheresheetpospheresheetpo
42
2
12-
2int
2int
10 x 61.5)5(4
001.02
001.0
10 x 85.8
001.0
4242
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9Electrostatics
Electric Flux Gauss Law
R = radius of the charge distribution r = radius of the sphere
dS Surface of Integration
Asphere = 4r = d
Page 193 from Supplied Reference Handbook
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10
Electrostatics
The Concept of Work Work (W) done by an external agent in moving a charge Q in an electric field from a point p1 to a point p2:
For a uniform field, the work done by moving a charge Q a distance d parallel to the uniform field:
VQWQEddFW
Note that equation, W = - QV, is always true for all fields. (V may not be easy to compute.)
E Q d V
+
-
ES +Q +W
-W W = 0 +Q
+QS +Q
(Nm or Joules)
Page 193 from Supplied Reference Handbook
(Joules) 21
p
pldEQW
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11
Electrostatics
Voltage Gradient points in the direction where the E-field increases or decreases most with voltage potential.
V = -W/Q
E Q
V
(V/m or N/C)
E
d
+ _
V
Page 193 from Supplied Reference Handbook
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12
Electrostatics
Example A source at zero potential emits electrons (mass: ) at negligible velocity. An open grid at 18 V is located 0.003 m from the source. An electron has a charge of C. At what velocity will the electrons pass through the grid?
Kg 10 x 11.9 -31
-1910 x 6022.1
18V
0.003m
QP
0V
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Electrostatics
Example/Contd.
2
21 T Handbook, FE of 59 Pg.in Equation Energy Kinetic mv
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14
Electrostatics
Current Electric current i(t) through a surface is defined as the rate of charge transport through that surface or i(t) = dq(t)/dt, which is a function of time t, since q(t) denotes instantaneous charge. A constant current i(t) is written as I, and the vector current density in amperes/m is defined as J.
J (Amps/m)
(C/s) or Amps
(Current direction matters)
( = Amps/m)
Page 193 from Supplied Reference Handbook
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15
Magnetostatics
Magnetism
(A/m)
(Wb/m or Teslas)
Right Hand Rule Generators
Left Hand Rule Motors
Page 193 from Supplied Reference Handbook
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16
Magnetostatics
Example
1 m (L) 30
B
|B|Sin
I
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17
Magnetostatics
Example What is the magnitude of the force on an electron moving at 0.1c in a uniform magnetic field of ?
T 10 x 0.5 -5
N 10 x 4.2)10 x m/s)(5.0 10 x C)(0.3 10 x 6022.1(
16-
5-819-
FTF
vBQF
B
v (-) electron velocity
magnetic field flux density
1 T = 1 Ns/Cm
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18
Magnetostatics
Induced Voltage v = the induced voltage = the average flux (Weber) enclosed by each turn N = number of loops L = distance AdBA
s
I
B
L ds/dt
ds/dt = velocity
s = distance that conductor travels.
B
Area
Page 193 from Supplied Reference Handbook
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19
Magnetostatics
Voltage The potential difference V between two points is the work per unit charge required to move the charge between points. For two parallel plates with potential difference V, separated by distance d, the strength of the E field between the plates is E = V/d directed from the + plate to the - plate.
+
-
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20
DC Circuits
Circuit Analysis
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21
DC Circuits
Kirchhoffs Laws (KCL, KVL)
The basic tools for solving DC circuit problems are Kirchhoffs laws: the current law and the voltage law, Ohms law, and the power relationship. The following configurations are typical:
RVRIVIP
IVRR
VIIRV
VVKVL
IIKCL
DROPSRISES
OUTIN
22
PowerResistive
LawsOhm'
LawssKirchhoff'
Voltage Drop (- V)
+
_ Voltage Drop (- V)
Voltage Rise (+V)
Voltage Rise (+V)
I1 I2
I3
I1 = I2 + I3
Page 194 from Supplied Reference Handbook
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22
DC Circuits KCL This law is also called Kirchhoff's first rule or Kirchhoff's nodal rule.
The principle of conservation of electric charge implies that at any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node or the algebraic sum of currents in a network of conductors meeting at a point is zero. (Assuming that current entering the junction is taken as positive and current leaving the junction is taken as negative). Recalling that current is a signed (positive or negative) quantity reflecting direction towards or away from a node, this principle can be stated as: n is the total number of branches with currents flowing towards or away from the node. The current entering any junction is equal to the current leaving that junction: i1 + i4 = i2 + i3 The law is based on the conservation of charge whereby the charge (measured in coulombs) is the product of the current (in amperes) and the time (which is measured in seconds).
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DC Circuits (KCL)
Example Consider the following circuit with the following parameters:
V1 = 15V V2 = 7V R1 = 20 R2 = 5 R3 = 10
Find current through R3 using Kirchhoff's Current Law.
The circuit above shows voltages at nodes a, b, c and d. We use node a as common node (ground if you like). Thus, Va = 0V. From Node b we get:
Vb = V1 = 15V From Node d we get:
Vd = V2 = 7V
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DC Circuits (KCL)
Example/Contd. Thus, we must solve Vc, in order to complete voltage definitions at all nodes. Vc will be found by applying KCL at node c and solving resulting equations follows: i1 = i2 + i3 Substituting values we get
VVV CC 857.1 Thus, A 65.0207
0 321
acacdbc V
RVV
RVV
RVV
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25
DC Circuits (KCL)
Example/Contd. Thus, now we can calculate current through R3 as follows:
IR3
AI
I
RVI
R
R
CR
186.010857.1
3
3
33
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26
DC Circuits
KVL This law is also called Kirchhoff's second law, or Kirchhoff's loop (or mesh) rule.
The principle of conservation of energy implies that the directed sum of the electrical potential
differences (voltage) around any closed circuit is zero.
Here, n is the total number of voltages measured. The voltages may also be complex:
The sum of all the voltages around the loop is equal to zero. v4 - v1 - v2 - v3 = 0
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DC Circuits (KVL)
Example
Consider the following circuit with the following parameters: V1 = 15V V2 = 7V R1 = 20 R2 = 5 R3 = 10 Find current through R3 using Kirchhoff's Voltage Law. We can see that there are two closed paths (loops) where we can apply KVL in, Loop 1 and 2 as shown in the circuit above.
From Loop 1 we get:
V1 VR3 VR1 = 0 From Loop 2 we get:
V2 VR2 VR3 = 0
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DC Circuits (KVL)
Example/Contd. The above results can further be simplified as follows: V1 (I1 I2) * R3 I1 * R1 = 0 (1) and
V2 I2 * R2 (I2 I1) * R3 = 0 (2)
By equating above (1) and (2) we can eliminate I2 and hence get the following: (3) We end up with the above three equations and now substitute the values given in the above equations and solve the variables. Only after we have arrived at a simplified equation then that we can substitute in values of Resistors, Voltages and Current.
32
3122 RR
RIVI
233213
323211 ))((
)(RRRRR
RVRRVI
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DC Circuits (KVL)
Example/Contd. The positive sign for I2 tells us that current I2 flows in the same direction to our initial assumed direction. Thus, now we can calculate current through R3 as follows: The negative sign for IR3 only tells us that current IR3 flows in the same direction as I2.
AVI
xAVI
AVI
VVI
03.115
4.15)510(
)1084.0(7
84.0350295
)10()105(*)2010()10()7()105()15(
2
2
1
21
IR3
AI
AAI
III
R
R
R
19.0
03.184.0
3
3
213
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30
DC Circuits
Series/Parallel Equivalent Circuits
Resistors in Series Resistors in Parallel
nRRRRRIf
RRIf
eq
2RR
321
eq21
Page 194 from Supplied Reference Handbook
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31
DC Circuits
Example What is the resistance, as seen from the battery, of the parallel and series resistors in this circuit? (A) 4 (B) 5 (C) 6 (D) 8 Two 8 resistors in parallel are 4 . Two 10 resistors in parallel are 5 . Thus, we have two 10 resistors in parallel are 5 . Therefore, the correct answer is (B).
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DC Circuits
Example
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33
DC Circuits
Node Voltage Circuit Analysis: 1. Convert all current sources to voltage sources. 2. Choose one node as reference (usually ground). 3. Identify unknown voltages at other nodes compared to reference. 4. Write Kirchhoffs current equation for all unknown nodes except reference node. 5. Write all currents in terms of voltage drops. 6. Write all voltage drops in terms of the node voltages.
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DC Circuits
Example Find the voltage potential at point A and the current i1.
A
VVVVi
VV
VVVVViii
A
A
AAA
35.72
3.35502
503.35
100
420
250
1
321
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35
DC Circuits
Loop Current Circuit Analysis
1. Select one less than the total number of loops. 2. Write Kirchhoffs voltage equation for each loop. 3. Use the simultaneous equations to solve for the current you want.
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DC Circuits
Example Find the current through the 0.5 resistor. The voltage sources around the left loop are equal to the voltage drops across the resistances. 20 V 19 V = 0.25 i1 + 0.4 (i1 i2) The same is true for the right loop. 19 V = 0.4 (i2 i1) + 0.5 i2 Solvetwo equations and two unknowns. 0.65 i1 0.4 i2 = 1 V 0.4 i1 + 0.9 i2 = 19 V i1 = 20 A i2 = 30 A The current through the 0.5 resistor is 30 A.
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DC Circuits
Voltage Divider The voltage across a resistor R in a loop with equivalent resistance Req with a voltage source V is In the general case, the voltage on impedance Zi in a loop with equivalent impedance Zeq with a Voltage source v is NOTE: Each symbol is a complex number in the general case.
VRRVeq
R
VZZ
Veq
ii
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DC Circuits
Example What is the voltage across the 6 resistor? (A) 5 V (B) 6 V (C) 8 V (D) 10 V Two 8 resistors in parallel equal 4 . The voltage across the 6 resistor is 6V. Therefore, the answer is (B).
VV 646
610
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39
DC Circuits
Current Divider The current through a resistor R in parallel with another resistance Rparallel and a current into the node of I is: (Resistance R does not appear explicitly. The denominator is the sum of the resistances in parallel.) In the general case, the current through impedance Z connected to a node in parallel with equivalent Impedance Zparallel with a current i into the node is: (The denominator is the sum of the impedances in parallel.) NOTE: Each symbol is a complex number in the general case. Procedure: 1. Identify the component you want the current through. 2. Simplify the circuit. 3. Determine the current into the node that is connected to the component of interest. 4. Allocate current in proportion to the reciprocal of resistance.
IRR
RIPARALLEL
PARALLELR )(
IZZ
ZIPARALLEL
PARALLELZ )(
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40
DC Circuits
Example What is the current through the 6 resistor? (A) 1/10 A (B) 1/3 A (C) 1/2 A (D) 1 A i = i1 + i2 Simplify the circuit. 3 in parallel with 6 = 2 2 in series with 4 = 6 Rparallel = 3 and Req = 3 + 6 = 9 Therefore, the answer is (B).
