Electric Circuits, 9th Edition P76 ppt

726 Complex Numbers A complex number in polar form can be put in rectangular form by writing ce je = c(cos0 + /sin 0) = c cos 0 + jc sin 0) = a + jb. (B.5) The transition from rectangular to polar form makes use of the geom- etry of the right triangle, namely, a + jb = Va 2 + b 2 )e je = ce j0 (B.6) where tan0 = b/a. (B.7) It is not obvious from Eq. B.7 in which quadrant the angle 0 lies. The ambi- guity can be resolved by a graphical representation of the complex number. b 0 s^ H J?\ C a Figure B.l • The graphical representation of a + jb when a and b are both positive. I l 5 36.87° 5 4 4+/3 = 5/36.87° (a) 143.13° ' \ 3 1 1 1 1 IX -4 J/l 1 1 i i i ; l. -4 + /3 = 5,/143.13° (b) i i 1 il U -4, -3 5 216.87° -4-/3 = 5/216,87 0 (c) i_L 0 1 1 1 1 1 I -3 i_ ~l 1 ZyS — 5 1 1 1 4 s 323. 1, 13° 4-/3 = 5/323.13° (d) Figure B.2 A The graphical representation of four complex numbers. B.2 The Graphical Representation of a Complex Number A complex number is represented graphically on a complex-number plane, which uses the horizontal axis for plotting the real component and the vertical axis for plotting the imaginary component. The angle of the complex number is measured counterclockwise from the positive real axis. The graphical plot of the complex number n = a + jb = c /0°, if we assume that a and b are both positive,is shown in Fig. B.l. This plot makes very clear the relationship between the rectangular and polar forms. Any point in the complex-number plane is uniquely defined by giving either its distance from each axis (that is, a and b) or its radial dis- tance from the origin (c) and the angle of the radial measurement 0. It follows from Fig. B.l that 0 is in the first quadrant when a and b are both positive, in the second quadrant when a is negative and b is positive, in the third quadrant when a and b are both negative, and in the fourth quadrant when a is positive and b is negative. These observations are illustrated in Fig. B.2, where we have plotted 4 + /3, -4 + /3, -4 - /3, and 4-/3. Note that we can also specify 0 as a clockwise angle from the positive real axis. Thus in Fig. B.2(c) we could also designate —4 - /3 as 5/-143.13°. In Fig. B.2(d) we observe that 5/323.13° = 5/-36.87°. It is customary to express 0 in terms of negative values when 0 lies in the third or fourth quadrant. The graphical interpretation of a complex number also shows the relationship between a complex number and its conjugate. The conjugate of a complex number is formed by reversing the sign of its imaginary component. Thus the conjugate of a + jb is a - jb, and the conjugate of —a + jb is — a - jb. When we write a complex number in polar form, we form its conjugate simply by reversing the sign of the angle 0. Therefore the conjugate of c/0° is c/-0°. The conjugate of a complex number is B.3 Arithmetic Operations 727 designated with an asterisk. In other words, n* is understood to be the /j, = -a+jb-c &•> conjugate of n. Figure B.3 shows two complex numbers and their conju- gates plotted on the complex-number plane. Note that conjugation simply reflects the complex numbers about the real axis. /j = — a+ib—v #-> ^. " d- ~ a ^^ -b- 112 — —u-jb=c -H2 n ^f ^¾^ «] = - (i- «+ ~jb = b = c 0, 1 a c -0 { B.3 Arithmetic Operations Addition (Subtraction) To add or subtract complex numbers, we must express the numbers in rec- tangular form. Addition involves adding the real parts of the complex numbers to form the real part of the sum, and the imaginary parts to form the imaginary part of the sum. Thus, if we are given Figure B.3 A The complex numbers n x and n 2 amd their conjugates n\ and «3. and then «! = 8 + /16 «2 = 12 - /3, n { + n 2 = (8 + 12) + /(16 - 3) = 20 + /13. Subtraction follows the same rule. Thus n 2 - «j = (12 -8)+ /(-3 - 16) = 4 - /19. If the numbers to be added or subtracted are given in polar form, they are first converted to rectangular form. For example, if «i = 10/53.13° and then and n 2 = 5/-135°, m + n 2 = 6 + /8 - 3.535 - /3.535 = (6 - 3.535) + /(8 - 3.535) = 2.465 + /4.465 = 5.10/61.10°, /11 - n 2 = 6 + /8- (-3.535 - /3.535) = 9.535 + /11.535 = 14.966 /50.42°. Multiplication (Division) Multiplication or division of complex numbers can be carried out with the numbers written in either rectangular or polar form. However, in most cases, the polar form is more convenient. As an example, let's find the product n x n 2 when /^ = 8 + /10 and n 2 = 5 - /4. Using the rectangular form, we have n x n 2 = (8 + /10)(5 - /4) = 40 - /32 + /50 + 40 = 80 + /18 = 82/12.68°. If we use the polar form, the multiplication n.