1. 1 Chapter 10 Sinusoidal Steady-State Analysis 電路學 ( 二 )

2. 2 Sinusoidal Steady-State Analysis Chapter 10 10.1 Introduction 10.2 Nodal Analysis 10.3 Mesh Analysis 10.4 Superposition Theorem 10.5 Source Transformation 10.6 Thevenin and Norton Equivalent Circuits

3. 3 Steps to Analyze AC Circuits: 1.Transform the circuit to the phasor or frequency domain. 2.Solve the problem using circuit techniques (nodal analysis, mesh analysis, superposition, etc.). 3.Transform the resulting phasor to the time domain. Time to Freq Solve variables in Freq Freq to Time 10.1 Introduction (1)

4. 4 10.2 Nodal Analysis (1) Example 1 Find i x in the circuit of figure below.

5. 5 10.2 Nodal Analysis (2) Example 2 Using nodal analysis, find v 1 and v 2 in the circuit of figure below. v 1 (t) = 11.32 sin(2t + 60.01  ) Vv 2 (t) = 33.02 sin(2t + 57.12  ) V Answer:

6. 6 10.2 Nodal Analysis (3) Example 3 Compute V 1 and V 2 in the circuit of figure below.

7. 7 10.3 Mesh Analysis (1) Example 3 Determine I o in the following figure using mesh analysis.

8. 8 10.3 Mesh Analysis (2) Example 4 Find I o in the following figure using mesh analysis.

9. 9 10.3 Mesh Analysis (3) Example 5 Solve for V o in the circuit using mesh analysis.

10. 10 10.4 Superposition Theorem (1) When a circuit has sources operating at different frequencies, The separate phasor circuit for each frequency must be solved independently, and The total response is the sum of time- domain responses of all the individual phasor circuits.

11. 11 10.4 Superposition Theorem (2) Example 6 Use the superposition theorem to find I o in the circuit.

12. 12 10.4 Superposition Theorem (3) Example 7 Find v o in the circuit using the superposition theorem.

13. 13 10.4 Superposition Theorem (4) Example 8 Calculate v o in the circuit of figure shown below using the superposition theorem. V o = 4.631 sin(5t – 81.12  ) + 1.051 cos(10t – 86.24  ) V

14. 14 10.5 Source Transformation (1)

15. 15 10.5 Source Transformation (2) Example 9 Find V x in the circuit of figure below using the method of source transformation.

16. 16 10.5 Source Transformation (3) Example 10 Find I o in the circuit of figure below using the concept of source transformation. I o = 3.288  99.46  A

17. 17 10.6 Thevenin and Norton Equivalent Circuits (1) Thevenin transform Norton transform

18. 18 10.6 Thevenin and Norton Equivalent Circuits (2) Example 11 Obtain the Thevenin equivalent at terminals a–b of the circuit below.

19. 19 10.6 Thevenin and Norton Equivalent Circuits (3) Example 12 Find the Thevenin equivalent at terminals a–b of the circuit below. Z th =12.4 – j3.2  V TH = 18.97  -51.57  V

20. 20 10.6 Thevenin and Norton Equivalent Circuits (4) Example 13 Obtain current I 0 in the figure using Norton’s theorem.

21. 21 10.7 Op Amp AC Circuit (1) An ideal op amp 1.No current enters either of its input terminals. 2.The voltage across its input terminals is zero.

22. 22 10.7 Op Amp AC Circuit (2) Example 14 Compute the closed-loop gain and phase shift for the circuit. Assume R 1 = R 2 = 10 k, C 1 = 2μF, C 2 = 1 μF, and  = 200 rad/s

23. 23 10.9 Applications (1) Capacitance Multiplier A circuit used in IC technology to produce a multiple of a small physical capacitance C when a large capacitance is needed.

24. 24 10.9 Applications (2) Oscillators A circuit produces an ac waveform as output when powered by a dc input. Barkhausen criteria: 1.The overall gain of the oscillator must be unity or greater. 2.The overall phase must be zero.

25. 25 10.9 Applications (3) Wein-bridge oscillator Used for generating sinusoids below 1 MHz

26. 26 10.9 Applications (4) 2nd criterion, V 2 and V o are in phase 1st criterion, overall gain  1 If R 1 = R 2 = R, C 1 = C 2 = C From noninverting amplifier R f = 2 R g

27. 27 4, 7, 11, 16, 20, 21, 23, 25, 33, 35, 40, 41, 43, 49, 51, 59, 65, 67, 73, 77, 79, 89, 90, 91