Solution Manual Electric Circuit Analysis and Design ch 15, Exercises of Electronic Circuits Analysis

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Electronic Circuit Analysis and Design, 2! edition Solutions Manual Chapter 15 Exercise Solutions E15.1 L dpole |] = = 6.0 dB fas * aE 1 (2) "RC yit as RO = aap = he = 159 10 1 2xfsaa — 2e(10*) 4-pote |] = eee = 7.24 BB La C =0.01 uF > 2= 159k je 10 Then E1s.4 Cy, = 0.03546 uF t Cz = 0.01392 uF Reg = Fcc Cy = 0.002024 uF a 7 3 * or fol = gm = Te a2 10 1 1 If C = 10 pF = fc = 20 kHz jTl= = 4(£ 1+{% E1S.$ faa Ww CQ 30 [T| = 0.124 or [7] = 18.1 dB Low-frequency gdias T= — 7 = — 3 _ fol 100 % 10°} (5 x 10777) E152 fuss = Fee Da(12 x 10") =— = foan = 6.63 kHe faa = Teng > RC = © ase =— = “6 E156 RE = Seip 7 1B 10 \ Let C= 0.001 wF =i aF => £=2i8k0 fo= Re Then Ry = 294k Ro = 3.44 Ba = 1.22 9 Re = 8,31 kD IT| = 0.01 = L L 1 6 RkO= = = 6.13 « 10 tehv3 2e(18 x 103),78 Let @ = 0.001 pF = 1 nF Thea R= 6.13 KO so Ry = BR = 49 kD B1S.7 fos 1 _ 1 °F ar VERG | 2xv/A(10*)(100 x 10-"7) = fp 265 ki Ra = 298 = 29(10°) => Ry = 290 kQ £15.8 —_ -—. f= apne 7°" ih l = = eee OE 0.02 pF C= ae * Se Ry = 2K, = 2(10) > Rp = 20 ER