EE201 Lecture 10 P 1 Linearity and Superposition Consider the circuit 6 vS 9 iS ix We can solve for the output ix using mesh analysis M1 vs 6i1 9 i1 i2 0 vs 15i1 9i2 0 M2 But i2 is i x i1 i2 ix vs 15 6is 15 EE201 Lecture 10 P 2 Solution 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 Step 1 Deactivate current source 6 vS 9 ix1 Solve for ix1 ix1 vs 15 EE201 Lecture 10 P 3 Step 2 Deactivate voltage source 6 9 iS ix2 Solve 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 This technique is called superposition Superposition works because internal voltages and currents depend linearly on the independent sources EE201 Lecture 10 Example P 4 1A i2 2 10V 1A 4 3A i1 12V I i 3 1 i5 i6 7V 5V 3 i4 2A Find I for this circuit using superposition 2A EE201 Lecture 10 P 5 Step 1 Short all voltage sources open all but one current source 2 1A 4 I 1 3 I1 1 6 1 6 1 4 1 0 4 A EE201 Lecture 10 P 6 Step 2 Short all voltage sources open all but one current source 1A 2 4 I 1 3 I2 1 8 1 8 1 2 1 0 2 A EE201 Lecture 10 P 7 Step 3 Short all voltage sources open all but one current source 2 4 I 1 2A 3 I3 1 1 1 1 1 9 2 1 8 A EE201 Lecture 10 P 8 Step 4 Short all voltage sources open all but one current source 2 4 I 1 3 2A I4 1 7 1 7 1 3 2 0 6 A EE201 Lecture 10 P 9 Step 5 Short all voltage sources open all but one current source 2 4 I 1 3A 3 I5 1 4 1 4 1 6 3 1 8 A EE201 Lecture 10 P 10 Step 6 Open all current sources 2 10V 4 12V I 1 7V 5V 3 KVL 5 1I6 10 2I6 4I6 12 3I6 7 0 10 10I6 0 I6 1A EE201 Lecture 10 P 11 Solution I I 1 I 2 I3 I 4 I5 I6 I 0 4 0 2 1 8 0 6 1 8 1 I 2 6 A see Lect 8 notes Principle of Linearity If x1 t and x2 t are two inputs excitations and y t is an output response then there exists coefficients A and B such that y 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 A are constants EE201 Lecture 10 P 12 Example Find A and B so that Vo A VS1 B VS2 6 24 VS1 VS2 12 Vo Step 1 Short VS2 and compute Vo1 Vo1 4 28 VS1 Step 2 Short VS1 and compute Vo2 Vo2 8 14 VS2 Therefore Vo 1 7 VS1 4 7 VS2 A 1 7 B 4 7 EE201 Lecture 10 P 13 Linearity cont and Source Transformations Last time the notion of linearity was introduced y t A1u1 A2u2 A3u3 Amum For any resistive circuit the outputs are linearly related to the inputs superposition Example 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 A2ib Va V ib A Vout V 5 0 3 0 2 6 EE201 Lecture 10 P 14 Solution Outputs are proportional to inputs Therefore Vo1 10 3 5 Vo1 6 V Vo2 8 6 2 Vo2 24 V Using superposition Vout Vo1 Vo2 30 V Now solve for coefficients vout A1va A2ib To find coefficients plug in numbers 3 V A1 5 A2 0 6 V A1 0 A2 2 Therefore vout 0 6 va 3 ib
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