Electric Circuits, 9th Edition P71 potx

676 The Fourier Transform Figure P17.34 ion -AMr- 1 H •o :62.5 mF 17.35 a) Use the Fourier transform method to find v a in PSPICE the circuit in Fig. PI 7.35 when MULTISIM i g = \Se m u(-t) - lSe- ]{)t u(t) A. b) Find v 0 (O~). c) Find v o (0 + ). d) Do the answers obtained in (b) and (c) make sense in terms of known circuit behavior? Explain. Figure P17.35 10 mF tt 25 0 + 17.36 When the input voltage to the system shown in Fig. P17.36 is \Su{i) V, the output voltage is v a = [10 + 30<T 20 ' - 40er 3Qf ]w(f)V. What is the output voltage if Vj = 15 sgn(f) V? Figure P17.36 Vt(t) (Input voltage) h{t) V a (t) (Output voltage) Section 17.8 17.37 It is given that F(<o) = e 0J u(-(o) + e~ w /*(w). a) Find/(0- b) Find the 1 f! energy associated with /(f) via time-domain integration. c) Repeat (b) using frequency-domain integration. d) Find the value of w t if /(f) has 90% of the energy in the frequency band 0 £ \<o\ s a> x . 17.38 The circuit shown in Fig. PI7.38 is driven by the current L = \2e~ m u(t)A. What percentage of the total 1 fi energy content in the output current i 0 lies in the frequency range 0 < \co\ < lOOrad/s? Figure PI7.38 s 25 n 500 mH 17.39 The input current signal in the circuit seen in Fig. P17.39 is i s = 30e~ 2/ u{t) fiA, t > 0 + . What percentage of the total 1 Cl energy content in the output signal lies in the frequency range 0 to 4 rad/s? Figure PI 7.39 1.25 fxF 17.40 The input voltage in the circuit in Fig. PI7.40 is v g = 30e _|f| V. a) Find v a (t). b) Sketch \V g (oo)\ for -5 < to < 5 rad/s. c) Sketch \V 0 (to)\ for -5 < w < 5 rad/s. d) Calculate the 1 ft energy content of v g . e) Calculate the 1 H energy content of v„. f) What percentage of the 1 O energy content in v g lies in the frequency range 0 ^ |w| ^ 2 rad/s? g) Repeat (f) for v (} . Figure PI7.40 125 mF 17.41 The amplitude spectrum of the input voltage to the high-pass RC filter in Fig. P17.41 is VM 200 , 100 rad/s < \w\ < 200 rad/s; Vj(co) = 0, elsewhere. Problems 677 a) Sketch |K,H| 2 for -300 < a> < 300 rad/s. b) Sketch \V 0 (a>)\ 2 for -300 < <o < 300 rad/s. c) Calculate the 1 Q, energy in the signal at the input of the filter. d) Calculate the 1 Q, energy in the signal at the out- put of the filter. Figure P17.41 0.5 (xF r—1(- v t 20 m 17.42 The input voltage to the high-pass RC filter circuit in Fig. P17.42 is Vi(t) = Ae-'"u(t). Let a denote the corner frequency of the filter, that is, a = 1/RC. a) What percentage of the energy in the signal at the output of the filter is associated with the fre- quency band 0 < \m\ < a if a = a? b) Repeat (a), given that a = V3a. c) Repeat (a), given that a = a/V3. Figure P17.42 -1(- C R + I. _." Hi I CHAPTER CONTENTS 18.1 The Terminal Equations p. 680 18.2 The Two-Port Parameters p. 681 18.3 Analysis of the Terminated Two-Port Circuit p. 689 18.4 Interconnected Two-Port Circuits p. 694 ^CHAPTER OBJECTIVES 1 Be able to calculate any set of two-port parameters with any of the following methods: • Circuit analysis; • Measurements made on a circuit; • Converting from another set of two-port parameters using Table 18.1. 2 Be able to analyze a terminated two-port circuit to find currents, voltages, impedances, and ratios of interest using Table 18.2. 3 Know how to analyze a cascade interconnection of two-port circuits. 678 Two-Port Circuits We have frequently focused on the behavior of a circuit at a specified pair of terminals. Recall that we introduced the Thevenin and Norton equivalent circuits solely to simplify circuit analysis relative to a pair of terminals. In analyzing some electri- cal systems, focusing on two pairs of terminals is also convenient. In particular, this is helpful when a signal is fed into one pair of terminals and then, after being processed by the system, is extracted at a second pair of terminals. Because the terminal pairs represent the points where signals are either fed in or extracted, they are referred to as the ports of the system. In this chapter, we limit the discussion to circuits that have one input and one output port. Figure 18.1 on page 680 illustrates the basic two- port building block. Use of this building block is subject to sev- eral restrictions. First, there can be no energy stored within the circuit. Second, there can be no independent sources within the circuit; dependent sources, however, are permitted. Third, the cur- rent into the port must equal the current out of the port; that is, i\ = i\ and / 2 = ii. Fourth, all external connections must be made to either the input port or the output port; no such connections are allowed between ports, that is, between terminals a and c, a and d, b and c, or b and d. These restrictions simply limit the range of cir- cuit problems to which the two-port formulation is applicable. The fundamental principle underlying two-port modeling of a system is that only the terminal variables (ij, V\, h* an< ^ ^¾) are °f interest. We have no interest in calculating the currents and volt- ages inside the circuit. We have already stressed terminal behavior in the analysis of operational amplifier circuits. In this chapter, we formalize that approach by introducing the two-port parameters. Practical Perspective Characterizing an Unknown Circuit Up to this point, whenever we wanted to create a model of a circuit, we needed to know what types of components make up the circuit, the values of those components, and the inter- connections among those components. But what if we want to model a circuit that is inside a "black box", where the com- ponents, their values, and their interconnections are hidden? In this chapter, we will discover that we can perform two simple experiments on such a black box to create a model that consists of just 4 values - the two-port parameter model for the circuit. We can then use the two-port parameter model to predict the behavior of the circuit once we have attached a power source to one of its ports and a load to the other port. In this example, suppose we have found a circuit, enclosed in a casing, with two wires extending from each side, as shown below. The casing is labeled "amplifier" and we want to determine whether or not it would be safe to use this amplifier to connect a music player modeled as a 2 V source to a speaker modeled as a 32 H resistor with a power rating of 100 W. 679 680 Two-Port Circuits '1 + Input port <"'i • a c* Circuit • K Am • D Q • l 2 + Output port '"': Figure 18.1 A The two-port building block. h + s-domain circuit h + Figure 18.2 A The 5-domain two-port basic building block. 18,1 The Terminal Equations In viewing a circuit as a two-port network, we are interested in relating the current and voltage at one port to the current and voltage at the other port. Figure 18.1 shows the reference polarities of the terminal voltages and the reference directions of the terminal currents. The references at each port are symmetric with respect to each other; that is, at each port the current is directed into the upper terminal, and each port voltage is a rise from the lower to the upper terminal. This symmetry makes it easier to generalize the analysis of a two-port network and is the reason for its uni- versal use in the literature. The most general description of