| Line 24: | Line 24: | ||
<math>y(z(t))</math> =?= <math>ay(x(t))</math> | <math>y(z(t))</math> =?= <math>ay(x(t))</math> | ||
| − | <math>3z(2t+1)</math> =?= <math>a*3x | + | <math>3z(2t+1)</math> =?= <math>a*3x(2t+1)</math> |
<math>3ax(2t+1) = 3ax(2t+1)</math> | <math>3ax(2t+1) = 3ax(2t+1)</math> | ||
Revision as of 19:10, 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
$ y(t) = 3x(2t+1) $
(i)
Let $ c(t) = a(t) + b(t) $
$ y(c(t)) $ =?= $ y(a(t)) + y(b(t)) $
$ 3c(2t+1) $ =?= $ 3a(2t+1) + 3b(2t+1) $
$ 3[a(2t+1) + b(2t +1)] $ =?= $ 3a(2t+1) + 3b(2t+1) $
$ 3a(2t+1) + 3b(2t+1) = 3a(2t+1) + 3b(2t+1) $
(ii)
Let $ z(t)=ax(t) $
$ y(z(t)) $ =?= $ ay(x(t)) $
$ 3z(2t+1) $ =?= $ a*3x(2t+1) $
$ 3ax(2t+1) = 3ax(2t+1) $
The system passes both requirements and is, therefore, linear.
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:
(ii)
$ y(ax(t)) $ =?= $ ay(x(t)) $
$ e^{ax(t)} \ne ae^{x(t)} $
In this case the system also fails the second requirement, but just failing one of the two is enough to prove that the system is non-linear.
