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246. Going in the ccw sense around the two secondary circuits, the sums of the voltage drops are (letting Z2 R j!L for Fig: 281: for Fig: 282: hence, by eq. (517) for Fig. 281, h21 I2 =I1 j!M=Z2 ; answer: then, by eq. (517), for Fig. 282, h21 I2 =I1 j !M=Z2 ; answer: The results simply show that, for the particular network of Figs. 281 and 282, the sign of h21 depends upon the sense in which the secondary turns are wound relative to the primary turns. 247. First verify that dh 0:1320. Next recall, from section 11.1, that two matrices can be equal only if all corresponding elements are equal. With this in mind, inspection of the fourth row of the conversion chart then gives the approximate answers g11 h22 =dh 0:0030 mhos g21 h21 =dh 197:0 g12 h12 =dh 0:0606 g22 h11 =dh 6439:4 ohms j!MI1 Z2 I2 0 j!MI1 Z2 I2 0
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248. One procedure is as follows. First solve eq. (507) for V2 , then substitute that value of V2 into eq. (506), which then becomes V1 h11 h12 h21 =h22 I1 h12 =h22 I2 which, since h11 h12 h21 =h22 h11 h22 h12 h21 =h22 dh=h22 , becomes V1 Next, by eq. (507), V2 h21 1 I1 I h22 2 h22 dh h I1 12 I2 h22 h22
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Or, in matrix form, the last two simultaneous equations become ! ! ! V1 dh=h22 h12 =h22 I1 V2 h21 =h22 1=h22 I2 Comparison of the above equation with eq. (519) shows it has to be true that ! ! dh=h22 h12 =h22 z11 z12 z21 z22 h21 =h22 1=h22 thus proving that the relationship given in the conversion table is correct. 249. From the rst row of the conversion chart we have that ! ! z11 z12 1=g11 g12 =g11 z21 z22 g21 =g11 dg=g11
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thus z11 1=g11 14:71 ohms z21 g21 =g11 3353 ohms 250. By eq. (511), g which can be written V1 I2 V1 I2 ! g ! z12 g12 =g11 1:074 ohms z22 dg=g11 8510 ohms !
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because g is a square matrix. Comparison of the last equation with eq. (508) shows that h g 1 as proposed. 251. First we have dh h11 h22 h12 h21 0:1320 Then, from the matrix conversion chart, the answers are z11 dh=h22 330 ohms z21 h21 =h22 65;000 ohms z12 h12 =h22 20 ohms z22 1=h22 2500 ohms
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252. First, from inspection of Fig. 286, note that (where e refers to the single equivalent two-port) V1e V1a V1b I1e I1a I1b Next, from eq. (521) we have for two-port a: V1a I2a for two-port b: " # " V2e V2a V2b I2e I2a I2b
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! "
h11a h21a
h12a h22a
I1a I2a
! x # # y
V1b I2b
h11b h21b h11b h21b
h12b h22b h12b h22b
#" #"
I1b V2b I1a V2a
Now note that the sum of eqs. (x) and (y) can be written as ! ! ! V1e h11a h11b h12a h12b I1e h21a h21b h22a h22b V2e I2e
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which, upon reference to eq. (521), shows that the h-parameters of the single equivalent two-port are equal to the sum of the h-parameters of the individual two-ports when the two-ports are connected in the series-parallel mode of Fig. 286. 253. Making use of eq. (536) and the conversion chart in section 11.8 we have, for three identical two-ports in parallel, ! ! ! 3y11 3y12 y11 y12 3z22 =dz 3z12 =dz 3 3z21 =dz 3y21 3y22 y21 y22 3z11 =dz and thus the answers are that parameters of the equivalent two-port, in terms of the z-parameters of the individual two-ports, are z11e 3z22 =dz; z12e 3z12 =dz; z21e 3z21 =dz; z22e 3z11 =dz
254. Since we re dealing with two identical two-ports, eq. (541) becomes ! ! ! ! a11 a12 a11 a12 Vo V1 I1 a21 a22 a21 a22 Io which, upon using the conversion chart in section 11.8, becomes ! ! ! ! V1 dh h11 Vo dh h11 1 2 h21 h22 1 I1 h22 1 Io Now make use of the procedure for matrix multiplication de ned in section 11.2; doing this gives the nal answer # ! " 2 ! V1 d h h11 h22 dhh11 h11 Vo I1 Io dhh22 h22 h11 h22 1 (From Fig. 288, note that, for two two-ports in cascade, Vo V4 V5 and Io I5 .) 255. Note that the answer to problem 254 is of the form ! ! ! ! V1 A B Vo c11 I1 Io c21 C D and therefore V1 I1 which says that V1 AVo BIo A B=ZL Vo and I1 CVo DIo C D=ZL Vo because, from Fig. 288, Io I7 Vo =ZL . Now solve the rst of the two simultaneous equations for Vo , then substitute the result in place of Vo in the second equation. Doing this gives us I1 C D=ZL V1 D CZL V1 A B=ZL B AZL ! AVo BIo CVo DIo !