1. ECEN3714 Network Analysis Lecture #27 23 March 2015 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 n Problems: 14.60 thru 14.62 n Quiz #7 this Friday n Exam #2 on 3 April

2. ECEN3714 Network Analysis Lecture #28 25 March 2015 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 n Review 12.14 Problems: Olde Quiz #7 Problems: Olde Quiz #7 n Quiz #7 this Friday n Exam #2 on 3 April

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 + (t) = v - (t) u a.k.a. "virtual ground" n Safe to assume no current enters Op Amp u If low Z outside paths exist

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. OpAmp Filters n 1st Order Low Pass n 1st Order High Pass n Band Pass u Hi Pass & Low Pass Filters Back-to-Back Hi Pass Break Point << Low Pass n Band Reject u Hi Pass & Low Pass Filters in Parallel Hi Pass Break Point >> Low Pass n Stacked Back-to-Back = steeper roll-offs u 3dB Break Points Change

9. 1st Order RC Low Pass Filter | H(ω) | ω 1 1 0.707

10. 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

11. 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

12. 2nd Order Butterworth Filter | H(ω) | ω 1 1 0.707 Butterworth has flatter passband. 2nd order Butterworth 2nd order Standard See also http://en.wikipedia.org/wiki/Butterworth_filter.

13. Some 5th Order Filters source: Wikipedia – Alessio Damato

14. 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

15. 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.

16. 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.

17. Generating a Square Wave... 0 1.5 -1.5 0 1.0 5 Hz + 15 Hz + 25 Hz + 35 Hz cos2*pi*5t - (1/3)cos2*pi*15t + (1/5)cos2*pi*25t - (1/7)cos2*pi*35t)