FE: Electric Circuits C.A. Gross EE1-1 FUNDAMENTALS OF ENGINEERING (FE) EXAMINATION REIEW 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 EE1-2 1
EE Review Problems 1. dc Circuits 2. Complex Numbers 3. ac Circuits 4. 3-phase Circuits 1 st Order Transients Control Signal Processing Electronics Digital Systems We will discuss these. We may discuss these as time permits FE: Electric Circuits C.A. Gross EE1-3 1. dc Circuits: Find all voltages, currents, and powers. FE: Electric Circuits C.A. Gross EE1-4 2
Solution The 8 and 7 resistors are in series: R18715 R1 and 1 are in parallel: 1 R2 1 1 1 R1 1R1 6 1 R1 FE: Electric Circuits C.A. Gross EE1-5 Solution 4 and R2 are in series: 4 R21 Rab L : ab 1 Ia 1 A R 1 ab 4 4Ia 4 ( L) 1 1 4 6 KL FE: Electric Circuits C.A. Gross EE1-6 3
Solution 1 6 Ic 6A L 1 1 KCL: I I I 1 6 4A b a c 8 8Ib 32 ( L) 7 7Ib 28 ( L) FE: Electric Circuits C.A. Gross EE1-7 Absorbed Powers... 2 R I W 2 4 a 4 1 4 2 R I W 2 1 c 1 6 36 2 R I W 2 7 b 7 4 112 2 R I W 2 8 b 8 4 128 Total Absorbed Power 1W Power Delivered by Source I 1 1 1W s a In General: P ABS = P DE (Tellegen's Theorem) FE: Electric Circuits C.A. Gross EE1-8 4
2. Complex Numbers 2 Consider x 2x5 x 21 2 4 1 1 12 1 2 2 2 ( 2) 4(1)(5) 2 16 The numbers "1 2 1" are called complex numbers 9 Summer 28 The I (j) operator Math Department... Define i 1 x 12i ECE Department Define j 1 x 1 j2 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 1 Summer 28 5
Polar Form Math Department... Z i Re a complex number R Z modulus of Z Z Z arg argument of radians ECE Department... Z Z a complex number Z Z magnitude of Z Z Z ang angle of 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 j2 1 real axis (X) 12 Summer 28 6
Conversions Retangular Polar Z X Y tan 2 2 Y X 1 Y Z Polar Retangular... X Zcos Y Zsin X 13 Summer 28 Example: Z 3 j4 X e Z 3 Y Im Z 4 4 5 14 Rect Polar... Z Summer 28 2 2 3 4 5 4 tan.9273 rad 53.1 3 1 3 7
Conjugate Z X jy Z Z* conjugate of Z X jy Z Example... (3 j4)* 3 j4 553.1 15 Summer 28 Addition (think rectangular) Aa jb A 3 j4553.1 Bc jd B 5 j12 1367.4 A B( a jb) ( c jd) ( ac) j( bd) A B (3 j4) (5 j12) (3 5) j(4 12) 8 j8 16 Summer 28 8
Multiplication (think polar) Aa jb A 3 j4553.1 Bc jd B 5 j12 1367.4 AB( A ) ( B ) AB( ) AB (5 53.1 ) (13 67.4 ) (5) (13) (53.1 67.4 ) 6514.3 17 Summer 28 Division (think polar) Aa jb A 3 j4553.1 Bc jd B 5 j12 1367.4 A A A ( ) B B B A 553.1 5 B 1363.4 13 53.1 67.4.384612.5 18 Summer 28 9
Multiplication (rectangular) ( ) ac bd j ad bc A B a jb c jd AB(3 j4) (5 j12) (15 48) j( 36 2) 63 j16 6514.3 19 Summer 28 EL EC 38 1 Addition (polar) 553.1 Complex number addition is the same as "vector addition"! 11.31 45 2 Summer 28 13 67.4 EL EC 38 1 1
3. ac Circuits v(t) - i(t) 8.663 mf 26.53 mh Find "everything" in the given circuit. vt ( ) 141.4cos(377 t) FE: Electric Circuits C.A. Gross EE1-21 Frequency, period vt ( ) 141.4cos(377 t) (radian) frequency 377 rad / s (cyclic) frequency f 6 Hz 2 1 1 Period 16.67 ms f 6 FE: Electric Circuits C.A. Gross EE1-22 11
