1. ECEN3713 Network Analysis Lecture #21 28 March 2006 Dr. George Scheets n Read Chapter 15.4 n Problems: 13.78, 15.5 – 15.7 n Thursday's Quiz u 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 Analysis Lecture #23 4 April 2006 Dr. George Scheets n Read Chapter 15.5 - End of Chapter n Problems: 15.21, 15.22, 15.24, 15.25 n Test on Thursday! Focus is on Material after last test u Chapters 14 & 15 u Initial Conditions u Chapters 12 & 13 may also show up

3. Op Amp Characteristics AvAv v p (t) v n (t) + - v out (t) = A v (v p (t)-v n (t)) n Zin? u In M ohms n H opamp (f) f 3dB ? u In XX or XXX MHz n Voltage gain Av? u On order of 10 4 - 10 6 Zin +Vcc -Vcc

4. Op Amps: No Feedback AvAv + - v out (t) = A v (v p (t)-v n (t)) n Output likely to hit rails u Unless tiny voltages +Vcc -Vcc v in (t)

5. Op Amps: Positive Feedback AvAv v in (t) + - v out (t) n Output likely to hit rails u May get stuck there

6. Op Amps: Negative Feedback AvAv v in (t) + - v out (t) n Safe to assume v p (t) = v n (t) n Safe to assume no current enters Op Amp u If low Z outside paths exist 0 v

7. Op Amps: Output Load AvAv v in (t) + - v out (t) n Ideally, load does not effect characteristics n Practically, load may effect characteristics u If Op Amp output can't source or sink enough current Z load

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

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

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

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

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

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

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

15. 1 & 2 Hz sinusoids 1000 samples = 1 second 02004006008001000 0 1.5  i x1 i 110 3  0i Suppose we need to maintain a phase relationship (low frequency sinusoid positive slope zero crossing same as the high frequency sinusoid's). samples

16. 1 & 2 Hz sinusoids 1000 samples = 1 second Both delayed by 30 degrees 02004006008001000 0 1.5  110 3  0i Delaying the two curves by the same phase angle loses the relationship.

17. 1 & 2 Hz sinusoids 1000 samples = 1 second 1 Hz delayed by 30, 2 Hz by 60 degrees Delaying the two curves by the same time keeps the relationship. θ low /freq low needs to = θ hi /freq hi. A transfer function with a linear phase plot θ out (f) = Kθ in (f) will maintain the proper relationship.