ECEN3714 Network Analysis Lecture #27 23 March 2015 Dr. George Scheets n Problems: thru n Quiz #7 this Friday n Exam #2 on 3 April

ECEN3714 Network Analysis Lecture #28 25 March 2015 Dr. George Scheets n Review Problems: Olde Quiz #7 Problems: Olde Quiz #7 n Quiz #7 this Friday n Exam #2 on 3 April

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 Zin +Vcc -Vcc

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)

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

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

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

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

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

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

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

2nd Order Butterworth Filter | H(ω) | ω Butterworth has flatter passband. 2nd order Butterworth 2nd order Standard See also

Some 5th Order Filters source: Wikipedia – Alessio Damato

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

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

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.

Generating a Square Wave 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)