1. ECEN3714 Network Analysis Lecture #16 18 February 2015 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 n Read 14.4 n Problems: Old Quiz #4 n Quiz #4 this Friday n Quiz #3 Results Hi = 9.0, Low = 0.5, Average = 5.50 Standard Deviation = 2.69

2. ECEN3714 Network Analysis Lecture #18 23 February 2015 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 n Read 14.5 n Problems: 13.55, 13.57, 14.2 n Quiz #5 this Friday

3. ECEN3714 Network Analysis Lecture #19 25 February 2015 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 n Read 14.6 n Problems: Old Quiz #5 n Quiz #5 this Friday n Quiz 4 Results Hi = 10, Low = 3, Average = 8.06 Standard Deviation = 1.97

4. Gustav Kirchoff n Born 1824 n Died 1887 n German Physicist n KCL & KVL formulated in 1852 Doctoral Dissertation n KVL: Σ(voltage drops around a closed loop) = 0 n KCL: Σ(current entering or exiting a node) = 0 n High side voltage on device end a current enters source: Wikipedia

5. V(s) = (100s + 10 6 )/(s + 5000) 2 v(t) = 500,000te -5000t + 100e -5000t Re Im x x -5000 t v(t) 100.001 Stability Issues: Location of poles on Real axis sets decay rate. Shape of curve indicates not pure exponential.

6. v(t) = 500,000te -5,000 t + 100e -5,000t ) t 500,000te -5,000t 100.001 t 100.001 100e -5,000t

7. Pulse in: u(t) – u(t-1)... t 1 10 x in (t) 1

8. Smeared pulse out: u(t)(1–e -0.5t ) - u(t-1)(1+e -0.5(t-1) ) t 1 10 x in (t) 1 t 0.632 10 1 x out (t)

9. V(s) = 4/(s [4Cs 2 +(1+4C)s +5]) Roots at [ –[1+4C] + {[1+4C] 2 – 80C} 0.5 ]/2 v out v in 1 Ω C 1 H 4 Ω Quiz 4B 2005 Will not oscillate when 0 < C < 13 mF Im x x -5 Re C = 0 Im x x -5.27 Re C = 3 mF x -79.04 Im x x -6.13 Re x -22.64 C = 9 mF Im x x -7.04 Re x -14.79 C = 12 mF

10. V(s) = 4/(s(4Cs 2 +(1+4C)s +5) v out v in 1 Ω C 1 H 4 Ω Quiz 4B Oscillates when 14 mF < C < 4.49 F Im x x -.75 Re x 1.39 C =.5 F Im x x -4.96 Re C = 28 mF x 4.47 Im x x -9.43 Re C = 14 mF.627 x C = 3 F Im x x -.54 Re x.35

11. V(s) = 4/(s(4Cs 2 +(1+4C)s +5) v out v in 1 Ω C 1 H 4 Ω Quiz 4B Will not oscillate when C > 4.49 F Im x x -.2 Re x -.83 C = 7.5 F Im x x -.36 Re C = 5 F x -.69 C = 4.48 F Im x x -.53 Re x.02 Im x x -.14 Re x -.88 C = 10 F

12. s = σ + jω… n Pole real (x axis) coordinate σ u Provides time constant F Time Waveform fades away if < 0 n Pole imaginary (y axis) coordinate ω u Frequency of oscillation F 0 = no oscillation n 1st Order Equation "s" u Has 1 pole n 2nd Order Equation "s 2 " u Has 2 poles n Etc.

13. y(t) = x(t) + y(t-1) → H(s) = 1/[1 – e –s ] σ Frequency Response σ = 0 axis This system, with a feedback loop, has an Infinite # of poles along The σ = 0 (jω) axis.

14. V(s) = 10s/(s 2 +2s+25) Im x x Re t 10.2 v(t) = 10.2e -t cos(4.899t +.0641π) Stability Issues: Underdamped Location of poles on Real axis sets decay rate. Location of poles on Imagninary axis sets oscillation rate.

15. V(s) = 10 5 /(s 2 + 2*10 5 s + 10 8 ) t v(t).5.002 v(t) =.5025(e -500t - e -199,500t ) Im xx -500 Re -199,500 Stability Issues: Overdamped No complex poles = no oscillation. Location of poles on real axis sets decay rate.

16. V(s) = (100s + 10 6 )/(s + 5000) 2 v(t) = 500,000te -5000t + 100e -5000t Re Im x x -5000 t v(t) 100.001 Stability Issues: Critically damped Location of poles on Real axis sets decay rate.

17. If system is under damped it's easy to get values of α & β α = decay rate & β = oscillation frequency Must have imaginary roots to oscillate.