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= [[:Category:Problem_solving|Practice Problem]]: the definition of a set= | = [[:Category:Problem_solving|Practice Problem]]: the definition of a set= | ||
---- | ---- | ||
| − | Does the following collection of signals form a set? | + | Does the following collection of signals form a set? '''(Revised)''' |
<math> | <math> | ||
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x_2(t) &= \cos t \\ | x_2(t) &= \cos t \\ | ||
x_3 (t) &= \sin \frac{t}{2} \\ | x_3 (t) &= \sin \frac{t}{2} \\ | ||
| − | x_4(t) & = \sin \left(t-\frac{\pi}{2} \right) | + | x_4(t) & = -\sin \left(t-\frac{\pi}{2} \right) |
\end{align} | \end{align} | ||
</math> | </math> | ||
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===Answer 1=== | ===Answer 1=== | ||
| − | It depends whether you consider the signals 'x(t) = | + | [[User:Green26|(alec green)]] |
| + | |||
| + | It depends whether you consider the signals '<math>x(t) = someFcn(t)</math>' 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." | From Wikipedia: "Every element of a set must be unique; no two members may be identical." | ||
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'''(a)''' a set | '''(a)''' a set | ||
| − | '''(b)''' not a set (eg x_1(0) = x_3(0)) | + | '''(b)''' not a set (eg <math>x_1(0) = x_3(0)</math>) |
'''(c)''' not a set (see below) | '''(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), | + | 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>. |
| + | :<span style="color:purple"> Instructor's comment: Actually, it's neither a) nor b) nor c). The question is whether the "signals" themselves are all distinct. Good thinking process though, and very well articulated. Keep up the good work! Anybody else wants to venture a guess? -pm </span> | ||
===Answer 2=== | ===Answer 2=== | ||
| − | + | These 4 elements do not form a set. Again the definition of a set is that it must contain unique elements. | |
| + | Signal 4 can be reduced to signal 2; therefore, these two signals are the same and thus not a set. The math is quite simple so I won't work it out, but if someone doesn't understand I would be happy to. | ||
| + | :<span style="color:purple"> Instructor's comment: Very good! Clearly explained, all the important elements are there. Answer is correct. -pm </span> | ||
| + | |||
| + | :<span style="color:blue"> TA: Just for your information, these 4 elements form what is called "multiset", in which the same element in the set can be appeared more than once. -wkh</span> | ||
===Answer 3=== | ===Answer 3=== | ||
Write it here. | Write it here. | ||
| + | |||
| + | The collection of those 4 are not a set because signal 2 and 4 are the same. The definition of the set in the first lecture is "a set is a collection of "distinct" objects called "elements"" | ||
---- | ---- | ||
[[2013_Spring_ECE_302_Boutin|Back to ECE302 Spring 2013 Prof. Boutin]] | [[2013_Spring_ECE_302_Boutin|Back to ECE302 Spring 2013 Prof. Boutin]] | ||
[[ECE302|Back to ECE302]] | [[ECE302|Back to ECE302]] | ||
Latest revision as of 13:45, 13 February 2013
Contents
Practice Problem: the definition of a set
Does the following collection of signals form a set? (Revised)
$ \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.
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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, it's neither a) nor b) nor c). The question is whether the "signals" themselves are all distinct. Good thinking process though, and very well articulated. Keep up the good work! Anybody else wants to venture a guess? -pm
Answer 2
These 4 elements do not form a set. Again the definition of a set is that it must contain unique elements. Signal 4 can be reduced to signal 2; therefore, these two signals are the same and thus not a set. The math is quite simple so I won't work it out, but if someone doesn't understand I would be happy to.
- Instructor's comment: Very good! Clearly explained, all the important elements are there. Answer is correct. -pm
- TA: Just for your information, these 4 elements form what is called "multiset", in which the same element in the set can be appeared more than once. -wkh
Answer 3
Write it here.
The collection of those 4 are not a set because signal 2 and 4 are the same. The definition of the set in the first lecture is "a set is a collection of "distinct" objects called "elements""
