(New page: == '''Problem 2''' == Although not difficult but I find these kind of problems the most confusing. An easy way to deal with such problems is to use the ''''box'''' method as taught by Mim...) |
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== '''Problem 2''' == | == '''Problem 2''' == | ||
| − | Although not difficult but I find these kind of problems the most confusing. | + | Although not difficult but I find these kind of problems the most confusing. |
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| + | Now we were asked to check whether the system defined by the equation '''<math>y(t) = x(1-t)</math>''' is time-invariant or not? | ||
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| + | One easy approach would be as follows | ||
| + | |||
| + | Now | ||
| + | <math>x(1-t) = x (-t+1))</math> | ||
| + | <br><math>x(t)</math> ⇒ '''DELAY''' ⇒ <math> x(t-to)</math> ⇒ '''SYSTEM''' ⇒ <math>x (-t-to+1))= x(1-t-to)</math><br> | ||
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| + | |||
| + | And,<br> | ||
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| + | |||
| + | <math>x(t)</math> ⇒ '''SYSTEM''' ⇒ <math> x(-t +1 )</math> ⇒ '''DELAY''' ⇒ <math>x (-(t-to)+1))=x(1-t+to)</math> | ||
| + | |||
| + | <br> | ||
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| + | These two outputs are not the same and hence the above system is not linear time invariant | ||
Latest revision as of 18:23, 14 October 2008
Problem 2
Although not difficult but I find these kind of problems the most confusing.
Now we were asked to check whether the system defined by the equation $ y(t) = x(1-t) $ is time-invariant or not?
One easy approach would be as follows
Now
$ x(1-t) = x (-t+1)) $
$ x(t) $ ⇒ DELAY ⇒ $ x(t-to) $ ⇒ SYSTEM ⇒ $ x (-t-to+1))= x(1-t-to) $
And,
$ x(t) $ ⇒ SYSTEM ⇒ $ x(-t +1 ) $ ⇒ DELAY ⇒ $ x (-(t-to)+1))=x(1-t+to) $
These two outputs are not the same and hence the above system is not linear time invariant
