FE: Electric Circuits
© C.A. Gross
EE1-
1
FUNDAMENTALS OF ENGINEERING
(FE) EXAMINATION REVIEW
ELECTRICAL
ENGINEERING
Charles A. Gross, Professor Emeritus
Electrical and Comp Engineering
Auburn University Broun 212
334.844.1812
gross@eng.auburn.edu
www.railway-technology.com
FE: Electric Circuits
© C.A. Gross
EE1-
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1
EE Review Problems
1.
2.
3.
4.
dc Circuits
Complex Numbers
ac Circuits
3-phase Circuits
1st Order Transients
Control
Signal Processing
Electronics
Digital Systems
FE: Electric Circuits
We will discuss
these.
We may discuss
these as time
permits
© C.A. Gross
EE1-
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EE1-
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1. dc Circuits:
Find all voltages, currents, and powers.
FE: Electric Circuits
© C.A. Gross
2
Solution
The 8 and 7 resistors are in series:
R1  8  7  15
R1 and 10 are in parallel:
1
R2 
1
1

10 R1
10  R1

 6
10  R1
FE: Electric Circuits
© C.A. Gross
EE1-
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Solution
4 and R2 are in series:
Rab  4  R 2  10
L :
Ia 
Vab 100

 10 A
Rab 10
V4  4  I a  40V
(L)
V10  100  40  60V
FE: Electric Circuits
 KVL 
© C.A. Gross
3
Solution
Ic 
V10 60

 6A
10 10
 L 
KCL :
I b  I a  I c  10  6  4 A
V8  8  I b  32V
(  L)
V7  7  I b  28V
(  L)
FE: Electric Circuits
© C.A. Gross
EE1-
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Absorbed Powers...
R4  I a2  4 10   400W
2
R10  I  10  6   360W
R7  I b2  7  4   112W
2
R8  I  8  4   128W
2
b
In General:
2
2
c
PABS = PDEV
(Tellegen's
2
Total Absorbed Power  1000W
Theorem)
Power Delivered by Source  Vs  I a  100 10   1000W
FE: Electric Circuits
© C.A. Gross
EE1-
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4
2. Complex Numbers
Consider
x2  2 x  5  0
2  (2) 2  4(1)(5) 2  16

x
2 1
2
4 1
 1
 1  2 1
2
The numbers "1  2 1" are called complex numbers
9
Summer 2008
The “I” (j) operator
Math Department....
ECE Department
Define i  1
Define j  1
x  1  j2
x  1  2i
We choose ECE notation! Terminology…
Rectangular Form.....
Z  X  jY  a complex number
X  e  Z   real part of Z
Y  Im  Z   imaginary part of Z
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5
Polar Form
Math Department.....
Z  R  e i  a complex number
R | Z | modulus of Z
  arg  Z   argument of Z  radians 
ECE Department.....
Z  Z   a complex number
Z | Z | magnitude of Z
  ang  Z   angle of Z  degrees 
11
The Argand Diagram
It is useful to plot complex numbers in a 2-D cartesian
space, creating the so-called Argand Diagram (Jean
Argand (1768-1822)).
“imaginary” axis (Y)
+2
Z  X  jY  1  j 2
“real” axis (X)
+1
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Summer 2008
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Conversions
Retangular  Polar
Z
Y
X2 Y2
Y 

X
  tan 1 
Z

Polar  Retangular.....
X  Z  cos  
X
Y  Z  sin  
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Summer 2008
Example:
Z  3  j4
X  e  Z   3
4
Y  Im  Z   4
Rect  Polar.....
5
0
Z  32  42  5
3
 4
 
Summer 2008
  tan 1    0.9273 rad  53.10
3
14
7
Conjugate
Z  X  jY  Z 
Z *  conjugate of Z  X  jY  Z   
Example...
(3  j 4)*  3  j 4  5  53.10
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Summer 2008
Addition (think rectangular)
A  a  jb  A  3  j 4  553.10
B  c  jd  B  5  j12  13  67.40
A  B  (a  jb)  (c  jd )
 (a  c )  j (b  d )
A  B  (3  j 4)  (5  j12)
 (3  5)  j (4  12)  8  j8
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Multiplication (think polar)
A  a  jb  A  3  j 4  553.10
B  c  jd  B  5  j12  13  67.40
A  B  ( A )  ( B )
 A  B(   )
A  B  (553.10 )  (13  67.40 )
 (5)  (13) (53.10  67.40 )  65  14.30
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Summer 2008
Division (think polar)
A  a  jb  A  3  j 4  553.10
B  c  jd  B  5  j12  13  67.40
A A  A 

