ELEC 202 Electric Circuit Analysis II Lecture 10(a) Complex Arithmetic and Rectangular/Polar Forms

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(a) Complex Arithmetic and Rectangular/Polar Forms THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Real vs. Complex Numbers j 1 The product of a real number & the operator j is an imaginary number. 2 3 j, j, j, 5.1 j 7 The sum of a real number & an imaginary number is a complex number, z. where the real part is denoted and the imaginary part is denoted a Re j j b z Re 2 4 2, Im 2 4 4 Im 24 j, 1 j z rectangular form 2

Addition & Subtraction Graphical addition & subtraction are performed like vector addition ( tip-to-tail ). M N 31j 22j M N 51j Algebraic addition & subtraction are performed piece-wise: M a b j 1 1 N a b j 2 2 M N a a b b j 1 2 1 2 3

Multiplication: Rectangular Form Multiplication may be accomplished in rectangular form z a b j 1 1 1 z a b j 2 2 2 z z a b j a b j 1 2 1 1 2 2 a a a b j a b j b b j 1 2 1 2 2 1 1 2 a1a2 a1b 2 a2b1 j b1b 2 a a b b a b a b j 1 2 1 2 1 2 2 1 2 M N 5 3 2 4 53j M N j j 2 24j 10 20 j 6 j 12 j 22 14 j but it is more easily accomplished in polar form. 4

Exponential Form e j cos jsin j z e z cos j z sin j z e a b j assume z is positive, real a b z z cos sin sin cos tan b a 2 2 2 2 2 2 a b z cos z sin 2 2 2 2 2 2 a b z cos sin z 2 2 a b z 5

Rectangular Exponential Form z a b Re Im z z tan b a z a b 2 2 M 43j z 2 2 4 3 5 1 tan 3 4 37 z = magnitude of z = phase/angle of z M 5e j37 6

Polar Form z a b j z e z j rectangular exponential polar Polar form is a shorthand for the exponential form. z = magnitude of z = phase/angle of z M j e 37 43 5 j 537 7

Multiplication: Polar Form Multiplication in polar form is carried out using exponentials z a b j z e 1 1 1 1 2 2 2 2 j z a b j z e 1 j 2 z z z e z e j 1 2 1 2 z z 1 2 1 2 z e z e 1 2 j j j 1 2 1 2 j z z, z z z z z z 1 1 1 2 2 1 1 2 1 2 1 2 M N 3 4 j 553 3 3 j 345 2 2 5 53 3 45 M N 1598 8

Division: Polar Form z z, z z 1 1 1 2 2 1 j1 j2 z e z z e 1 1 1 j2 j2 z e z e z 2 2 2 e j j 1 2 z z z 11 1 1 2 22 z2 M N 6 8 j 1053 5 5 j 545 2 2 10 53 5 45 M N 28 9

Complex Conjugate The complex conjugate of z is denoted z and if z a b j then z a b j M 31j The conjugate of z is the same number, except that the imaginary part is negated. M 31j Graphically, the complex conjugate of z is the mirror image of z across the Real axis. 10

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(b) Complex Arithmetic Examples and Matlab Scripts THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Example: Complex Addition 9 2 j A 31j A 2 6 j A >> i1 = 9 + 2*j; >> i2 = -3 + j; >> i3 = -2 + 6*j; >> i = i1 + i2 + i3 i = 4.0000 + 9.0000i 12

Example: Complex Addition 9 2 j A 31j A 2 6 j A 4 9 j A >> i1 = 9 + 2*j; >> i2 = -3 + j; >> i3 = -2 + 6*j; >> i = i1 + i2 + i3 i = 4.0000 + 9.0000i 13

Example: Complex Multiplication Find v x i in rectangular form: v i 7 3 j m 5 4 j ma 7 3 5 4 v i j j 35 15 j 28 j 12 j 47 13 j μw 2 v 29 j, i 3 5 j A 2 9 3 5 v i j j 6 27 j10 j 45 5117 j W >> = 2 + 9*j; >> I = -3 + 5*j; >> p = * I p = -51.0000-17.0000i 14

Example: Complex Multiplication Find v x i in rectangular form: v i 7 3 j m 5 4 j ma 7 3 5 4 v i j j 35 15 j 28 j 12 j 47 13 j μw 2 v 29 j, i 3 5 j A 2 9 3 5 v i j j 6 27 j10 j 45 5117 j W >> = 2 + 9*j; >> I = -3 + 5*j; >> p = * I p = -51.0000-17.0000i 15

Example: Complex Arithmetic Determine the quantity v a v b in polar form if v n = 0. v a v b 110 0 110120 v n 110 240 v c >> v_a = 110*exp(j*0); >> v_b = 110*exp(j*-2*pi/3); >> v = v_a - v_b; >> abs(v) ans = 190.5256 >> angle(v)*180/pi ans = 30.0000 16