The ac Circuit To solve the problem, we convert the circuit into an "ac circuit": RLC,, elements Z(impedance) vi, sources, I(phasors) R : Z R j8 j R L: ZL j L j(.377)(26.53) j1 C: 1 1 ZC j j4 jc.377(.663) FE: Electric Circuits C.A. Gross EE1-23 The Phasor vt () cos( t ) MAX To convert to a phasor... For example.. MAX 2 vt ( ) 141.4cos(377 t) MAX 1 2 o FE: Electric Circuits C.A. Gross EE1-24 12
The "ac circuit" 1 I R C L( ac): I Z 1 8 j1 j4 136.9 o L o it ( ) 14.14 cos(377t 36.9 ) FE: Electric Circuits C.A. Gross EE1-25 Solving for voltages o o Z I (8)(136.9 ) 836.9 R R o v ( t) 113.1cos(377t 36.9 ) R o o Z I ( j4)(136.9 ) 4126.9 C C o v ( t) 56.57 cos(377t 126.9 ) C o o Z I ( j1)(136.9 ) 153.1 L L o v ( t) 141.4 cos(377t 53.1 ) L FE: Electric Circuits C.A. Gross EE1-26 13
Absorbed powers S I* P jq o o S I* 836.9 (136.9 )* 8 j R R o o S I* 4126.9 (136.9 )* j4 C C o o S I* 153.1 (136.9 )* j1 L L S S S S 8 j6 P TOT R C L TOT 8 watts; Q 6 vars; S S 1A TOT TOT TOT FE: Electric Circuits C.A. Gross EE1-27 Delivered power o o S I* 1 (136.9 )* 8 j6 S S S S 8 j6 P Q S S S P Q TOT TOT TOT 8 watts 6 vars In General: P ABS = P DE Q ABS = Q DE (Tellegen's Theorem) FE: Electric Circuits C.A. Gross EE1-28 14
The Power Triangle S 8 j6 S = 1 A = 36.9 Q = 6 var 1 o P = 8 W I 136.9 o P power factor pf cos( ).8 lagging S FE: Electric Circuits C.A. Gross EE1-29 Leading, Lagging Concepts Leading Case I Lagging Case I Q < Q > FE: Electric Circuits C.A. Gross EE1-3 15
A Lagging pf Example 7.2 k R jx R 13.68 jx j43.2 FE: Electric Circuits C.A. Gross EE1-31 Currents I R I I R I L R jx 7.2 69.44 A R 13.68 7.2 IL j166.7 A jx j43.2 I I I R I 69.44 j166.7 I 18.667.38 L A FE: Electric Circuits C.A. Gross EE1-32 16
Powers pf cos cos 67.36.3845 I R I R jx S * 5 R IR kw j I L 13 ka 12 k var S I * j12 kvar S I* S S S L L S R L S 5 j12 1367.38 5 kw FE: Electric Circuits C.A. Gross EE1-33 Add Capacitance I I R I L jx C IC j125 I C I 69.44 R 7.2 I C j125 A jx j57.6 I I I I R L C I 69.44 j166.7 j125 8131 A IL j166.7 FE: Electric Circuits C.A. Gross EE1-34 17
Powers pf cos cos 31.8575 9 kvar 12 kvar S * 9kvar C IC j S I* S S S SS 5 j12 j9 S 583.131 ka S R L C S 583.1 ka 5 kw 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 After Note that: I 18.6 A; pf.3845 I 81 A; pf.8575 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 EE1-36 18
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 C.A. Gross EE1-37 The Power Factor Correction problem Utility Load pf correcting capacitance An Electric Utility supplies 7.2 k to a customer whose load is 7.2 k 13 ka @ pf =.3845 lagging. The utility offers the customer a reduced rate if he will correct ( improve or raise ) his pf to.8575. Determine the requisite capacitance. FE: Electric Circuits C.A. Gross EE1-38 19
