10: Sine waves and phasors
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- Vivien Holland
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1 E. Analysis of Circuis (26-924) Phasors: /
2 Sine Waves For inducors and capaciorsi = C dv d andv = Ldi d differeniae i() and v() when analysing circuis conaining hem. so we need o E. Analysis of Circuis (26-924) Phasors: 2 /
3 Sine Waves For inducors and capaciorsi = C dv d andv = Ldi d differeniae i() and v() when analysing circuis conaining hem. Usually differeniaion changes he shape of a waveform. so we need o E. Analysis of Circuis (26-924) Phasors: 2 /
4 Sine Waves For inducors and capaciorsi = C dv d andv = Ldi d differeniae i() and v() when analysing circuis conaining hem. Usually differeniaion changes he shape of a waveform. For bounded waveforms here is only one excepion: so we need o v() = sin dv d = cos E. Analysis of Circuis (26-924) Phasors: 2 /
5 Sine Waves For inducors and capaciorsi = C dv d andv = Ldi d differeniae i() and v() when analysing circuis conaining hem. Usually differeniaion changes he shape of a waveform. For bounded waveforms here is only one excepion: so we need o v() = sin dv d = cos v() dv/d E. Analysis of Circuis (26-924) Phasors: 2 /
6 Sine Waves For inducors and capaciorsi = C dv d andv = Ldi d differeniae i() and v() when analysing circuis conaining hem. Usually differeniaion changes he shape of a waveform. For bounded waveforms here is only one excepion: so we need o v() = sin dv d = cos same shape bu wih a ime shif. v() dv/d E. Analysis of Circuis (26-924) Phasors: 2 /
7 Sine Waves For inducors and capaciorsi = C dv d andv = Ldi d differeniae i() and v() when analysing circuis conaining hem. Usually differeniaion changes he shape of a waveform. For bounded waveforms here is only one excepion: so we need o v() = sin dv d = cos same shape bu wih a ime shif. sin complees one full period every imeincreases by2π. v() dv/d E. Analysis of Circuis (26-924) Phasors: 2 /
8 Sine Waves For inducors and capaciorsi = C dv d andv = Ldi d differeniae i() and v() when analysing circuis conaining hem. Usually differeniaion changes he shape of a waveform. For bounded waveforms here is only one excepion: so we need o v() = sin dv d = cos same shape bu wih a ime shif. sin complees one full period every imeincreases by2π. v() dv/d sin2πf makesf complee repeiions every imeincreases by ; his gives a frequency off cycles per second, orf Hz. E. Analysis of Circuis (26-924) Phasors: 2 /
9 Sine Waves For inducors and capaciorsi = C dv d andv = Ldi d differeniae i() and v() when analysing circuis conaining hem. Usually differeniaion changes he shape of a waveform. For bounded waveforms here is only one excepion: so we need o v() = sin dv d = cos same shape bu wih a ime shif. sin complees one full period every imeincreases by2π. v() dv/d sin2πf makesf complee repeiions every imeincreases by ; his gives a frequency off cycles per second, orf Hz. We ofen use he angular frequency, ω = 2πf insead. ω is measured in radians per second. E.g. 5Hz 34rad.s. E. Analysis of Circuis (26-924) Phasors: 2 /
10 Roaing Rod A useful way o hink of a cosine wave is as he projecion of a roaing rod ono he horizonal axis. For a uni-lengh rod, he projecion has lengh cos θ. E. Analysis of Circuis (26-924) Phasors: 3 /
11 Roaing Rod A useful way o hink of a cosine wave is as he projecion of a roaing rod ono he horizonal axis. For a uni-lengh rod, he projecion has lengh cos θ. If he rod is roaing a a speed off revoluions per second, hen θ increases uniformly wih ime: θ = 2πf. E. Analysis of Circuis (26-924) Phasors: 3 /
12 Roaing Rod A useful way o hink of a cosine wave is as he projecion of a roaing rod ono he horizonal axis. For a uni-lengh rod, he projecion has lengh cos θ. If he rod is roaing a a speed off revoluions per second, hen θ increases uniformly wih ime: θ = 2πf. The only difference beweencos andsin is he saring posiion of he rod: E. Analysis of Circuis (26-924) Phasors: 3 /