AVi 166
AAi31
63312
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41
DC Circuits
Superposition Theorem The net current/voltage is the sum of the current/voltage caused by each current/voltage source. Procedure: 1. Short all voltage sources, and open all current sources, then turn on only one source at a time. 2. Simplify the circuit to get the current/voltage of interest. 3. Repeat until all sources have been used. 4. Add the results for the answer.
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42
DC Circuits
Example Using the Superposition Theorem, determine the current through the center leg of the circuit. Short the 20 V source. Short the 50 V source. Ieq = 25 A + 5 A = 30 A
AVI 25250
AVI 5420
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43
DC Circuits
Thevenin Equivalent Circuit
Page 194 from Supplied Reference Handbook
NTHSC
OCeq RRI
VR
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44
DC Circuits
Example Find the Thevenin equivalent voltage and resistance of the circuit as seen by the 10 resistor. The Thevenin resistance is the same as the Norton resistance in the previous example, which is 1.33 with the 10 resistor open-circuited, apply the Kirchhoff voltage law around the loop and find VTH = 40V.
2 4
33.1424 x 2
THR
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45
DC Circuits
Example/Contd. ( 50 V 20 V) = I(2 + 4 ) I = 5 A V = 50 V I(2 ) = 50 V (5 A)(2 ) = 40 V
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46
DC Circuits
Norton Equivalent
Page 194 from Supplied Reference Handbook
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47
DC Circuits
Example Find the Norton equivalent current and resistance of the circuit as seen by the 10 resistor. With the 10 resistor open circuited, and the voltage sources shorted, the circuit is 4.0 and 2.0 in parallel. With the 10 resistor shorted, the circuit looks just like the previous superposition example. IN = 30A
33.1424 x 2
NR 4 2
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48
DC Circuits
Maximum Power Transfer DC Circuit Maximum power is transferred when the load resistance is the same as the Thevenin Equivalent circuit resistance, Rs = RL AC Circuit Maximum power is transferred when the load impedance is the complex conjugate of the Thevenin equivalent circuit impedance, Zs = x + jy and ZL = x jy = Zs*
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49
DC Circuits
Example A load resistance is to be connected to the circuit below. (a) What value of load resistance will provide maximum power transfer to the load? (b) what is the value of that maximum power? (a) The maximum power theorem tells us that the load resistance should be, RL = RTH = 0.8
Ohms. (b) We can find the maximum power by finding the current through the load. Essentially the
11.2V source is across two 0.8 ohm resistors in series. Thus, the current is, I = 11.2V/1.6 ohm = 7.0 A. Thus, the power is WohmsAP 4.78)6.1()7( 2
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50
DC Circuits
Capacitors
Q = CV
C = Q/V
V = Q/C
Pages 194, 195 from Supplied Reference Handbook
(Joules) 222
Cvenergy
v(t)][varying )()(22
C CCC
CC
vqC
q
tCvtq
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51
DC Circuits
Example
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52
DC Circuits
Example
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53
DC Circuits
Inductors
2
2LLiEnergy
L = Inductance
N = # of loops
= Flux
I = Current
(initial current) (current value from 0 to t)
21
21
:parallelin inductors 2For
LLLLLeq
Page 195 from Supplied Reference Handbook
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54
DC Circuits
Example
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55
DC Circuits
Transient Response
RC Transients
Constant TimeCircuit RC
(t = t = t2 t1)
Capacitor Voltage (V is battery voltage if present)
seconds) in 36.8V 100V 100V, V (If
VR
Page 196 from Supplied Reference Handbook
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56
DC Circuits
Example
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57
DC Circuits
Example/Contd.
i (t) actual
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58
DC Circuits
Example/Contd.
stssts
sst
e
V
VeVetvc
sst
sst
sst
C
73.430.2657.3
10ln1
658.35ln
10lnln58.35ln
10ln34.58Veln
equation theof sidesboth of logarithm natural theTake
101058.34)( )(
16
1s6s-t
16
16
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59
DC Circuits
RL Transients
Constant TimeCircuit RL
VR(t)
VL(t)
Page 196 from Supplied Reference Handbook
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60
DC Circuits
Example Find the voltage at point A at the instant the switch is closed. The switch has been open for a long time, and there is no initial current in the inductor.
(Inductor behaves like a short at t >> 0)
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61
DC Circuits
Example Consider the following circuit with R1= 3 , R2= 4 , R3= 2 , C = 1 F, and E = 12 volts. Assume that the switch has been in the close position for a long time and at t = 0 the switch is open. Write the equation for vc(t) and ic(t) for t > 0. Calculate vc(t) and ic(t) at t = 3 sec and at t = 6 sec.
ampseRRtv
ti
voltsevtv
VVERRR
RRv
FCRRRC
RCt
CC
RCt
cC
C
32
321
32
32
)()(
)(
81296)0(
sec616)(
Ic(t)
amptivoltstvamptivoltstv
ampsetivoltsetv
C
C
C
C
stC
stC
49.0sec)6(94.2sec)6(81.0sec)3(
85.4sec)3(33.1)(
8)(sec6
6/
6/
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62
DC Circuits
Example Consider the following circuit with R1= 6 k, R2= 12 k, L = 2 H, and E = 12 volts. Assume that the switch has been in the close position for a long time and at t = 0 the switch is opened. Write the equation for iL(t) and vL(t) for t > 0. Calculate iL(t) and vL(t) at t = 100 sec and at t = 200 sec.
