Question

Q 3. Time 250 the input Impedance is defined as 2-VA find the input impedance of the part below when R - 16., 2010-10-20 and 5000 hPereport your newer the male posted all angles are in the range of negative 180 degrees to positive 180 degrees C R w R www 12 The value of Z- 2

Answer 1

Ig VO Vo www RI www R₂ C2 Zin Given, Ri= 16kr R2 = 20532 , 0,= lonF, 42 = 200 F W: 5000 rad/s By applying by virtual KCL at V and ground concept Vx-vo Vx-vo 0 T Vx-vg R + R2 jwa) ,الدز By kcL at V. = - Vo (jwGR2) Vo Vo-Vx -> Vo-Vx R₂ lowe, - Vx Yo (i+ Rajax(s)) .

from vo (tj w Q₂ (2) avg voljW6,862). Vo (jWGR)Ciwa) X2 Ri (vo (14) WRZCx) + Vo (jwca) R, + Vo (-w64R24 R.)) = vg we know that, - I Vy Vg -vx Ri vg (It j w R2 (2) It jwR2C2 + jWC, R, - W² CG RR, R jwR, C2 - W² C₂C, R, R2 R. (1+ j w ļC2 + 3 W GR, - w² CzC, P, R2) - Ig » [ date to the room - w gg, RiR2) Zin= Vg Ig R. (1+j w Rz (z tj w Rg (2 - ; WR, G - W² CC, R, R₂ 다 16x103 (1+; (5000) (20*10° x 20x104) +; (5000) (16x16x20x109) (5000)? (20x169) (10x109) (16*103 *20*103)] j (5000) (16x103) (20x109) – (50001 (20x10*)(10x109) (16 220° x 20x109)

1642031 1+j2 +j1.6 -1.6) jl.6+(-1.6) 104 (-0.6+)3.6) (-1+;) Exio} (-1+;6) (tj) 6x103 (-1+36) (-1-;) 1+1 3x100 (1 +3 +63 +6) 3x103 (7-5;) 25.8x103 4- 35-53° 2 L2= -35.53° .. 121= 25.8kr