Basic Electric Circuits Linearity And Superposition Lesson 9

Basic Electric Circuits Linearity And Superposition Lesson 9

Basic Electric Circuits Linearity And Superposition Lesson 9 Basic Electric Circuits Linearity and Superposition: Linearity. Basically, a mathematical equation is said to be linear if the following properties hold. homogenity additivity What does this mean? We first look at the property of homogenity. 1 Basic Electric Circuits Linearity : Homogeneity. Homogenity requires that if the input (excitation) of a system (equation) is multiplied by a constant, then the output should be obtained by multiplying by the same constant to obtain the correct solution.

Sometimes equations that we think are linear, turn out not be be linear because they fail the homogenity property. We next consider such an example. 2 Basic Electric Circuits Linearity : Homogeneity (scaling). Illustration: Does homogenity hold for the following equation? Given, Eq 9.1 y = 4x If x = 1, y = 4. If we double x to x = 2 and substitute this value into Eq 9.1 we get y = 8. Now for homogenity to hold, scaling should hold for y. that is, y has a value of 4 when x = 1. If we increase x by a factor of 2 when we should be able to multiply y by the same factor and get the same answer and when we substitute into the right side of the equation for x = 2. 3 Basic Electric Circuits

Linearity : Homogeneity (scaling). Illustration: Does homogenity hold for the following equation? Given, y = 4x + 2 Eq 9.2 If x = 1, then y = 6. If we double x to x=2, then y = 10. Now, since we doubled x we should be able to double the value that y had when x = 1 and get y = 10. In this case we get y = (2)(6) = 12, which obviously is not 10, so homogenity does not hold. 4 We conclude that Eq 9.2 is not a linear equation. In some ways that goes against the gain of what we have been taught about linear equations. Basic Electric Circuits Linearity : Homogeneity (scaling). Many of us were brought-up to think that if plotting an equation yields a straight line, then the equation is linear. From the following illustrations we have; h o m o g e n ity

d o e s n o t h o ld h o m o g e n ity h o ld s y y Lnear 0 5 x 2 N o t L in e a r 0 x

Basic Electric Circuits Linearity : Additivity Property. The additivity property is equivalent to the statement that the response of a system to a sum of inputs is the same as the responses of the system when each input is applied separately and the individual responses summed (added together). This can be explained by considering the following illustrations. 6 Basic Electric Circuits Linearity : Additivity Property. Illustration: Given, y = 4x. Let x = x1, then y1 = 4x1 Let x = x2, then y2 = 4x2 Then y = y1 + y2 = 4x1 + 4x2 Also, we note, y = f(x1 + x2) = 4(x1 + x2) = 4x1 + 4x2

7 Since Equations (9.3) and (9.4) are identical, the additivity property holds. Eq 9.3 Eq 9.4 Basic Electric Circuits Linearity : Additivity Property. Illustration: Given, y = 4x + 2. Let x = x1, then y1 = 4x1 + 2 Let x = x2, then y2 = 4x2 + 2 Then y = y1 + y2 = 4x1+2 + 4x2+2 = 4(x1+x2) + 4 Also, we note, y = f(x1 + x2) = 4(x1 + x2) + 2 Since Equations (9.5) and (9.6) are not identical, 8 the additivity property does not hold.

Eq 9.5 Eq 9.6 Basic Electric Circuits Linearity : Example 9.1: Given the circuit shown in Figure 9.1. Use the concept of linearity (homogeneity or scaling) to find the current I0. I0 11 V S = 90 V +_ 6 12

Figure 9.1: Circuit for Example 9.1. 9 Assume I0 = 1 A. Work back to find that this gives VS = 45 V. But since VS = 90 V this means the true I0 = 2 A. Basic Electric Circuits Linearity : Example 9.2: In the circuit shown below it is known that I0 = 4 A when IS = 6 A. Find I0 when IS = 18 A. R IS I0 x 4

2 Figure 9.2: Circuit for Example 9.2 10 Since IS NEW = 3xIS OLD we conclude I0 NEW = 3xI0 OLD. Thus, I0 NEW = 3x4 = 12 A. Basic Electric Circuits Linearity : Question. For Example 9.2, one might ask, how do we know the circuit is linear? That is a good question. To answer, we assume a circuit of the same form and determine if we get a linear equation between the output current and the input current. What must be shown for the circuit below? R IS I0 X

R 1 R 2 Figure 9.3: Circuit for investigating linearity. Basic Electric Circuits Linearity : Question, continued. We use the current splitting rule (current division) to write the following equation. ( I S )( R1 ) I0 ( K )( I s ) ( R1 R2 ) Eq 9.7 The equation is of the same form of y = mx, which

we saw was linear. Therefore, if R1 and R2 are constants then the circuit is linear. 12 Basic Electric Circuits Superposition : One might read (hear) the following regarding superposition. (1) A system is linear if superposition holds. (2) Superposition holds if a system is linear. This sounds a little like the saying of which comes first, the chicken or the egg. Of the two statements, I believe one should remember that if a system is linear then superposition applies (holds). 14 Basic Electric Circuits Superposition: Characterization of Superposition. Let inputs f1 and f2 be applied to a system y such that, y = k1f1 + k2f2 Where k1 and k2 are constants of the systems. Let f1 act alone so that, y = y1 = k1f1 Let f2 act alone so that, y = y2 = k2f2

The property of superposition states that if f1 and f2 Are applied together, the output y will be, y = y1 + y2 = k1f1 + k2f2 13 Basic Electric Circuits Superposition: Illustration using a circuit. Consider the circuit below that contains two voltage sources. R V2 _ + 1 I V 1 R +_

R 2 3 Figure 9.4: Circuit to illustrate superposition We assume that V1and V2 acting together produce current I. 15 Basic Electric Circuits Superposition: Illustration using a circuit. R V 1 +_ R

1 I1 R R V 1 _ + I2 R 2 3 V1 produces current I1

2 R 2 3 V2 produces current I2 Superposition states that the current, I, produced by both sources acting together (Fig 9.4) is the same as the sum of the currents, I1 + I2, where I1 is produced by V1 and I2 is produced by V2. 16 Basic Electric Circuits Superposition: Example 9.3. Given the circuit below. Demonstration by solution that superposition holds. V 2

_ B 3 + IT V A + V _ V

A = 10 V, V B = 5 V, V C C = 15 V Figure 9.5: Circuit for Example 9.3 With all sources acting: IL = 6 A 17 _ + Basic Electric Circuits Superposition: Example 9.3. Given the circuit below.

Demonstration by solution that superposition holds. 2 _ A + V IT V 3 B + V _

V A = 10 V, V B = 5 V, V C = 15 V With VA + VB acting,, VC = 0: IA+B = 3 A With VC acting, VA + VB = 0: IC = 3 A

We see that superposition holds. 18 C _ + Basic Electric Circuits Superposition: Example 9.4. Given the circuit below. Find the current I by using superposition. 12 I IS = 3 A 6 + _ V S

= 54 V Figure 9.5: Circuit for Example 9.4. First, deactivate the source IS and find I in the 6 resistor. Second, deactivate the source VS and find I in the 6 resistor. 19 Sum the two currents for the total current. Basic Electric Circuits Superposition: Example 9.4. Given the circuit below. Find the current I by using superposition. 12 IV s 6 IVs = 3 A 20 + _

V S = 54 V Basic Electric Circuits Superposition: Example 9.4. Given the circuit below. Find the current I by using superposition. Is IS = 3 A 12 6 3 x12 IS 2 A (3 12) Total current I: I = IS + Ivs = 5 A 21

CIRCUITS End of Lesson 9 Linearity and Superposition

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