13 Roaing Rod A useful way o hink of a cosine wave is as he projecion of a roaing rod ono he horizonal axis. For a uni-lengh rod, he projecion has lengh cos θ. If he rod is roaing a a speed off revoluions per second, hen θ increases uniformly wih ime: θ = 2πf. The only difference beweencos andsin is he saring posiion of he rod: v = cos2πf E. Analysis of Circuis (26-924) Phasors: 3 /
14 Roaing Rod A useful way o hink of a cosine wave is as he projecion of a roaing rod ono he horizonal axis. For a uni-lengh rod, he projecion has lengh cos θ. If he rod is roaing a a speed off revoluions per second, hen θ increases uniformly wih ime: θ = 2πf. The only difference beweencos andsin is he saring posiion of he rod: v = cos2πf v = sin2πf E. Analysis of Circuis (26-924) Phasors: 3 /
15 Roaing Rod A useful way o hink of a cosine wave is as he projecion of a roaing rod ono he horizonal axis. For a uni-lengh rod, he projecion has lengh cos θ. If he rod is roaing a a speed off revoluions per second, hen θ increases uniformly wih ime: θ = 2πf. The only difference beweencos andsin is he saring posiion of he rod: v = cos2πf v = sin2πf = cos ( 2πf π 2 ) E. Analysis of Circuis (26-924) Phasors: 3 /
16 Roaing Rod A useful way o hink of a cosine wave is as he projecion of a roaing rod ono he horizonal axis. For a uni-lengh rod, he projecion has lengh cos θ. If he rod is roaing a a speed off revoluions per second, hen θ increases uniformly wih ime: θ = 2πf. The only difference beweencos andsin is he saring posiion of he rod: v = cos2πf v = sin2πf = cos ( 2πf π 2 ) sin2πf lags cos2πf by9 (or π 2 radians) because is peaks occurs 4 of a cycle laer (equivalenly cos leads sin). E. Analysis of Circuis (26-924) Phasors: 3 /
17 Phasors If he rod has lenghaand sars a an angleφhen he projecion ono he horizonal axis is Acos(2πf+φ) E. Analysis of Circuis (26-924) Phasors: 4 /
18 Phasors If he rod has lenghaand sars a an angleφhen he projecion ono he horizonal axis is Acos(2πf+φ) = Acosφcos2πf Asinφsin2πf E. Analysis of Circuis (26-924) Phasors: 4 /
19 Phasors If he rod has lenghaand sars a an angleφhen he projecion ono he horizonal axis is Acos(2πf+φ) = Acosφcos2πf Asinφsin2πf = Xcos2πf Y sin2πf E. Analysis of Circuis (26-924) Phasors: 4 /
20 Phasors If he rod has lenghaand sars a an angleφhen he projecion ono he horizonal axis is Acos(2πf+φ) = Acosφcos2πf Asinφsin2πf = Xcos2πf Y sin2πf A ime =, he ip of he rod has coordinaes (X, Y) = (Acosφ, Asinφ). E. Analysis of Circuis (26-924) Phasors: 4 /
21 Phasors If he rod has lenghaand sars a an angleφhen he projecion ono he horizonal axis is Acos(2πf+φ) = Acosφcos2πf Asinφsin2πf = Xcos2πf Y sin2πf A ime =, he ip of he rod has coordinaes (X, Y) = (Acosφ, Asinφ). If we hink of he plane as an Argand Diagram (or complex plane), hen he complex numberx +jy corresponding o he ip of he rod a = is called a phasor. E. Analysis of Circuis (26-924) Phasors: 4 /
22 Phasors If he rod has lenghaand sars a an angleφhen he projecion ono he horizonal axis is Acos(2πf+φ) = Acosφcos2πf Asinφsin2πf = Xcos2πf Y sin2πf A ime =, he ip of he rod has coordinaes (X, Y) = (Acosφ, Asinφ). If we hink of he plane as an Argand Diagram (or complex plane), hen he complex numberx +jy corresponding o he ip of he rod a = is called a phasor. The magniude of he phasor,a = X 2 +Y 2, gives he ampliude (peak value) of he sine wave. E. Analysis of Circuis (26-924) Phasors: 4 /
23 Phasors If he rod has lenghaand sars a an angleφhen he projecion ono he horizonal axis is Acos(2πf+φ) = Acosφcos2πf Asinφsin2πf = Xcos2πf Y sin2πf A ime =, he ip of he rod has coordinaes (X, Y) = (Acosφ, Asinφ). If we hink of he plane as an Argand Diagram (or complex plane), hen he complex numberx +jy corresponding o he ip of he rod a = is called a phasor. The magniude of he phasor,a = X 2 +Y 2, gives he ampliude (peak value) of he sine wave., gives he phase shif relaive The argumen of he phasor,φ = arcan Y X ocos2πf. Ifφ >, i is leading and ifφ <, i is lagging relaive ocos2πf. E. Analysis of Circuis (26-924) Phasors: 4 /
