Resolution of Homework 8 on Circuit Analysis and Design | ECE 225, Assignments of Electrical Circuit Analysis

Download Resolution of Homework 8 on Circuit Analysis and Design | ECE 225 and more Assignments Electrical Circuit Analysis in PDF only on Docsity! ECE225 HOMEWORK #8 – SOLUTIONS SPRING 2009 Problem 8.1 (a) G = 1000 =⇒ G = 20 log10 103 = 60 log10 10 = 60 dB G = 0.01 =⇒ G = 20 log10 10−2 = − 40 log10 10 = − 40 dB G = 6 =⇒ G = 20 log10 6 ∼= 15.6 dB G = 1√ 2 =⇒ G = 20 log10 2(−1/2) = − 10 log10 2 = − 3.01dB (b) G = −60 dB = 20 log10 G =⇒ G = 10−3 = 0.001 G = 6 dB = 20 log10 G =⇒ G = 100.3 ∼= 2 G = 14 dB = 20 log10 G =⇒ G = 100.7 ∼= 5 G = 3 dB = 20 log10 G =⇒ G = 100.15 = 1.413 ∼= √ 2 Problem 8.2 (a) H(s) = 100(s + 4) (s + 2)(s + 8) = 100× 4(1 + s/4) 2(1 + s/2)× 4(1 + s/4) = 25(s + s/4) (1 + s/2)(1 + s/8) (b) H̄(jω) = 25(1 + jω/4) (1 + jω/2)(1 + jω/8) = 25 √ 1 + (ω/4)2√ 1 + (ω/2)2 √ 1 + (ω/8)2︸ ︷︷ ︸ 6 tan−1 ω 4 − tan−1 ω 2 − tan−1 ω 8︸ ︷︷ ︸ = H(ω) θ(ω) (c) H(0) = 25× 1 1× 1 = 25 H̄(j3) = 25 √ 1 + (3/4)2√ 1 + (3/2)2 √ 1 + (3/8)2 6 tan−1 3 4 − tan−1 3 2 − tan−1 3 8 = 25 √ 1.5625√ 3.15 √ 1.140625 6 36.9o − 56.3o − 20.6o ∼= 16.5 6 − 40.0o H̄(j20) = 25 √ 1 + (20/4)2√ 1 + (20/2)2 √ 1 + (20/8)2 6 tan−1 20 4 − tan−1 20 2 − tan−1 20 8 = 25 √ 26√ 101 √ 7.25 6 78.7o − 84.3o − 68.2o ∼= 4.71 6 − 73.8o (d) See attached Bode magnitude and phase plots. (e) H ∼= 13.75 dB =⇒ H(20) ∼= 1013.75/20 = 4.87 θ ∼= −75o 1 Problem 8.2 (Continued) 10 20 30 40 43 8 80 100200.1 0.2 0.3 0.4 0.8 13.75 0 1 2 10 ω −20 dB/dec 0.4 0.8 30 40 −45 deg/dec −45 deg/dec −45 deg/dec −75 −15 0 30 15 −45 degθ ( ) 100800.3 −20 dB/dec 28 22.5 H (dB) 0.1 0.2 1 2 10 20 ω −30 −60 −90 3 4 8 Problem 8.3 1 2 10 20 ω 1 2 10 ω 10 20 30 40 50 43 8 0.2 H (dB) 0.1 80 −75 −120 −105 −40 dB/dec −12 dB/oct 14 −6 dB/oct −45 deg/dec 45 deg/dec 45 deg/dec −45 deg/dec 0.1 2 0 θ ( )deg −10 −20 80 10020 3 4 80.3 0.4 0.2 0.3 0.4 0.8 0.8 30 40 −90 −20 dB/dec −20 dB/dec 100 2