Example: Complex Arithmetic Determine the quantity v a v b in polar form if v n = 0. v a v b 110 0 110120 v n v a v 1100 110 120 b 110cos0 110sin 0 190.5 30 j j 110 cos 120 110sin 120 110 j0 110 1 2 j 110 3 2 165 j 55 3 2 2 165 55 3 tan 1 55 3 165 17 110 240 >> v_a = 110*exp(j*0); v c >> v_b = 110*exp(j*-2*pi/3); >> v = v_a - v_b; >> abs(v) >> angle(v)*180/pi ans = 190.5256 ans = 30.0000

Example: Complex Division Determine the ratio of L to I L : I L L L L IL IL + L 3 3 3 j m 1 j 3 ma I L >> _L = -3*sqrt(3)+3*j; >> I_L = 1+j*sqrt(3); >> Z = _L / I_L Z = -0.0000 + 3.0000i >> abs(z) ans = 3.0000 >> angle(z)*180/pi ans = 90.0000 18

Example: Complex Division Determine the ratio of L to I L : I L L L L IL IL + L 3 3 3 j m 1 j 3 ma I L I L L 3 3 3 j m 1 j 3 ma 2 2 1 2 2 1 3 tan 1 3 3 3 3 tan 1 3 6150 2 60 390 >> _L = -3*sqrt(3)+3*j; >> I_L = 1+j*sqrt(3); >> Z = _L / I_L Z = -0.0000 + 3.0000i >> abs(z) ans = 3.0000 >> angle(z)*180/pi ans = 90.0000 19

Example: Complex Conjugate Write the quantity x I* in polar form, given 35 j I 6 7 j ma >> = 3-5*j; >> I = 6 + 7*j; >> S = * conj(i) S = -17.0000-51.0000i >> abs(s) ans = 53.7587 >> angle(s)*180/pi ans = -108.4349 20

Example: Complex Conjugate Write the quantity x I* in polar form, given I 3 5 j6 7 j 18 30 j 21j 35 17 51j mw I 17 51 tan 51 17 I 2 2 1 53.8 108 mw 34 59 85 49 35 j I 6 7 j ma >> = 3-5*j; >> I = 6 + 7*j; >> S = * conj(i) S = -17.0000-51.0000i >> abs(s) ans = 53.7587 >> angle(s)*180/pi ans = -108.4349 3485 59 49 53.8 108 mw 21

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(c) Sinusoids and Sinusoidal Steady-State THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Alternating Current (Sinusoidal) v t sin t, 0 m 0 0 + vt A A B B m = amplitude (in olts), 0 = phase (in radians) = frequency (in radians/second) T = period (in seconds) f = frequency (in cycles/second) = 1/T = / 2 sin t cos t 2 m 0 m 0 23

Review of Sinusoids sin, sin v t t v t t 1 m 2 m v 2 leads v 1 by v 1 lags v 2 by 24

Sinusoids & Exponential Form sin, sin v t t v t t 1 m 2 m e j cos jsin sin j t e cos t j t m m m Re jt me mcos t Im jt me msin t 25

RL Circuit with a Sinusoidal Source cos 0 t i d R i t i t 0 cos t dt L L -- oscillates forever -- never settles to a DC value (e.g. zero) It s possible that the solution is of the form i t I cos t 0 Substituting i(t) into the differential equation R I 0sin t I0cos t cos t L L 0 Solving for I 0 and substituting back into i(t) yields 0 1 cos t tan i t R L 2 2 2 L R amplitude scaling, phase shift 26

transient RL Circuit with a Sinusoidal Source v s i L v s L 400 μh 5t 5 5 1μs R 2kΩ The RL circuit s transient response is negligible after 5t. The remaining response is sinusoidal. i L 27

transient RC Circuit with a Sinusoidal Source v s + v C v s 5t 5 RC 5 2 kω 100 pf 1μs The RC circuit s transient response is negligible after 5t. The remaining response is sinusoidal. v C 28

transient RLC Circuit with a Sinusoidal Source + v C i s t cos sin v t e t t c 1 d 2 d 3 1 1 rad 3125 2RC 210 16 μf s t s 5 1.6 ms The RLC circuit s transient response is negligible after t s. The remaining response is sinusoidal. v C 29

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(d) Phasors THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Phasor Notation I cos m t 2 Im 2 cos m t 1 m 1 j t 1 j1 j1 cos t Re e Re e e e m 1 m m m m 1 jt jt j t 2 j2 j2 I cos t Re I e Re I e e I e I m 2 m m m m 2 Assume all voltages & currents oscillate with frequency = 2 f Pick off the amplitude & phase for each v/i ; write each in polar form. 31 phasor notation (longer derivation provided in textbook section 10.3)

Example: Phasor vs. Time Domain Let all phasors be referenced to a cosine (zero phase). Convert the following time-domain functions to the phasor domain : (a) 40cos(t + 30 ) m, = 100 rad/s (b) 25cos(t 75 ) A, = 400 rad/s (c) 70cos(2 f + 45 ), f = 20 MHz (d) 36sin(2 f + 110 ) ma, f = 8 MHz (a) (b) (c) (d) t t sin cos 2 32

