Introduction to Electric Circuits

226 Two-port networks With the output port short circuited so that V2 - 0, VI=BI2 and 11=DI2 The short circuit input impedance i ...

11 Insertion loss 227 terminals of ZI~, and terminating the input port with Z~I results in an impedance looking into the outpu ...

228 Two-port networks 1300f~ o I I o r~ 880t2 Figure 9.21 0 0 75f~ ~] Figure 9.22 75f~ , Solution The network is now as shown in ...

20 log (IbRu/IaRL) = 20 log (Vb/Va) Thus the insertion loss is 20 log (0.5V1/O.O479V1) = 20.37 dB 12 Propagation coefficient (Y ...

230 Two-port networks 9.13 SELF-ASSESSMENT TEST 1 Define a two-port network. 2 Give two examples of a two-port network. 3 Give a ...

14 Problems 231 9.14 PROBLEMS 1 Viewed from the input port a four-terminal network consists of a series impedance of 300 II fo ...

232 Two-port networks 13 Each series arm of a symmetrical low-pass filter consists of a pure inductor of 0.018 H and the shunt b ...

10 Duals and analogues 10.1 DUALS OF CIRCUIT ELEMENTS We have seen that in linear circuit theory there is an intimate relationsh ...

234 Duals and analogues Solution The energy stored in the capacitor is given by W- (CV2)/2. The dual of capacitance is inductanc ...

10.2 Dual circuits 235 I = V(Y1 + Yz + Y3 + Y4) and the circuit described by this equation is shown in Fig. 10.2(b), which is th ...

236 Duals and analogues of a short circuit current in parallel with an admittance Ysc. The voltage across the load admittance YL ...

10.2 Dual circuits 237 'T G (a) l TC ,,, L G I J IT c l L (b) Figure 10.5 'T L H 1 L (c) (c) This circuit shows an ideal transfo ...

238 Duals and analogues 10.3 ANALOGUES If an equation describing the operation of a physical system is identical in form to one ...

10.3 Ana/ogues 239 Solution The magnetic field analogues of D and E are, respectively, B and H so that the analogous energy equa ...

240 Duals and analogues Applying KVL to the outer loop gives 11R1 = 12R2 (10.10) Equation (10.10) could of course been obtained ...

Table 10.4 10.4 Self-assessment test 241 Mechanical Electrical analogue Force source, P Current source, I Velocity, v Voltage, V ...

Answers to self-assessment tests and problems Chapter 1 Self-assessment test second; newton; kilogram. 2. coulomb; newton; volt ...

Answers 243 14 = 44 mA. 7. (a) 6012, (b) 8 A, 4 A, 4 A. 8. 0.332 fL 66.4 fL 0.664 ~. 9. (a) 2 f~, (b) 112.5 W. 10. (a) 1.4 V, 3. ...

244 Answers (a) 38.5 Hz, (b) 50.23 Hz, (c) 1 A, 0.1 A. 9. (a) 15.9 kHz, (b) 20, (c) 5 kHz, (d) 13.4 kHz, 18.4 kHz. 10. (a) 33.7 ...

Answers 245 and each successive negative voltage becoming numerically larger. i(t) = (2/to) exp (-at) sin tot where a = r/EL = ...