PF Correction: the solution 1. Draw the load power triangle. 13 ka @ pf =.3845 lagging. S S pf.3845 cos 67.38 LOAD LOAD S 1367.38 5 j12 12 kvar 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. 67.38 5 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.8575 lagging. pf.8575 cos 31 Closing the switch (inserting the capacitors) S 5 j12 jq 5 j 12 Q S C C Let QX 12 QC Therefore 5 31 SS jqx SS ka Q X Tan Then Tan 31.6 5 FE: Electric Circuits C.A. Gross EE1-4 2
PF Correction: the solution Q X.6 Q 3 kvar X 5 Q C 12 Q 9 kvar The new source power triangle Install 9 kvar of 7.2 k Capacitors X 9 kvar 3 kvar 12 kvar 5 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 1 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 I a I b I c I n If I I I I then I 3I a b c n We need a total of A A A 3A = 6A conductors. 22
But what if the currents are not in phase? Suppose I I I I12 I I12 Then a b c I I I I I I12 I12 n a b c 1.5.866.5.866 In I j j j In I 1..5.5) j(..866.866 I( j) Now we only need a total of A A A = 3A conductors! A 5% savings! The 3-Phase Situation FE: Electric Circuits C.A. Gross EE1-46 23
"Balanced" voltage means equal in magnitude, 12 o separated in phase v ( t) cos( t) 2 cos( t) an max v t t t bn( ) max cos( 12 ) 2 cos( 12 ) v t t t cn ( ) max cos( 12 ) 2 cos( 12 ) an bn cn 12 12 FE: Electric Circuits C.A. Gross EE1-47 The Line oltages By KL ab an bn 1 3 ab 12 1 j j 33 2 2 bc ca 39 315 ab bc ca L 3 FE: Electric Circuits C.A. Gross EE1-48 24
An Example When a power engineer says the primary distribution voltage is 12 k he/she means 12.47 k ab bc ca L ab bc ca 12.473 12.479 12.4715 k k k L an bn cn 7.2 k 3 an bn cn 7.2 k 7.212 7.212 k k FE: Electric Circuits C.A. Gross EE1-49 bc 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 12! To demonstrate FE: Electric Circuits C.A. Gross EE1-5 bc 25
Recall the pf Correction Problem An Electric Utility supplies 7.2 k to a single-phase customer whose load is 7.2 k 13 ka @ pf =.3845 lagging. 7.2 k 14 j43.2 I IR IL 18167 A 13 ka 12 kvar 5 kw FE: Electric Circuits C.A. Gross EE1-51 bc The pf Correction Problem in the 3-phase case An Electric Utility supplies 12.47 k to a 3-phase customer whose load is 12.47 k 39 ka @ pf =.3845 lagging. an I a 7.2 k 18167 A 14 j43.2 39 ka 3 times bigger! 36 kvar 15 kw FE: Electric Circuits C.A. Gross EE1-52 26
If we want all the s, I s, and S s an bn cn I I I a b c 7.2 k 7.212 7.212 18167 181187 18153 k k A A A ab bc ca 12.473 12.479 12.4715 k k k S 5 j12 ka a S 5 j12 ka b S 5 j12 ka c FE: Electric Circuits C.A. Gross EE1- PF Correction: the 3ph solution Install 27 kvar of Capacitance. The circuitry in the 3- phase case is a bit more complicated. There are two possibilities. 15 kw 27 kvar 9 kvar 36 kvar FE: Electric Circuits C.A. Gross EE1-54 27
