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Because none of the above periodic functions are injective (ie multiple distinct inputs (t) may result in same output (x), like <math>x_1(0) = x_1(pi) = 0</math>), <math>\{x_1(t), x_2(t), x_3(t), x_4(t)\}</math> does not comprise a set, nor do <math>\{x_1(t)\}</math>, <math>\{x_2(t)\}</math>, <math>\{x_3(t)\}</math>, or <math>\{x_4(t)\}</math>.
 
Because none of the above periodic functions are injective (ie multiple distinct inputs (t) may result in same output (x), like <math>x_1(0) = x_1(pi) = 0</math>), <math>\{x_1(t), x_2(t), x_3(t), x_4(t)\}</math> does not comprise a set, nor do <math>\{x_1(t)\}</math>, <math>\{x_2(t)\}</math>, <math>\{x_3(t)\}</math>, or <math>\{x_4(t)\}</math>.
  
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<span style="color:purple"> Instructor's comment: Actually, </span>
 
===Answer 2===
 
===Answer 2===
 
Write it here.
 
Write it here.

Revision as of 15:41, 7 January 2013

Practice Problem: the definition of a set


Does the following collection of signals form a set?

$ \begin{align} x_1(t) &= \sin t \\ x_2(t) &= \cos t \\ x_3 (t) &= \sin \frac{t}{2} \\ x_4(t) & = \sin \left(t-\frac{\pi}{2} \right) \end{align} $

Justify your answer.


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Answer 1

It depends whether you consider the signals '$ x(t) = someFcn(t) $' as (a) character strings, (b) input/output pairs (t,x), or (c) the outputs (x) for all valid inputs (t). I assume that case (c) was intended for consideration here.

From Wikipedia: "Every element of a set must be unique; no two members may be identical."

(a) a set

(b) not a set (eg $ x_1(0) = x_3(0) $)

(c) not a set (see below)

Because none of the above periodic functions are injective (ie multiple distinct inputs (t) may result in same output (x), like $ x_1(0) = x_1(pi) = 0 $), $ \{x_1(t), x_2(t), x_3(t), x_4(t)\} $ does not comprise a set, nor do $ \{x_1(t)\} $, $ \{x_2(t)\} $, $ \{x_3(t)\} $, or $ \{x_4(t)\} $.

Instructor's comment: Actually,

Answer 2

Write it here.

Answer 3

Write it here.


Back to ECE302 Spring 2013 Prof. Boutin

Back to ECE302

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