the two-port network is carried out in the s domain. For purely resistive networks, the analysis reduces to solving resistive circuits. Sinusoidal steady-state problems can be solved either by first finding the appropriate ^-domain expressions and then replacing s with jo), or by direct analysis in the frequency domain. Here, we write all equations in the s domain; resistive networks and sinusoidal steady-state solutions become special cases. Figure 18.2 shows the basic building block in terms of the s-domain variables I\,Vy, / 2 , and V 2 . Of these four terminal variables, only two are independent. Thus for any circuit, once we specify two of the variables, we can find the two remaining unknowns. For example, knowing V{ and V 2 and the circuit within the box, we can determine I { and I 2 . Thus we can describe a two- port network with just two simultaneous equations. However, there are six different ways in which to combine the four variables: V2 = Zuh + Zizhi (18.1) (18.2) V t = a n V 2 - a l2 I 2 , h = rt 21^2 — rt 22-^2» (18.3) V 2 = buYi - b l2 I h h = bnYx ~ ^22 A '•> (18.4) V t = h^I x + h 12 V 2 , h = h 2X U + ^22^2; (18.5) h = g\\V\ + gnh> V 2 = &i Vi + g 22 I 2 . (18.6) These six sets of equations may also be considered as three pairs of mutually inverse relations. The first set, Eqs. 18.1, gives the input and out- put voltages as functions of the input and output currents. The second set, Eqs. 18.2, gives the inverse relationship, that is, the input and output cur- rents as functions of the input and output voltages. Equations 18.3 and 18.4 are inverse relations, as are Eqs. 18.5 and 18.6. The coefficients of the current and/or voltage variables on the right- hand side of Eqs. 18.1-18.6 are called the parameters of the two-port cir- cuit. Thus, when using Eqs. 18.1, we refer to the z parameters of the circuit. Similarly, we refer to the y parameters, the a parameters, the b parameters, the h parameters, and the g parameters of the network. 18.2 The Two-Port Parameters 681 18.2 The Two-Port Parameters We can determine the parameters for any circuit by computation or meas- urement. The computation or measurement to be made comes directly from the parameter equations. For example, suppose that the problem is to find the z parameters for a circuit. From Eqs. 18.1, Zu Zn z 2 \ Z?2 Vi h h h h ft, /,=0 n, /,=0 ft, /-»=0 ft. (18.7) (18.8) (18.9) (18.10) /, =n Equations 18.7-18.10 reveal that the four z parameters can be described as follows: • Z\\ is the impedance seen looking into port 1 when port 2 is open. • Zi2 is a transfer impedance. It is the ratio of the port 1 voltage to the port 2 current when port 1 is open. • in is a transfer impedance. It is the ratio of the port 2 voltage to the port 1 current when port 2 is open. • Z22 is the impedance seen looking into port 2 when port 1 is open. Therefore the impedance parameters may be either calculated or measured by first opening port 2 and determining the ratios V\/I\ and V2/I], and then opening port 1 and determining the ratios V|// 2 and Vjjl^. Example 18.1 illustrates the determination of the z parameters for a resis- tive circuit. Example 18.1 Finding the z Parameters of a Two-Port Circuit and therefore Find the z parameters for the circuit shown in Fig. 18.3. Figure 18.3 • The circuit for Example 18.1. Solution The circuit is purely resistive, so the s-domain cir- cuit is also purely resistive. With port 2 open, that is, h = 0, the resistance seen looking into port 1 is the 20 ft resistor in parallel with the series combination of the 5 and 15 ft resistors. Therefore Zn = /^=0 (20)(20) 40 ion. When / 2 is zero, V 2 is 1/,= 15 + 5 (15) = ().75V b Z 2 \ h 0.75¼ = 7.5 ft. / 2 =o Vi/10 When I] is zero, the resistance seen looking into port 2 is the 15 ft resistor in parallel with the series combination of the 5 and 20 X2 resistors. Therefore V, Zll ~ /,=0 (15)(25) 40 = 9.375 ft. Kjis When port 1 is open, I x is zero and the voltage V> = V, -(20) = 0.8K 2 . 