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| − | + | = Practice problem on time-invariance of a CT system = | |
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| − | =Practice problem on time-invariance of a CT system= | + | |
| − | == Question == | + | == Question == |
| − | + | ||
| + | <span class="texhtml">''Y''(''t'') = ''x''(''t'' − 1) − ''x''(1 − ''t'')</span> | ||
| + | |||
| + | <br> It is Time Invariant? Justify. | ||
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| − | + | == Answer == | |
| − | + | No. | |
| − | < | + | <span class="texhtml">''S''<sub>1</sub> = ''Y''(''t'') = ''x''(''t'' − 1) − ''x''(1 − ''t'')</span> |
| − | < | + | <span class="texhtml">''S''<sub>2</sub> = ''Y''(''t'') = ''x''(''t'' − ''t''<sub>''o''</sub>)</span> |
| − | < | + | <span class="texhtml">''x''(''t'') − > ''S''1 − > ''S''2 − > ''x''(''t'' − ''t''<sub>''o''</sub> − 1) − ''x''(1 − ''t'' + ''t''<sub>''o''</sub>)</span> |
| + | |||
| + | <span class="texhtml">''x''(''t'') − > ''S''2 − > ''S''1 − > ''x''(''t'' − ''t''<sub>''o''</sub> − 1) − ''x''(1 − ''t'' − ''t''<sub>''o''</sub>)</span> | ||
| + | |||
| + | <span class="texhtml">''x''(''t'' − ''t''<sub>''o''</sub> − 1) − ''x''(1 − ''t'' + ''t''<sub>''o''</sub>) = / = ''x''(''t'' − ''t''<sub>''o''</sub> − 1) − ''x''(1 − ''t'' − ''t''<sub>''o''</sub>)</span> | ||
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| − | == Comments== | + | |
| − | Please comment on this answer. Are there any mistakes? Is it clear? Could it be improved? | + | == Comments == |
| + | |||
| + | Please comment on this answer. Are there any mistakes? Is it clear? Could it be improved? | ||
| + | |||
---- | ---- | ||
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| − | ==Comment 2== | + | == Comment 1 == |
| − | Write it here | + | |
| + | Actually this is where I'm unsure on HW3 Q1b. It would definitely be improved and more clear if the student showed intermediate steps. | ||
| + | |||
| + | x(t) −> [S1] −> y(t) = x(t - 1) - x(1 - t) --> [S2] −> z<sub>1</sub>(t) = y(t - t<sub>o</sub>) = x((t - t<sub>o</sub>) -1) - x(1 - (t - t<sub>o</sub>)) | ||
| + | |||
| + | --> z<sub>1</sub>(t) = x(t − t<sub>o</sub> − 1) − x(1 − t + t<sub>o</sub>)<br>x(t) −> [S2] −> y(t) = x(t - t<sub>o</sub>) --> [S1] −> z<sub>2</sub>(t) = y(t - 1) - y(1 - t) = x((t - 1) - t<sub>o</sub>) - x((1 - t) - t<sub>o</sub>)<sub></sub> | ||
| + | |||
| + | --> z<sub>2</sub>(t) = x(t − t<sub>o</sub> − 1) − x(1 − t − t<sub>o</sub>) | ||
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| + | Comment 2 | ||
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| + | Write it here | ||
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| − | [[ | + | |
| + | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
| + | |||
| + | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
Revision as of 21:41, 2 February 2011
Contents
Practice problem on time-invariance of a CT system
Question
Y(t) = x(t − 1) − x(1 − t)
It is Time Invariant? Justify.
Answer
No.
S1 = Y(t) = x(t − 1) − x(1 − t)
S2 = Y(t) = x(t − to)
x(t) − > S1 − > S2 − > x(t − to − 1) − x(1 − t + to)
x(t) − > S2 − > S1 − > x(t − to − 1) − x(1 − t − to)
x(t − to − 1) − x(1 − t + to) = / = x(t − to − 1) − x(1 − t − to)
Comments
Please comment on this answer. Are there any mistakes? Is it clear? Could it be improved?
Comment 1
Actually this is where I'm unsure on HW3 Q1b. It would definitely be improved and more clear if the student showed intermediate steps.
x(t) −> [S1] −> y(t) = x(t - 1) - x(1 - t) --> [S2] −> z1(t) = y(t - to) = x((t - to) -1) - x(1 - (t - to))
--> z1(t) = x(t − to − 1) − x(1 − t + to)
x(t) −> [S2] −> y(t) = x(t - to) --> [S1] −> z2(t) = y(t - 1) - y(1 - t) = x((t - 1) - to) - x((1 - t) - to)
--> z2(t) = x(t − to − 1) − x(1 − t − to)
Comment 2
Write it here
