ECEN3713 Network Analysis Lecture #15 15 February 2016 Dr ECEN3713 Network Analysis Lecture #15 15 February 2016 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 Read 13.7 Problems: 13.8, 10, & 11 Friday, Quiz #4 Initial Conditions (i.e. everything ≠ 0 when t < 0)
ECEN3713 Network Analysis Lecture #16 17 February 2016 Dr ECEN3713 Network Analysis Lecture #16 17 February 2016 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 Problems: 13.15 & 13.78; Old Quiz #5 Friday, Quiz #4 Initial Conditions (i.e. everything ≠ 0 when t < 0) Exam 1 Results Hi = 100, Low = 53, Average = 80.73, Standard Deviation = 10.25 A > 89, B > 79, C > 68, D > 57
OSU IEEE February General Meeting 5:30-6:30 pm, Wednesday, 17 February ES201b Reps from Zeeco will present Dinner will be served + 3 points extra credit All are invited
V(s) = (100s + 106)/(s + 5000)2 v(t) = 500,000te-5000t + 100e-5000t Stability Issues: Location of poles on Real axis sets decay rate. t v(t) 100 .001 Shape of curve indicates not pure exponential. Re Im x -5000
v(t) = 500,000te-5,000t + 100e-5,000t) 500,000te-5,000t 100e-5,000t .001 100e-5,000t 100 t .001
s = σ + jω… Pole real (x axis) coordinate σ Provides time constant Time Waveform fades away if < 0 Pole imaginary (y axis) coordinate ω Frequency of oscillation 0 = no oscillation 1st Order Equation "s" Has 1 pole 2nd Order Equation "s2" Has 2 poles Etc.
Pulse in: u(t) – u(t-1)... xin(t) 1 t 1 10
Pulse & RC Low Pass Filter y(t) = u(t)(1–e-t/[RC]) - u(t-1)(1-e-(t-1)/[RC]) 10 xin(t)
e-αs term in Laplace Domain In numerator? Indicates a delay in the time domain Ignore it for Partial Fraction Expansion Find undelayed time domain equation Then delay it by α seconds In denominator? Indicates a math error
V(s) = 4/(s(4Cs2 +(1+4C)s +5) Will not oscillate when 0 < C < 13 mF 1 H vin Im x -5 Re C = 0 Im x -5.27 Re C = 3 mF -79.04 1 Ω C vout 4 Ω Quiz 4B 2005 Im x -7.04 Re -14.79 C = 12 mF Im x -6.13 Re -22.64 C = 9 mF
V(s) = 4/(s(4Cs2 +(1+4C)s +5) Oscillate when 14 mF < C < 4.49 F vin Im x -9.43 Re C = 14 mF .627 Im x -4.96 Re C = 28 mF 4.47 1 Ω C vout 4 Ω Quiz 4B C = 3 F Im x -.54 Re .35 Im x -.75 Re 1.39 C = .5 F
V(s) = 4/(s(4Cs2 +(1+4C)s +5) Will not oscillate when C > 4.49 F Im vin Im x -.36 Re C = 5 F -.69 1 Ω C = 4.48 F Im x -.53 Re .02 C vout 4 Ω Quiz 4B Im x -.14 Re -.88 C = 10 F Im x -.2 Re -.83 C = 7.5 F
H(s) = 1/[1 – e–Δs]; s = σ + jω Δ = 1 second Frequency Response H(jω) = 1/[1 – e–jω] Plot of jω axis; σ = 0 1st & 2nd Quadrant Blows Up at Integer Multiples of 1 Hz (2π rads/sec) jω σ
Need Voltage Gain at ω = 5.5 rad/sec? Plot of jω axis; σ = 0 Frequency Response H(j5.5) = 1/[1 – e–j5.5] [1 - e–j5.5] = [1 - e–j2π0.8754] = [1 – e–j315.1◦] = [1 – ej44.9◦] = [1 – (0.7083 + j0.7059)] = [0.2917 - j0.7059] = 0.7638∟-67.55◦ H(j5.5) = 1.309∟67.55◦ Voltage Gain = 1.309, 67.55 degree phase shift