    (   )
B B  B 
A  553.10   5 

    53.10   67.40 
0 
B  13  63.4   13 


 0.3846120.50
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Multiplication (rectangular)
A  B  (a  jb)   c  jd 
  ac  bd   j  ad  bc 
A  B  (3  j 4)  (5  j12)
 (15  48)  j ( 36  20)
 63  j16  65  14.30
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Summer 2008
EL
EC
38
10
Addition (polar)
553.10
Complex number
addition is the
same as "vector
addition"!
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Summer 2008
11.31  450
EL
13  67.40EC
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10
10
3. ac Circuits
i(t)
Find "everything"
in the given
circuit.
8
+
v(t)
-
0.663 mF
26.53 mH
v (t )  141.4cos(377 t ) V
FE: Electric Circuits
© C.A. Gross
EE1- 21
Frequency, period
v (t )  141.4cos(377 t ) V
(radian) frequency    377 rad / s
(cyclic) frequency  f 
Period 
FE: Electric Circuits

 60 Hz
2
1
1

 16.67 ms
f 60
© C.A. Gross
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The ac Circuit
To solve the problem, we convert the circuit
into an "ac circuit":
R, L, C elements  Z (impedance)
v , i sources  V , I (phasors)
R : Z R  R  j0  8  j0
L : Z L  0  j L  0  j (0.377)(26.53)  0  j10
C : ZC  0 
1
1
 0 j
 0  j4
0.377(0.663)
j C
FE: Electric Circuits
© C.A. Gross
EE1- 23
The Phasor
v (t )  VMAX cos( t   )
To convert to a phasor...
For example..
V
FE: Electric Circuits
V
VMAX
2

v (t )  141.4cos(377 t )
VMAX
2
  100 0o
© C.A. Gross
EE1- 24
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The "ac circuit"
I
L(ac ) :

VR

1000

I 



VC


FE: Electric Circuits
100
8  j10  j 4
 10  36.9o

VL
V
Z
i (t )  14.14  cos(377 t  36.9o )
© C.A. Gross
EE1- 25
Solving for voltages
VR  Z R  I  (8)(10  36.9o )  80  36.9o V
v R (t )  113.1  cos(377 t  36.9o )
VC  Z C  I  ( j 4)(10  36.9o )  40  126.9o V
vC (t )  56.57  cos(377t  126.9o )
VL  Z L  I  ( j10)(10  36.9o )  10053.1o V
v L (t )  141.4  cos(377 t  53.1o )
FE: Electric Circuits
© C.A. Gross
EE1- 26
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Absorbed powers
S  V  I *  P  jQ
S R  VR  I *  80  36.9o (10  36.9o )*  800  j 0
SC  VC  I *  40  126.9o (10  36.9o )*  0  j 400
S L  VL  I *  10053.1o (10  36.9o )*  0  j1000
STOT  S R  SC  S L  800  j 600
PTOT  800 watts;
QTOT  600 var s;
STOT | STOT | 1000 VA
FE: Electric Circuits
© C.A. Gross
EE1- 27
Delivered power
S S  VS  I *  100 0o (10  36.9o )*  800  j 600
S S  STOT  800  j 600
PS  PTOT  800 watts
QS  QTOT  600 var s
In General: PABS = PDEV
QABS = QDEV
(Tellegen's Theorem)
FE: Electric Circuits
© C.A. Gross
EE1- 28
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The Power Triangle S  800  j 600
S = 1000 VA
Q = 600 var
 = 36.90
V  1000o
I  10  36.9o
P = 800 W
power factor  pf 
FE: Electric Circuits
P
 cos( )  0.8 lagging
S
© C.A. Gross
EE1- 29
Leading, Lagging Concepts
Leading Case
I
Q<0
V
Lagging Case
V
Q>0
I
FE: Electric Circuits
© C.A. Gross
EE1- 30
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A Lagging pf Example

V  7.200 kV

jX
R
jX  j 43.2 
R  103.68 
FE: Electric Circuits
© C.A. Gross
EE1- 31
Currents

I
V
IR
V

R
IR 
IL 
IL
jX
V
7.2

 69.44 A
R 103.68
I  IR  IL
V
7.2

  j166.7 A
jX j 43.20
FE: Electric Circuits
I  69.44  j166.7
I  180.6  67.380 A
© C.A. Gross
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pf  cos    cos  67.360   0.3845
Powers

I
IL
IR
V

R
1300 kVA
jX
1200 k var
S R  V  I R *  500 kW  j 0
S L  V  I L *  0  j1200 k var
SS  V  I *  S R  S L
500 kW
SS  500  j1200  130067.380
FE: Electric Circuits
EE1- 33
© C.A. Gross
Add Capacitance
I
I C  j125

 jX C
V
IR

IC 
IL
IC
I R  69.44
V
7.2

 j125 A
 jX  j 57.6
I L   j166.7
I  I R  I L  IC
I  69.44  j166.7  j125  81  310 A
FE: Electric Circuits
© C.A. Gross
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Powers
pf  cos    cos  310   0.8575
900 kvar
1200 kvar
SC  V  I C *  0  j 900 kvar
583.1 kVA
SS  V  I *  S R  S L  SC
SS  500  j1200  j 900
500 kW
SS  583.1310 kVA
FE: Electric Circuits
© C.A. Gross
EE1- 35
Observations
By adding capacitance to a lagging pf (inductive) load, we
have significantly reduced the source current., without
changing P!
Before
I  180.6 A;
pf  0.3845
I  81 A;
pf  0.8575
After
Note that:
low pf, high current;
high pf, low current;
If we consider the “source” in the example to represent an
Electric Utility, this reduction in current is of major practical
importance, since the utility losses are proportional to the
square of the current.
FE: Electric Circuits
© C.A. Gross
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Observations
That is, by adding capacitance the utility losses have
been reduced by almost a factor of 5! Since this results
in significant savings to the utility, it has an incentive to
induce its customers to operate at high pf.
This leads to the “Power Factor Correction” problem, which
is a classic in electric power engineering and is extremely
likely to be on the FE exam.
We will be using the same numerical data as we did in the previous example.
Pretty clever, eh’ what?
FE: Electric Circuits
EE1- 37
© C.A. Gross
The Power Factor Correction problem
Utility
Load
pf correcting
capacitance
An Electric Utility supplies 7.2 kV to a customer whose load
is 7.2 kV 1300 kVA @ pf = 0.3845 lagging. The utility offers
the customer a reduced rate if he will “correct” (“improve”
or “raise”) his pf to 0.8575. Determine the requisite
capacitance.
FE: Electric Circuits
© C.A. Gross
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PF Correction: the solution
1. Draw the load power triangle.
1300 kVA @ pf = 0.3845 lagging.
pf  0.3845  cos  
  67.380
S LOAD  S   130067.38
1200 kvar
0
S LOAD  500  j1200
Because the pf is lagging, the load is
inductive, and Q is positive. Therefore we
must add negative Q to reduce the total,
which means we must add capacitance.
67.380
500 kW
FE: Electric Circuits
© C.A. Gross
EE1- 39
PF Correction: the solution
2. We need to modify the source complex power so
that the pf rises to 0.8575 lagging.
pf  0.8575  cos  
  310
Closing the switch (inserting the capacitors)
SS  500  j1200  jQC  500  j 1200  QC 
Let QX  1200  QC
Therefore SS  500  jQX  S S 310 kVA
Then Tan   
FE: Electric Circuits
QX
 Tan  310   0.6
500
© C.A. Gross
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PF Correction: the solution
QX
 0.6
500
QX  300 kvar
QC  1200  Q X  900 kvar
900 kvar
The new source power triangle
1200 kvar
Install 900 kvar of
7.2 kV Capacitors
300 kvar
500 kW
FE: Electric Circuits
© C.A. Gross
EE1- 41
4. Three-phase ac Circuits
Although essentially all types of EE’s use ac circuit analysis
to some degree, the overwhelming majority of applications
are in the high energy (“power”) field.
It happens that if power levels are above about 10 kW, it is
more practical and efficient to arrange ac circuits in a
“polyphase” configuration. Although any number of
“phases” are possible, “3-phase” is almost exclusively used
in high power applications, since it is the simplest case that
achieves most of the advantage of polyphase.
It is virtually certain that some 3-phase problems will appear
on the FE and PE examinations, which is why 3-phase
merits our attention.
21
A single-phase ac circuit
a
+
n
Generator
Line
Load
“a” is the “phase” conductor
“n” is the “neutral” conductor
For a given load, the phase a conductor must have a crosssectional area “A”, large enough to carry the requisite
current. Since the neutral carries the return current, we need
a total of “2A worth” of conductors.
Tripling the capacity
+
+
+
Ia
Ib
Ic
In
If I a  I b  I c  I  then I n  3I 
We need a total of A + A +A + 3A = 6A conductors.
22
But what if the currents are not in phase?
Suppose I a  I 00 I b  I  1200 I c  I   1200
Then
I n  I a  I b  I c  I 00  I   1200  I   1200
I n  I 1  j0   0.5  j 0.866   0.5  j0.866 
I n  I 1.0  0.5  0.5)  j(0.0  0.866  0.866   I (0  j 0)  0
Now we only need a total of A + A +A + 0 = 3A conductors!
A 50% savings!
The 3-Phase Situation
FE: Electric Circuits
© C.A. Gross
EE1- 46
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"Balanced" voltage means equal in
magnitude, 120o separated in phase
v an ( t )  V max cos( t ) 
2  V  cos( t )
vbn ( t )  V max cos( t  120 0 ) 
2  V  cos( t  120 0 )
vcn ( t )  V max cos( t  120 0 ) 
2  V  cos( t  120 0 )
Van  V 00
Vbn  V   1200
Vcn  V   1200
FE: Electric Circuits
© C.A. Gross
EE1- 47
The “Line” Voltages
By KVL Vab  Van  Vbn

 1
3 
0
Vab  V 00  V   1200  V 1  j 0     j
   V 330
2
2


 
Vbc  V 3  900
Vca  V 31500
Vab  Vbc  Vca  VL  V 3
FE: Electric Circuits
© C.A. Gross
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When a power engineer says “the
primary distribution voltage is 12 kV”
he/she means…
An Example
Vab  12.47300 kV
Vbc  12.47  900 kV
Vab  Vbc  Vca  VL  12.47 kV
Vca  12.47  1500 kV
Van  Vbn  Vcn 
Van  7.2 00 kV
VL
 7.2 kV
3
FE: Electric Circuits
Vbn  7.2  1200 kV
Vcn  7.2  1200 kV
© C.A. Gross
bc
EE1- 49
An Important Insight….
All balanced three-phase problems can be
solved by focusing on a-phase, solving the
single-phase (a-n) problem, and using 3-phase
symmetry to deal with b-n and c-n values!
This involves judicious use of
the factors 3, 3, and 1200!
To demonstrate…
FE: Electric Circuits
© C.A. Gross
bc
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25
Recall the pf Correction Problem
An Electric Utility supplies 7.2 kV to a single-phase customer
whose load is 7.2 kV 1300 kVA @ pf = 0.3845 lagging.

V  7.200 kV

104 
j 43.2 
1300 kVA
1200 kvar
I  I R  I L  181  67 A
0
FE: Electric Circuits
© C.A. Gross
500 kW
bc
EE1- 51
The pf Correction Problem in the 3-phase case
An Electric Utility supplies 12.47 kV to a 3-phase customer
whose load is 12.47 kV 3900 kVA @ pf = 0.3845 lagging.

Van  7.200 kV
104 
j 43.2 

3900 kVA
3600 kvar
I a  181  670 A
3 times bigger!
1500 kW
FE: Electric Circuits
© C.A. Gross
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If we want all the V’s, I’s, and S’s
Van  7.2 00 kV
Vab  12.47300 kV
Vbn  7.2  1200 kV
Vbc  12.47  900 kV
Vcn  7.2  1200 kV
Vca  12.47  1500 kV
I a  181  670 A
Sa  500  j1200 kVA
I b  181  187 A
Sb  500  j1200 kVA
I c  181  53 A
Sc  500  j1200 kVA
0
0
FE: Electric Circuits
© C.A. Gross
EE1-
PF Correction: the 3ph solution
Install 2700 kvar of
Capacitance.
2700 kvar
3600 kvar
The circuitry in the 3phase case is a bit
more complicated.
There are two
possibilities….
900 kvar
1500 kW
FE: Electric Circuits
© C.A. Gross
EE1- 54
27
The wye connection….
S
O
U
R
C
E
a
L
O
A
D
b
c
n
Install three 900 kvar
7.2 kV wye-connected
Capacitors.
FE: Electric Circuits
© C.A. Gross
EE1- 55
The delta connection….
S
O
U
R
C
E
a
L
O
A
D
b
c
n
Install three 900 kvar 12.47 kV delta-connected
Capacitors.
FE: Electric Circuits
© C.A. Gross
EE1- 56
28
Z  3 ZY
wye-delta connections
wye case
2700
Qan 
 900 kvar
3
Ia 
delta case
Qan 
Qan 900

 125 A
Van 7.2
Zan  ZY 
CY 
Iab 
Van
 57.6 
Ia
2700
 900 kvar
3
Qab 900

 72.17 A
Vab 12.47
Zab  Z 
1
 46.05  F
 ZY
C 
FE: Electric Circuits
12.47
 172.8 
72.17
1
 15.35 F
 Z
© C.A. Gross
EE1- 57
1st Order Transients…..
i t
Network B
Network A

contains
dc sources,
resistors,
one switch
vt

contains
one energy
storage
element
(L or C)
The problem….(1) solve for v and/or i @ t < 0; (2) switch is
switched @ t = 0; (3) solve for v and/or i for t > 0
FE: Electric Circuits
© C.A. Gross
EE1- 58
29
The inductive case
vL  L 
di L
dt
L's are SHORTS to dc
iL(t) cannot change in zero time
i L (0 )  i L (0)  i L (0 )
FE: Electric Circuits
© C.A. Gross
EE1- 59
An Example…
vL  L 
di L
dt
FE: Electric Circuits
i L (0 )  i L (0)  i L (0 )
© C.A. Gross
EE1- 60
30
Solution....
 constant 
vC (t )  vC ( )  constant 
t  0:
vC (t )  vC (0)
t :
0 t  :
vC (t )  vC ( )  vC (0)  vC ( )   e  t /
  Rab  C
Our job is to determine
vC (0); vC ( ); and  Rab  C
FE: Electric Circuits
© C.A. Gross
EE1- 61
Solution....
For a
capacitor:
iC  C 
dvC
dt
C's are OPENS to dc
vC(t) cannot change in zero time
Therefore, if the circuit is switched at t = 0:
vC (0 )  vC (0)  vC (0 )
FE: Electric Circuits
© C.A. Gross
EE1- 62
31
Solution: T < 0; switch and "C" OPEN

+
120V
-


A
vC  0  
a
0
+
vC
b
120
12   48 V
12  6  12
FE: Electric Circuits
© C.A. Gross
EE1- 63
Solution: T > 0; switch CLOSED

+
120V
-


a
+
vC
-
A
vC    
iC
b
Rab 
6  12
4
6  12
0.2 F
  Rab C
 4(0.2)  0.8 s
120
12   80 V
0  6  12
FE: Electric Circuits
© C.A. Gross
EE1- 64
32
Solution....
vC (0)  48 vC ( )  80
t  0:
vC (t )  80   48  80   e 1.25 t
vC (t )  80  32  e 1.25 t
FE: Electric Circuits
© C.A. Gross
EE1- 65
3. 1st Order Transients: RL
0.4 H
b. The switch is closed at t = 0. Find and
plot iL (t).
FE: Electric Circuits
© C.A. Gross
EE1- 66
33
Solution....
t  0:
i L (t )  i L (0)
i L (t )  i L ( )   i L (0)  i L ( )  e  t /
t  0:

L
Rab
Our job is to determine
i L (0); i L ( ); and  L / Rab
FE: Electric Circuits
© C.A. Gross
EE1- 67
Solution....
For an
inductor:
vL  L 
di L
dt
L's are SHORTS to dc
iL(t) cannot change in zero time
i L (0 )  i L (0)  i L (0 )
FE: Electric Circuits
© C.A. Gross
EE1- 68
34
Solution: T < 0; switch OPEN; L SHORT

+
120V
-

a
+
vC
-

A
iL  0  
iL
b
120
 6.667 A
12  6
FE: Electric Circuits
© C.A. Gross
EE1- 69
Solution: T > 0; switch CLOSED

+
120V
-

a
iL
+
vL
-

A
b
120
iL    
 20 A
060
FE: Electric Circuits
© C.A. Gross
Rab 
6  12
4
6  12
0.4 H

L
0.4

4
Rab
 0.1 s
EE1- 70
35
Solution....
t  0:
i L (t )  6.667
t  0:
i L (t )  20   6.667  20   e  t /
i L (t )  20  13.33  e 10 t
FE: Electric Circuits
© C.A. Gross
EE1- 71
5. Control
Given the following feedback control system:
R(s)
+
C(s)
G(s)
H(s)
G ( s) 
1
( s  1)( s  4)
FE: Electric Circuits
© C.A. Gross
H ( s)  K
EE1- 72
36
a. Write the closed loop transfer function in
rational form
1
C
G
( s  1)( s  4)


K
R 1  GH 1 
( s  1)( s  4)
C
1
1

 2
R ( s  1)( s  4)  K s  3s  ( K  4)
FE: Electric Circuits
© C.A. Gross
EE1- 73
b. Write the characteristic equation
s 2  3s  ( K  4)  0
c. What is the system order?
2
d. For K = 0, where are the poles located?
s 2  3s  4   s  1   s  4   0
s = +1; s = - 4
e. For K = 0, is the system stable?
FE: Electric Circuits
© C.A. Gross
NO
EE1- 74
37
s 2  3s  ( K  4)  0
f. Complete the table
Roots of the CE are poles of the CLTF
K
poles
0
4
damping
 4 .0 0,  1 .0 0
 3 .0 0, 0 .0 0
u n stab le
o ver
5
 2 .6 2,  0 .3 8 2
o ver
6 .2 5  1 .5 0,  1 .5 0
critical
1 0 .2 5  1 .5  j 2,  1 .5  j 2 u n d er
FE: Electric Circuits
© C.A. Gross
EE1- 75
f. Sketch the root locus
j
g. Find the range on K
for system stability.

-4
+1
If K = 4:
s 2  3s  0   s    s  3   0
Poles at s = 0; s = - 3
Therefore for K>4, poles are in LH s-plane
and system is stable.
K≥4
FE: Electric Circuits
© C.A. Gross
EE1- 76
38
h. Find K for critical damping
CE :
s 2  3s  ( K  4)  0
Solving the CE: s 
3  9  4( K  4)
2
Critical damping occurs when the poles are
real and equal
9  4( K  4)  0
K  4  9 / 4;
K  4  2.25  6.25
FE: Electric Circuits
© C.A. Gross
EE1- 77
6. Signal Processing
a. periodic time-domain functions have
continuous discrete
frequency spectra.
(circle the correct adjective)
b. aperiodic time-domain functions have
continuous discrete
frequency spectra.
(circle the correct adjective)
FE: Electric Circuits
© C.A. Gross
EE1- 78
39
c. Matching
d.
Laplace Transform
d
Fourier Transform
Fourier Series
a
Convolution integral b
Inverse FT
e
N
 D exp( jn t)
a. x(t ) 
n
n  N
t
b. y(t ) 
 x( )  h(t  )  d

e. x(t ) 
1
2
0
FE: Electric Circuits
0

 x(t )  e
c. X ( j ) 

d. X ( s )   x(t )  e  st  dt
c


 j t
 X ( j )  e
 dt
 j t
 d

© C.A. Gross
EE1- 79
c. Matching
d.
Z-Transform
d
DFT
Inverse ZT
a
Discrete Convolution b
Inverse DFT
e
b.
y[ k ] 
c
a. X ( j) 

