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<math>e^{a(t)}e^{b(t)} \ne e^{a(t)} + e^{b(t)}</math>
 
<math>e^{a(t)}e^{b(t)} \ne e^{a(t)} + e^{b(t)}</math>
  
and
+
Proving that the system fails the first requirement is enough to prove that the system is non-linear, but, continuing anyway:

Revision as of 17:40, 8 September 2008

Definition

A system is a linear system if (i) the output produced by first summing any two inputs and then putting the result through the system is identical to the output produced by first putting both signals through the system separately and then summing the results and (ii) the output produced by first multiplying an input signal by a constant and then putting the result through the system is identical to the output produced by first putting the original signal through the system and then multiplying the result by the constant.

Example 1: Linear System

Example 2: Non-Linear System

$ y(t) = e^{x(t)} $

(i)

$ y(a(t) + b(t)) $ =?= $ y(a(t)) + y(b(t)) $

$ e^{a(t) + b(t)} $ =?= $ e^{a(t)} + e^{b(t)} $

$ e^{a(t)}e^{b(t)} \ne e^{a(t)} + e^{b(t)} $

Proving that the system fails the first requirement is enough to prove that the system is non-linear, but, continuing anyway:

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva