FUNDAMENTALS OF ENGINEERING (FE ... - Auburn University

FE: Electric Circuits

© C.A. Gross

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FUNDAMENTALS OF ENGINEERING (FE) EXAMINATION REVIEW

ELECTRICAL ENGINEERING Charles A. Gross, Professor Emeritus Electrical and Comp Engineering Auburn University Broun 212 334.844.1812 [email protected] www.railway-technology.com

FE: Electric Circuits

© C.A. Gross

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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

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1. dc Circuits:

Find all voltages, currents, and powers. FE: Electric Circuits

© C.A. Gross

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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

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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

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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

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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

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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

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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 10

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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)

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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   13

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Example:

Z  3  j4

X  e  Z   3

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Y  Im  Z   4 Rect  Polar.....

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0

Z  32  42  5

3

 4   Summer 2008

  tan 1    0.9273 rad  53.10 3 14

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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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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 16

Summer 2008

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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 17

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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 18

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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 19

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 38 10

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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

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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

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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

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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

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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

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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

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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

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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

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Leading, Lagging Concepts Leading Case

I

Q0

I

FE: Electric Circuits

© C.A. Gross

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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

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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

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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

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© 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

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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

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© 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

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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

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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.

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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.

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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

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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

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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

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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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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

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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

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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

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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

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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

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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

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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 [email protected]

Good Evening... FE: Electric Circuits

© C.A. Gross

EE1- 92

46