voltseRRitv
mAeiti
mAmAkki
RRRi
mAkV
RREi
mkH
RRL
RL
t
LL
t
LL
EL
E
))(0()(
)0(
13186)0(
3412
||
sec11.0182
21
21
1
21
21
voltstvmAtivoltstv
mAtivoltsetv
mAeti
L
L
L
L
stL
stL
97.2sec)200(16.0sec)200(
73.0sec)100(41.0sec)100(
18)(
1)(sec110msec 11.0
110/
110/
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63
DC Circuits (Sampling Theorem)
Sampling is the process of converting a signal (for example, a function of continuous time or space) into a numeric sequence (a function of discrete time or space). A sample refers to a value or set of values at a point in time and/or space. If a function x(t) contains no frequencies higher than B Hertz, it is completely determined by giving its ordinates at a series of points spaced 1/(2B) seconds apart. The theorem shows that a band-limited analog signal that has been sampled can be perfectly reconstructed from an infinite sequence of samples if the sampling rate exceeds 2B samples per second, where B is the highest frequency in the original signal. If a signal contains a component at exactly B hertz, then samples spaced at exactly 1/(2B) seconds do not completely determine the signal. A signal or function is band-limited if it contains no energy at frequencies higher than some band-limit or bandwidth B. A signal that is band-limited is constrained in how rapidly it changes in time, and therefore how much detail it can convey in an interval of time. The sampling theorem asserts that the uniformly spaced discrete samples are a complete representation of the signal if this bandwidth is less than half the sampling rate.
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64
DC Circuits
Sampling Frequency Determines the sampling rate to reproduce accurately in the discrete time system. Sampling Rate or Frequency: {where fN is the maximum frequency component of g(t) and fN is the Nyquist frequency} Sampling Frequency: (Sample rate to recover signal g(t) exactly from samples)
NS ff 2
TS = 1/2fN
Spectrum of band-limited signal g(t)
NS ff 2
SS Tf 1
g(t)
Page 110 from Supplied Reference Handbook
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65
DC Circuits
Sampled Messages A low-pass message m(t) can be exactly reconstructed from uniformly spaced samples taken at a sampling frequency of fs = 1/Ts fs 2fN where M(f ) = 0 for f > fN The frequency fN is called the Nyquist frequency. Sampled messages are typically transmitted by some form of pulse modulation. The minimum bandwidth B required for transmission of the modulated message is inversely proportional to the pulse length . B 1 / Frequently, for approximate analysis B 1 / is used as the minimum bandwidth of a pulse of length .
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66
DC Circuits
Therefore, a band-limited signal can be reconstructed exactly if it is sampled at a rate at least twice the maximum frequency component in it. T(t) is the sampling signal with fs = 1/Ts > 2fN
Original Signal Spectrum G()
Spectrum T()
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67
DC Circuits
Sampled Signal s = 2N Ts = 1/2fN
Spectrum Gs() Sampled Signal gs(t)
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68
DC Circuits
Aliasing What happens if we sample the signal at a frequency that is lower that the Nyquist rate? When the signal is converted back into a continuous time signal, it will exhibit a phenomenon called aliasing. Aliasing is the presence of unwanted components in the reconstructed signal. These components were not present when the original signal was sampled. In addition, some of the frequencies in the original signal may be lost in the reconstructed signal. Aliasing occurs because signal frequencies can overlap if the sampling frequency is too low. Frequencies "fold" around half the sampling frequency - which is why this frequency is often referred to as the folding frequency.
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69
DC Circuits
Aliasing Aliasing is a phenomenon where the high frequency components of the sampled signal interfere with each other because of inadequate sampling s < 2m. Aliasing leads to distortion in recovered signal. This is the reason why sampling frequency should be at least twice the bandwidth of the signal.
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70
DC Circuits
Oversampling In practice, signals are oversampled, where fs is significantly higher than Nyquist rate to avoid aliasing.
Oversampled signal avoids aliasing
fs >> 2fN
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71
DC Circuits
Example
or fs = 1/Ts
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72
DC Circuits
Example/Contd. The sampling frequency must be greater than the Nyquist rate for accurate reproduction. fs 2fN The greatest frequency that can be reproduced at this sampling rate is Therefore, the correct is (C).
HzHzHzf
f SN66
6
10 x 16 10 x 7.16210 x 33
2
Page 110 from Supplied Reference Handbook
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73
DC Circuits
Example What is the minimum sampling frequency that can be used to avoid aliasing with the following analog signal? x(t) = Cos 100t + Sin 200t + Sin 60t (A) 30Hz (B) 60Hz (C) 200Hz (D) 400Hz Solution: To avoid aliasing, the analog signal must be sampled at least at the Nyquist rate, which must be at least twice the highest frequency contained in the signal. The highest frequency contained in the given signal is 200 rad/s or 100 Hz. Therefore, the minimum frequency that can be used to avoid aliasing is 200 Hz.
= 2f
f = /2
f = 200/2 Hz
= 100 Hz
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74
DC Circuits
Instrumentation Transducers Transducer sensitivity the ratio of change in electrical signal magnitude to the change in magnitude of the physical parameter being measured.
Page 110 from Supplied Reference Handbook
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75
DC Circuits
Instrumentation Resistance Temperature Detector (RTD) a device used to relate change in resistance to change in temperature. Typically made from platinum, the controlling equation for an RTD is given by: where: RT is the resistance of the RTD at temperature T (measured in C) R0 is the resistance of the RTD at the reference temperature T0 (usually 0 C) is the temperature coefficient of the RTD
)(1 00 TTRRT
Page 110 from Supplied Reference Handbook
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76
DC Circuits
Example
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77
DC Circuits
Example/Contd. The error in the measurement is: Error = 383.1 C - 400 C = -16.9 C
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78
DC Circuits
Resistance Example A cube with an edge length of 0.01 m has resistivity of 0.01 m. What is the resistance from one side to the opposite side? (A) 0.0001 (B) 0.001 (C) 0.1 (D) 1 Therefore, the answer is (D).
)( ALR
1)01.0(
)01.0)(01.0(2m
mmALR
0.01m
0.01m
0.01m
A = (0.01) m
= 0.01 m
Page 194 from Supplied Reference Handbook
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79
DC Circuits
Strain Gauges Metal or semiconductor foils that change resistance linearly with the strain. Strain Gauge a device whose electrical resistance varies in proportion to the amount of strain in the device. Gauge Factor (GF) the ratio of fractional change in electrical resistance to the fractional change in length (strain): Where: R is the nominal resistance of the strain gauge at nominal length L. R is the change in resistance due to the change in length L. is the normal strain sensed by the gauge. The gauge factor for metallic strain gauges is typically around 2.
RRLLRRGF /
//
Page 110 from Supplied Reference Handbook
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80
DC Circuits
Example
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81
DC Circuits
Wheatstone Bridges/Instrumentation
Page 110 from Supplied Reference Handbook
balanced is bridge theand 0V If 03241 VRRRR
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82
DC Circuits
Example
R1 = 10.00k + 0.24k = 10.24k
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83
DC Circuits Measurement Uncertainty/Instrumentation
Kline-McClintock Equation: A method for estimating the uncertainty in a function that depends on more than one measurement. Suppose that a calculated result R depends on measurements whose value are x1 w1, x2 w2, x3 w3, etc, where R = f(x1, x2, x3,xn), xi is the measured value, and wi is the uncertainty in that value. The uncertainty in R, wR, can be estimated using the Kline- McClintock equation:
Page 111 from Supplied Reference Handbook
22
22
2
11 ...
!
!
!
!
!
!
nnR x
fwxfw
xfww
-
84
DC Circuits
Example
-
85
III. AC Circuits
Circuit Analysis
-
86
AC Circuits
Frequency and Period of a Sinusoidal Waveform Frequency (Linear and Angular) Period = T
Tf
Tf "
" 22 2
1
Page 195 from Supplied Reference Handbook
-
87
AC Circuits
Average Value Average and Effective Values
T
t XMAX
2
31
:Sawtoothor TriangularFor
MAXAVEMAXRMS
XXXX
Page 195 from Supplied Reference Handbook
-
88
AC Circuits
Average Value Example
AmpstqI
dtdqti
sC
)(
FsA 96,485 electrons
of mole one of charge of Magnitude Faraday 1Constant sFaraday' x eCapacitancq
Handbook)in 19 (Pg. 485,96
:Constant sFaraday'
FsA
VMAX = 170V
(q)
-
89
AC Circuits
Effective or Root-Mean-Square (RMS) Value
VRMS %)70( 707.0
707.02
1
MAXRMS VV
*
*
* Not in Handbook
T=1/f
T
t
Page 195 from Supplied Reference Handbook
-
90
AC Circuits
Effective or Root-Mean-Square (RMS) Value Example
120.2V = 170V
= 3610 W 3.61kW
-
91
AC Circuits
Effective or Root-Mean-Square (RMS) Value Example
-
92
AC Circuits
Effective or Root-Mean-Square (RMS) Value Example
-
93
AC Circuits
Effective or Root-Mean-Square (RMS) Value Example
VRMS VRMS = VDC = steady-state voltage
-
94
AC Circuits
Phasor Transforms Phase Angle = as the angle between the reference and voltage as the angle between the reference and current If the phase angle is positive, the signal is called lagging or inductive. If the phase angle is negative, the signal is called leading of capacitive. If the phase angle is zero, the signal is called in phase.
t
Ref
i
v
(+) (-)
ELI the ICE man
If = + Lagging (Inductive)
If = - Leading (Capacitive)
-
95
AC Circuits
Impedance Definitions
XL
XC
R 90
-90
0
0
901j:Inductor
9011:Capacitor
#
# jj
Page 195 from Supplied Reference Handbook
-
96
AC Circuits
Example
VMAXSin (wt + )
160V0 w = 400 rads/sec
ZT
R jwL 1/jwC
ZT = ZR + ZL + ZC + ZE
An Element
ZE = RCos + jRSin
Z = R jX ZT = |Z|
VL
i(t) = V/Z = |i| = Tan (x/y)
VL
-
97
AC Circuits
Example/Contd.
j
Z = R + jwL j/wC + 940
R = RCos + jRSin
)F 10 6x
-
98
AC Circuits
Example/Contd.
(Voltage Across Inductor)
-
99
AC Circuits
Example
i1 i2
1800 1800
- j4
i1 =
i2 =
- j0.25 = 0.25-90 - j = 1- 90
-
100
AC Circuits
Complex Power Lagging (Inductive) Complex Power Triangle
Cos = P/S
I = a + jb
I* = a - jb
KW = p.f. x KVA
S = P + jQ
= kW/kVA
Page 197 from Supplied Reference Handbook
-
101
AC Circuits Complex Power Example
RIRVVIP 2
2
3714.02.68..2.68387736001440
0
0
$
#
CosfpVarjWS
-
102
AC Circuits
Complex Power
S = P + jQ = I* V Inductive
ELI Q = - 3600 VAR
If S were Capacitive
ICE
P = 1440 W
S = 3877 VA
-
103
AC Circuits
Resonance
CL
LCf
00
00
1
)(Resonance 12
"
"
"
R L C
R C L
Z
RLRC
= w0 (RC/2)
Z
= w0 (R/2L) Resistance = Reactance
PARALLELSERIES Q
Q 1
X = XL + XC
Page 196 from Supplied Reference Handbook
-
104
AC Circuits
Example
Z
At Resonance:
XL = - XC
Therefore, Z = R
-
105
AC Circuits
Example/Contd.
= W0 (R/L)
At Resonance VC = - VL
Peak Voltage across each component.
VL
VC
- 90
+90
-
106
AC Circuits
Example
52
10NEWR
2
2
121
1212 12
OLDNEW
NEWOLD
OLDNEW
RR
RRRR
RCRC
BWBW
(ROLD)
(RNEW)
10kHz 20 kHz BW1 BW2
-
107
AC Circuits
Analog-to-Digital Conversion Voltage Resolution The range from a high voltage, VH, and a low voltage, VL, is divided up into the 2 ranges For example, if all the bits are 1 then the discrete value is somewhere between VH and VH v. To calculate the discrete value from the digital value use
Page 110 from Supplied Reference Handbook
number of bits Voltage Resolution =
Discrete Value =
Typical values for n:
4, 8, 10, 12, 16 bits
Key Design Parameter
N: { 0, 2 - 1 }
digital value VH
VL
Highest measurable voltage = VH - VL
-
108
AC Circuits
Analog-to-Digital Conversion Example
LHVn
LH
HLnLH
V
VVVV
VVVV
2 221
2
-
109
AC Circuits
Transformer Equivalent Circuit with Secondary Impedance
IS
VS
Zeq
P
S
S
P
SSPP
SECPRI
II
VV
IVIVPP
PPRI
PSEC
Page 198 from Supplied Reference Handbook
-
110
AC Circuits
Transformer Example
V2
-
111
AC Circuits
Transformer
Example
3 + j1 (50/100)(12 + j4)
10050
2
1
NNa
Ideal Transformer has no losses
= 6 + j2
-
112
AC Circuits
Rotating Machines
Angular Velocity = = 2w/p = 4f/p w = (p)/2
= 120w/2p = 9.5 x
Synchronous Speed (RPM):
p = # of poles n = rotor operating speed 120
ABC = BCA = CAB
Page 198 from Supplied Reference Handbook
-
113
AC Circuits
Example Armature: The structure (rotor or stator) where a voltage is induced by a magnetic field. (Motor armature: rotor; Generator armature: stator.)
= 125.67 rads/s
pnS " 2
602
f = p/4 p = 4f
1 pair = 2 poles
-
114
AC Circuits
Rotating Machines
snn
snn
nns
speedssynchronounspeedoperatingn
S
S
S
S
1
1
1
Page 198 from Supplied Reference Handbook
-
115
AC Circuits
Rotating Machines Example
p = 4 poles
60 Hz = 1800 RPM
50 Hz = 1500 RPM 5%
-
116
AC Circuits
Rotating Machines
(Kf is a design constant)
Lf
If
Rf
Vf
(Webers)
Stator
Rotor
Motor
Page 198 from Supplied Reference Handbook
-
117
AC Circuits
Rotating Machines DC Generators
A device that produces DC potential
B
VOUT
VRMS Vpeak
or VMAX
%)64(6366.02
%)70(707.02
MAXMAX
AVG
MAXMAX
RMS
VVV
VVV
-
118
AC Circuits
Rotating Machines DC Generators
Va = Ka n
Power MechanicalxTPTorque T
Speedngular ConstantDesign
peedOperatingFlux
m
AK
Sn
a
%
(Heat Loss)
Pa = IaRa + IaEg = PHEAT + Pm
Page 198 from Supplied Reference Handbook
-
119
AC Circuits
Rotating Machines Example
RPMRPMAA
RPMxn
n
nn
nKnK
nn
II
a
a
150018001012
15006
5180056
1012 1800
1012
VV
2
2
2
1
2
1
2
1
2
1
2
1
%
%
-
120
AC Circuits
Rotating Machines
Example/Contd.
RPMxn
n
nn
II
15006
5180056
1012 1800
1012
2
2
2
1
2
1
-
121
AC Circuits
Rotating Machines
Va = Kan Pm = Ta x = VaIa
Handbook)in 198 (Pg.2
1
2
1
%
%
TT
= 2ns/60
Ph = Heat Loss
(Nm)
-
122
AC Circuits
Rotating Machines
Example
-
123
IV. Computers
Terminology, Spreadsheets, and
Structured Programming
-
124
Computers
Terminology CPU stands for Central Processing Unit. This is where all the computer's data processing is handled - all the
data manipulation, calculations and formatting data for output. When you buy a computer you will become more focused on the CPU and its capabilities.
The execution of the instructions within the computer system is extremely fast and is measured in cycles of time and referred to as megahertz. For this reason the MHz of a computers processor is sometimes referred to as its clock speed. This pulsing is expressed as "MHz" e.g. 2400Mhz.
The CPU pulsing is turning electrical current on and off. An electrical circuit can be either open or closed, and the power is either running through or not; that is, it is either on or off. This two-state situation is called binary, and the two states are controlled by binary digits or bits. The two bits of a computer are zero and one. For you to talk to your computer your message must be translated into binary form, a series of ones and zeros.