Example: Phasor vs. Time Domain Let all phasors be referenced to a cosine (zero phase). Convert the following time-domain functions to the phasor domain : (a) 40cos(t + 30 ) m, = 100 rad/s (b) 25cos(t 75 ) A, = 400 rad/s (c) 70cos(2 f + 45 ), f = 20 MHz (d) 36sin(2 f + 110 ) ma, f = 8 MHz (a) 4030 m (b) 25 75 A (c) 7045 (d) 3620 ma t t sin cos 2 33

Example: Phasors and oltage/current Let = 2000 rad/s with phasors be referenced to a cosine (zero phase). Determine the instantaneous value, at t = 1 ms, of the current corresponding to this phasor: 20 + 10j A. omega = 2000; I = 20 + 10*j; T = 2*pi/omega; delta_t = T/1000; t = 0:delta_t:2*T; i_t = abs(i)*cos(omega*t + angle(i)); plot(t,i_t) grid axis([0 2*T -Inf Inf]) ylabel('current (A)') xlabel('time (s)') 34

Example: Phasors and oltage/current Let = 2000 rad/s with phasors be referenced to a cosine (zero phase). Determine the instantaneous value, at t = 1 ms, of the current corresponding to this phasor: 20 + 10j A. 20 10 j 20 10 tan 10 20 22.426.6 A 2 2 1 omega = 2000; I = 20 + 10*j; t 1 ms 22.4cos 2.464 i t i 22.4 cos 2000 0.464 A 17.5 A T = 2*pi/omega; delta_t = T/1000; t = 0:delta_t:2*T; i_t = abs(i)*cos(omega*t + angle(i)); plot(t,i_t) grid axis([0 2*T -Inf Inf]) ylabel('current (A)') xlabel('time (s)') 35

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(e) Phasors & Ohm s Law, KL/KCL THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Phasor oltage vs. Current: R, L, C it I t v t cos m t 1 m cos 2 v t Ri t For this equation to be true, I R and m m 1 2 d vlt L il t dt m Im 1 2 90, L 2 1 90, C I m d ict C vc t dt For this equation to be true, For this equation to be true, m v and i are v leads i in phase by 90 I = jc i leads v by 90 = R I = jl I 37

Example: KCL, Phasor Domain Determine v(t). 6 i t 8cos 2 10 t 30 ma s 70 10 mh 2 nf + vt 38

Example: KCL, Phasor Domain Determine v(t). 6 i t 8cos 2 10 t 30 ma s 70 10 mh 2 nf + vt Convert to phasor form I 8 s 30 ma Employ the appropriate Kirchhoff Law(s) I R I L I C 70 10 mh 2 nf + I s jc 0 R jl 6 9 830 j 2 10 210 0 70 j 2 10 10 10 6 6 39

Example: KCL, Phasor Domain Determine v(t). 6 i t 8cos 2 10 t 30 ma s 70 10 mh 2 nf + vt Convert between rectangular & polar forms as necessary 6 9 830 j 2 10 210 0 70 j 2 10 10 10 6 6 1 1 j 0.0126 830 70 j 62.8 0.0143 0.0159 j 0.0126 j 830 0.0143 0.0033 j 830 0.0147 13 830 40

Example: KCL, Phasor Domain Determine v(t). 6 i t 8cos 2 10 t 30 ma s 70 10 mh 2 nf + vt omega = 2*pi*10^6; I = 8*exp(j*30*pi/180); R = 70; L = 10e-6; C = 2e-9; Y = (1/R + 1/(j*omega*L) + j*omega*c); = I / Y; abs() angle()*180/pi ans = 545.2174 ans = 43.1941 41

Example: KCL, Phasor Domain Determine v(t). 6 i t 8cos 2 10 t 30 ma s 70 10 mh 2 nf + vt Convert between rectangular & polar forms as necessary 0.0147 13 830 830 ma 54443 m 0.0147 13 Convert from phasors to time domain 6 t v t 544cos 2 10 43 m omega = 2*pi*10^6; I = 8*exp(j*30*pi/180); R = 70; L = 10e-6; C = 2e-9; Y = (1/R + 1/(j*omega*L) + j*omega*c); = I / Y; abs() angle()*180/pi ans = 545.2174 ans = 43.1941 42

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(f) Impedance THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Impedance Impedance, Z is the ratio of phasor voltage to phasor current for an electrical element or network. like resistance, but it is complex -- It is a measure of an element/network s ability to impede current flow. RI Z Z For a resistor, R I -- current and voltage are always in-phase -- there is no frequency dependence R For an inductor, jl I Z L jl -- voltage always leads current by 90 -- at higher frequencies, less current is passed (for constant ) For a capacitor, I jc Z 1 44 C jc j C -- current always leads voltage by 90 -- at higher frequencies, more current is passed (for constant )

Impedance vs. Frequency 2 f R = 100 Z Z R R R R Z 0 R Z L Z L jl Z 90 L L = 3 mh L Z C Z C Z 1 jc C 1 C C = 300 pf 90 45