24 Phasor Examples V =,f = 5Hz E. Analysis of Circuis (26-924) Phasors: 5 /
25 Phasor Examples V =,f = 5Hz v() = cos2πf E. Analysis of Circuis (26-924) Phasors: 5 /
26 Phasor Examples V =,f = 5Hz v() = cos2πf V = j E. Analysis of Circuis (26-924) Phasors: 5 /
27 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf E. Analysis of Circuis (26-924) Phasors: 5 /
28 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j E. Analysis of Circuis (26-924) Phasors: 5 /
29 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j v() = cos2πf+.5sin2πf E. Analysis of Circuis (26-924) Phasors: 5 /
30 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j =.2 53 v() = cos2πf+.5sin2πf E. Analysis of Circuis (26-924) Phasors: 5 /
31 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j =.2 53 v() = cos2πf+.5sin2πf =.2cos(2πf 2.68) E. Analysis of Circuis (26-924) Phasors: 5 /
32 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j =.2 53 v() = cos2πf+.5sin2πf =.2cos(2πf 2.68) V = X +jy E. Analysis of Circuis (26-924) Phasors: 5 /
33 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j =.2 53 v() = cos2πf+.5sin2πf =.2cos(2πf 2.68) V = X +jy v() = Xcos2πf Y sin2πf E. Analysis of Circuis (26-924) Phasors: 5 /
34 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j =.2 53 v() = cos2πf+.5sin2πf =.2cos(2πf 2.68) V = X +jy v() = Xcos2πf Y sin2πf Beware minus sign E. Analysis of Circuis (26-924) Phasors: 5 /
35 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j =.2 53 v() = cos2πf+.5sin2πf =.2cos(2πf 2.68) V = X +jy v() = Xcos2πf Y sin2πf Beware minus sign. V = A φ E. Analysis of Circuis (26-924) Phasors: 5 /
36 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j =.2 53 v() = cos2πf+.5sin2πf =.2cos(2πf 2.68) V = X +jy v() = Xcos2πf Y sin2πf Beware minus sign V = A φ v() = Acos(2πf+φ) E. Analysis of Circuis (26-924) Phasors: 5 /
37 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j =.2 53 v() = cos2πf+.5sin2πf =.2cos(2πf 2.68) V = X +jy v() = Xcos2πf Y sin2πf Beware minus sign V = A φ = Ae jφ v() = Acos(2πf+φ) E. Analysis of Circuis (26-924) Phasors: 5 /
38 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j =.2 53 v() = cos2πf+.5sin2πf =.2cos(2πf 2.68) V = X +jy v() = Xcos2πf Y sin2πf Beware minus sign V = A φ = Ae jφ v() = Acos(2πf+φ) A phasor represens an enire waveform (encompassing all ime) as a single complex number. We assume he frequency, f, is known. E. Analysis of Circuis (26-924) Phasors: 5 /
39 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j =.2 53 v() = cos2πf+.5sin2πf =.2cos(2πf 2.68) V = X +jy v() = Xcos2πf Y sin2πf Beware minus sign V = A φ = Ae jφ v() = Acos(2πf+φ) A phasor represens an enire waveform (encompassing all ime) as a single complex number. We assume he frequency, f, is known. A phasor is no ime-varying, so we use a capial leer: V. A waveform is ime-varying, so we use a small leer: v(). E. Analysis of Circuis (26-924) Phasors: 5 /
40 Phasor Examples V =,f = 5Hz v() = cos2πf V = j v() = sin2πf V =.5j =.2 53 v() = cos2πf+.5sin2πf =.2cos(2πf 2.68) V = X +jy v() = Xcos2πf Y sin2πf Beware minus sign V = A φ = Ae jφ v() = Acos(2πf+φ) A phasor represens an enire waveform (encompassing all ime) as a single complex number. We assume he frequency, f, is known. A phasor is no ime-varying, so we use a capial leer: V. A waveform is ime-varying, so we use a small leer: v(). Casio: Pol(X,Y) A,φ,Rec(A,φ) X,Y. Saved X &Y mems. E. Analysis of Circuis (26-924) Phasors: 5 /
41 Phasor arihmeic Phasors V = P +jq Waveforms v() = P cosω Qsinω whereω = 2πf. E. Analysis of Circuis (26-924) Phasors: 6 /
42 Phasor arihmeic Phasors V = P +jq Waveforms v() = P cosω Qsinω whereω = 2πf. a v() E. Analysis of Circuis (26-924) Phasors: 6 /