5 + 20 With port 1 open, the current into port 2 is V 2 Hence 2|2 ll 9.375* 0.8V 2 /i=0 l/ 2 /9.375 = 7.5 ft. Equations 18.7-18.10 and Example 18.1 show why the parameters in Eqs. 18.1 are called the z parameters. Each parameter is the ratio of a volt- age to a current and therefore is an impedance with the dimension of ohms. We use the same process to determine the remaining port parameters, which are either calculated or measured. A port parameter is obtained by either opening or shorting a port. Moreover, a port parameter is an imped- ance, an admittance, or a dimensionless ratio. The dimensionless ratio is the ratio of either two voltages or two currents. Equations 18.11-18.15 summarize these observations. yu yn h s, 1/2=0 s, v 2 =a yn yn v. V, v,=o S. 1/(=0 (18.11) a n V, A=0 a n o, 1/-,=0 «21 Vi /-,=0 a 2 2 K-,=0 (18.12) '1] Vi /,=0 '12 /1 a 1/,=0 ^ = u /,=0 '22 Vi=0 (18.13) An = 7- *1 a 1/,=0 /*1? = /,=0 /bl = 1/,=0 *22 tt /,=0 (18.14) £ll Vi s. /-,=0 #12 = 1/,=0 &1 /,=0 V, g22 = a. K,=0 (18.15) The two-port parameters are also described in relation to the reciprocal sets of equations. The impedance and admittance parameters are grouped into the immittance parameters. The term immittance denotes a quantity 18.2 The Two-Port Parameters 683 that is either an impedance or an admittance. Tlie a and b parameters are called the transmission parameters because they describe the voltage and current at one end of the two-port network in terms of the voltage and cur- rent at the other end. Tlie immittance and transmission parameters are the natural choices for relating the port variables. In other words, they relate either voltage to current variables or input to output variables. The h and g parameters relate cross-variables, that is, an input voltage and output cur- rent to an output voltage and input current. Therefore the h and g parame- ters are called hybrid parameters. Example 18.2 illustrates how a set of measurements made at the ter- minals of a two-port circuit can be used to calculate the a parameters. Example 18.2 Finding the a Parameters from Measurements The following measurements pertain to a two-port circuit operating in the sinusoidal steady state. With port 2 open, a voltage equal to 150 cos 4000/ V is applied to port 1. The current into port 1 is 25 cos (4000/ - 45°) A, and the port 2 voltage is 100cos (4000/ + 15°) V. With port 2 short-circuited, a voltage equal to 30cos4000r V is applied to port 1. The current into port 1 is 1.5 cos (4000/ + 30") A, and the current into port 2 is 0.25 cos (4000/ + 150°) A. Find the a parameters that can describe the sinusoidal steady-state behavior of the circuit. Solution The first set of measurements gives V, = 150 /Q a V, I, = 25/-45° A, V 2 = 100/15° V, I, = 0A. From Eqs. 18.12, «ii a 2 \ h 150/0 C 25/-45 1 = 1.5/-15% = 0.25/-60°S. /2=0 100/15° The second set of measurements gives V! = 30/0° V, Ij = 1.5 /30°A, V 2 = 0V, I 2 = 0.25/150° A. Therefore a n = rt 2 i = y, i 2 i, -30/0 c „ 2=0 0.25/150° -1.5/30° i/ := o ~ 0.25/150° : 120/30° O, 6/60°. I/ASSESSMENT PROBLEMS Objective 1—Be able to calculate any set of two-port parameters 18.1 Find the v parameters for the circuit in Fig. 18.3. Answer: y n = 0.25 S, 18.3 Vl2 = y 2] = ^ 2= 15 S - -0.2 S, 18.2 Find the g and h parameters