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Matrix Algebra. Networks
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Now try the following problems. As you may note, some of the problems would be as easy, or easier, to work without using matrices. But this is, of course, beside the point, as our object here is to provide practice in thinking in terms of matrices. Matrix algebra is a shorthand method of manipulating systems of simultaneous equations; its real value becomes evident in the analysis of complex networks represented by such equations. It allows us to study systems of interconnected blocks of elements without having to write out the mass of individual equations associated with the system. Digital computer programs for solving matrix equations are available, and are used to provide actual numerical answers if this is required. Problem 257 For eq. (545) show that I1 Z22 Z V1 Z12 Z V2 dz Z11 Z22 Z12 Z21 Z
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where dz Z11 Z22 Z12 Z21 Problem 258 Write eq. (545) in terms of the h-parameters of the transistor. Problem 259 Can eq. (504), in section 11.6, be applied directly to eq. (543) Problem 260 Solve eq. (548) for the matrix Vo Io by taking the inverse of the coe cient matrix. (Note: the above will be easy if you take advantage of the special formula for nding the inverse of a 2 2 matrix given in the solution to problem 256.) Next, in the basic Fig. 277, suppose a load impedance of ZL ohms is connected to the output terminals, as in Fig. 307, and that the PROBLEM is to nd the value of the output load current IL . !
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Fig. 307
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* At junction point a in Fig. 307, by Kirchho s current law, I2 IL 0; that is, I2 IL .
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CHAPTER 11 Matrix Algebra. Networks
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Problem 261 For the network of Fig. 307 inside the box, suppose it is found that h11 400 ohms h21 20 h12 0:100 h22 0:002 mhos
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Given that V1 12 volts, nd the value of load current if ZL RL 150 ohms. (Answer: IL 0:2927 amps) Problem 262 Rework problem 261, this time beginning with the matrix equation (514) in section 11.6. Problem 263 Two identical transistors, operating in common-emitter mode, are connected in cascade as shown in Fig. 308.
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Fig. 308
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In the gure, let it be given that the transistor h-parameter values are h11 1000 ohms h21 40 h12 0:004 h22 0:0005 mhos
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If it is given that R 500 ohms and RL 900 ohms, nd the output voltage VL if the input voltage V1 is 0.001 volt. (Again, as in problem 262, let us begin with eq. (514) in section 11.6.) (Answer: 0.3196 volts) Problem 264 Write the set of simultaneous eqs. (455), (456), and (457), in Chap. 10, in the form of a single matrix equation. Problem 265 Note that the answer to problem 264 says that 3 1 2 3 2 3 2 A A1 1 1 1 6 7 6 7 6 7 2 4 A2 5 4 a a 1 5 4 B 5 C a2 a 1 A0 As an exercise in matrix manipulation, verify that the above expression does produce eqs. (460), (463), and (466) in Chap. 10. Our nal example, which follows, will provide further practice in matrix manipulation and will also bring to light an interesting fact concerning power in unbalanced three-phase systems. In doing this we ll freely make use of our previous work in three-phase theory
Matrix Algebra. Networks
Fig. 309
in sections 10.7 through 10.11. Let us begin with the three-phase generator depicted in Fig. 309. In the gure, Va , Vb , and Vc represent the rms values of three unbalanced phase voltages, with Ia , Ib , and Ic representing the rms values of the three corresponding unbalanced phase currents (also the line currents here), as shown. In the work here we wish to concentrate our attention on the POWER produced in the unbalanced condition, ESPECIALLY in regard to expressing the power in terms of the SYMMETRICAL COMPONENTS of the unbalanced system. To begin, let PT denote the total true power produced by the above unbalanced generator. From inspection of the gure it s clear that PT is equal to the SUM OF THE POWERS produced by the three individual phases of the generator; thus (see note 29 in Appendix) in terms of the actual phase voltages and currents the value of PT is equal to the sum of the real parts (srp) in the expression " " " PT srp : Va Ia Vb Ib Vc Ic 549
" " " in which the overscore in Ia ; Ib ; Ic denotes the CONJUGATE of the quantity represented by the letter. (It s understood that the Vs and Is are, in general, complex numbers.) Note that eq. (549) is expressed in terms of the actual phase voltages and currents; but we, however, wish to express the power in terms of the SYMMETRICAL COMPONENTS of the phase voltages and currents. To do this, let us start by writing eq. (549) in matrix notation; thus 2 3 " Ia 6" 7 PT srp: Va Vb Vc 4 Ib 5 550 " Ic Let us now rst work on the above current matrix, as follows. From inspection of Fig. 309 we have Ia I1 I2 Io Ib aI1 a2 I2 Io Ic a2 I1 aI2 Io Now take the CONJUGATES of the above equations. Remembering that the conjugate of the sum of a number of complex numbers is the sum of the conjugates and that the
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