\n 2 becomes n 1 n 2 = (12.81 /51.34° )(6.40 /-38.66° ) = 82/12.68° = 80 + /18. The first step in dividing two complex numbers in rectangular form is to multiply the numerator and denominator by the conjugate of the denomi- nator. This reduces the denominator to a real number. We then divide the real number into the new numerator. As an example, let's find the value of n\/n 2 , where rt\ = 6 + /3 and n 2 = 3 - /1. We have «1 n 2 6 + /3 3 - /1 (6 + (3- 18 + /6 + /9 - 9 + 1 15 + /15 10 = 2.12 /45° /3)(3 + /1)(3 + -3 = 1.5 + /1.5 /1) /1) In polar form, the division of n x by n 2 is n { 6.71 /26.57° n 2 3.16/-18.43° = 1.5 + /1.5. 2.12 /45' B.4 Useful Identities In working with complex numbers and quantities, the following identities are very useful: ± / 2 = + 1, (B.8) (-/)(/) = U (B.9) / = ^7, (B.10) ff */»/2 = ± /. (B.i2) Given that n = a + jb = c/0°, it follows that nn = a 2 + b z = <r, (B.13) « + n = 2«, (B.14) n - n* = jib, (B.15) «/w* = 1/20°. (B.16) B.5 The Integer Power of a Complex Number To raise a complex number to an integer power k, it is easier to first write the complex number in polar form. Thus n k = (a + jb) k = (cei°) k = c k e jk0 = c k (coskd + j sinkO). For example, (2e /12 °) 5 = 2V 6() ° = 32e mr = 16 +)27.71, and (3 + /4) 4 = (5e^ y ) 4 = 5V m52 ° = 625^ 212,52 ° = -527 - /336. B.6 The Roots of a Complex Number To find the /cth root of a complex number, we must recognize that we are solving the equation x k -ce' 6 = 0, (B.17) which is an equation of the kth degree and therefore has k roots. To find the k roots, we first note that ce je = ce mi*) = ce m-w = > (B 18) It follows from Eqs. B.17 and B.18 that Xj = (ce'°y /k = c yk eW\ X 2 = [ ce W+2iryil/k = c l/k e j{fi+2v)/k^ X$ = [ce' i0+47T) } ]/k = c Vk e J(0+47r)/k^ We continue the process outlined by Eqs. B.19, B.20, and B.21 until the roots start repeating. This will happen when the multiple of n is equal to 2k. For example, let's find the four roots of 81tf y6{) °. We have Xt = 8lVV6°/4 = 3e' ls °, X2 = 81^/^+360)/4 = ^/1(^ x 3 = 8W m+m V 4 = 3e' 195 '\ x = g l l/4 e /(6()+ t«H0)/4 = 2e j2 ^'\ Xj = 81 l/4^(60 + 1440)/4 = 3t ,/375^ = 3^ Here, x$ is the same as X\, so the roots have started to repeat. Therefore we know the four roots of 81*? ; are the values given by X\, X 2 , X 3 , and X4. It is worth noting that the roots of a complex number lie on a circle in the complex-number plane. The radius of the circle is c !///< . The roots are uniformly distributed around the circle, the angle between adjacent roots being equal to lir/k radians, or 360/k degrees. The four roots of 81 e ' 6(r are shown plotted in Fig. B.4. 3. 105° / / 1 1 1• 1 1 3 195°^ "s. N _ \ \ fc3 15° 1 i 1 1 1 - I / ~ 3. 285° Figure B.4 • The four roots of %\e m ". (B.19) (B.20) (B.21) Appendix _ More on Magnetically { Coupled Coils and Ideal Transformers C.l Equivalent Circuits for Magnetically Coupled Coils At times, it is convenient to model magnetically coupled coils with an equivalent circuit that does not involve magnetic coupling. Consider the two magnetically coupled coils shown in Fig. C.l. The resistances Ri and R 2 represent the winding resistance of each coil. The goal is to replace the magnetically coupled coils inside the shaded area with a set of inductors that are not magnetically coupled. Before deriving the equivalent circuits, we must point out an important restriction: The voltage between terminals b and d must be zero. In other words, if terminals b and d can be shorted together without disturbing the voltages and currents in the original cir- cuit, the equivalent circuits derived in the material that follows can be used to model the coils. This restriction is imposed because, while the equivalent circuits we develop both have four terminals, two of those four terminals are shorted together. Thus, the same requirement is placed on the original circuits. We begin developing the circuit models by writing the two equations that relate the terminal voltages i?i and v 2 to the terminal currents i x