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Phasor Analysis Appendix A Phasor Analysis A.l Vector Representation of Sinusoids: The Concept of Phasors Consider the harmonic, or sinusoidal, time functions u(t) = urnax sin wt; (A.l) v(t) = vrnax sin(wt - a). These functions may represent steady-state voltages, currents, velocities, or any other physical variables. Graphically, the functions may be depicted (Figure A.I) as the vertical projections of the rotating vectors U and V. These vectors have the lengths urnax and vrnax ' respectively, and rotate with the angular velocity, w, [rad/s]. The algebraic sum wet) = u(t) + vet) (A.2) is also sinusoidal and can be represented by the projection of a third rotating vec­ tor, W. This vector is obtained as the vectorial sum of the vectors U and V as shown in Figure A. I. It is clear that analysis of sinusoidally varying quantities can be carried out by a geometric study of vectors. The studies can be further simplified by "freezing" the rotating vectors in a given position. Such "time-frozen" vectors are often referred to as phasors. One may think of a phasor as a snapshot of a rotating vector. Example A-I Find the sum w of the two sinusoids u(t) = 6 sinwt, (A.Ll) v(t) = 5 sin(wt - 37°). Solution: The two vectors U and V are shown in an x-y coordinate system (Fig­ ure A.2). For simplicity they have been "frozen" at the instant when the U vector 405 406 Appendix A Phasor Analysis FigureA.l y u x ---- -- v ---- w FigureA.2 A.1 Vector Representation of Sinusoids: The Concept of Phasors 407 coincides with the x axis. According to our definition, the rotating vectors have been turned into phasors. The U phasor, which coincides with the x axis, is referred to as the reference phasor. If the sum of the two phasors U and V is the phasor, W we first determine its com­ ponents, Wx and Wy along the x and y axes, respectively, and obtain Wx = 6 + 5 cos 37° = 9.993, (A.1.2) Wy = 0 - 5 sin37° = -3.009. Iwi is given by Iwi = \1'9.993 2 + 3.009 2 = 10.436. (A. 1.3) The angle I W of phasor, W relative to the reference phasor is ~= tan-I ( -Wy) = tan-I (-3.009) = -16.76°. (A. 1.4) Wx 9.993 For the time function, wet), we can write wet) = 10.436 sin(wt - 16.76°). (A. 1.5) ExampleA.2 A sinusoidal current, i = imax sin wt [A], (A.2.1) is injected into a circuit consisting of a resistor, R, in series with an inductor, L, and a capacitor, C. Describe the steady-state voltages v R , v L , and Vc defined in Figure A.3. Solution: The voltage across the resistor, according to Ohm's law, equation (3.23) is V R = Ri = Rimax sin wt [V]. (A.2.2) The voltage across the inductor, from equation (3.70), is i L C -=----A.M.J\r--.rmnI~---1~ - y vc , y v FigureA.3 408 Appendix A Phasor Analysis di vL = L- [V], dt = wLimax cos wt [V], (A.2.3) = wLimax sin (wt + 90°) [V]. The voltage across the capacitor, from equation (3.11) is 1 1 Vc = CQ = C f idt = -1 f i sin wt dt = - -imax cos wt (A.2.4) C max wC = imax sin (wt - 90°) [V]. wC Summary: The resistor voltage vR is a sinusoid and in phase with the current i. The inductor voltage vL and capacitor voltage Vc are likewise sinusoidal but, respectively, leading and lagging the current by 90°. If we represent the above time variables by the phasors VR , VL , Vc ' and I, respectively, we obtain the pha­ sor diagram shown in Figure A.4. y I x FigureA.4 A.l Vector Representation of Sinusoids: The Concept of Phasors 409 ExampleA.3 For the circuit in Example A2, we have the following numerical values: imax = 10 [A], R = 20 [0], w = 377 [rad/s] (60 Hz), (A3.1) L = 0.040 [H], e = 300 [JLF]' Find the amplitudes of the voltages across R, L, and e, and also the total voltage across the circuit. Solution: From equation (A2.2) we get, vR max = Rimax = 20 . 10 = 200 [V]. (A3.2) From equation (A2.3), vLmax = wLimax = 377 . 0.040 . 10 = 150.8 [V]. (A3.3) From equation (A2.4), v = imax = 10 = 88.4 [V]. (A3.4) C max we 377 . 300 . 