The CPU is located usually around the center of the motherboard, and under a giant fan employed to keep it cool. The motherboard is the main piece of circuitry of your computer. It houses the CPU, the ROM and RAM, a variety of computer cards for receiving signals from other input devices, the power supply, hard drive and so on.
The term CPU is sometimes used to refer to the case that houses the computers components, but technically the CPU is only one chip inside the computers case. The case is sometimes described as the "tower" which means the case stands on the floor on its narrow side. Whereas a "desktop" machine sits on your desk and generally has the monitor placed on it.
-
125
Computers
Terminology
RAM RAM means Random Access Memory" (or sometimes "ready access memory"). It is like a temporary notepad where your computer sends information it has processed before writing it to disk, or where instructions from other input devices (keyboard, mouse, floppy drive) are stored. The words "random access" indicate that memory locations in RAM are accessible in any order unlike sequential access of a data cassette tape, for example.
ROM ROM means Read Only Memory. Your computers ROM is a chip on the motherboard that stores a permanent set of start-up instructions for your computer. The familiar term for ROM is "BIOS", short for "basic input output system". ROM is sometimes referred to as "firmware" (as opposed to software) because it is permanently housed inside the hardware. Instructions in ROM remain intact when the power is off. Generally you can not alter the information in ROM; it is fixed at the time of manufacture. Some manufacturers provide updates to the ROM of their chips. The process of installing these updates is called "flashing the bios". This is only recommended for advanced users and only after all other avenues of repair are explored. DRAM Dynamic random-access memory (DRAM) is a type of random-access memory that stores each bit of data in a separate capacitor within an integrated circuit. The capacitor can be either charged or discharged; these two states are taken to represent the two values of a bit, conventionally called 0 and 1.
-
126
Computers
Terminology Bit, Bytes and Binary A bit stands for Binary Digit. A byte is 8 bits. A kilobyte (kB) is 1,024 bytes. A megabyte (MB) is approximately 1,024,000 bytes. A gigabyte (GB) 1,024,000,000 bytes - thats a lot of bits!!! A bit is the smallest information parcel on a computer. A computer file can be made up of many hundreds of bytes and many thousands of bits. The bits are like switches, they are represented by 0s and 1s. Characters on a keyboard are therefore represented by 1s and 0s, so a "P" may be 01010000 to a computer - it knows when it sees those on/off instructions to output a "P". Hardware Hardware is the term given to all the physical parts of a computer system. Hardware includes the monitor, the keyboard, the mouse, the main case which houses the RAM, CPU and the motherboard. Operating System Every computer needs a set of programs called the operating system to run the system and make all the other programs work. The word processor, database or spreadsheet programs can not operate unless the operating system is present. Programs written for one operating system will not work on a different operating system. Common operating systems include: Windows 7, Vista, Macintosh OSX, and Linux. There are many versions aimed at different uses.
-
127
Computers
Terminology Hard Drive A computers hard disk drive is like an audio CD, except the computer can read and write to it. In other words, a computer can take data off its hard drive (to process it in the CPU or place it in RAM to work with) or it can record the results of the work it does back to the disk, this is called "writing to disk". The abbreviation HDD means "hard disk drive". If the HDD is opened up (not recommended as this usually voids the manufacturer's warranty), a pancake stack would be found on double sided disks. Each side of each disk has an arm that holds a "head". The head is responsible for writing data to the disk. When one handles a disk, carry a computer, or laptop, one should be careful not to bang the case as this can force the heads to collide with the surface of the disk and potentially damage it. Hard disks are "formatted" to make them writable. Through this process "clusters", "sectors" and a "file allocation table" are created. With these mechanisms the computer writes information to the disk and can retrieve it later.
-
128
Computers
Terminology
access time - The performance of a hard drive or other storage device - how long it takes to locate a file. application - a program in which you do your work. ASCII (pronounced ask-key ) - American Standard Code for Information Interchange. a commonly used data
format for exchanging information between computers or programs. bit - the smallest piece of information used by the computer. Derived from "binary digit". In computer
language, either a one (1) or a zero (0). backup - a copy of a file or disk you make for archiving purposes. boot - to start up a computer. bug - a programming error that causes a program to behave in an unexpected way. bus - an electronic pathway through which data is transmitted between components in a computer. byte - a piece of computer information made up of eight bits. cartridge drive - a storage device, like a hard drive, in which the medium is a cartridge that can be removed. Chooser - A desk accessory used to select a printer, or other external device, or to log onto a network. Clock Rate (MHz) - The instruction processing speed of a computer measured in millions of cycles per second
(i.e., 200 MHz). compiler - a program the converts programming code into a form that can be used by a computer. compression - a technique that reduces the size of a saved file by elimination or encoding redundancies (i.e.,
JPEG, MPEG, LZW, etc.) CPU - the Central Processing Unit. The processing chip that is the "brains" of a computer. crash - a system malfunction in which the computer stops working and has to be restarted. daisy chaining - the act of stringing devices together in series (such as Small Computer System Interface
(SCSI). database - an electronic list of information that can be sorted and/or searched.
