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| + | '''Part A''' | ||
| + | ---- | ||
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We know that a time invariant system is one whose output does not depend explicitly on time. | We know that a time invariant system is one whose output does not depend explicitly on time. | ||
'''Now is the given system time invariant?''' | '''Now is the given system time invariant?''' | ||
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As we can see they are not the same. Thus the system is time variant. | As we can see they are not the same. Thus the system is time variant. | ||
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| + | '''Part B''' | ||
| + | ---- | ||
| + | What input x[n] would yield the output Y[n]=u[n-1]? | ||
| + | |||
| + | We are told that the system is linear.. Thus from the data that is given to us, | ||
| + | An input u[n] should yield an ouput Y[n]=u[n-1]. | ||
Latest revision as of 10:53, 12 September 2008
Part A
We know that a time invariant system is one whose output does not depend explicitly on time.
Now is the given system time invariant?
Well the answer is no, It is not time invariant. Here's why.
Input $ {x_k[n]}=\delta[n-k] $
Output
$ {y_k[n]}={(k+1)^2\delta[n-(k+1)]} $
Let us consider a particular case of, input, $ {x_1[n]}=\delta[n-1] $ and output $ {y_1[n]}=4\delta[n-2] $ and a time delay of 5 units.
Now, $ \delta[n-1] $$ {\xrightarrow{time delay}} $$ \delta[n-6] $$ {\xrightarrow{system}} $$ {49\delta[n-7]} $
When we put it through the system first, This is what we get,
$ \delta[n-1] $$ {\xrightarrow{system}} $$ {4\delta[n-2]} $$ {\xrightarrow{time delay}} $$ {4\delta[n-7]} $
As we can see they are not the same. Thus the system is time variant.
Part B
What input x[n] would yield the output Y[n]=u[n-1]?
We are told that the system is linear.. Thus from the data that is given to us, An input u[n] should yield an ouput Y[n]=u[n-1].
