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ECEN3713 Network Analysis Lecture #11 14 February 2006 Dr. George Scheets

ECEN3713 Network Analysis Lecture #11 14 February 2006 Dr. George Scheets. Read 13.4 – 13.5, 13.7 Problems: 13.9 – 13.12 Exam #1, Thursday 16 February Chapter 12 & 13.1 - 13.5, no initial conditions. v(t). 100. Im. t. .001. x. x. -5000. Re.

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ECEN3713 Network Analysis Lecture #11 14 February 2006 Dr. George Scheets

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  1. ECEN3713 Network AnalysisLecture #11 14 February 2006Dr. George Scheets • Read 13.4 – 13.5, 13.7 • Problems: 13.9 – 13.12 • Exam #1, Thursday 16 February • Chapter 12 & 13.1 - 13.5, no initial conditions

  2. v(t) 100 Im t .001 x x -5000 Re 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. Shape of curve indicates not pure exponential.

  3. v(t) = 500,000te-5,000t + 100e-5,000t) 500,000te-5,000t 100 t .001 100e-5,000t 100 t .001

  4. Pulse in: u(t) – u(t-1)... xin(t) 1 t 1 10

  5. Smeared pulse out: u(t)(1–e-0.5t) - u(t-1)(1+e-0.5(t-1)) xin(t) 1 t 1 10 xout(t) 0.632 t 10 1

  6. Im Im C = 3 mF C = 0 -79.04 x x x x x Re Re -5 -5.27 Im Im C = 12 mF C = 9 mF -14.79 -22.64 x x x x x x Re Re -7.04 -6.13 V(s) = 4/(s(4Cs2 +(1+4C)s +5) Will not oscillate when 0 < C < 13 mF 1 H vin 1 Ω C vout 4 Ω Quiz 4B 2005

  7. Im Im C = 28 mF C = 14 mF x 4.47 .627 -4.96 x x x x Re Re x -9.43 Im Im C = 3 F C = .5 F x 1.39 .35 x x x x x -.75 Re Re -.54 V(s) = 4/(s(4Cs2 +(1+4C)s +5) Oscillate when 14 mF < C < 4.49 F 1 H vin 1 Ω C vout 4 Ω Quiz 4B

  8. Im Im C = 4.48 F C = 5 F .02 x x -.36 x x x x Re -.69 -.53 Re Im Im C = 10 F C = 7.5 F -.88 -.83 x x x x x x Re -.14 Re -.2 V(s) = 4/(s(4Cs2 +(1+4C)s +5) Will not oscillate when C > 4.49 F 1 H vin 1 Ω C vout 4 Ω Quiz 4B

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