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Thus, <math>\,y(t) = \pi(x(t))</math> is a linear system. | Thus, <math>\,y(t) = \pi(x(t))</math> is a linear system. | ||
| + | |||
| + | == Example of a Non Linear System == | ||
| + | ::<math>\,x1(t) = sin(t) </math> | ||
| + | ::<math>\,x2(t) = cos(t) </math> | ||
| + | |||
| + | ::<math>\,y1(t) = ln(x1(t)) = ln(sin(t))</math> | ||
| + | ::<math>\,y2(t) = ln(x2(t)) = ln(cos(t))</math> | ||
| + | |||
| + | ::<math>\,ay1(t) + by2(t) = a*ln(sin(t)) + b*ln(cos(t)) \neq ln(ax1(t) + bx2(t))</math> | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | Thus, <math>\,y(t) = ln(x(t))</math> is not a linear system. | ||
Latest revision as of 05:55, 10 September 2008
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.
