Linearity and Super Position

Linearity

Basically, a mathematical equation is said to be linear if the following properties hold.

•  homogenity

•  additivity

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.

Illustration:  Does homogenity hold for the following equation?

 y = 4x  Eq 9.1

 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.  

Homogeneity (scaling).

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.

Illustration: Given, y = 4x Eq 9.4

Let x = x1, then y1 = 4x1 Let x = x2, 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.

Super Position

For a circuit having multiple independent sources, the voltage (or current through) an element is equal to the algebraic sum of the voltages (or currents) due to each independent source acting one at a time.

Two things to remember: 

• Consider one independent source at a time while other independent sources are turned off. 
• Replace voltage source with short circuit and current source with open circuit.
• Dependent sources are left intact because they are controlled by circuit variables.

Given the circuit below.Find the current I by using superposition.

First, deactivate the source IS and find I in the 6 W resistor. Second, deactivate the source VS and find I in the 6 W resistor. Sum the two currents for the total current.

IVs =  3 A

Total current I:  I = IS + Ivs = 5 A