Question

Consider the two-port network shown with n=3, R=9k and C=6?F.

(50 points) Consider the two-port network shown with n - 3, R - 9kS2 and C -6pF. a. Determine the hybrid [h] and impedance [z] parameters (as functions of s; hint: each hybrid parameter is a first order transfer function) 1000 ? 10.333333 0.3333 6.667E-17 166666.676 | ? | 500000/s 500000/s |?| 9000/s

Answer 1

(a) Consider the circuit. 1:3 9 k22 76 LF 3 & c a The voltage across the capacitor is equal to the input voltage, V. V, =(9 k)12 +3V, I= 31, + 1 1. =-31, +6x10°s, V,=6x1065* +6x10-67 V. – 166666.671, + 5000001 S Substitute the expression for V, in the equation for V. V2 =(9 k)], +3V (166666.67_ 500000 2+3 -1,- -12 +(9 k). +3(196 7., 590) _500, +(9k + 15000) v, - 500000 1, +(9000 1500000), Compare the two equations with the equations for Z-parameters. V = ZuI+2212 V2 = Z211, +Z2212 166666.67 vo 2, - 500000 212 = - O 2 - 500000 - 2 222 = 9000 + 1500000

Consider the equations for h-parameters. V = h,1 +h2V2 12 = h, 1+hV, VA 12 11 11,50 V, = 0 represents a short circuit at the load terminals. 3V. 1 9 k 3k 1 =-31, +6x10-SV, 1, = - = - V. 1, ==3( +6x10“sv, = (10- +6x10Ⓡs), 3k) -10° +6x10^s hist10 166666.67 S +166.67 1. - - -V = 1 = 166666.67 s +166.67 3 k ܚܙܐ 55.55 5+166.67 1, = 0 represents an open-circuit at the input terminals. 11 =-31, +6x10-65V, =0 312 =3x106SV, V2 = (9 k)1, +372 = V=V, -(9k)12 V, =(9 k)(3x10-65V)+3V, V 1 V, 3+0.027s has 37.04 5+111.11 12 = 3x10-6 sv. 1, = (3x10^s)[V, –(9 k)13] (1+0.0275)/2 = 3x10-ⓇsV h. - IL 3x10“s V2(1+0.0275) h, -1.11x104 S +37.04

12 = 3x10-6 sv. 12 =(3x10s)[V> –(9 k)12] (1+0.027s)), = 3x10-65V, h - 123x1065 V2 Ip-0 (1+0.0275) h, _ 1.11x104 S+ 37.04 (b) Consider the circuit. + Vs + 1 The source current is, is = 5 cos (300t) mA Determine the current through the capacitor. ie= - = jQCvi = j(300)(6x10-4)= jl.8v, mA joc Determine the current at the primary of the transformer. i, ri, +ic =5+ jl.8v, mA iz = -1;=1.667 + j0.6v, ma Write the voltage equation at the secondary of the transformer. V. = iz (9k)+3v; =(1.667 + j0.6v;)(9)+3v; =15+(3 + j5.4) v; - V. -15 8-15 v = -=1.1332119.050 V ? 3+ j5.4 3+ j5.4 v= 1.133 cos (300t +119.059) V; t>0 in =1.667 + j0.6v, mA =1.667 + j0.6x1.133 Z119.05° =1.1222-17.1° mA iz(t)=1.122 cos (300t – 17.1°) mA