Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10EE201 Lecture 31 P. 1PhasorsPhasors: Complex number representations of sinusoidal signals at fixed frequency, .Complex number = Aejt+ = A(ejt ej)But if is constant: Aej ≡ A ImReFor A=1ej = cos+ j sinsin cos Unit circlee j = e j e-j = cos 2 + sin 2 =1EE201 Lecture 31 P. 2A completely specifies the quantity A cos (t+) for known .Phasor voltage : V = Vm Phasor current: I = Im Denoted by bold type in text.v(t)= -15 sin (t - 60)v(t)= 15 cos (t + 30)i(t) = 25 cos (t + 45)V = 15 30 I = 25 45 Ohm’s Law Relationships for SSSExists for resistors, inductors & capacitors.General form: V = z (j) I z (j)- impedance (units: ohms)EE201 Lecture 31 P. 3Impedance concept for resistorsRV R= z R (j) I R V R= R I RiR(t)=IRe j(t+)IRvR(t)=RIRe j(t+)VR+_vR(t)z R (j) =RResistor impedance is independent of frequencyImpedance concept for inductorsiL(t) = ILe j(t+)ILvL(t) = VLe j(t+)VL+_EE201 Lecture 31 P. 4From I-V relationship for inductorsVLe j(t+) = L [ ILe j(t+)]ddtVLe j(t+) = jL ILe j(t+)zL(j) = jLWhen = 0, inductor behaves as a short.When = , inductor behaves as an openConsequence of multiplying IL by jIL= IL(cos + jsin )VL = jL IL = L IL(-sin + j cos )Taking dot period: -sin cos + sin cos=0Therefore, VL and IL are orthogonal.VL= jL ILEE201 Lecture 31 P. 5ImReVLILVL leads IL by 90IL lags VL by 90Impedance concept for capacitors+_vc(t) = Vce j(t+)VciC(t) = ICe j(t+)ICEE201 Lecture 31 P. 6ICe j(t+) = C [ VCe j(t+)]ddtICe j(t+ ) = jC Vce j(t+)ICe j = jC Vce j IC= jC Vczc(j) = 1/jCWhen = 0, capacitor behaves as an open.When = , capacitor behaves as a short.Consequences of multiplying Ic by 1/(jC)Note : 1 -jj C C=VC= Ic= zc(jw) IC1jCEE201 Lecture 31 P. 7ImReVCI CVC lags IC by 90IC = IC e j = IC(cos + j sin ) I C = (sin - j cos )Therefore, VC and IC are perpendicular.-jCICCIC sin IC cos EE201 Lecture 31 P. 8KCL and KVL for Phasors i2(t) = 4 cos (25t - 60 ) AFind i3(t)I1 = 2 30 = 2(cos 30 + jsin 30) = 1.73 + j1.0I2 = 4 – 60 = 4[cos (-60) +jsin (-60)] =2 - j3.46I3 = I1 – I2 = - 0.27 + j 4.46i1(t)i2(t)i3(t)i1(t) = 2 cos (25t + 30 ) AEE201 Lecture 31 P. 9(-0.27)2+(4.46)2 = 4.47KVL: Find v3(t) by phasor methods.i3(t) = 4.47cos (25t + 93.4 ) = tan –1 (4.46/-0.27) = 93.4 I3 = 4.47 93.4Converting to real sinusoids (i.e. cos function),-+ -+ -+-+v3(t)v2(t)v1(t)vs(t) = 10sin(t)EE201 Lecture 31 P. 10v1(t) = 5 cos (t + 45 ) v2(t) = 6 cos (t - 45 )Convert real sinusoids to phasorsv1(t) = 5 cos (t + 45) V1 = 5 45 =3.53 + j3.53v2(t) = 6 cos (t – 45) V2 = 6 – 45 = 4.24 – j4.24vs(t) = 10 sin (t) 10 cos(t – 90 )Vs = – j10 = 10 –90o Using KVL, V3 = Vs – V1– V2V3 = – j10 – (3.53 + j3.53) – (4.24 – j 4.24)V3 = – 7.77 – j 9.29 = 12.11 – 130v3(t) = 12.11 cos (t – 130) in
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