Lesson 9 Linearity and Superposition

8/13/2019 Lesson 9 Linearity and Superposition

1/23

Basic Electric Circuits

Linearity And

Superposition

Lesson 9


2/23

Basic Electric CircuitsLinearity 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


3/23

Basic Electric CircuitsLinearity : 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 homogenityproperty. We next consider such an example.

2


4/23

Basic Electric CircuitsLinearity : Homogeneity (scaling).

Illustration: Does homogenity hold for the following equation?

Given,

y = 4x Eq 9.1

If x = 1, y = 4. If we double x to x = 2 and substitutethis 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


5/23


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.

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.4


6/23


y y

x x0 0

2

homogenity

holds

homogenity

does not hold

Lnear Not Linear

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;

5


7/23

Basic Electric CircuitsLinearity : 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 appliedseparately and the individual responses

summed (added together).

This can be explained by considering the

following illustrations.

6


8/23


Illustration: Given, y = 4x.

Let x = x1, then y1= 4x1Let x = x

2

, then y2

= 4x2

Then y = y1+ y2 = 4x1+ 4x2 Eq 9.3

Also, we note,

y = f(x1+ x2) = 4(x1+ x2) = 4x1+ 4x2 Eq 9.4

Since Equations (9.3) and (9.4) are identical,

the additivity property holds.

7


9/23


Illustration: Given, y = 4x + 2.

Let x = x1, then y1= 4x1 + 2Let x = x2, then y2= 4x2+ 2

Then y = y1+ y2 = 4x1+2 + 4x2+2 = 4(x1+x2) + 4 Eq 9.5

Also, we note,

y = f(x1+ x2) = 4(x1+ x2) + 2 Eq 9.6

Since Equations (9.5) and (9.6) are not identical,

the additivity property does not hold.

8


10/23

Basic Electric CircuitsLinearity : Example 9.1: Given the circuit shown

in Figure 9.1. Use the concept of linearity(homogeneity or scaling) to find the current I0.

6 12

11

90 V +_

I0

VS =

Figure 9.1: Circuit for Example 9.1.

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.9


11/23

Basic Electric CircuitsLinearity : Example 9.2: In the circuit shown

below it is known that I0= 4 A when IS= 6 A. FindI0when IS= 18 A.

4 2

Rx I0

IS

Figure 9.2: Circuit for Example 9.2

Since IS NEW= 3xIS OLDwe conclude I0 NEW =3xI0 OLD.

Thus, I0 NEW= 3x4 = 12 A.

10


12/23

Basic Electric CircuitsLinearity : 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?

RX

R1 R2

I0

IS

Figure 9.3: Circuit for investigating linearity.


13/23

Basic Electric CircuitsLinearity : Question, continued.

We use the current splitting rule (current division)

to write the following equation.

10

1 2

( )( )

( )( )( )

S

s

I R

I K IR R

Eq 9.7

The equation is of the same form of y = mx, which

we saw was linear. Therefore, if R1and R2 are constants

then the circuit is linear.

12


14/23

Basic Electric CircuitsSuperposition :

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


15/23

Basic Electric CircuitsSuperposition: Characterization of Superposition.

Let inputs f1 and f2 be applied to a system y such that,

y = k1f1+ k2f2

Where k1and k2are constants of the systems.Let f1act alone so that, y = y1= k1f1Let f2 act alone so that, y = y2= k2f2

The property of superposition states that if f1and f2

Are applied together, the output y will be,

y = y1+ y2= k1f1+ k2f2

13


16/23

Basic Electric CircuitsSuperposition: Illustration using a circuit.

Consider the circuit below that contains two voltage sources.

V1

V2R1

R2

R3

+

+_

_

I

Figure 9.4: Circuit to illustrate superposition

We assume that V1and V2 acting together

produce current I.

15


17/23

Basic Electric CircuitsSuperposition: Illustration using a circuit.

V1

R1

R2

R3

+_

I1

V2

R1

R2

R3

I2

+_

V1produces current I1 V2produces current I2

Superposition states that the current, I, produced by bothsources acting together (Fig 9.4) is the same as the sum of

the currents, I1+ I2, where I1is produced by V1and I2is

produced by V2.

16


18/23

Basic Electric CircuitsSuperposition: Example 9.3. Given the circuit below.

Demonstration by solution that superposition holds.

2 3

VA

VB

VC+

+

+

_

__

VA= 10 V, VB= 5 V, VC= 15 V

IT

Figure 9.5: Circuit for Example 9.3

With all sources acting: IL= 6A

17


19/23


Demonstration by solution that superposition holds.2 3

VA

VB

VC+

+

+

_

__

VA= 10 V, VB= 5 V, VC= 15 V

IT

With VA+ VBacting, VC= 0: IA+B= 3 A,

With VCacting, VA+ VB= 0: IC= 3 A

We see that superposition holds.

18


20/23


Find the current I by using superposition.

I

6 VS= 54 VIS = 3 A

12

+_

Figure 9.5: Circuit for Example 9.4.First, deactivate the source ISand find I in the 6 resistor.

Second, deactivate the source VSand find I in the 6 resistor.

Sum the two currents for the total current.19


21/23



IVs

6 VS= 54 V

12

+_

IVs = 3 A

20


22/23



Is

6 IS = 3 A

12

3 12 2(3 12)

S

xI A

Total current I: I = IS+ Ivs= 5 A

21


23/23

Documents

Lesson 9 Linearity and Superposition