The wye connection. S O U R C E a b c n L O A D Install three 9 kvar 7.2 k wye-connected Capacitors. FE: Electric Circuits C.A. Gross EE1-55 The delta connection. S O U R C E a b c n L O A D Install three 9 kvar 12.47 k delta-connected Capacitors. FE: Electric Circuits C.A. Gross EE1-56 28
wye-delta connections Z 3 ZY wye case 27 Qan 9 kvar 3 I a Z C Qan 9 125 A 7.2 an Y an an ZY 57.6 I 1 Z Y a 46.5 F I Z C delta case 27 Qan 9 kvar 3 ab ab Qab 9 72.17 A 12.47 ab 12.47 Z 172.8 72.17 1 15.35 F Z FE: Electric Circuits C.A. Gross EE1-57 1 st Order Transients.. Network A contains dc sources, resistors, one switch it v t Network B contains one energy storage element (L or C) The problem.(1) solve for v and/or i @ t < ; (2) switch is switched @ t = ; (3) solve for v and/or i for t > FE: Electric Circuits C.A. Gross EE1-58 29
The inductive case di L vl L dt L's are SHORTS to dc i L (t) cannot change in zero time i ( ) i () i ( ) L L L FE: Electric Circuits C.A. Gross EE1-59 An Example di L vl L dt i ( ) i () i ( ) L L L FE: Electric Circuits C.A. Gross EE1-6 3
Solution... t : v ( t) v () constant C t : v ( t) v ( ) constant C C C t : vc( t) vc( ) vc() vc( ) e R C ab Our job is to determine v (); v ( ); and R C C C ab t / FE: Electric Circuits C.A. Gross EE1-61 Solution... For a capacitor: dv C ic C dt C's are OPENS to dc v C (t) cannot change in zero time Therefore, if the circuit is switched at t = : v ( ) v () v ( ) C C C FE: Electric Circuits C.A. Gross EE1-62 31
Solution: T < ; switch and "C" OPEN a vc 12 - A v C - 12 12 48 12 6 12 b FE: Electric Circuits C.A. Gross EE1-63 Solution: T > ; switch CLOSED vc 12 - A 12 612 12 8 a v C - b i C R ab.2 F 612 4 612 RabC 4(.2).8 s FE: Electric Circuits C.A. Gross EE1-64 32
Solution... v C () 48 v ( ) 8 C 1.25t 1.25 t : v ( t) 8 48 8 e C v () t 832e C t FE: Electric Circuits C.A. Gross EE1-65 3. 1 st Order Transients: RL.4 H b. The switch is closed at t =. Find and plot i L (t). FE: Electric Circuits C.A. Gross EE1-66 33
Solution... t : i ( t) i () L L t / t : i ( t) i ( ) i () i ( ) e L R ab L L L L Our job is to determine i (); i ( ); and L/ R L L ab FE: Electric Circuits C.A. Gross EE1-67 Solution... For an inductor: di L vl L dt L's are SHORTS to dc i L (t) cannot change in zero time i ( ) i () i ( ) L L L FE: Electric Circuits C.A. Gross EE1-68 34
Solution: T < ; switch OPEN; L SHORT a i L 12 - A il 12 6.667 12 6 A v C - b FE: Electric Circuits C.A. Gross EE1-69 Solution: T > ; switch CLOSED a i L R ab 612 4 612 12 - A il 12 2 A 6 v L - b.4 H L R ab.1 s.4 4 FE: Electric Circuits C.A. Gross EE1-7 35
Solution... t : i ( t) 6.667 L t / t : i ( t) 2 6.667 2 e L i ( t) 2 13.33e L 1t FE: Electric Circuits C.A. Gross EE1-71 5. Control Given the following feedback control system: R(s) - G(s) H(s) Gs 1 () () ( s 1)( s 4) Hs K C(s) FE: Electric Circuits C.A. Gross EE1-72 36