Impedance vs. Frequency freq = 2e6:1e3:10e6; omega = 2*pi*freq; f 2 MHz 10 MHz, 2 f R = 100; Z_R = R*ones(1,length(omega)); R 100, ZR R L = 3e-6; Z_L = j*omega*l; L 3 μh, ZL jl C = 300e-12; Z_C = 1./(j*omega*C); C 300 pf, Z 1 C jc figure(1) subplot(2,1,1) plot(freq/10^6,abs(z_r)) axis([-inf Inf 0 300]) set(gca,'xtick',[2:1:10]) ylabel(' Z_R (ohms)') title('{\bf Resistor Impedance... vs. Frequency}') subplot(2,1,2) plot(freq/10^6,phase(z_r)*180/pi) axis([-inf Inf -180 180]) set(gca,'xtick',[2:1:10]) ylabel('phase[ Z_R ] (degrees)') xlabel('frequency (MHz)') figure(2) subplot(2,1,1) plot(freq/10^6,abs(z_l)) axis([-inf Inf 0 300]) set(gca,'xtick',[2:1:10]) ylabel(' Z_L (ohms)') title('{\bf Inductor Impedance... vs. Frequency}') subplot(2,1,2) plot(freq/10^6,phase(z_l)*180/pi) axis([-inf Inf -180 180]) set(gca,'xtick',[2:1:10]) ylabel('phase[ Z_L ] (degrees)') xlabel('frequency (MHz)') figure(3) subplot(2,1,1) plot(freq/10^6,abs(z_c)) axis([-inf Inf 0 300]) set(gca,'xtick',[2:1:10]) ylabel(' Z_C (ohms)') title('{\bf Capacitor Impedance... vs. Frequency}') subplot(2,1,2) plot(freq/10^6,phase(z_c)*180/pi) axis([-inf Inf -180 180]) set(gca,'xtick',[2:1:10]) ylabel('phase[ Z_C ] (degrees)') xlabel('frequency (MHz)') 46

Impedances in Series & Parallel I I N n1 s Z n s Z 1 Z 2 Z 3 1 2 3 Impedances in series are combined like resistors in series. R s N R n1 n N n1 I 1 s Z n Impedances in parallel are combined like resistors in parallel. Z N N Z s N Z n1 n N 1 1 R p n1 R n N 1 1 Z p n1 Z n 47

Example: Equivalent Impedance Determine the equivalent impedance of the network at terminals A B if = 5 rad/s. A 4 33.3 mf 1.6 H 2 100 mf B 48

Example: Equivalent Impedance Determine the equivalent impedance of the network at terminals A B if = 5 rad/s. A 4 33.3 mf 1.6 H 2 100 mf B Convert all resistances, inductances, capacitances into impedances A 6j 8j Z Z C Z L R j R jl C B 4 2 2j 49

Example: Equivalent Impedance Combine impedances in series & parallel, starting away from A B and working towards A B Z 1 2 2j 4 90 2 45 1 j 2 2j 8 45 Z2 Z1 8j 6j 1 j 8 j 6 j 1 j Z 1 50

Example: Equivalent Impedance Combine impedances in series & parallel, starting away from A B and working towards A B Z AB Z Z 2 2 4 1 j 4 4 1 j 4 4 4 j 3245.0 5 j 2611.4 1.1 33.6 Z 2 If a test source were applied at terminals A B : Itest 10 A then the voltage across A B would be 51 test 10 1.133.6 1.133.6

Example: Equivalent Impedance Determine the equivalent impedance of the network at terminals A B if = 5 rad/s. A 4 33.3 mf 1.6 H 2 100 mf B omega = 5; C1 = 100e-3; C2 = 33.3e-3; R1 = 2; R2 = 4; L1 = 1.6; Z_C1 = 1/(j*omega*C1); Z_C2 = 1/(j*omega*C2); Z_R1 = R1; Z_R2 = R2; Z_L1 = j*omega*l1; Z_eq1 = Z_C1*Z_R1/(Z_C1 + Z_R1); Z_eq2 = Z_eq1 + Z_L1 + Z_C2; Z_ab = Z_R2*Z_eq2 / (Z_R2 + Z_eq2) Z_ab = 0.9217 + 0.6120i abs(z_ab) ans = 1.1063 phase(z_ab)*180/pi ans = 33.5837 52

Example: Phasor Analysis (a) Determine v C (t). (b) Plot v C (t) and the source versus time for 2 cycles. 3 cos(2 10 6 t + 20 ) 1 k + v C 1 nf 53

Example: Phasor Analysis (a) Determine v C (t). (b) Plot v C (t) and the source versus time for 2 cycles. 3 cos(2 10 6 t + 20 ) 1 k + v C 1 nf v S v C 54

Example: Phasor Analysis (a) Determine v C (t). (b) Plot v C (t) and the source versus time for 2 cycles. 3 cos(2 10 6 t + 20 ) 1 k + v C 1 nf Convert the circuit from the time domain to the phasor domain. ZC j C j j 6 9 2 10 10 160 Ω 320 I 1 k + C 160j Use KL/KCL to solve for /I in the phasor domain. I j 160 ji 320 I 1000 160 0 C 55

