Electric Circuits, 9th Edition P72 ppt

686 Two-Port Circuits Example 18.3 Finding h Parameters from Measurements and Table 18.1 Two sets of measurements are made on a two-port resistive circuit. The first set is made with port 2 open, and the second set is made with port 2 short-circuited. Hie results are as follows: Port 2 Open V, = 10 mV /i = 10/xA V, = -40 V Port 2 Short-Circuited l/j = 24 mV /j = 20 /AA /•> = 1 raA Find the /; parameters of the circuit. Solution We can find h u and h 2 \ directly from the short- circuit test: k u = Vi V\ = () 24 x 10 20 X 10" -3 = 1.2 kO, h->i = ~r v 2 =o 10"" 3 20 X 10 -IT 50. The parameters h l2 and h 2 i cannot be obtained directly from the open-circuit test. However, a check of Eqs. 18.7-18.15 indicates that the four a parameters can be derived from the test data. Therefore, h n and h 22 can be obtained through the conversion table. Specifically, A« «22 «21 h 22 = a i 7 The a parameters are flu = Yi h /-, = 0 /, = 0 10 X IP" 3 -40 10 X 10~ 6 -40 = -0.25 X 10~ 3 , -0.25 X 10" 6 S, «12 = Ha - h V-,=() 1/,=0 24 X 10" 3 10" 3 = -24 11, 20 x IP" 6 10~ 3 = -20 X 10~\ The numerical value of Aa is Art = «n«22 — a U a 21 = 5 X 10~ 6 - 6 X 10~ 6 = -10 -6 . Tli us /*12 = A« «22 10 -6 -20 X 10" = 5 X 10 -5 , «21 «22 = «22 -0.25 x icr* m -20 X 10~ 3 /ASSESSMENT PROBLEM Objective 1—Be able to calculate any set of two-port parameters 18.4 The following measurements were made on a two-port resistive circuit: With port 1 open, V 2 = 15 V, V 1 = 10 V, and h = 30 A; with port 1 short-circuited, V 2 = 10 V, 1 2 = 4 A, and /] = -5 A. Calculate the z parameters. Answer: z n «12 (4/15) ft; (1/3) ft; Z21 = -1.6 ft; Z22 = 0.5 ft. NOTE: Also try Chapter Problem 18.13. 18.2 The Two-Port Parameters 687 Reciprocal Two-Port Circuits If a two-port circuit is reciprocal, the following relationships exist among the port parameters: Zl2 ~ z 2h >'i2 = yai< rt ll a 22 — a 12^21 = Art = 1, ^11^22 ~~ ^12¾] = A/; = 1, /i ]2 = -/^21 (18.28) (18.29) (18.30) (18.31) (18.32) gl2 •gl\- (18.33) A two-port circuit is reciprocal if the interchange of an ideal voltage source at one port with an ideal ammeter at the other port produces the same ammeter reading. Consider, for example, the resistive circuit shown in Fig. 18.4. When a voltage source of 15 V is applied to port ad, it produces a current of 1.75 A in the ammeter at port cd.Thc ammeter current is eas- ily determined once we know the voltage V hd . Thus 60 30 20 (18.34) and V b(i = 5 V. Therefore 5 15 / = — + — = 1.75 A. 20 10 (18.35) If the voltage source and ammeter are interchanged, the ammeter will still read 1.75 A. We verify this by solving the circuit shown in Fig. 18.5: 60 30 20 (18.36) From Eq. 18.36, V^ = 7.5 V. The current / at) equals 7.5 15 U = — + — = 1.75 A. 'ad 30 10 (18.37) A two-port circuit is also reciprocal if the interchange of an ideal cur- rent source at one port with an ideal voltmeter at the other port produces 10(1 -AW- 30 O b 20 0 - / Wv f -VW- 15 VI 6()() /'( /j Ammeter d d Figure 18.4 A A reciprocal two-port circuit ion 30 n b 20 n c AM* f VW •—•- Ammeter ( / yl 60 n 15 V Figure 18.5 A The circuit shown in Fig. 18.4, with the voltage source and ammeter interchanged. 