43 Phasor arihmeic Phasors V = P +jq Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω E. Analysis of Circuis (26-924) Phasors: 6 /
44 Phasor arihmeic Phasors V = P +jq av Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω E. Analysis of Circuis (26-924) Phasors: 6 /
45 Phasor arihmeic Phasors V = P +jq av Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω v ()+v 2 () E. Analysis of Circuis (26-924) Phasors: 6 /
46 Phasor arihmeic Phasors V = P +jq av V +V 2 Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω v ()+v 2 () E. Analysis of Circuis (26-924) Phasors: 6 /
47 Phasor arihmeic Phasors V = P +jq av V +V 2 Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω v ()+v 2 () Adding or scaling is he same for waveforms and phasors. E. Analysis of Circuis (26-924) Phasors: 6 /
48 Phasor arihmeic Phasors V = P +jq av V +V 2 Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω v ()+v 2 () Adding or scaling is he same for waveforms and phasors. dv d = ωp sinω ωqcosω E. Analysis of Circuis (26-924) Phasors: 6 /
49 Phasor arihmeic Phasors V = P +jq av V +V 2 Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω v ()+v 2 () Adding or scaling is he same for waveforms and phasors. dv d = ωp sinω ωqcosω = ( ωq)cosω (ωp)sinω E. Analysis of Circuis (26-924) Phasors: 6 /
50 Phasor arihmeic Phasors V = P +jq av V +V 2 Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω v ()+v 2 () Adding or scaling is he same for waveforms and phasors. V = ( ωq)+j(ωp) dv d = ωp sinω ωqcosω = ( ωq)cosω (ωp)sinω E. Analysis of Circuis (26-924) Phasors: 6 /
51 Phasor arihmeic Phasors V = P +jq av V +V 2 Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω v ()+v 2 () Adding or scaling is he same for waveforms and phasors. V = ( ωq)+j(ωp) = jω(p +jq) dv d = ωp sinω ωqcosω = ( ωq)cosω (ωp)sinω E. Analysis of Circuis (26-924) Phasors: 6 /
52 Phasor arihmeic Phasors V = P +jq av V +V 2 Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω v ()+v 2 () Adding or scaling is he same for waveforms and phasors. V = ( ωq)+j(ωp) = jω(p +jq) = jωv dv d = ωp sinω ωqcosω = ( ωq)cosω (ωp)sinω E. Analysis of Circuis (26-924) Phasors: 6 /
53 Phasor arihmeic Phasors V = P +jq av V +V 2 Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω v ()+v 2 () Adding or scaling is he same for waveforms and phasors. V = ( ωq)+j(ωp) = jω(p +jq) = jωv dv d = ωp sinω ωqcosω = ( ωq)cosω (ωp)sinω Differeniaing waveforms corresponds o muliplying phasors by jω. E. Analysis of Circuis (26-924) Phasors: 6 /
54 Phasor arihmeic Phasors V = P +jq av V +V 2 Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω v ()+v 2 () Adding or scaling is he same for waveforms and phasors. V = ( ωq)+j(ωp) = jω(p +jq) = jωv dv d = ωp sinω ωqcosω = ( ωq)cosω (ωp)sinω Differeniaing waveforms corresponds o muliplying phasors by jω. E. Analysis of Circuis (26-924) Phasors: 6 /
55 Phasor arihmeic Phasors V = P +jq av V +V 2 Waveforms v() = P cosω Qsinω whereω = 2πf. a v() = ap cosω aqsinω v ()+v 2 () Adding or scaling is he same for waveforms and phasors. V = ( ωq)+j(ωp) = jω(p +jq) = jωv dv d = ωp sinω ωqcosω = ( ωq)cosω (ωp)sinω Differeniaing waveforms corresponds o muliplying phasors by jω. Roae ani-clockwise 9 and scale byω = 2πf. E. Analysis of Circuis (26-924) Phasors: 6 /
56 Complex Impedances Resisor: v() = Ri() E. Analysis of Circuis (26-924) Phasors: 7 /
57 Complex Impedances Resisor: v() = Ri() V = RI E. Analysis of Circuis (26-924) Phasors: 7 /
58 Complex Impedances Resisor: v() = Ri() V = RI V I = R E. Analysis of Circuis (26-924) Phasors: 7 /
59 Complex Impedances Resisor: v() = Ri() V = RI V I = R Inducor: v() = L di d E. Analysis of Circuis (26-924) Phasors: 7 /
60 Complex Impedances Resisor: v() = Ri() V = RI V I = R Inducor: v() = L di d V = jωli E. Analysis of Circuis (26-924) Phasors: 7 /
61 Complex Impedances Resisor: v() = Ri() V = RI V I = R Inducor: v() = L di d V = jωli V I = jωl E. Analysis of Circuis (26-924) Phasors: 7 /