Example: Phasor Analysis (a) Determine v C (t). (b) Plot v C (t) and the source versus time for 2 cycles. 320 I 1 k + C 160j Perform complex algebra to find /I C 160 j 160 90 320 320 1000 160 j 1000 160 tan 160 1000 160 90 320 0.472 60.9 1013 9.1 2 2 1 Convert back to the time domain 56 6 v t 472cos 2 10 t 60.9 m C

Example: Plotting Sinusoids, Matlab (a) Determine v C (t). (b) Plot v C (t) and the source versus time for 2 cycles. 3 cos(2 10 6 t + 20 ) 1 k + v C 1 nf omega = 2*pi*10^6; T = 2*pi/omega; delta_t = T/100; t = 0 : delta_t : 2*T - delta_t ; v_s = 3.000 * cos( omega*t + 20.0*pi/180 ); v_c = 0.472 * cos( omega*t - 60.9*pi/180 ); figure(1) plot(t/10^-6, v_s, 'r--',... t/10^-6, v_c, 'b-', 'LineWidth', 2) ylabel('oltage ()') xlabel('time (\mus)') legend('v_s','v_c') grid 6 v t 472cos 2 10 t 60.9 m C 57

Example: Transient AC Circuit, PSpice Amplitude is 472 m as determined by written analysis. 6 v t 472cos 2 10 t 60.9 m C 58

Example: Frequency Response, Matlab Plot the amplitude of v C (t) versus f 0 for 500 khz < f 0 < 2 MHz. Z C 9 9 10 j j C j 10 Ω 3 cos(2 10 6 t + 20 ) 1 k + v C 1 nf C 9 ZC 10 j 320 320 9 1000 Z 1000 10 j C _s = 3 * exp( j*20*pi/180 ); R = 1000; C = 1e-9; f = 500e3 : 1e3 : 2e6 ; 6 6 for f 10 Hz ( 210 rad s)... 6 v t 472cos 2 10 t 60.9 m C omega = 2 * pi * f; Z_C = -j./ (omega * C); _C = _s * Z_C./ (R + Z_C); plot(f/10^3, abs(_c), 'LineWidth', 2) ylabel(' _C (volts)') xlabel('frequency (khz)') grid Frequency (khz) 59

Example: Frequency Response, PSpice Plot the amplitude of v C (t) versus f 0 for 500 khz < f 0 < 2 MHz. AC part, SOURCE library 60

Example: Frequency Response, PSpice Plot the amplitude of v C (t) versus f 0 for 500 khz < f 0 < 2 MHz. AC part, SOURCE library 6 6 for f 10 Hz ( 210 rad s)... 6 v t 472cos 2 10 t 60.9 m C 61

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(g) Nodal & Mesh Analysis in the Phasor Domain THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Nodal Analysis with Phasors v 1 v 2 v 3 1 2 3 0 0 Analysis Steps SAME as with DC circuits. Now use complex arithmetic. (1) Choose a reference node (usually ground or the bottom node) to have a voltage of zero. (2) Assign a unique voltage variable to each node that is not the reference (v 1, v 2, v 3, v N 1 ). (3) For independent & dependent voltage sources, identify a supernode and write the voltage across the supernode in terms of node voltages. Write a KCL equation at all N 1 nodes including the supernode (and not the reference, or a supernode which includes the reference). (4) Solve the N 1 node equations + source equations simultaneously. 63

Example: Nodal Analysis, Phasors Write a valid matrix equation whose solution includes the phasor form of v(t). 0.5i A + v(t) i A 64

Example: Nodal Analysis, Phasors Write a valid matrix equation whose solution includes the phasor form of v(t). 0.5i A v 2 Convert to phasor form L 3 2070 10 1.4 3 Z jl j j Z j C j 20 25010 0.2 j C v 1 v 3 v 4 i A + v(t) Identify supernode(s) 1 2 0.5I A Identify v sources next to the reference 9, 9 j 1 4 Write KCL equations 0.2 j 1.4 j 0.2 j 3 3 1 3 2 3 4 3 0 Write equations governing dep. src. ctrl I A 3 3 65 1 0.5I A 0 0.2 j 2 1.4 j 3 4 0.2 j I A 9 0 9 90 0 +

Example: Nodal Analysis, Phasors Write a valid matrix equation whose solution includes the phasor form of v(t). 0.5I A 2 0.2 j 1.4 j 0.2 j 3 3 1 3 2 3 4 3 0 9, 9 j 1 2 0.5I A 1 4 Rearrange into matrix form j j 1 5 j 1 2 5 j 5 j 3 5 j 4 0 1.4 1.4 3 I A 3 3 1 0.2 j 1.4 j 3 4 0.2 j I A 9 0 9 90 0 + 1 4 9 9j 1 2 0.5I A 0 3 0 3 I A 2 j 1 5 j 9.3 j 5 j 0 1 0 1.4 3 9 1 0 0 0 0 3 9 j 0 0 0 1 0 4 0 1 1 0 0 0.5 I A 0 0 0 1 3 0 1 66 8.35.5 t v t 8.3cos 20 5.5

