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ECEN3713 Network Analysis Lecture #17 7 March 2006 Dr. George Scheets

ECEN3713 Network Analysis Lecture #17 7 March 2006 Dr. George Scheets. Quiz 5 Results: Hi = 5, Low = 2, Ave. = 3.55, StanDev = 1.13 Read Chapter 14.5 – 14.6 Problems: 14.20, 14.21, 14.23, 14.24 Thursday's Quiz Chapter 14.1 – 14.6

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ECEN3713 Network Analysis Lecture #17 7 March 2006 Dr. George Scheets

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  1. ECEN3713 Network AnalysisLecture #17 7 March 2006Dr. George Scheets Quiz 5 Results: Hi = 5, Low = 2, Ave. = 3.55, StanDev = 1.13 • Read Chapter 14.5 – 14.6 • Problems: 14.20, 14.21, 14.23, 14.24 • Thursday's Quiz • Chapter 14.1 – 14.6 Thursday's AssignmentProblems: 14.26, 14.28 – 14.30

  2. ECEN3713 Network AnalysisLecture #19 21 March 2006Dr. George Scheets Quiz 6 Results: Hi = 10, Low = 2.0, Ave. = 5.63, StanDev = 2.42 • Read Chapter 15.1 – 15.3 • Problems: 15.1, 15.2, 13.45, 13.47 • Thursday's Quiz • Chapter 14.1 – 14.5, 15.1 – 15.3 Thursday's AssignmentProblems: 15.3, 15.4, 13.48, 13.52 Exam #2: 6 April

  3. Generating a Square Wave... 5 Hz+ 15 Hz + 25 Hz + 35 Hz 1.5 0 -1.5 1.0 0 cos2*pi*5t - (1/3)cos2*pi*15t + (1/5)cos2*pi*25t - (1/7)cos2*pi*35t) 5 cycle per second square wave generated using 4 sinusoids

  4. Generating a Square Wave... 1.5 0 -1.5 1.0 0 5 cycle per second square wave generated using 100 sinusoids. Max frequency at (N*10 – 5) Hz = 995 Hz

  5. 5 Hz square wave after Single Pole Low Pass Filtering159.2 Hz half power frequency 1.5 0 -1.5 1.0 0 Not much visible change. Blue = Error = Output(t) – Input(t)

  6. 5 Hz square wave after Single Pole High Pass Filtering159.2 Hz half power frequency 1.5 0 -1.5 1.0 0

  7. Integrator: H(jω) = -1/jωRC |H(ω)| 10 ω 0 100 1000

  8. 5 Hz Square Wave In... 1.5 0 -1.5 1.0 0 This one made up of 100 sinusoids. Fundamental frequency of 5 Hz & next 99 harmonics (15, 25, ..., 995 Hz).

  9. 5 Hz Triangle Out... 1.5 0 -1.5 1.0 0 50 -50

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