688 Two-Port Circuits /, »- + V\ • 2 a 2 b —(i— 2 a /, "*— + V"; • + + V, (a) (b) /l + 2 a 2 C Vi • 2 b —it— 2 a U + V 2 • /, + xC\ 2 a /•> ^/ + (c) (d) Figure 18.6 A Four examples of symmetric two-port circuits, (a) A symmetric tee. (b) A symmetric pi. (c) A symmetric bridged tee. (d) A symmetric lattice. the same voltmeter reading. For a reciprocal two-port circuit, only three calculations or measurements are needed to determine a set of parameters. A reciprocal two-port circuit is symmetric if its ports can be inter- changed without disturbing the values of the terminal currents and volt- ages. Figure 18.6 shows four examples of symmetric two-port circuits. In such circuits, the following additional relationships exist among the port parameters: Z\\ = z 22 , (18.38) v ii = to a u = a 22 , ^11^22 _ ^12½ = A/* = 1, 8n822 - Sngn = &g = 1- (18.39) (18.40) (18.41) (18.42) (18.43) For a symmetric reciprocal network, only two calculations or meas- urements are necessary to determine all the two-port parameters. ^ASSESSMENT PROBLEM Objective 1—Be able to calculate any set of two-port parameters 18.5 The following measurements were made on a resistive two-port network that is symmetric and reciprocal: With port 2 open, V\ = 95 V and I { = 5 A; with a short circuit across port 2, V l = 11.52 V and I 2 = -2.72 A. Calculate the Z parameters of the two-port network. Answer: z\\ = z 22 = 19 O, zn = £21 = 17 ft. NOTE: Also try Chapter Problem 18.14. 18.3 Analysis of the Terminated Two-Port Circuit 689 18.3 Analysis of the Terminated Two-Port Circuit In the typical application of a two-port model, the circuit is driven at port 1 and loaded at port 2. Figure 18.7 shows the s-domain circuit diagram for a typically terminated two-port model. Here, Z ? represents the internal impedance of the source, V g the internal voltage of the source, and Z L the load impedance. Analysis of this circuit involves expressing the terminal currents and voltages as functions of the two-port parameters, V s , Z g , and Z L . Six characteristics of the terminated two-port circuit define its termi- nal behavior: the input impedance Z m = V\jl\, or the admittance Y- m = l\/V\ the output current h theThevenin voltage and impedance (K Th , Z Th ) with respect to port 2 the current gain I 2 /Ii the voltage gain V 2 /V\ the voltage gain VtfV n /l i> z g i- Two-port model of a network h + ? i. Figure 18.7 • A terminated two-port model. The Six Characteristics in Terms of the z Parameters To illustrate how these six characteristics are derived, we develop the expressions using the z parameters to model the two-port portion of the circuit. Table 18.2 summarizes the expressions involving the y, a, b, h, and g parameters. The derivation of any one of the desired expressions involves the algebraic manipulation of the two-port equations along with the two con- straint equations imposed by the terminations. If we use the z-parameter equations, the four that describe the circuit in Fig. 18.7 are V\ = Zii/i + 212^21 V 2 - Z2\h + v t = v s - i { z g , V 2 =-I 2 Z L . (18.44) (18.45) (18.46) (18.47) Equations 18.46 and 18.47 describe the constraints imposed by the terminations. To find the impedance seen looking into port 1, that is, Z m = V x /I h we proceed as follows. In Eq. 18.45 we replace V 2 with -I 2 Z L and solve the resulting expression for / 2 : h = -z 2] h Z L + z 22 (18.48) 690 Two-Port Circuits TABLE 18.2 Terminated Two-Port Equations z Parameters -^12^21 Z m - Z\\ I, = -^22 + Z/ -221V. («n + Z^izn + Z,) - z n z 2 [ zn + z « Zxh — Z?? ~ £12¾ :„ + Z g *2 — ^21 /] ~ Z 2 2 + Zl VI z 2 \Z L V, z u Z L + Az ^2 Z 2 \Z L Vf. (Zll + 2,)(¾ + 2/.) - <12*21 y Parameters y m = vn - i, = y\ 2 y 2 \Z L 1 + y 22 Z L VnV n V Th = 1 + y 22 Z L + y n Z g + &yZ n Z L Z-Th = >^ 2 + A}>Z(, 1 + >')iZ, V22 + A>'Z S ^2 _ >'21 1 { y n + AyZ 7 , V2 _ -y*z L V\ 1 + y 2 2^ y 2 yiiZ L V K y\2}'2\Z R Z L -(1 + y u Z R ){\ + j^Z L ) a Parameters a\\Z L + «12 Z in = /, = fl 2 ]Z/_ + «22 • v * a\\Z[ + « 12 + a 2 \Z^Z. h + ^22^ V„ 1½. = Zlh — « n + a n Z s «12 + «22^ fl,, + «21 Zj. h = -1 /] «21Z/, + ^22 V2 _ Z L V x a u Z L + « 12 V2 Z; ^ ( a ll + a 2\Z R )Z L + «12 + ^22¾ b Parameters k&Zi + /?i2 Z, n = 6 21 Z/. + b n h -V^b buZf, + b 2[ ZgZ L + b 22 Z L + b [2 hx + hiZg _h u Z R + b x2 b 2 \Z R + b 22 h _ -A/J /l 6l ] + &2lZ L V, hbZ L V\ b\ 2 + buZi V 2 U?Z L V s b n + 6 n Z s + /> 22 Z,. + 6 2 iZ fi Z/, h Parameters Zin = /'ll ~ hvM\Z L 1 + /J 22 Z ; . (1 + h 22 Z,)(h u + Z s ) - h n h 2{ Z L -/'2.V, "Th " Z'lh / 2 h Vi h 22 Z„ + A/? Z K + i h&Zft A21 1 + /z 2: -/l 2 ] + A/i sZ/. Z/. Vx AhZ L + h u V 2 -h 2 {Z L V, (Mi + Z ? )(l + h 12 Z,) - h l2 h 2l Z L g Parameters S12S21 2i-> + Z £22 ~»~ ^/. /0 '£21 K« VTH = (1 + guZ n )(g22 + Zi) - gng2iZf, 1 + SiiZg ZJU - 822 /, gl282lZ s I + gaZ K •&i /1 guZ L + Ag ^2 _ g2lZ L V\ 812 + Z/. ^2 8i\Z L V„ (1 + gllZ s )(g 22 + Z L ) - gi 2 g2lZ s We then substitute this equation into Eq. 18.44 and solve for Z m : Z in = z u 7^-, (18.49) Z22 + Z L To find the terminal current h, we first solve Eq. 18.44 for I\ after replacing V x with the right-hand side of Eq. 18.46. The result is h = ' 7 • (18.50) Z\\ -r A, We now substitute Eq. 18.50 into Eq. 18.48 and solve the resulting equa- tion for / 2 : -z 2l V R (z n + 2^)(¾ + Z L ) - z 12 z 2 ] The Thevenin voltage with respect to port 2 equals V 2 when U = 0. With I 2 = 0, Eqs. 18.44 and 18.45 combine to yield ^(/,=0 = Z 2 l/l = %— • (18.52) Zll But V, = V^ - I\Z g , and /j = V g /(Z g + Z\\Y therefore substituting the results into Eq. 18.52 yields the open-circuit value of V 2 : V 2 |/,-o = Vvk = 7 I V K - (18.53) The Thevenin, or output, impedance is the ratio V 2 /I 2 when V M is replaced by a short circuit. When V„ is zero, Eq. 18.46 reduces to V, = -7,Z f . (18.54) Substituting Eq. 18.54 into Eq. 18.44 gives — Sl2'2 A = f^T- ( 18 - 55 ) We now use Eq. 18.55 to replace ^ in Eq. 18.45, with the result that Z\lZ2\ = Z Th = z 22 —= (18.56) V H =i) Zu -r & R Tire current gain I 2 /I\ comes directly from Eq. 18.48: h _ ~Zi\ h Z L + z 22 (18.57) To derive the expression for the voltage gain V 2 fV\, we start by replac- ing I 2 in Eq. 18.45 with its value from Eq. 18.47; thus V 2 = Z 2 ]I] + Z 22 -7T 1 - • (18.58) 692 Two-Port Circuits or Next we solve Eq. 18.44 for /, as a function of V\ and V 2 \ Zuh — V\ - z\2 v x z v y 2 Z]l Zll^L (18.59) We now replace I { in Eq. 18.58 with Eq. 18.59 and solve the resulting expression for V 2 /V\: Vi ZnZ, \**L Z\\Z L + Z\\Z22 ~ Z\ 2 Z 2 \ = z 2l Z L z n Z L + Az' (18.60) To derive the voltage ratio V 2 /Vg, we first combine Eqs. 18.44, 18.46, and 18.47 to find /, as a function of V 2 and K: Z12V2 V g Zdzu + Z R ) + Z„ (18.61) We now use Eqs. 18.61 and 18.47 in conjunction with Eq. 18.45 to derive an expression involving only V 2 and V g ; that is, Vi Z22 v* Z L (zn + Z g ) z n + Z g Z L which we can manipulate to get the desired voltage ratio: V 2 z 2 \Z L n (z n + Zg)(z 22 + Z L ) - Z12Z21 (18.62) (18.63) The first entries in Table 18.2 summarize the expressions for these six attributes of the terminated two-port circuit. Also listed are the corre- sponding expressions in terms of the y, a, b, h, and g parameters. Example 18.4 illustrates the usefulness of the relationships listed in Table 18.2. Example 18.4 Analyzing a Terminated Two-Port Circuit The