 x[n]  e
 jn
n 
k

x[ n ]  h[ n  k ]
n  
N 1
c. X k   x[ n]  e  j 2 kn / N
n0

d. X ( z )   x[ n]  z  n
n0
FE: Electric Circuits
e. x[n] 
© C.A. Gross
1
N
N 1
X
k 0
k
 e  j 2 kn / N
EE1- 80
40
7. Electronics

e (t )

v
v
e
e (t )  169.7  sin( t )
T
a. Darken the conducting diodes at time T
FE: Electric Circuits
© C.A. Gross
EE1- 81
b. Given the "OP Amp" circuit
Ideal OpAmp....
•infinite input resistance
• zero input voltage
•infinite gain
• zero output resistance
FE: Electric Circuits
© C.A. Gross
EE1- 82
41
Find the output voltage.
vi  5 V
Ri  10 k 
R f  50 k 
KCL :
 Rf 
v0   
  vi
 Ri 
 50 
v0      5  25 V
 10 
FE: Electric Circuits
vi
v
 0 0
Ri R f
© C.A. Gross
EE1- 83
c. Find the output voltage.
40k
0 +
5 +
5 +
20k
10k
KCL :
v
v1 v2 v3


 0 0
R1 R2 R3 R f
10k
+
+
vo LOAD
-
FE: Electric Circuits
© C.A. Gross
EE1- 84
42
"SUMMER"
Solution:
0
5
5 v

  0 0
40 20 10 10
40k
0 +
5 +
5 +
20k
10k
10k
 10 
 10 
 10 
v0      5     5     0
 10 
 20 
 40 
+
+
vo LOAD
v0  7.5 V
-
FE: Electric Circuits
© C.A. Gross
EE1- 85
8. Digital Systems Logic Gates
FE: Electric Circuits
© C.A. Gross
EE1- 86
43
a. Complete the Truth Table
C
A
0
Half Adder (HA)
AND
S
B
0
XOR
A
B
0
0
1
1
0
1
0
1
C
0
0
0
1
A + B = CS
S
0
1
1
0
FE: Electric Circuits
0 + 0 = 00
0 + 1 = 01
1 + 0 = 01
1 + 1 = 10
© C.A. Gross
EE1- 87
b. Complete the indicated row in the TT
A1
0
0
1
1
0
0
1
1
B1 C2 S1
0 0
0
0 1
1
0 1
0
1 0
1
0 1
0
1 0
1
1 0
0
1 1
1
FE: Electric Circuits
A
11
C0
A
HA S 1
B
01
HA
1C1
OR
1
C2
1
C
0
C1
0
0
0
0
1
1
1
1
S1
Full Adder (FA)
© C.A. Gross
EE1- 88
44
c. Indicate the inputs and outputs to perform
the given sum in a 4-bit adder
1 1 1
0 1 1
1 1 0
0 0
A3 B3 C2
A2 B2 C1
A1 B1 C0
A0 B0
FA
FA
FA
HA
C3
S3
S2
S1
S0
1
1
0
0
0
FE: Electric Circuits
© C.A. Gross
1010
+1110
11000
EE1- 89
d. Design a D/A Converter to accomodate 3-bit
digital inputs (5 volt logic)
Resolution: 3-bits
(23 = 8 levels;
10 V scale)
Example...
Convert "110"
to analog
FE: Electric Circuits
Digital
Analog (V)
000
0.00
001
1.25
010
2.50
011
3.75
100
5.00
101
6.25
110
7.50
111
8.75
Binary Word: ABC
(A msb; C lsb)
© C.A. Gross
EE1- 90
45
d. Finished Design ABC = 110
40k
0 C+
5 B+
5 A+
20k
10k
10
10
10
5
5
0
10
20
40
 7.50
v0  
10k
+
+
vo LOAD
-
FE: Electric Circuits
© C.A. Gross
EE1- 91
Good Luck on the Exam!
If I can help with any ECE material, come
see me (7:30 - 11:00; 1:15 - 2:30)
Charles A. Gross, Professor Emeritus
Electrical and Comp Engineering
Auburn University Broun 212
334.844.1812
gross@eng.auburn.edu
Good Evening...
FE: Electric Circuits
© C.A. Gross
EE1- 92
46