62 Complex Impedances Resisor: v() = Ri() V = RI V I = R Inducor: v() = L di d V = jωli V I = jωl Capacior: i() = C dv d E. Analysis of Circuis (26-924) Phasors: 7 /
63 Complex Impedances Resisor: v() = Ri() V = RI V I = R Inducor: v() = L di d V = jωli V I = jωl Capacior: i() = C dv d I = jωcv E. Analysis of Circuis (26-924) Phasors: 7 /
64 Complex Impedances Resisor: v() = Ri() V = RI V I = R Inducor: v() = L di d V = jωli V I = jωl Capacior: i() = C dv d I = jωcv V I = jωc E. Analysis of Circuis (26-924) Phasors: 7 /
65 Complex Impedances Resisor: v() = Ri() V = RI V I = R Inducor: v() = L di d V = jωli V I = jωl Capacior: i() = C dv d I = jωcv V I = jωc For all hree componens, phasors obey Ohm s law if we use he complex impedances jωl and jωc as he resisance of an inducor or capacior. E. Analysis of Circuis (26-924) Phasors: 7 /
66 Complex Impedances Resisor: v() = Ri() V = RI V I = R Inducor: v() = L di d V = jωli V I = jωl Capacior: i() = C dv d I = jωcv V I = jωc For all hree componens, phasors obey Ohm s law if we use he complex impedances jωl and jωc as he resisance of an inducor or capacior. If all sources in a circui are sine waves having he same frequency, we can do circui analysis exacly as before by using complex impedances. E. Analysis of Circuis (26-924) Phasors: 7 /
67 Phasor Analysis Givenv = sinω whereω = 2π, findv C (). v (ms) E. Analysis of Circuis (26-924) Phasors: 8 /
68 Phasor Analysis Givenv = sinω whereω = 2π, findv C (). () Find capacior complex impedance Z = jωc = 6.28j 4 = 592j v (ms) E. Analysis of Circuis (26-924) Phasors: 8 /
69 Phasor Analysis Givenv = sinω whereω = 2π, findv C (). () Find capacior complex impedance Z = jωc = 6.28j 4 = 592j (2) Solve circui wih phasors V C = V Z R+Z v (ms) E. Analysis of Circuis (26-924) Phasors: 8 /
70 Phasor Analysis Givenv = sinω whereω = 2π, findv C (). () Find capacior complex impedance Z = jωc = 6.28j 4 = 592j (2) Solve circui wih phasors V C = V Z R+Z = j 592j 592j (ms) v E. Analysis of Circuis (26-924) Phasors: 8 /
71 Phasor Analysis Givenv = sinω whereω = 2π, findv C (). () Find capacior complex impedance Z = jωc = 6.28j 4 = 592j (2) Solve circui wih phasors V C = V Z v R+Z = j 592j 592j = j = (ms) E. Analysis of Circuis (26-924) Phasors: 8 /
72 Phasor Analysis Givenv = sinω whereω = 2π, findv C (). () Find capacior complex impedance Z = jωc = 6.28j 4 = 592j (2) Solve circui wih phasors V C = V Z R+Z = j 592j 592j = j = v C = 8.47cos(ω 22 ) C (ms) v v C E. Analysis of Circuis (26-924) Phasors: 8 /
73 Phasor Analysis Givenv = sinω whereω = 2π, findv C (). () Find capacior complex impedance Z = jωc = 6.28j 4 = 592j (2) Solve circui wih phasors V C = V Z R+Z = j 592j 592j = j = v C = 8.47cos(ω 22 ) C (ms) v v C (3) Draw a phasor diagram showing KVL: V = j V C = j V R = V V C = j = E. Analysis of Circuis (26-924) Phasors: 8 /
74 Phasor Analysis Givenv = sinω whereω = 2π, findv C (). () Find capacior complex impedance Z = jωc = 6.28j 4 = 592j (2) Solve circui wih phasors V C = V Z R+Z = j 592j 592j = j = v C = 8.47cos(ω 22 ) C R vr (ms) v v C (3) Draw a phasor diagram showing KVL: V = j V C = j V R = V V C = j = E. Analysis of Circuis (26-924) Phasors: 8 /
75 Phasor Analysis Givenv = sinω whereω = 2π, findv C (). () Find capacior complex impedance Z = jωc = 6.28j 4 = 592j (2) Solve circui wih phasors V C = V Z R+Z = j 592j 592j = j = v C = 8.47cos(ω 22 ) C R vr (ms) v v C (3) Draw a phasor diagram showing KVL: V = j V C = j V R = V V C = j = Phasors add like vecors E. Analysis of Circuis (26-924) Phasors: 8 /
76 CIVIL Capaciors: i = C dv d Inducors: v = L di d I leadsv V leadsi E. Analysis of Circuis (26-924) Phasors: 9 /
77 CIVIL Capaciors: i = C dv d Inducors: v = L di d I leadsv V leadsi Mnemonic: CIVIL = In a capaciori leadv buv leadsi in an inducor. E. Analysis of Circuis (26-924) Phasors: 9 /
78 CIVIL Capaciors: i = C dv d Inducors: v = L di d I leadsv V leadsi Mnemonic: CIVIL = In a capaciori leadv buv leadsi in an inducor. COMPLEX ARITHMETIC TRICKS: E. Analysis of Circuis (26-924) Phasors: 9 /