for the circuit in Fig. 18.3. Answer: g u = 0.1 S; g n = -0.75; &i = °-75; # 22 = 3.75 H; k n = 4(1; h l2 = 0.8; h 2 \ = -0.8; /* 22 = 0.1067 S. The following measurements were made on a two-port resistive circuit. With 50 mV applied to port 1 and port 2 open, the current into port 1 is 5 /xA, and the voltage across port 2 is 200 mV. With port 1 short-circuited and 10 mV applied to port 2, the current into port 1 is 2 ^tA, and the current into port 2 is 0.5 ^A. Find the g parameters of the network. Answer: g n = 0.1 mS; gu = 4; &i = 4; gr> = 20 kn. NOTE: Also try Chapter Problems 18.2,18.3, and 18.8. 684 Two-Port Circuits Relationships Among the Two-Port Parameters Because the six sets of equations relate to the same variables, the parame- ters associated with any pair of equations must be related to the parameters of all the other pairs. In other words, if we know one set of parameters, we can derive all the other sets from the known set. Because of the amount of algebra involved in these derivations, we merely list the results in Table 18.1. TABLE 18.1 Parameter Conversion Table ^11 Z\2 Zl\ yn yu 3 ; 21 >22 «11 rtt-> «21 «22 bii b\2 V22 = Ay = yn Ay -yi\ Ay yn Ay Z 22 = Az~ = Zu Az Z 2 \ Az _ Zu 3 Az - ill — *21 _ Az _ Z2\ 1 Z21 Zl2 Zn Z22 zu _ Az _ Z\2 flu _ ^22 _ Ah _ 1 «21 h\ h n g\\ Art 1 h[ 2 gu «21 b 2 \ h 2 2 gn 1 Ab h 2 \ gii «21 hi h 22 gn «22 ^it 1 kg «21 ^21 ^22 g\\ «22 ^11 1 Ag «12 b l2 /'11 §22 Art 1 h l2 gu «12 ^12 ^11 S22 1_ ._ _ A/> _ /jn_ _ _#2i «12 b\2 fhl £22 «11 b 22 Ah 1 «12 b\2 'Mi S22 y22 _ ^22 _ __&l _ _1_ m A6 h 2] g 2] _J_ _ bl - _ h VL _ Sn y 21 Ab h 2l &, Ay = h L= h n _ gn y 2l A6 /z 2 i &i y n 6 n 1 Ag y 2 i A/? /z 21 £21 _yu _ «22 _ J_ _ Ag y 12 Art h l2 gn 1_ _ «12 _ fh± _ _gn yu Art h n gn b 2 \ = •>22 1 _ "^ _ * _ *-iz _ "ii _ &11 h-t-y = h 2l = 1 Zn Zu = Zu Az _ 222 £12 _ £22 _£21 Z22 1 *22 1 Z\\ _zn Zu zn = zu Az_ Zn yu yn _ }'n 1 yn _yn yn _ yn yu Ay = yn Ay = y 22 = yn yn _yn y 22 1 > ; 22 _ £21 _ ^22 _ gn Art h l2 gu flu Ah 1 Art h l2 gn «12 ^12 g22 «22 b n Ag _ Art _ J_ _ gi2 «22 ^11 Ag 1_ _ _Ab_ _ _gn_ a 22 b n Ag «21 _ &21 __ gn_ «22 t> n Ag «21 _ ^21 _ ^22 rt n b 22 Ah Art 1 h l2 «11 b 2 2 Ah 1 _ Ab h 2 \ «11 ^22 A/i «12 ^12 ^11 «11 b 12 Ah h 22 - gn = gi2 = gn = g22 = Az = ZyZ22 ~~ ^12-^21 Ay = yny22 - yuyn Art = rt n «22 - «12«21 Ab = bub 22 ~ b n b 2 \ Ah = /Zn/l22 _ ^12^21 Ag = gng22 " gl2g21 Although we do not derive all the relationships listed in Table 18.1, we do derive those between the z and y parameters and between the z and a parameters. These derivations illustrate the general process involved in relating one set of parameters to another. To find the z parameters as functions of the y parameters, we first solve Eqs. 18.2 for V t and V 2 . We then compare the coefficients of I\ and I 2 in the resulting expressions to the coefficients of I Y and / 2 in Eqs. 18.1. From Eqs. 18.2, V\ v 7 /l h yw yi\ yw 3¾] yn >'22 y\i yz2 h h yn j yiz T A/ 1 " Ay' 2 ' (18.16) Ay A/ 1 + Ay' 2 " (18.17) Comparing Eqs. 18.16 and 18.17 with Eqs. 18.1 shows ^22 Z\\ = zn = z 2 \ = Z22 = Ay' >>12 Ay' yn V Ay (18.18) (18.19) (18.20) (18.21) To find the z parameters as functions of the a parameters, we rearrange Eqs. 18.3 in the form of Eqs. 18.1 and then compare coefficients. From the second equation in Eqs. 18.3, 1 #97 «21 «21 (18.22) Therefore, substituting Eq. 18.22 into the first equation of Eqs. 18.3 yields Vi = — /, + -^ ~ «12 / 2 - (18.23) "21 V «21 / From Eq. 18.23, Z\\ = Z\2 = «11 «21 Aw «21 From Eq. 18.22, ^21 = «21 ^22 «22 «21 (18.24) (18.25) (18.26) (18.27) Example 18.3 illustrates the usefulness of the parameter conversion table.