-
129
Computers
Terminology driver - a file on a computer which tells it how to communicate with an add-on piece of equipment (like a
printer). Ethernet - a protocol for fast communication and file transfer across a network. expansion slot - a connector inside the computer which allows one to plug in a printed circuit board that
provides new or enhanced features. file - the generic word for an application, document, control panel or other computer data. folder - an electronic subdirectory which contains files. gig - a gigabyte = 1024 megabytes. hard drive - a large capacity storage device made of multiple disks housed in a rigid case. head crash - a hard disk crash caused by the heads coming in contact with the spinning disk(s). high density disk - a 1.4 MB floppy disk. installer - software used to install a program on a hard drive. kilobyte - 1024 bytes. Measurements
*a bit = one binary digit (1 or 0) * 1 byte (or Octet) = 8 bits
*1024 bytes = one kilobyte *K = kilobyte *Kb = kilobit *MB = megabyte *Mb = megabit *MB/s = megabytes per second *Mb/s = megabits per second *bps = bits per second i.e., 155 Mb/s = 19.38 MB/s
megabyte - 1024 kilobytes. memory - the temporary holding area where data is stored while it is being used or changed; the amount of
RAM a computer has installed. nibble - a piece of computer information made up of four bits. It is equal to half a byte.
-
130
Computers
Terminology operating system - the system software that controls the computer. RAM - acronym for Random-Access Memory. RISC - acronym for Reduced Instruction Set Computing; the smaller set of commands used by the PowerPC
and Power Mac. ROM - acronym for Read Only Memory; memory that can only be read from and not written to. server - a central computer dedicated to sending and receiving data from other computers (on a network). software - files on disk that contain instructions for a computer. 32 bit addressing - a feature that allows the Mac to recognize and use more than 8MB of memory. vaporware - "software" advertised, and sometimes sold, that does not yet exist in a releasable for. virtual memory - using part of your hard drive as though it were "RAM". WORM - acronym for Write Once-Read Many; an optical disk that can only be written to once (like a CD-
ROM).
-
131
Computers
Spreadsheets A B C D
1
2
3
4
Functions
=SUM(A4:D4), INT(n) = m where m n, SIN(), PMT(rate, nper, pv, fv, n) where rate = interest, nper = total # of payments for loan, n = 0 (end of period) or 1
(beginning of period).
Page 109 from Supplied Reference Handbook
-
132
Computers
Spreadsheets
Example
C3 = $A$4 + B$2 + B2
C4 = $A$4 + B$2 + B3
C5 = $A$4 + B$2 + B4
D5 = $A$4 + C$2 + C4
A B C D E F
1
2
3
4
5
$A$4
Page 109 from Supplied Reference Handbook
-
133
Computers
Spreadsheets Example
A
4
5
6
7
6
A4 + $A$4
A5 + $A$4
A6 + $A$4
12
18
24
Relative Addressing
Absolute Addressing
-
134
Computers
Algorithm Flowcharts and Symbols
(To a different page)
(Connects to algorithm or subroutine)
-
135
Computers
Algorithm Flowcharts
Example
to main program
(= is replaced by)
X = 4
T = 5
-
136
Computers
Structured Programming Sequence of operations from left to right in the following hierarchy: Brackets and Parentheses, then Exponentiation, then multiplication and division, then addition and subtraction.
- 6 5
/ 4* 3
**or ^ 2) ( ] [ 1
Hierarchy lOperationa
th
th
th
rd
nd
st
-
137
Computers
Structured Programming Example
12224
3 - 56 4
x
-
138
Computers
Structured Programming
Example
7X 8X 4T 734X 4 82X 4T 824 X 4 2
4222X 4 2 2 T 3 4T X
&
'
&
-
139
Computers
Structured Programming
Example (Contd.)
-
140
Computers
Structured Programming
Example
6X 615X 514X61 - 7T 71 - 8 T
Pass 2 Pass 1 ndst
return to main program
-
141
Computers
Structured Programming Example
-
142
Computers
Structured Programming Example
8X ENDWHILE
826X 624XXT 42 - 6T 62 - 8T
No Yes Yes Pass 3 Pass 2 Pass1 rdndst
'
-
143
Computers
Structured Programming
Example (Contd.)
-
144
Computers
Structured Programming Example
-
145
Computers
Structured Programming Example (Contd.)
-
146
Computers
Structured Programming
Example What is most nearly Z after structured programming is executed?
If not DO
-
147
Computers
Structured Programming
Example (Contd.)
3.7429
IY Z
413I 312 I Endwhile 292)(2 25Y 25 5)(50Y 211I 2X 5X 3X
3I No, 3I Yes, 3I Yes, 3I Yes,Pass4 Pass3 Pass2 Pass1 thrdndst
(
(
&
-
148
V. Ethics and Business Practices
Professional Engineering Ethics
-
149
Ethics and Business Practices
Hints for Exam Questions Page 121 from Supplied Reference Handbook
-
150
Ethics and Business Practices
Obligations to Society Page 121 from Supplied Reference Handbook
-
151
Ethics and Business Practices
Obligation to Society
-
152
Ethics and Business Practices
Obligations to Employers and Clients
-
153
Ethics and Business Practices
Obligations to Employers and Clients
-
154
Ethics and Business Practices
Obligations to Employers and Clients
Example
-
155
Ethics and Business Practices
Obligations to Employers and Clients
Example
-
156
Ethics and Business Practices
Obligations to Employers and Clients
-
157
Ethics and Business Practices
Obligations to Other Registrants
-
158
Ethics and Business Practices
Obligations to Other Registrants
-
159
Ethics and Business Practices
Expert Witness
-
160
Ethics and Business Practices
Expert Witness
Example
-
161
Ethics and Business Practices
Consultants
-
162
Ethics and Business Practices
Consultants
Example