79 CIVIL Capaciors: i = C dv d Inducors: v = L di d I leadsv V leadsi Mnemonic: CIVIL = In a capaciori leadv buv leadsi in an inducor. COMPLEX ARITHMETIC TRICKS: ()j j = j j = E. Analysis of Circuis (26-924) Phasors: 9 /
80 CIVIL Capaciors: i = C dv d Inducors: v = L di d I leadsv V leadsi Mnemonic: CIVIL = In a capaciori leadv buv leadsi in an inducor. COMPLEX ARITHMETIC TRICKS: ()j j = j j = (2) j = j E. Analysis of Circuis (26-924) Phasors: 9 /
81 CIVIL Capaciors: i = C dv d Inducors: v = L di d I leadsv V leadsi Mnemonic: CIVIL = In a capaciori leadv buv leadsi in an inducor. COMPLEX ARITHMETIC TRICKS: ()j j = j j = (2) j = j (3)a+jb = r θ = re jθ wherer = a 2 +b 2 andθ = arcan b a (±8 ifa < ) E. Analysis of Circuis (26-924) Phasors: 9 /
82 CIVIL Capaciors: i = C dv d Inducors: v = L di d I leadsv V leadsi Mnemonic: CIVIL = In a capaciori leadv buv leadsi in an inducor. COMPLEX ARITHMETIC TRICKS: ()j j = j j = (2) j = j (3)a+jb = r θ = re jθ wherer = a 2 +b 2 andθ = arcan b a (±8 ifa < ) (4)r θ = re jθ = (rcosθ)+j(rsinθ) E. Analysis of Circuis (26-924) Phasors: 9 /
83 CIVIL Capaciors: i = C dv d Inducors: v = L di d I leadsv V leadsi Mnemonic: CIVIL = In a capaciori leadv buv leadsi in an inducor. COMPLEX ARITHMETIC TRICKS: ()j j = j j = (2) j = j (3)a+jb = r θ = re jθ wherer = a 2 +b 2 andθ = arcan b a (±8 ifa < ) (4)r θ = re jθ = (rcosθ)+j(rsinθ) (5)a θ b φ = ab (θ +φ) and a θ (θ φ). b φ = a b Muliplicaion and division are much easier in polar form. E. Analysis of Circuis (26-924) Phasors: 9 /
84 CIVIL Capaciors: i = C dv d Inducors: v = L di d I leadsv V leadsi Mnemonic: CIVIL = In a capaciori leadv buv leadsi in an inducor. COMPLEX ARITHMETIC TRICKS: ()j j = j j = (2) j = j (3)a+jb = r θ = re jθ wherer = a 2 +b 2 andθ = arcan b a (±8 ifa < ) (4)r θ = re jθ = (rcosθ)+j(rsinθ) (5)a θ b φ = ab (θ +φ) and a θ (θ φ). b φ = a b Muliplicaion and division are much easier in polar form. (6) All scienific calculaors will conver recangular o/from polar form. E. Analysis of Circuis (26-924) Phasors: 9 /
85 CIVIL Capaciors: i = C dv d Inducors: v = L di d I leadsv V leadsi Mnemonic: CIVIL = In a capaciori leadv buv leadsi in an inducor. COMPLEX ARITHMETIC TRICKS: ()j j = j j = (2) j = j (3)a+jb = r θ = re jθ wherer = a 2 +b 2 andθ = arcan b a (±8 ifa < ) (4)r θ = re jθ = (rcosθ)+j(rsinθ) (5)a θ b φ = ab (θ +φ) and a θ (θ φ). b φ = a b Muliplicaion and division are much easier in polar form. (6) All scienific calculaors will conver recangular o/from polar form. Casio fx-99 (available in all exams excep Mahs) will do complex arihmeic (+,,,,x 2, x, x,x ) in CMPLX mode. Learn how o use his: i will save los of ime and errors. E. Analysis of Circuis (26-924) Phasors: 9 /
86 Impedance and For any nework (resisors+capaciors+inducors): E. Analysis of Circuis (26-924) Phasors: /
87 Impedance and For any nework (resisors+capaciors+inducors): () Impedance = Resisance + j Reacance Z = R+jX (Ω) E. Analysis of Circuis (26-924) Phasors: /
88 Impedance and For any nework (resisors+capaciors+inducors): () Impedance = Resisance + j Reacance Z = R+jX (Ω) Z 2 = R 2 +X 2 Z = arcan X R E. Analysis of Circuis (26-924) Phasors: /
89 Impedance and For any nework (resisors+capaciors+inducors): () Impedance = Resisance + j Reacance Z = R+jX (Ω) Z 2 = R 2 +X 2 (2) = Y = Z Impedance = G+jB Siemens (S) Z = arcan X R = Conducance +j Suscepance E. Analysis of Circuis (26-924) Phasors: /
90 Impedance and For any nework (resisors+capaciors+inducors): () Impedance = Resisance + j Reacance Z = R+jX (Ω) Z 2 = R 2 +X 2 (2) = Impedance Z = arcan X R = Conducance +j Suscepance Y = Z = G+jB Siemens (S) Y 2 = = G 2 +B 2 Y = Z = arcan B Z 2 G E. Analysis of Circuis (26-924) Phasors: /
