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ECEN3713 Network Analysis Lecture #21 28 March 2006 Dr. George Scheets

ECEN3713 Network Analysis Lecture #21 28 March 2006 Dr. George Scheets. Read Chapter 15.4 Problems: 13.78, 15.5 – 15.7 Thursday's Quiz Series or Parallel Combinations of Active Filters Thursday's Assignment Problems 15.8, 15.10, 15.11, 15.20 Quiz 7 kicked back for a regrade.

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ECEN3713 Network Analysis Lecture #21 28 March 2006 Dr. George Scheets

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  1. ECEN3713 Network AnalysisLecture #21 28 March 2006Dr. George Scheets • Read Chapter 15.4 • Problems: 13.78, 15.5 – 15.7 • Thursday's Quiz • Series or Parallel Combinations of Active Filters Thursday's Assignment Problems 15.8, 15.10, 15.11, 15.20 Quiz 7 kicked back for a regrade.

  2. ECEN3713 Network AnalysisLecture #23 4 April 2006Dr. George Scheets • Read Chapter 15.5 - End of Chapter • Problems: 15.21, 15.22, 15.24, 15.25 • Test on Thursday!Focus is on Material after last test • Chapters 14 & 15 • Initial Conditions • Chapters 12 & 13 may also show up

  3. Op Amp Characteristics +Vcc vp(t) Av Zin + • Zin? • In M ohms • Hopamp(f) f3dB? • In XX or XXX MHz • Voltage gain Av? • On order of 104 - 106 vout(t) = Av(vp(t)-vn(t)) vn(t) - -Vcc

  4. Op Amps: No Feedback +Vcc • Output likely to hit rails • Unless tiny voltages Av + vout(t) = Av(vp(t)-vn(t)) vin(t) - -Vcc

  5. Op Amps: Positive Feedback • Output likely to hit rails • May get stuck there Av vout(t) + vin(t) -

  6. Op Amps: Negative Feedback • Safe to assume vp(t) = vn(t) • Safe to assume no current enters Op Amp • If low Z outside paths exist Av vout(t) - 0 v vin(t) +

  7. Op Amps: Output Load • Ideally, load does not effect characteristics • Practically, load may effect characteristics • If Op Amp output can't source or sink enough current Av vout(t) - vin(t) + Zload

  8. Differentiator: H(jω) = -jωRC |H(ω)| 1 ω 0 100 1000

  9. 5 Hz Square Wave In... 1.5 0 -1.5 1.0 0 This curve made up of 100 sinusoids. Fundamental frequency of 5 Hz & next 99 harmonics (15, 25, ..., 995 Hz).

  10. Spikes Out... 1.5 0 -1.5 1.0 0

  11. 1st Order RC Low Pass Filter |H(ω)| 1 0.707 1 ω

  12. 2nd Order Low Pass Filter(Two back-to-back 1st order active filters) |H(ω)| 3dB break point changes. 1 0.707 1st order 2nd order 1 ω

  13. Scaled 2nd Order Low Pass Filter(Two back-to-back 1st order active filters) |H(ω)| 2nd order filter has faster roll-off. 1 0.707 1st order 2nd order 1 ω

  14. 2nd Order Butterworth Filter |H(ω)| Butterworth has flatter passband. 1 0.707 2nd order Butterworth 2nd order Standard 1 ω

  15. 1 & 2 Hz sinusoids1000 samples = 1 second 1.5 i 0 x1 i - 1.5 samples 0 200 400 600 800 1000 0 i 3 1 ´ 10 Suppose we need to maintain a phase relationship (low frequency sinusoid positive slope zero crossing same as the high frequency sinusoid's).

  16. 1 & 2 Hz sinusoids1000 samples = 1 secondBoth delayed by 30 degrees 1.5 0 - 1.5 0 200 400 600 800 1000 0 i 3 1 ´ 10 Delaying the two curves by the same phase angle loses the relationship.

  17. 1.5 0 - 1.5 0 200 400 600 800 1000 0 i 3 1 ´ 10 1 & 2 Hz sinusoids1000 samples = 1 second1 Hz delayed by 30, 2 Hz by 60 degrees Delaying the two curves by the same time keeps the relationship. θlow/freqlow needs to = θhi/freqhi. A transfer function with a linear phase plot θout(f) = Kθin(f) will maintain the proper relationship.

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