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<math>y_1(t) + y_2(t) = a\times t - b + 5 + c\times t + d + 5 = (a + c)t + d - b + 10</math> | <math>y_1(t) + y_2(t) = a\times t - b + 5 + c\times t + d + 5 = (a + c)t + d - b + 10</math> | ||
<math>x(a\times t-b) + x(c\times t+d) = a\times t - b + 5 + c\times t + d + 5 = (a + c)t + d - b + 10</math> | <math>x(a\times t-b) + x(c\times t+d) = a\times t - b + 5 + c\times t + d + 5 = (a + c)t + d - b + 10</math> | ||
| + | Therefore: <math>x(a\times t-b) + x(c\times t+d) = y_1(t) + y_2(t)</math> | ||
Revision as of 15:29, 12 September 2008
A linear system is one in which the magnitude of the output is proportional to the magnitude of the input. If an input x(t) produces an output y(t) then the input ax(t) should produce the output ay(t) for some number a. In order for a system to be linear it must also be able to take combined inputs and produce their combined outputs. This means that if an input x1(t) produces the output y1(t) and x2(t) yields y2(t) then the combined inputs x1(t) + x2(t) should yield the combined output y1(t) + y2(t).
Linear system: $ x(t) = t + 5 $
Proof: Given that a, b, c, and d are any numbers real or complex we have
$ x(a\times t-b) = at - b + 5 = y_1(t) $ $ x(c\times t+d) = ct + d + 5 = y_2(t) $ $ y_1(t) + y_2(t) = a\times t - b + 5 + c\times t + d + 5 = (a + c)t + d - b + 10 $ $ x(a\times t-b) + x(c\times t+d) = a\times t - b + 5 + c\times t + d + 5 = (a + c)t + d - b + 10 $ Therefore: $ x(a\times t-b) + x(c\times t+d) = y_1(t) + y_2(t) $