91 Impedance and For any nework (resisors+capaciors+inducors): () Impedance = Resisance + j Reacance Z = R+jX (Ω) Z 2 = R 2 +X 2 (2) = Impedance Y = Z = G+jB Siemens (S) Y 2 = = G 2 +B 2 Z 2 Z = arcan X R = Conducance +j Suscepance Y = Z = arcan B G Noe: Y = G+jB = Z E. Analysis of Circuis (26-924) Phasors: /
92 Impedance and For any nework (resisors+capaciors+inducors): () Impedance = Resisance + j Reacance Z = R+jX (Ω) Z 2 = R 2 +X 2 (2) = Impedance Y = Z = G+jB Siemens (S) Y 2 = = G 2 +B 2 Z 2 Z = arcan X R = Conducance +j Suscepance Y = Z = arcan B G Noe: Y = G+jB = Z = R+jX E. Analysis of Circuis (26-924) Phasors: /
93 Impedance and For any nework (resisors+capaciors+inducors): () Impedance = Resisance + j Reacance Z = R+jX (Ω) Z 2 = R 2 +X 2 (2) = Impedance Y = Z = G+jB Siemens (S) Y 2 = = G 2 +B 2 Z 2 Z = arcan X R = Conducance +j Suscepance Y = Z = arcan B G Noe: Y = G+jB = Z = R+jX = R R 2 +X 2 +j X R 2 +X 2 E. Analysis of Circuis (26-924) Phasors: /
94 Impedance and For any nework (resisors+capaciors+inducors): () Impedance = Resisance + j Reacance Z = R+jX (Ω) Z 2 = R 2 +X 2 (2) = Impedance Y = Z = G+jB Siemens (S) Y 2 = = G 2 +B 2 Z 2 Z = arcan X R = Conducance +j Suscepance Y = Z = arcan B G Noe: Y = G+jB = Z = R+jX = R R 2 +X 2 +j X R 2 +X 2 So G = R R 2 +X 2 = R Z 2 B = X R 2 +X 2 = X Z 2 E. Analysis of Circuis (26-924) Phasors: /
95 Impedance and For any nework (resisors+capaciors+inducors): () Impedance = Resisance + j Reacance Z = R+jX (Ω) Z 2 = R 2 +X 2 (2) = Impedance Y = Z = G+jB Siemens (S) Y 2 = = G 2 +B 2 Z 2 Z = arcan X R = Conducance +j Suscepance Y = Z = arcan B G Noe: Y = G+jB = Z = R+jX = R R 2 +X 2 +j X R 2 +X 2 So G = R R 2 +X 2 = R Z 2 B = X R 2 +X 2 = X Z 2 Beware: G R unlessx =. E. Analysis of Circuis (26-924) Phasors: /
96 Summary Sine waves are he only bounded signals whose shape is unchanged by differeniaion. E. Analysis of Circuis (26-924) Phasors: /
97 Summary Sine waves are he only bounded signals whose shape is unchanged by differeniaion. Think of a sine wave as he projecion of a roaing rod ono he horizonal (or real) axis. A phasor is a complex number represening he lengh and posiion of he rod a ime =. E. Analysis of Circuis (26-924) Phasors: /
98 Summary Sine waves are he only bounded signals whose shape is unchanged by differeniaion. Think of a sine wave as he projecion of a roaing rod ono he horizonal (or real) axis. A phasor is a complex number represening he lengh and posiion of he rod a ime =. IfV = a+jb = r θ = re jθ, hen v() = acosω bsinω = rcos(ω+θ) = R ( Ve jω) E. Analysis of Circuis (26-924) Phasors: /
99 Summary Sine waves are he only bounded signals whose shape is unchanged by differeniaion. Think of a sine wave as he projecion of a roaing rod ono he horizonal (or real) axis. A phasor is a complex number represening he lengh and posiion of he rod a ime =. IfV = a+jb = r θ = re jθ, hen v() = acosω bsinω = rcos(ω+θ) = R ( Ve jω) The angular frequency ω = 2πf is assumed known. E. Analysis of Circuis (26-924) Phasors: /
100 Summary Sine waves are he only bounded signals whose shape is unchanged by differeniaion. Think of a sine wave as he projecion of a roaing rod ono he horizonal (or real) axis. A phasor is a complex number represening he lengh and posiion of he rod a ime =. IfV = a+jb = r θ = re jθ, hen v() = acosω bsinω = rcos(ω+θ) = R ( Ve jω) The angular frequency ω = 2πf is assumed known. If all sources in a linear circui are sine waves having he same frequency, we can use phasors for circui analysis: E. Analysis of Circuis (26-924) Phasors: /
101 Summary Sine waves are he only bounded signals whose shape is unchanged by differeniaion. Think of a sine wave as he projecion of a roaing rod ono he horizonal (or real) axis. A phasor is a complex number represening he lengh and posiion of he rod a ime =. IfV = a+jb = r θ = re jθ, hen v() = acosω bsinω = rcos(ω+θ) = R ( Ve jω) The angular frequency ω = 2πf is assumed known. If all sources in a linear circui are sine waves having he same frequency, we can use phasors for circui analysis: Use complex impedances: jωl and jωc E. Analysis of Circuis (26-924) Phasors: /