and i 2 . For the given references and polarity dots, di\ diz i-h U-r + M~r dt dt (C.l) and Vi du dh dt dt (C.2) «1 +- 'l a + "i * L* Ll ) M «1 ••2 V 2 Figure C.l • The circuit used to develop an equivalent circuit for magnetically coupled coils. The T-Equivalent Circuit To arrive at an equivalent circuit for these two magnetically coupled coils, we seek an arrangement of inductors that can be described by a set of equations equivalent to Eqs. C.l and C.2. The key to finding the arrange- ment is to regard Eqs. C.l and C.2 as mesh-current equations with i y and i 2 as the mesh variables. Then we need one mesh with a total inductance of L\ H and a second mesh with a total inductance of L 2 H. Furthermore, the two meshes must have a common inductance of M H. The T-arrangement of coils shown in Fig. C.2 satisfies these requirements. Figure C.2 • The T-equivalent circuit for the magneti- cally coupled coils of Fig. C.l. 731 732 More on Magnetically Coupled Coils and Ideal Transformers You should verify that the equations relating v Y and v 2 to /, and i 2 reduce to Eqs. C.l and C.2. Note the absence of magnetic coupling between the inductors and the zero voltage between b and d. The ^-Equivalent Circuit We can derive a 7r-equivalent circuit for the magnetically coupled coils shown in Fig. C.l.This derivation is based on solving Eqs. C.l and C.2 for the derivatives dijdt and di^jdt and then regarding the resulting expres- sions as a pair of node-voltage equations. Using Cramer's method for solv- ing simultaneous equations, we obtain expressions for di\jdt and di 2 /dt: di\ dt V\ v 2 u M M L 2 M L 2 LiL 1^2 M Vl M L X L 2 - M •v 2 ; (C.3) di 2 dt M v 2 •M UU - M 2 UL <\^2 1^2 w Vi + Li L,L>, - M l jVl (C.4) Now we solve for /, and i 2 by multiplying both sides of Eqs. C.3 and C.4 by dt and then integrating: k = *i(0) + L X L 2 - M 2 J{) V { dT M L X L 2 - M l h v 2 dr (C.5) and ' 2 (0) UU x / v\dr + r / M 2 J {) L X L 2 -M 2 k v 2 dT. (C.6) If we regard v x and v 2 as node voltages, Eqs. C.5 and C.6 describe a circuit of the form shown in Fig. C.3. All that remains to be done in deriving the 7r-equivalent circuit is to find L A , L B , and L c as functions of L h L 2 , and M. We easily do so by writ- ing the equations for / t and i 2 in Fig. C.3 and then comparing them with Eqs. C.5 and C.6. Thus Figure C.3 • The circuit used to derive the 7r-equivalent circuit for magnetically coupled coils. C.l Equivalent Circuits for Magnetically Coupled Coils 733 1 f 1 /*' ii = /j(0) + — / Vidr + — I («! - v 2 )dr LA JO L B JO and 1 /"' 1 /' * 2 = «2(0) + -^- v 2 dT + — I (v 2 - v x )dr I-C ./0 L3 JO = «' 2 (0) + 7" / M* +(7- + 7-] Then M L B L^ - M 2' (C.9) L A L 2 -M L X L 2 - M 2 ' (CIO) L c Li /V/ LiL 1^2 Mr (til) When we incorporate Eqs. C.9-C.11 into the circuit shown in Fig. C.3, the ^-equivalent circuit for the magnetically coupled coils shown in Fig. C.l is as shown in Fig. C.4. Note that the initial values of iy and i 2 are explicit in the ^-equivalent circuit but implicit in the T-equivalent circuit. We are focusing on the sinu- soidal steady-state behavior of circuits containing mutual inductance, so we can assume that the initial values of ij and i 2 are zero. We can thus eliminate the current sources in the ^-equivalent circuit, and the circuit shown in Fig. C.4 simplifies to the one shown in Fig. C.5. The mutual inductance carries its own algebraic sign in the T- and ^-equivalent circuits. In other words, if the magnetic polarity of the cou- pled coils is reversed from that given in Fig. C.l, the algebraic sign of M Figure C.4 A The 7r-equivalent circuit for the magnetically coupled coils of Fig. C.l. Figure C.5 • The ^-equivalent circuit used for sinusoidal steady-state analysis. 