1、Chapter 10Sinusoidal Steady-state Analysis,要求深刻理解與熟練掌握的重點(diǎn)內(nèi)容有： 正弦穩(wěn)態(tài)電路的分析。 難點(diǎn)：相量圖表示、提高功率因數(shù)、功率分析,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,10-7 Op Amp AC Circuits,10-10 Summary and Review,1

2、0.1 Introduction,Steps to Analyze ac Circuits:,Transform the circuit to the phasor or frequency domain. Solve the problem using circuit techniques (nodal analysis, mesh analysis, superposition, etc.). Transform the resulting phasor to the time domain.,oscillator 振蕩器,transistor 晶體管,Capacitance multip

3、lier 電容倍增器,10.2 Nodal Analysis,The basis of nodal analysis is Kirchhoffs current law. Since KCL is valid for phasors, as demonstrated in Section 9.6 , we can analyze ac circuits by nodal analysis.,DC Circuit,AC Circuit,Example： Find the time-domain node voltages v1 (t) and v2(t) in the circuit shown

4、 in Fig. a.,A frequency-domain circuit for which node voltages V1 and V2 are identified.,Fig. a,.,.,Solution：,left node :,right node:,Combining terms, we have:,Solving, we find that V1 = 2.24 -63.4o V, V2 = 47.47 116.6o V,Converting to the time domain :,v1(t) = 2.24*1.414 cos ( t 63.4o) V v2(t) = 47

5、.7 *1.414 cos ( t + 116.6o) V,.,.,Note that the value of would have to be known in order to compute the impedance values given on the circuit diagram. Also, both sources must be operating at the same frequency.,20cos4t V,10,ix,0.1F,1H,2ix,0.5H,Example： Find ix in the circuit of Fig.10.1 using nodal

6、analysis.,We first convert the circuit to the frequency domain:,20cos4t,Solution：,Node 1,Node 2,The equations can be put in matrix form as,Transforming this to the time domain,10.3 Mesh analysis,Kirchhoffs voltage law (KVL) forms the basis of mesh analysis. The validity of KVL for ac circuits was sh

7、own in section 9.6 and is illustrated in the following examples.,Mesh-Current,DC Circuit,AC Circuit,Example :Use phasors and mesh analysis on the circuit of Fig. 10.63 to find .,Solution:,Example :Use phasor analysis to determine the three mesh currents i1(t), i2(t), and i3(t) in the circuit of Fig.

8、 10.70,Solution:,We define an additional clockwise mesh current i4(t) flowing in the upper right-hand mesh.,The inductor is replaced by a j0.004 W impedance :,The 750 mF capacitor is replaced by a j/ 0.0015 W impedance,The 1000 mF capacitor is replaced by a j/ 2 W impedance.,And ZR =1 W,We replace t

9、he left voltage source with a 6 -13o V source, and the right voltage source with a 6 0o V source.,Thus ,Mesh-Current equation is,(1 j/ 0.0015) I1 ( j/ 0.0015 )I2 I3 +0 = 6 -13o 1,j/ 0.0015 I1 + j0.004 I2 j0.004 I4 = 0.005 I1 2,-I1 + (1 j/ 2) I3 + j0.5 I4 = -6 0o 3,-j0.004 I2 + j0.5 I3 + (j0.004 j0.5

10、) I4 = 0 4,Solving, we find that I1 = 0.00144 -51.5o A, I2 = 233.6 39.65o A, and I3 = 6.64 173.5o A,Converting to the time domain,i1(t) = 1.44 cos (2t 51.5o) mA i2(t) = 233.6 cos (2t + 39.65o) A i3(t) = 6.64 cos (2t + 173.5o) A,10.4 Superposition theorem,When a circuit has sources operating at diffe

11、rent frequencies, one must add the responses due to the individual frequencies in the time domain.,Example： Find v0 in the circuit in Fig.10.13 using the superposition theorem.,We let v0=v1+v2+v3,where v1 is due to the 5V dc voltage source, v2 is due to the 10cos2t V voltage source, and v3 is due to

12、 the 2sin5t A current source.,Solution：,To find v1, we set to zero all sources except the 5V dc source. The equivalent circuit is as shown in Fig.10.14(a).,v1=-1V,To find v2, we set to zero both the 5V source and the 2sin5t current source and transform the circuit to the frequency domain.,In the tim

13、e domain,To obtain v3, we set the voltage sources to zero and transform the circuit to the frequency domain.,By current division,In the time domain,10.5 Source transformation,10.6 Thevenin and Norton equivalent circuits,Thevenins and Nortons theorems are applied to ac circuits in the same way as the

14、y are to dc circuits. The only additional effort arises from the need to manipulate complex numbers.,1,1,ZTh,+ _,1,1,Linear circuit,The frequency-domain version of a Thevenin equivalent circuit is depicted in Fig.(a), where a linear circuit is replaced by a voltage source in series with an impedance

15、.,The frequency-domain version of a Norton equivalent circuit is depicted in Fig.(b), where a linear circuit is replaced by a current source in parallel with an impedance.,1,1,Linear circuit,Example： Find the Thevenin equivalent of the circuit in Fig.10.25 as seen from terminals a-b.,1. To find ,we

16、apply KCL at node 1 ,1,+ _,Solution：,Thus , the Thevenin voltage is,2. Find ZTh.,3. The Thevenin equivalent of the circuit.,ZTh,+ _,a,b,10.7 Op amp ac circuits,As usual, we will assume ideal op amps.,As discussed in Chapter 5, the key to analyzing op amp circuits is to keep two important properties of an ideal op amp in mind :,No current enters either of its input terminals. The voltage across its input terminals is zero.,Example： Determine v0(t) for the op amp circuit in Fig.10.31(a) if vS=3cos1000t V.,V0,We first transform the circuit to the frequency domain, as shown in Fig.(b), wh