Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14EE201 Lecture 10 P. 1Linearity and SuperpositionConsider the circuit:9ixvSiS+_6We can solve for the ‘output’ ix using mesh analysis.M1: vs - 6i1 - 9(i1 - i2) = 0 vs - 15i1 + 9i2 = 0M2: i2 = -isBut, ix = i1 - i2ix = (vs / 15) + (6is / 15)EE201 Lecture 10 P. 2Solution, ix = (vs / 15) + (6is / 15) is linearly related to the ‘inputs’ vs and is. Solve for output ix by considering the effect of each input separately.9ix1vS+_6Step 1: Deactivate current sourceSolve for ix1.ix1 = vs / 15EE201 Lecture 10 P. 3Step 2: Deactivate voltage sourceSolve for ix2.ix2 = [(1 / 9) / (5 / 18)] is = (2/5) is Step 3: Add ix1 and ix2 to obtain solution. ix = (vs / 15) + (2/5) is ix = (vs / 15) + (6is / 15)9ix2iS6This technique is called superposition. Superposition works because internal voltages and currents depend linearly on the independent sources.EE201 Lecture 10 P. 4_+_+_+_+2143i1i2i310V1A2A2A3Ai4i5i612V5V7VIFind I for this circuit using superposition.Example:1AEE201 Lecture 10 P. 521431AIStep 1: Short all voltage sources, open all but one current source.I1 = - [(1/6) / ((1/6)+(1/4))] 1 = - 0.4 AEE201 Lecture 10 P. 621431AIStep 2: Short all voltage sources, open all but one current source.I2 = - [(1/8) / ((1/8)+(1/2))] 1 = - 0.2 AEE201 Lecture 10 P. 721432AIStep 3: Short all voltage sources, open all but one current source.I3 = - [(1/1) / ((1/1)+(1/9))] 2 = - 1.8 AEE201 Lecture 10 P. 821432AIStep 4: Short all voltage sources, open all but one current source.I4 = [(1/7) / ((1/7)+(1/3))] 2 = 0.6 AEE201 Lecture 10 P. 921433AIStep 5: Short all voltage sources, open all but one current source.I5 = - [(1/4) / ((1/4)+(1/6))] 3 = - 1.8 AEE201 Lecture 10 P. 10_+_+_+_+214310V12V5V7VIStep 6: Open all current sources.KVL: 5 - 1I6 + 10 - 2I6 - 4I6 - 12 - 3I6 + 7 = 0 10 - 10I6 = 0 I6 = 1AEE201 Lecture 10 P. 11Solution: I = I1 + I2 + I3 + I4 + I5 + I6I = -0.4 - 0.2 - 1.8 + 0.6 - 1.8 + 1I = -2.6 A (see Lect. 8 notes)Principle of LinearityIf x1(t) and x2(t) are two inputs (excitations) and y(t) is an output (response), then there exists coefficients A and B such thaty(t) = A x1(t) + B x2(t)Note: inputs and outputs are voltages or currents.For general case, y(t) = A1x1 + A2x2 + A3x3 + ... + Amxm where xj are voltage or current values from independent sources, and Aj are constants.EE201 Lecture 10 P. 12Example: Find A and B so that Vo = A VS1 + B VS224VS16+_+_VS212_+VoStep 1: Short VS2 and compute Vo1.Vo1 = (4 / 28) VS1Step 2: Short VS1 and compute Vo2.Vo2 = (8 / 14) VS2Therefore,Vo = (1/7) VS1 + (4/7) VS2A = (1/7); B = (4/7)EE201 Lecture 10 P. 13Linearity (cont.) and Source TransformationsLast time, the notion of linearity was introduced.y(t) = A1u1 + A2u2 + A3u3 + ... + AmumFor any resistive circuit, the outputs are linearly related to the inputs (superposition).Va (V)ib (A)Vout (V)500236Example: The table contains measurement data from a linear circuit. Find the output voltage when va = 10 V and ib = 8 A. Also, determine the coefficients A1 and A2 for vout = A1va + A2ibEE201 Lecture 10 P. 14Solution: Outputs are proportional to inputs.Therefore, Vo1 / 10 = (3 / 5) Vo1 = 6 V Vo2 / 8 = (6 / 2) Vo2 = 24 VUsing superpositionVout = Vo1 + Vo2 = 30 VNow, solve for coefficients.vout = A1va + A2ibTo find coefficients, plug in numbers, 3 V = A1(5) + A2 (0) 6 V = A1(0) + A2 (2)Therefore, vout = 0.6 va + 3
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