734 More on Magnetically Coupled Coils and Ideal Transformers reverses. A reversal in magnetic polarity requires moving one polarity dot without changing the reference polarities of the terminal currents and voltages. Example C.l illustrates the application of theT-equivalent circuit. Example C.l a) Use the T-equivalent circuit for the magnetically coupled coils shown in Fig. C.6 to find the phasor currents I| and I 2 . The source frequency is 400 rad/s. b) Repeat (a), but with the polarity dot on the sec- ondary winding moved to the lower terminal. Solution a) For the polarity dots shown in Fig. C.6, M carries a value of +3 H in the T-equivalent circuit. Therefore the three inductances in the equiva- lent circuit are L { - M = 9 - 3 = 6 H; L 2 - M = 4 - 3 = 1 H; M = 3 H. Figure C.7 shows the T-equivalent circuit, and Fig. C.8 shows the frequency-domain equivalent circuit at a frequency of 400 rad/s. Figure C.9 shows the frequency-domain circuit for the original system. Here the magnetically coupled coils are modeled by the circuit shown in Fig. C.8. To find the phasor currents I] and I 2 , we first find the node voltage across the 1200 O inductive reac- tance. If we use the lower node as the reference, the single node-voltage equation is 300 + 900 - /2100 = 0. 700 + y'2500 /1200 Solving for V yields V = 136 - /8 = 136.24/-3.37° V(rms). Then 300 -(136- /8) 700 + /2500 63.25 /-71.57° mA (rms) 500 a /loo a _TVYY>_ II 300/0 Q V a 200 a /1200 a 4. o I • loo a 800 a A/W 6H 1 H |3H Figure C.7 A The T-equivalent circuit for the magnetically coupled coils in Example C.l. /2400 /400 :/1200 Figure C.8 • The frequency-domain model of the equivalent circuit at 400 rad/s. 500 a / loo a 200 a /2400 a /400 a 100 a 6 3()0,()° V /I200a Figure C.9 A The circuit of Fig. C.6, with the magnetically coupled coils replaced by their T-equivalent circuit. and I, = 136 - /8 900 - /2100 59.63 /63.43° mA (rms). Vi /3600 a b) When the polarity dot is moved to the lower ter- minal of the secondary coil, M carries a value of -3 H in the T-equivalent circuit. Before carrying out the solution with the new T-equivalent cir- cuit, we note that reversing the algebraic sign of M has no effect on the solution for Ij and shifts I 2 by 180°.Therefore we anticipate that /2500 a Figure C.6 A The frequency-domain equivalent circuit for Example C.l. and Ij = 63.25/-71.57° mA (rms) I 2 = 59.63 /-116.57° mA (rms). We now proceed to find these solutions by using the new T-equivalent circuit. With M = -3 H, the three inductances in the equiv- alent circuit are Lj - M = 9 - (-3) = 12 H; L 2 - M = 4- (-3) = 7H; M = -3H. At an operating frequency of 400 rad/s, the frequency-domain equivalent circuit requires two inductors and a capacitor, as shown in Fig. CIO. The resulting frequency-domain circuit for the original system appears in Fig. C.ll. As before, we first find the node voltage across the center branch, which in this case is a capacitive reactance of — /'1200 H. If we use the lower node as reference, the node-voltage equation is V - 300 + + 700 + /4900 -/1200 900 + /300 Solving for V gives V = -8 - /56 = 56.57 /-98.13° V (rms). C.2 The Need for Ideal Transformers in the Equivalent Circuits 735 Then 300 -(-8- /56) h = and 700 + /4900 = 63.25 /-71.57° mA (rms) -8 - /56 900 + /300 = 59.63 /-116.57° mA (rms). /4800 fl /2800 0 -/1200O Figure CIO • The frequency-domain equivalent circuit for M = -3 H and a> = 400 rad/s. 500 n /loo n 200 n /48ooa /28oon 1000 r^/ 3 ° ( M jc I V -/120012: 800 O -/25()0 il Figure C.ll • The frequency-domain equivalent circuit for Example C.l(b). C.2 The Need for Ideal Transformers in the Equivalent Circuits The inductors in the T- and 77-equivalent circuits of magnetically cou- pled coils can have negative values. For example, if L\ = 3 mH, L 2 = 12 mH, and M = 5 mH, the T-equivalent circuit requires an induc- tor of —2 mH, and the 7r-equivalent circuit requires an inductor of -5.5 mH. These negative inductance values are not troublesome when you are using the equivalent circuits in computations. However, if you are to build the equivalent circuits with circuit components, the negative inductors can be bothersome. The reason is that whenever the frequency of the sinusoidal source changes, you must change the capacitor used to simulate the negative reactance. For example, at a frequency of 50 krad/s, a -2 mH inductor has an impedance of -/100 fi.This imped- ance can be modeled with a capacitor having a capacitance of 0.2 /xF. If the frequency changes to 25 krad/s, the -2 mH inductor impedance changes to -/50 il. At 25 krad/s, this requires a capacitor with a capaci- tance of 0.8 /xF. Obviously, in a situation where the frequency is varied . area with a set of inductors that are not magnetically coupled. Before deriving the equivalent circuits, we must point out an important restriction: The voltage between terminals b and d must