Example: Nodal Analysis, Phasors Write a valid matrix equation whose solution includes the phasor form of v(t). 0.5I A 2 0.2 j 1.4 j 0.2 j 3 3 1 3 2 3 4 3 0 9, 9 j 1 4 I A I 1 2 0.5 A 3 3 1 0.2 j 1.4 j 3 4 0.2 j I A 9 0 9 90 0 + >> A = [ -5*j j/1.4 9.29*j+1/3-5*j 0 ; 1 0 0 0 0 ; 0 0 0 1 0 ; 1-1 0 0-0.5 ; 0 0-1/3 0 1 ]; >> B = [0 ; 9 ; -9*j ; 0 ; 0 ]; >> x = A^-1 * B >> = x(2); >> abs() >> angle() * 180/pi x = ans = ans = 9.0000-0.0000i 8.2703 + 0.7913i 4.3784-4.7477i 0.0000-9.0000i 1.4595-1.5826i 8.3080 5.4653 8.35.5 67

Mesh Analysis with Phasors Analysis Steps SAME as with DC circuits. Now use complex arithmetic. (1) Draw a mesh current for each mesh. (2) Identify supermeshes. i 1 i 2 (3) Write KL around each supermesh, then KL for each mesh that is not part of a supermesh. (4) Express additional unknowns (e.g. dependent-source /I) in terms of mesh currents. (5) Solve the simultaneous equations. I 1 I 2 68

Example: Mesh Analysis, Phasors Write a valid matrix equation whose solution includes the phasor form of v(t). 0.5i A + v(t) i A 69

Example: Mesh Analysis, Phasors Write a valid matrix equation whose solution includes the phasor form of v(t). 0.5i A Convert to phasor form 3 3 Z j 20 7010 1.4 j L Z j 20 250 10 0.2 j C i 2 i 3 + v(t) Identify supermesh(es) i 1 i 4 i A Write KL equations 0.5I A j I1 I2 I1 I4 I A j I2 I3 j I2 I1 3 2 3 3 4 I I j I I j 9 0.2 3 0 0.5 1.4 0.2 0 1.4 j I I 5I 9 j 0.2 j I I 0 3 0.2 9 0 4 1 4 3 Write equations governing I A I1I4 dep. src. ctrl 9 0 I 2 1.4 j I 3 0.2 j 0.2 j I 1 I 4 I A 9j + 70

Example: Mesh Analysis, Phasors Write a valid matrix equation whose solution includes the phasor form of v(t). 0.5I A j I1 I2 I1 I4 I A j I2 I3 j I2 I1 3 2 3 3 4 I I j I I j 9 0.2 3 0 0.5 1.4 0.2 0 1.4 j I I 5I 9 j 0.2 j I I 0 3 0.2 9 0 4 1 4 3 I I I 5I A 1 4 3 9 0 0.2 j I 2 1.4 j I 3 0.2 j I 1 I 4 I A 9j + Rearrange into matrix form j I1 j I2 I4 1 2 3 A 2 3 4 0.2 3 0.2 3 9 0.2 j I 1.2 j I 1.4 j I 0.5 I 0 1.4 j I 1.2 j 5 I 0.2 j I 9 j 3 I 0.2 j I 3 0.2 j I 9 j 1 3 4 I I I 0 1 4 5I 0 3 A 0.2 j 3 0.2 j 0 3 0 0 I1 9 0.2 j 1.2 j 1.4 j 0 0.5 0 I 0 2 0 1.4 j 1.2 j 5 0.2 j 0 0 I 3 9 j 3 0 0.2 j 30.2 j 0 0 I4 9 j 1 0 0 1 1 0 I A 0 0 0 5 0 0 1 0 8.35.5 vt 8.3cos 20t 5.5 71

Example: Mesh Analysis, Phasors Write a valid matrix equation whose solution includes the phasor form of v(t). 0.5I A j I1 I2 I1 I4 I A j I2 I3 j I2 I1 3 2 3 3 4 I I j I I j 9 0.2 3 0 0.5 1.4 0.2 0 1.4 j I I 5I 9 j 0.2 j I I 0 3 0.2 9 0 4 1 4 3 I I I 5I A 1 4 3 9 0 0.2 j I 2 1.4 j I 3 0.2 j I 1 I 4 I A 9j + A = [ -0.2*j+3 0.2*j 0-3 0 0 ; 0.2*j 1.2*j -1.4*j 0 0.5 0 ; 0-1.4*j 1.2*j+5 0.2*j 0 0 ; -3 0 0.2*j 3-0.2*j 0 0 ; 1 0 0-1 -1 0 ; 0 0-5 0 0 1 ]; B = [ 9 ; 0 ; -9*j ; 9*j ; 0 ; 0 ]; x = A^-1 * B = x(6); abs() angle() * 180/pi 72 x = -18.1368 +20.4776i 5.6121-2.6198i 1.6540 + 0.1583i -19.5970 +22.0609i 1.4602-1.5833i 8.2699 + 0.7916i ans = ans = 8.3077 5.4679 8.35.5