10-6 The total voltage v across the circuit in Figure A3 is the sum of the individual voltages. By representing v by its phasor V, we obtain the latter simply by vecto­ rial addition (Figure AS) of the phasors VR , VL , and Vc. We get y v Vc FigureA.5 410 Appendix A Phasor Analysis Ivl = Y2002 + (150.8 - 88.4)2 = 209.5 [V]. (A.3.5) I 150.8 - 88.4) fl: = tan- ( = l7.33°. (A.3.6) 200 For the total voltage we can write, v = 209.5 sin(wt + 17.33°) [V]. (A.3.7) A.2 Phasor Representation Using Complex Numbers Complex algebra is a most valuable analytical tool for the study of phasors. We first give a very brief expose of complex algebra followed by examples. A.2.1 Complex Numbers: Definition Figure A.6 shows a complex number plane. The complex number z is defined by its coordinates, x andjy, along the real and imaginary axes, respectively. We write the complex number z in the cartesian form: z = x + jy. (A.3) The factor j is defined by (A.4) We say that x is the "real part of z," and y the "imaginary part." The distance Iz I from the origin to the coordinate point z is referred to as the modulus or mag­ nitude of z. jy-axis ~~------------~z Izl x x-axis Figure A.6 A.2 Phasor Representation Using Complex Numbers 411 We have (A.5) The angular orientation I..!:. of z in reference to the real axis is obtained from I..!:. = tan- 1 C) (A. 6) From Figure A.6 we have x = Izl cos I..!:., (A.7) y = Iz I sin I..!:.. Euler's theorem 1 states that ej ~ = cos I..!:. + j sin I..!:. (A.8) The complex number z can be written in the alternate polar form: z = x + jy, Iz I cos I..!:. + j Iz I sin I..!:., (A.9) Izl(cosl..!:. + jsinl..!:.) = Izle j / z• A.2.2 Complex Algebra Addition, subtraction, multiplication, and division of complex numbers are defined in terms of the same operations valid for real numbers. We give one example of each operation. ExampleA.4 Given the two complex numbers z\ = 3 + jl and Z2 = 4 - j2, (A.4.1) I This theorem can be proved as follows: From the series x 2 x 3 X4 eX = 1 + x + 2! + 3! + 4! + ... , we obtain . ~2 .~3 ~4 e1<P = 1 + J~ --- J - + - + ... 2! 3! 4! = cos~ + jsin~ 412 Appendix A Phasor Analysis find: (i) ZI + Z2' (ii) ZI - Z2' (iii) ZI Z2' (iv) ZI/Z2' Solution: (i) The sum of the complex numbers is obtained as follows: ZI + Z2 = (3 + jl) + (4 - j2) (A.4.2) = (3 + 4) + j(l - 2) = 7 - jl. Note that the real and imaginary parts are added separately. Note also (Figure A.7) that if each complex number is associated with a phasor, then the complex sum of the numbers corresponds to a vectorial addition of the phasors. (ii) The difference: ZI - Z2 = (3 + j1) - (4 - j2) (AA.3) =(3-4)+j(1 +2)= -1 +j3. (iii) The product: Z,Z2 = (3 + j1)(4 - j2) = 3 ·4 + (j 1)( - j2) + j(1 ·4 - 3 . 2) (A.4A) = 14 - j2. jy t ---x Figure A.7 A.2 Phasor Representation Using Complex Numbers 413 (iv) The quotient: Zl _ 3 + j1 _ (3 + jl)(4 + j2) Z2 - 4 - j2 - (4 - j2)(4 + j2) 3·4 + (j I) (j2) + j(1. 4 + 2·3) (A.4.5) 16 + 4 10 10 = 20 + j 20 = 0.5 + jO.5. By using the polar fonn of complex numbers, we obtain (v) The product: ZlZ2 = (3 + j1)(4 - j2) = 3.162ejI8 .435° ·4.472e-j26.5W = 14.140e-j8.13Qo = 14.140(0.990 - jO.141) (A.4.6) = 14.00 - j2.oo (as before). (vi) The quotient: 3 162ej 18.4350 Zl = . _ j45° Z2 4.472e-j26.5W - 0.707e , (A.4.7) = 0.5 + jO.5 (as before). ExampleA.5 Use complex algebra to solve Example A.3. Solution: If we express the voltage phasors, VR , VL , and Vc as complex num­ bers we get VR = 200 [V]. VL = j 150.8 [V]. (A.5.1) Vc = - j88.4 [V]. The applied voltage represented by the phasor V is then obtained by the applica­ tionofKVL: V = VR + VL + Vc = 200 + j150.8 - j88.4, (A.5.2) = 200 + j62.4 = 209.5ejI7.33° [V]. When the total voltage is expressed as a function of time, we get vet) = 209.5 sin(wt + 17.33°) [V]. (A.5.3) Compare this to equation (A.3.7). 414 Appendix A Phasor Analysis A.3 Impedances Consider the phasor P and the complex number z. The product zp can be written as zp = Iz kilo Ip ki.e = Iz lip k(il+ i.e). (A 10) We make the following observation: Multiplication by z results in a magnitude amplification of Iz Iand a phase advancement by L.!:. degrees. In particular, when a phasor is multiplied by the factor j, it is rotated counter­ clockwise through the angle 90° with an unchanged magnitude.
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