102 Summary Sine waves are he only bounded signals whose shape is unchanged by differeniaion. Think of a sine wave as he projecion of a roaing rod ono he horizonal (or real) axis. A phasor is a complex number represening he lengh and posiion of he rod a ime =. IfV = a+jb = r θ = re jθ, hen v() = acosω bsinω = rcos(ω+θ) = R ( Ve jω) The angular frequency ω = 2πf is assumed known. If all sources in a linear circui are sine waves having he same frequency, we can use phasors for circui analysis: Use complex impedances: jωl and jωc Mnemonic: CIVIL ells you wheheri leadsv or vice versa ( leads means reaches is peak before ). E. Analysis of Circuis (26-924) Phasors: /
103 Summary Sine waves are he only bounded signals whose shape is unchanged by differeniaion. Think of a sine wave as he projecion of a roaing rod ono he horizonal (or real) axis. A phasor is a complex number represening he lengh and posiion of he rod a ime =. IfV = a+jb = r θ = re jθ, hen v() = acosω bsinω = rcos(ω+θ) = R ( Ve jω) The angular frequency ω = 2πf is assumed known. If all sources in a linear circui are sine waves having he same frequency, we can use phasors for circui analysis: Use complex impedances: jωl and jωc Mnemonic: CIVIL ells you wheheri leadsv or vice versa ( leads means reaches is peak before ). Phasors eliminae ime from equaions E. Analysis of Circuis (26-924) Phasors: /
104 Summary Sine waves are he only bounded signals whose shape is unchanged by differeniaion. Think of a sine wave as he projecion of a roaing rod ono he horizonal (or real) axis. A phasor is a complex number represening he lengh and posiion of he rod a ime =. IfV = a+jb = r θ = re jθ, hen v() = acosω bsinω = rcos(ω+θ) = R ( Ve jω) The angular frequency ω = 2πf is assumed known. If all sources in a linear circui are sine waves having he same frequency, we can use phasors for circui analysis: Use complex impedances: jωl and jωc Mnemonic: CIVIL ells you wheheri leadsv or vice versa ( leads means reaches is peak before ). Phasors eliminae ime from equaions, convers simulaneous differenial equaions ino simulaneous linear equaions. E. Analysis of Circuis (26-924) Phasors: /
105 Summary Sine waves are he only bounded signals whose shape is unchanged by differeniaion. Think of a sine wave as he projecion of a roaing rod ono he horizonal (or real) axis. A phasor is a complex number represening he lengh and posiion of he rod a ime =. IfV = a+jb = r θ = re jθ, hen v() = acosω bsinω = rcos(ω+θ) = R ( Ve jω) The angular frequency ω = 2πf is assumed known. If all sources in a linear circui are sine waves having he same frequency, we can use phasors for circui analysis: Use complex impedances: jωl and jωc Mnemonic: CIVIL ells you wheheri leadsv or vice versa ( leads means reaches is peak before ). Phasors eliminae ime from equaions, convers simulaneous differenial equaions ino simulaneous linear equaions. Needs complex numbers bu worh i. E. Analysis of Circuis (26-924) Phasors: /
106 Summary Sine waves are he only bounded signals whose shape is unchanged by differeniaion. Think of a sine wave as he projecion of a roaing rod ono he horizonal (or real) axis. A phasor is a complex number represening he lengh and posiion of he rod a ime =. IfV = a+jb = r θ = re jθ, hen v() = acosω bsinω = rcos(ω+θ) = R ( Ve jω) The angular frequency ω = 2πf is assumed known. If all sources in a linear circui are sine waves having he same frequency, we can use phasors for circui analysis: Use complex impedances: jωl and jωc Mnemonic: CIVIL ells you wheheri leadsv or vice versa ( leads means reaches is peak before ). Phasors eliminae ime from equaions, convers simulaneous differenial equaions ino simulaneous linear equaions. Needs complex numbers bu worh i. see Hay Chaper E. Analysis of Circuis (26-924) Phasors: /
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