1. ECEN3714 Network Analysis Lecture #30 30 March 2015 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 Problems: Olde Quiz #8 Problems: Olde Quiz #8 n Exam #2 this Friday

2. ECEN3714 Network Analysis Lecture #31 1 April 2015 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 Problems: Olde Exams (#2) Problems: Olde Exams (#2) n Exam #2 this Friday n Quiz #7 Results Hi = 10, Low = 2.0, Average = 6.22 Standard Deviation = 2.70

3. 2015 OSU ECE Spring Banquet n Hosted by Student Branch of IEEE n Wednesday, 15 April, at Meditations n Doors open at 5:30 pm, meal at 6:00 pm n Cash Bar n Sign up in ES202 to reserve your seat(s) u $5 if pay in advance (< 8 April) & resume submitted to ieee@okstate.edu u $10 on 9 & 10 April. (A $20 value.) n Speakers: Ron Sinnes (Level3, Director IP) & Eric Miller (VYVX, Sports Manager) n Dress is Business Casual n Many door prizes available! u Known best: 8.4" Samsung Galaxy Tab S, two Fluke Multimeters n All are invited! Sponsored in part by:

4. Quiz 7 - + 10 μF v out 0.1 H 50 Ω n Suppose you needed to find the input current i in (t) n What can be ignored? i in (t) 10 μF 0.1 H 50 Ω i in (t) 1 MΩ 10 μF 0.1 H 50 Ω i in (t)

5. Quiz 7 - + 10 μF v out 0.1 H 50 Ω n Suppose you needed to find the input impedance Z in n What can be ignored? Z in 10 μF 0.1 H 50 Ω 1 MΩ 10 μF 0.1 H 50 Ω Z in

6. Quiz 7 - + 10 μF v out 0.1 H 50 Ω n Suppose you needed to find the voltage transfer function H(s) = V out (s)/V in (s) n What can be ignored? v in - + 10 μF v out 0.1 H 50 Ω v in 10 μF v out 0.1 H 50 Ω v in

7. H(f) for Quiz #7

8. H(f) for Quiz #7 w/o Inductor

9. H(f) for Quiz #7 w/small Inductor

10. The Ideal Filter n Frequency Response u H(f) = K1; over some frequency range u = 0; elsewhere n Phase Response u θ(f) = K2*f; over same frequency range Will delay all frequencies by same time amount n Input with energy 100% in frequency range? u Amplitude may be changed u May be shifted in time u General shape unchanged

11. 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) H(f) = 100 & θ(f) = 0; 0 < f < 50 H(f) = 100 & θ(f) = 18f degrees; 0 < f < 50

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

13. Single Real Pole, Two Real Poles 1/(jω+3), 1/(jω+3) 2 |H(ω)|

14. Single Real Pole, Two Real Poles 1/(jω+3), 3/(jω+3) 2 |H(ω)| Note: 2 nd order system has sharper roll-off. Also, 3 dB break point has moved.

15. Complex Conjugate Poles, |real| = 0 H(s) = 1/(s 2 + 100) = 1/[(s + j10)(s – j10)] H(jω) = 1/(ω 2 – 100) |H(ω)|

16. Complex Conjugate Poles, real < 0 H(s) = 1/(s 2 + 4s + 104) = 1/[(s + 2 + j10)(s + 2 – j10)] H(jω) = 1/[104 – ω 2 +j4ω] |H(ω)|

17. H(f) for Quiz #7 w/small Inductor

18. Complex Conjugate Poles, real < 0 H(s) = 1/(s 2 + 10s + 125) = 1/[(s + 5 + j10)(s + 5 – j10)] H(jω) = 1/[125 – ω 2 +j10ω] |H(ω)|

19. Let t o = α & f( ) = h( ) h(t – α)u(t – α); α > 0 h(t – α); α > 0, t > α