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(h) Thevenin Equivalence in AC Circuits THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Review of Thevenin Equivalents allows us to replace a large, complicated circuit with a much simpler 2-element series/parallel circuit the simpler circuit allows for rapid calculations of, I, P that the original circuit can deliver to a load helps us to choose the best value of load resistance to maximize the power delivered (e.g. from an amplifier, to a speaker) 74

Review of Thevenin Equivalents To determine TH... YES only ind. src.? NO source transformation find TH = OC find TH = OC To determine Z TH... YES only ind. src.? NO deactivate src., find Z TH = Z eq source transformation find I SC, use Z TH = OC /I SC insert test source, find Z TH = test /I test NO TH = 0? YES find I SC, use Z TH = OC /I SC insert test source, find Z TH = test /I test insert test source, find Z TH = test /I test 75

Review of Thevenin Equivalents Determine the Thevenin equivalent of Network A using open-circuit voltage and short-circuit current. I SC + OC I Z SC OC TH 6 6 12 12 8 = 6 3 9 1 7 8 4 A 1 2 1 7 9 OC 9 I SC 76 8 89 TH

Review of Thevenin Equivalents Determine the Thevenin equivalent of Network A by using a test source. + OC 8 TH I test 2 I test test test I test test 2 7 I test test 9 Z TH 77

Example: Thevenin & Sinusoids Determine the phasor voltage difference 1 2 (with j10 ). Use the Thevenin equivalent at 1, 2 (without j10 ). 78

Example: Thevenin & Sinusoids Determine the phasor voltage difference 1 2 (with j10 ). Use the Thevenin equivalent at 1, 2 (without j10 ). A B j j 4 2 j 0.5 j2 4 j 4 2 j j 2 6 3 j TH OC A B 10 4 2 0.5 90 2 4 ZTH 4 2 j 2 4 j 6 2 j 1 2 6 3j 10 j 6 2 j10 j 3 6 j 6.7 63.4 79

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(i) Phasors & Superposition THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Linearity 1 x t linear network N y t x t network y t 1 2 linear N 2 x = input or source or stimulus x t t N y t t 1 1 1 1 y = output or response A x t 1 1 1 t N A y t t N 1 1 1 A x t t A x t t 1 1 1 2 2 2 N A y t t A y t t 1 1 1 2 2 2 81

Superposition: oltage Sources Determine the current i using superposition. deactivate all but 1 solve sum 16 32 2 k 4 2 k i 16 4 4 ma i 32 4 8 ma 8 i i i i i 4 8 2 1 15 ma 60 i 8 4 2 ma i 4 4 1mA 82 i 60 4 15 ma

Superposition: Current Sources Determine the voltage v using superposition. 6 A 3 12 A 6 24 A deactivate all but 1 solve sum v v 6 0 3 6 v 12 v v 12 0 3 6 v 24 v v 24 0 3 6 v 48 83 v 12 24 48 60

Example: Superposition & Sinusoids Determine the phasor voltage difference 1 2. Use superposition. 84

Example: Superposition & Sinusoids Determine the phasor voltage difference 1 2. Use superposition. 1 2 1 2 1 2 1 2 1 2 4 2j 1 2 1 10 j 2 4 j 4 2 j 2 6 j 2 4j 1 2 0.5 j 10 j 1 2 j 2 4 j 4 12 j 1 2 3 6 j 9 36 tan 2 6.7 63.4 1 85

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(x,1) Thevenin Equivalent Example THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Example: Thevenin & Sinusoids, #2 Determine the Thevenin equivalent circuit with respect to terminals a b. 87

Example: Thevenin & Sinusoids, #2 Determine the Thevenin equivalent circuit with respect to terminals a b. Z TH 5 j10 5 j 5 j 10 5 j Zeq 5 j 10 5 j 25 50 j 2.5 5 j 10 Y I Z TH RL OC s YRL YC 1 10 5 j j j j 330 10 5 1 10 5 1 5 13.0 10.6 j RL 88

Example: Thevenin & Sinusoids, #2 Determine the Thevenin equivalent circuit with respect to terminals a b. Z I TH OC SC 13.0 10.6 3 30 2.5 5 j 2.5 5 j 13.0 10.6 j 89

Example: Thevenin & Sinusoids, #2 Determine the Thevenin equivalent impedance with respect to terminals a b. ZTH 2.5 5 j 90

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(j) Phasors & Source Transformation THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Review of Source Transformation If these two circuits provide the same v/i characteristics at their outputs (v L, i L ), the two circuits are equivalent. s I s Imax s Rs max s R s R p 92 I condition for equivalence s s I max I I R s max s p

Source Transformation & Phasors If these two circuits provide the same /I characteristics at their outputs ( L, I L ), the two circuits are equivalent. Zs I L I L s L ZL I s Z p L ZL I Z max s s max s Z s Z p 93 I condition for equivalence s s I max I I Z s max s p

Example: Src. Transform. & Sinusoids Determine the phasor voltage difference 1 2. Use source transformation(s). 94

Example: Src. Transform. & Sinusoids Determine the phasor voltage difference 1 2. Use source transformation(s). S1 Z 1 Z 2 S2 1 2 S1 S2 10 j 10 j Z Z 1 2 Z Z 1 2 4 2j 2 4j j S1 10 4 2 4 2 j j j j j S1 0.5 90 2 4 0.5 2 4 2 1 2 4 2 j j 2 10 j 10 j 4 2 j 2 4 j 10 j 6 3 j 3 6 j 8j 6 6.7 63.4 95

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(k,1) PSpice for AC Circuits THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

Example #1: Written Analysis Solve for v(t). 0.5i A From nodal analysis 0.2 j 1.4 j 0.2 j 3 3 1 3 3 4 3 0 9, 9 j 1 2 0.5I A 1 4 I A 3 3 i A + v(t) j 1 5 j 9.3 j 5 j 0 1 0 1.4 3 9 1 0 0 0 0 3 9 j 0 0 0 1 0 4 0 1 1 0 0 0.5 I A 0 0 0 1 3 0 1 8.35.5 t v t 8.3cos 20 5.5 97 1 0.5I A 0.2 j 2 1.4 j 3 4 0.2 j I A 9 0 9 90 0 +

Example #1: Matlab Solve for v(t). 0.5I A 2 0.2 j 1.4 j 0.2 j 3 3 1 3 2 3 4 3 0 9, 9 j 1 4 Rearrange into matrix form I A I 1 2 0.5 A 3 3 1 0.2 j 1.4 j 3 4 0.2 j I A 9 0 9 90 + >> A = [ -5*j j/1.4 9.29*j+1/3-5*j 0 ; 1 0 0 0 0 ; 0 0 0 1 0 ; 1-1 0 0-0.5 ; 0 0-1/3 0 1 ]; >> B = [0 ; 9 ; -9*j ; 0 ; 0 ]; >> x = A^-1 * B >> = x(2); >> abs() >> angle() * 180/pi 98 x = ans = ans = 9.0000-0.0000i 8.2703 + 0.7913i 4.3784-4.7477i 0.0000-9.0000i 1.4595-1.5826i 8.3080 5.4653 8.35.5 t v t 8.3cos 20 5.5

Example #1: PSpice Plot v(t) using PSpice. Amplitude is 8.4 as determined by written analysis. 99 8.35.5 t v t 8.3cos 20 5.5

Example #2: Written Analysis Determine the phasor voltage difference 1 2. Confirm this answer using PSpice. + Z 1 S1 Z 2 S2 + 1 2 S1 S2 10 j 10 j Z Z 1 2 Z Z 1 2 4 2j 2 4j j S1 10 4 2 4 2 j j j j j S1 0.5 90 2 4 0.5 2 4 2 1 2 4 2 j j 2 10 j 10 j 4 2 j 2 4 j 10 j 6 3 j 3 6 j 8j 6 6.7 63.4 100

Example #2: PSpice Determine the phasor voltage difference 1 2. Confirm this answer using PSpice. IAC part, SOURCE library 1 2 6.7 63.4 101

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(k,2) Op Amps in AC Circuits THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

103 Example #3: Phasors & Op Amps Sketch out / in vs. for 1 rad/s < < 100 krad/s. Determine v out (t) for v in (t) = cos(4x10 4 t). 125 nf 2k 200

104 Example #3: Phasors & Op Amps Sketch out / in vs. for 1 rad/s < < 100 krad/s. Determine v out (t) for v in (t) = cos(4x10 4 t). 125 nf 2k out in Z f Rf 1 jc f Rf 1 jc f Z R R R 1 jc i i i f f Rf 1 Rf C f 4000 10 R j1 R C j4000 i f f 200 out in 10 4000 4000 2 2 out 4 4 410 10 4 10 3 in 7.1 135 j410 4000 j1 4 vout t 7.1cos 4 10 t 135 4000

Dr. Gregory J. Mazzaro Spring 2016 ELEC 202 Electric Circuit Analysis II Lecture 10(k,3) AC Thevenin Equivalent w/ a Dependent Source THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA 171 Moultrie Street, Charleston, SC 29409

106 Example #4: Thevenin, Dependent Src Determine the Thevenin equivalent of this circuit at terminals A B. A B

107 Example #4: Thevenin, Dependent Src Determine the Thevenin equivalent of this circuit at terminals A B. A j j25ia OC 1030 30I 5I 25 I 0 a a a B I a 1030 0.2875 25 j25 OC j 25 0.2875 7.07 15 SC j j I 1030 30 I I 5I 25 I 0 a a a 25 50I 0 a SC Z TH 7.07 15 0.106 2 66.7 13 A 25 j25 30 Ia 1030 j25 50 SC 0 ISC 106 2 ma I 7.0715 B