Linearity

A system is said to be linear if it satisfies the properties of scaling and superposition. Thus, the following holds true for all linear systems:

Suppose there are two inputs
$ \,x1(t) $
$ \,x2(t) $
with outputs
$ \,y1(t) = C\left\{x1(t)\right\} $
$ \,y2(t) = C\left\{x2(t)\right\} $
A linear system must satisfy the condition
$ \,ay1(t) + by2(t) = C\left\{ax1(t) + bx2(t)\right\} $

Example of a Linear System

$ \,x1(t) = sin(t) $
$ \,x2(t) = cos(t) $
$ \,y1(t) = \pi\left\{x1(t)\right\} = \pi(sin(t)) $
$ \,y2(t) = \pi\left\{x2(t)\right\} = \pi(cos(t)) $
$ \,ay1(t) + by2(t) = a*\pi*sin(t) + b*\pi*cos(t) = \pi\left\{asin(t) + bcos(t)\right\} = \pi\left\{ax1(t) + bx2(t)\right\} $


Thus, $ \,y(t) = \pi(x(t)) $ is a linear system.

Example of a Non Linear System

$ \,x1(t) = sin(t) $
$ \,x2(t) = cos(t) $
$ \,y1(t) = ln(x1(t)) = ln(sin(t)) $
$ \,y2(t) = ln(x2(t)) = ln(cos(t)) $
$ \,ay1(t) + by2(t) = a*ln(sin(t)) + b*ln(cos(t)) \neq ln(ax1(t) + bx2(t)) $



Thus, $ \,y(t) = ln(x(t)) $ is not a linear system.

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn