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CHAPTER 8

CHAPTER 8. NETWORKS 1: 0909201-01 11 December 2002 – Lecture 8b ROWAN UNIVERSITY College of Engineering Professor Peter Mark Jansson, PP PE DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING Autumn Semester 2002. admin. hw 7 due today, hw 8 due at final

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CHAPTER 8

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  1. CHAPTER 8 NETWORKS 1: 0909201-01 11 December 2002 – Lecture 8b ROWAN UNIVERSITY College of Engineering Professor Peter Mark Jansson, PP PE DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING Autumn Semester 2002

  2. admin • hw 7 due today, hw 8 due at final • hw 8 posted end of today • last lab 6 due by end of day monday • final exam: Next Wed 18 Dec 10:15am • take – home portion • Pick up at end of day from office window • Yellow and Blue folders (by Lab Section)

  3. networks I • Today’s learning objectives – • understand first order circuits • build your knowledge of the concept of complete response • learn how Thevenin and Norton equivalents help simplify analysis of first order circuits • learn how to calculate the natural (transient) response and forced (steady-state) response

  4. new concepts from ch. 8 • response of first-order circuits • to a constant input • the complete response • stability of first order circuits • response of first-order circuits • to a nonconstant (sinusoidal) source

  5. What does First Order mean? • circuits that contain capacitors and inductors can be defined by differential equations • circuits with ONLY ONE capacitor OR ONLY ONE inductor can be defined by a first order differential equation • such circuits are called First Order Circuits

  6. complete response? Complete response = transient response + steady state response For circuits we will review today: both – constant source p.315 and sinusoidal source p. 304… Complete response = natural response + forced response

  7. finding the CR of 1st Ord. Cir • Find the forced response before the disturbance. Evaluate at t = t(0-) to determine initial conditions • Find forced response (steady state) after the disturbance t= t(0+) • Add the natural response (Ke-t/) to the new forced response. Use initial conditions to calculate K

  8. simplifying for analysis • Using Thevenin and Norton Equivalent circuits can greatly simplify the analysis of first order circuits • We use a Thevenin with a Capacitor • and a Norton with an Inductor

  9. Rt + v(t) - C Voc + – Thevenin Equivalent at t=0+ i(t) + -

  10. + v(t) - Isc Rt i(t) L Norton equivalent at t=0+

  11. t = 0 R1 R2 + v(t) - R3 vs C + – 1st ORDER CIRCUITS WITH CONSTANT INPUT

  12. Example (before switch closes) • If vs = 4V, R1 = 20kohms, • R2 = 20 kohms • R3 = 40 kohms • What is v(0-) ?

  13. as the switch closes… • THREE PERIODS emerge….. • 1. system change (switch closure) • 2. (immediately after) capacitor or inductor in system will store / release energy (adjust and/or oscillate) as system moves its new level of steady state (a.k.a. transient or natural response) …. WHY??? • 3. new steady state is then achieved (a.k.a. the forced response)

  14. Rt + v(t) - C Voc + – Thevenin Equivalent at t=0+ i(t) + - KVL

  15. SOLUTION OF 1st ORDER EQUATION

  16. SOLUTION CONTINUED

  17. SOLUTION CONTINUED

  18. so complete response is… complete response = v(t) = forced response (steady state) = Voc + natural response (transient) = (v(0-) –Voc) * e -t/RC)

  19. Example • 8.3-1, p. 315

  20. + – WITH AN INDUCTOR t = 0 R1 R2 R3 i(t) vs L Why ?

  21. + v(t) - Isc Rt i(t) L Norton equivalent at t=0+ Why ? KCL

  22. SOLUTION

  23. so complete response is… complete response = i(t) = forced response (steady state) = Isc + natural response (transient) = (i(0-) –isc) * e -tR/L)

  24. Example • 8.3-2, p. 316

  25. for more practice: Exercises • 8.3-1, p. 321 • 8.3-2, p. 321

  26. forced response summary

  27. Example • 8.7-2, p. 336

  28. HANDY CHART ELEMENTCURRENT VOLTAGE

  29. IMPORTANT CONCEPTS FROM CHAPTER 8 • determining Initial Conditions • determining T or N equivalent to simplify • setting up differential equations • solving for v(t) or i(t)

  30. Don’t forget HW 8 (test review) • Course Evals……

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