two-port circuit shown in Fig. 18.8 is described in terms of its b parameters, the values of which are b u = -20, b l2 = -3000X1, b 2] = -2 mS, b 22 = -0.2. a) Find the phasor voltage V 2 . b) Find the average power delivered to the 5 kH load. c) Find the average power delivered to the input port. 500Z0 0 .1500 0 • -vw + T> : ^1 [b] + v : 1 5 kft d) Find the load impedance for maximum average power transfer. e) Find the maximum average power delivered to the load in (d). Solution a) To find V 2 , we have two choices from the entries in Table 18.2. We may choose to find I 2 and then find V 2 from the relationship V 2 = -1 2 Z L , or we may find the voltage gain V 2 /V g and calculate V 2 from the gain. Let's use the latter approach. For the 6-parameter values given, we have &b = (-20)(-0.2) - (-3000)(-2 x 10" 3 ) = 4-6 = -2. Figure 18.8 • The circuit for Example 18.4. 18.3 Analysis of the Terminated Two-Port Circuit 693 From Table 18.2, V 2 &bZ, Y g b ]2 + b n Z g + b 22 Z L + b 2l Z„Z L The average power delivered to the input port is 0.78947 2 Px = -(13333) = 41.55 W. (-2)(5000) -3000 + (-20)500 + (-0.2)5000 + [-2 X 10 10 ~ 19' Then, 3 (500)(5000)] d) The load impedance for maximum power trans- fer equals the conjugate of the Thevenin imped- ance seen looking into port 2. From Table 18.2, bnZg + b u V, = 500 = 263.16/0° V. b) The average power delivered to the 5000 ft load is 263.16 2 P> = 2(5000) 6.93 W. c) To find the average power delivered to the input port, we first find the input impedance Z in . From Table 18.2, 7 = b 22 Z L + b 12 b 2 ]Z L + b n (-0.2)(5000) - 3000 " -2 X 10 _3 (5000) - 20 400 3 Now i! follows directly: 500 133.33 ft. Ii = 500 + 133.33 = 789.47 mA. -Th b 2l Z g + b 22 (-20)(500) - 3000 (-2 X 10~ 3 )(500) - 0.2 13,000 1.2 = 10,833.33 ft. Therefore Z L = Zj h = 10,833.33 ft. e) To find the maximum average power delivered to Z L , we first find V 2 from the voltage-gain expression V 2 /V / ,. When Z L is 10,833.33 ft, this gain is V 2 Thus and 0.8333. V 2 = (0.8333)(500) = 416.67 V, P L (maximum) 1 416.67 2 2 10,833.33 8.01 W. •ASSESSMENT PROBLEM Objective 2—Be able to analyze a terminated two-port circuit to find currents, voltages, and ratios of interest 18.6 The a parameters of the two-port network shown are a n = 5 X 10~ 4 , a l2 = 10 ft, a 2l = 10~ 6 S, and a 22 = -3 X 10~ 2 . The net- work is driven by a sinusoidal voltage source having a maximum amplitude of 50 mV and an internal impedance of 100 + /0 ft. It is termi- nated in a resistive load of 5 kft. a) Calculate the average power delivered to the load resistor. b) Calculate the load resistance for maximum average power. Answer: (a) 62.5 mW; c) Calculate the maximum average power (J°) 70/6 kft; delivered to the resistor in (b). (c) 74.4 mW. /l 1 z* + /4\ ( ) K ' T - Two-port model of a network + Vi /- z L NOTE: Also try Chapter Problems 18.29,18.30, and 18.34. 18.4 Interconnected Two-Port Circuits Synthesizing a large, complex system is usually made easier by first designing subsections of the system. Interconnecting these simpler, easier-to-design units then completes the system. If the subsections are modeled by two-port circuits, synthesis involves the analysis of interconnected two-port circuits. Two-port circuits may be interconnected five ways: (1) in cascade, (2) in series, (3) in parallel, (4) in series-parallel, and (5) in parallel-series. Figure 18.9 depicts these five basic interconnections. We analyze and illustrate only the cascade connection in this section. However, if the four other connections meet certain requirements, we can obtain the parameters that describe the interconnected circuits by simply adding the individual network parameters. In particular, the z parameters describe the series connection, the y parameters the parallel connection, the h parameters the series-parallel connection, and the g parameters the parallel-series connection. 