a. Write the closed loop transfer function in rational form 1 C G ( s1)( s4) R 1 GH K 1 ( s 1)( s 4) C 1 1 R s s K s s K 2 ( 1)( 4) 3 ( 4) FE: Electric Circuits C.A. Gross EE1-73 b. Write the characteristic equation 2 s s K 3 ( 4) c. What is the system order? 2 d. For K =, where are the poles located? 2 s 3s4 s1 s4 s = 1; s = - 4 e. For K =, is the system stable? NO FE: Electric Circuits C.A. Gross EE1-74 37
f. Complete the table 2 s s K 3 ( 4) Roots of the CE are poles of the CLTF K poles damping 4., 1. unstable 4 3.,. over 5 2.62,.382 over 6.25 1.5, 1.5 critical 1.25 1.5 j2, 1.5 j2 under FE: Electric Circuits C.A. Gross EE1-75 f. Sketch the root locus -4 j 1 g. Find the range on K for system stability. If K = 4: 2 s 3s s s3 Poles at s = ; 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 s K 2 : 3 ( 4) 3 9 4( K 4) Solving the CE: s 2 Critical damping occurs when the poles are real and equal 9 4( K 4) K 49/4; K 42.256.25 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 Fourier Series Inverse FT t b. yt ( ) x( ) ht ( ) d st d. X( s) x( t) e dt d a e Fourier Transform c Convolution integral b N a. x( t) Dn exp( jn t) nn jt c. X( j) x( t) e dt 1 jt e. x( t) X( j) e d 2 FE: Electric Circuits C.A. Gross EE1-79 c. Matching d. Z-Transform d Inverse ZT a Inverse DFT e d. X( z) x[ n] z k b. y[ k] x[ n] h[ n k] n n n DFT c Discrete Convolution b a. X( j) x[ n] e N 1 n n c. X x[ n] e k N 1 1 e. xn [ ] Xk e N k jn j2 kn/ N j2 kn/ N FE: Electric Circuits C.A. Gross EE1-8 4
7. Electronics et () v v e et ( ) 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. v 5 525 1 v 5 i R 1 k R i f KCL v 5 k v R i : R f Ri i v R v i f FE: Electric Circuits C.A. Gross EE1-83 c. Find the output voltage. 5 4k 2k 1k KCL : v v v v R R R R 1 2 3 1 2 3 f 5 1k - v o LOAD - FE: Electric Circuits C.A. Gross EE1-84 42
Solution: "SUMMER" 5 5 4k 2k 1k 1k - v o 5 5 v 4 2 1 1 v LOAD 1 1 1 5 5 1 2 4 v 7.5 - 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 A C AND Half Adder (HA) B S XOR A B C S A B = CS = 1 1 1 = 1 1 1 1 = 1 1 1 1 1 1 = 1 FE: Electric Circuits C.A. Gross EE1-87 b. Complete the indicated row in the TT C 1 A 1 B 1 C 2 S 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 A C A1 1 1 HA S 1 B OR C 2 1 1 C HA 1C 1 Full Adder (FA) S 1 FE: Electric Circuits 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 1 1 1 1 A 3 B 3 C 2 A 2 B 2 C 1 A 1 B 1 C A B FA FA FA HA 11 111 11 C 3 S 3 S 2 S 1 S 1 1 FE: Electric Circuits C.A. Gross EE1-89 d. Design a D/A Converter to accomodate 3-bit digital inputs (5 volt logic) Resolution: 3-bits (2 3 = 8 levels; 1 scale) Example... Convert "11" to analog Digital Analog (). 1 1.25 1 2.5 11 3.75 1 5. 11 6.25 11 7.5 111 8.75 Binary Word: ABC (A msb; C lsb) FE: Electric Circuits C.A. Gross EE1-9 45
d. Finished Design ABC = 11 5 C B 4k 2k 1k 1 1 1 v 5 5 1 2 4 7.5 5 A 1k - v o 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 (7:3-11:; 1:15-2:3) 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