1. ECEN3714 Network Analysis Lecture #21 2 March 2015 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 n Read 14.7 n Problems: 14.5, 7, & 55 n Quiz #6 this Friday n Quiz #5 Results Hi = 10, Low = 4.0, Average = 7.89 Standard Deviation = 2.02

2. ECEN3714 Network Analysis Lecture #24 9 March 2015 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 n Problems: 14.11, 22,37 n No Quizzes or Labs week prior to Spring Break n Exam #2 on 3 April n Note: Lab Practicum dropped u 10 Labs & 1 Design Problem u Still worth 220 points total in the end ≈ 1/3 of grade

3. ECEN3714 Network Analysis Lecture #25 11 March 2015 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 n Problems: 14.51, 52, 57 n Next Quiz is week after Spring Break n Exam #2 on 3 April

4. ECEN3714 Network Analysis Lecture #26 13 March 2015 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 n Problems: 14.58, 59, 72 n Next Quiz is week after Spring Break n Exam #2 on 3 April

5. Transforms X(s) = x(t) e -st dt 0-0- ∞ Laplace s = σ +jω ∞ X(f) = x(t) e -j2πft dt -∞-∞ Fourier

6. Got a Laplace Transform? Re(s) = σ Im(s) = jω |V(s)| The Fourier Transform of x(t) is on the jω axis.* *Provided x(t) = 0; t < 0

7. y(t) = x(t) + y(t-1): H(s) = 1/[1 – e –s ] σ = 0 Frequency Response σ = 0 axis Re(s) = σ Im(s) = jω |V(s)| ω = 0

8. 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) 5 cycle per second square wave generated using 4 sinusoids

9. Generating a Square Wave... 5 cycle per second square wave generated using 100 sinusoids. Max frequency at (N*10 – 5) Hz = 995 Hz 0 1.5 -1.5 0 1.0

10. 5 Hz square wave after Single Pole Low Pass Filtering 159.2 Hz half power frequency 0 1.5 -1.5 0 1.0 Not much visible change. Blue = Error = Output(t) – Input(t)

11. Generating a Square Wave... 5 cycle per second square wave generated using 100 sinusoids. Max frequency at (N*10 – 5) Hz = 995 Hz 0 1.5 -1.5 0 1.0

12. 5 Hz square wave after Single Pole High Pass Filtering 159.2 Hz half power frequency 0 1.5 -1.5 0 1.0

13. Filters n Low Pass, High Pass, & Band Pass u All are LTI if made from R's, C's, and/or L's n LTI Systems have an impulse response h(t) u δ(t) in? h(t) is output. n If LTI, for any input x(t), In Time Domain: x(t)☺h(t) = y(t) In Laplace Domain: X(s)H(s) = Y(s) In Frequency Domain: X(jω)H(jω) = Y(jω) n If not LTI, h(t) does not exist n We'll hit ☺ in more detail later

14. OpAmps n High Gain Devices u Typically, Voltage Gain > 10,00 u Vout(t) = [Vp(t) – Vn(t)]*Voltage Gain n High Input Impedance u Zin typically > 1 MΩ

15. Integrator: H(jω) = -1/jωRC ω |H(ω)| 10 1000 0 100

16. 5 Hz Square Wave In... 0 1.5 -1.5 0 1.0 This one made up of 100 sinusoids. Fundamental frequency of 5 Hz & next 99 harmonics (15, 25,..., 995 Hz).

17. 5 Hz Triangle Out... 0 1.5 -1.5 0 1.0 -50 50

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

19. 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).

20. Spikes Out... 0 1.5 -1.5 0 1.0