1 The cascade connection is important because it occurs frequently in the modeling of large systems. Unlike the other four basic interconnec- tions, there are no restrictions on using the parameters of the individual two-port circuits to obtain the parameters of the interconnected circuits. The a parameters are best suited for describing the cascade connection. We analyze the cascade connection by using the circuit shown in Fig. 18.10, where a single prime denotes a parameters in the first circuit and a double prime denotes a parameters in the second circuit. The output 1 2 (a) 1 2 (b) (c) 1 -• •- 2 1 ^ ^ 2 (d) (e) Figure 18.9 • The five basic interconnections of two-port circuits. (a) Cascade, (b) Series, (c) Parallel, (d) Series-parallel, (e) Parallel-series. 1 A detailed discussion of Ihese four interconnections is presented in Henry Ruston and Joseph Bordogna, Electric Networks: Functions, Filters, Analysis (New York: McGraw-Hill. 1966). ch. 4. 18.4 Interconnected Two-Port Circuits 695 voltage and current of the first circuit are labeled V 2 and / 2 , and the input voltage and current of the second circuit are labeled V\ and I\. The prob- lem is to derive the ^-parameter equations that relate V 2 and I 2 to Vj and / lt In other words, we seek the pair of equations L\ — (121*2 ^22-^2^ /, m —*• + Circuit 1 a 21 a 22 + + r 2 v, Circuit 2 «"ll «"l2 »21 ^ 22 /, + ^2 (18.64) (18 65) Fl '9 ure 18 « 10 A A cascade connection. where the a parameters are given explicitly in terms of the a parameters of the individual circuits. We begin the derivation by noting from Fig. 18.10 that VI = a\iV 2 - a\ 2 I' 2 , (18.66) The interconnection means that V 2 — V\ and V 2 •- these constraints into Eqs. 18.66 and Eqs. 18.67 yields (18.67) •l\. Substituting Vi = a' u V\ + a' l2 I' h (18.68) /, = a ' 21 v\ + a' 22 I\. (18.69) The voltage V\ and the current l\ are related to V 2 and I 2 through the a parameters of the second circuit: V\ = a" n V 2 - a n n I z , (18.70) /1 = a 2 \V 2 - ahh- (18.71) We substitute Eqs. 18.70 and 18.71 into Eqs. 18.68 and 18.69 to generate the relationships between V h /j and V 2 ,1 2 : Vi = (flnfln + a' u a 2 \)V 2 -{a' n a'{ 2 + a\ 2 a 22 )I 2 , (18.72) /] = («21«11 + «22«2l)^2-(«21«f2 + (lVflh)h- (18.73) By comparing Eqs. 18.72 and 18.73 to Eqs. 18.64 and 18.65, we get the desired expressions for the a parameters of the interconnected networks, namely, a u = a' u a'{ : + a\ 2 a 2 \. (18.74) «12 = flii fl i2 + (iiiah- (18.75) «21 = 021«?] + "22"21- (18.76) «22 = rt 21 fl 12 + rt 22«22- (18.77) . V 2 • /, + xC 2 a /•> ^/ + (c) (d) Figure 18.6 A Four examples of symmetric two-port circuits, (a) A symmetric tee. (b) A symmetric pi. (c) A symmetric bridged tee. (d) A symmetric. currents and volt- ages. Figure 18.6 shows four examples of symmetric two-port circuits. In such circuits, the following additional relationships exist among the port parameters: Z\ = z 22 ,. easier-to-design units then completes the system. If the subsections are modeled by two-port circuits, synthesis involves the analysis of interconnected two-port circuits. Two-port circuits