(New page: == Part A == This system is not time invariant because it depends on the time shift of the function. This System goes in the order of: X1(t) --> delay --> System --> multiplication ...)
 
(Part B)
 
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== Part B ==
 
== Part B ==
 +
 +
With only inputing
 +
 +
x[n] = u[n]
 +
 +
Then the coefficient of the output is just the square of the time time delay value, which this time delay would be "1".
 +
 +
So u^2 is just (1)^2 = 1
 +
 +
So the output of the signal is:
 +
 +
y[n] = (u * 1)[n - 1] = u[n - 1]

Latest revision as of 13:16, 11 September 2008

Part A

This system is not time invariant because it depends on the time shift of the function. This System goes in the order of:

X1(t) --> delay --> System --> multiplication and addition --> Y(t)

However, if it goes in this order:

X1(t) --> multiplication and addition --> system --> delay --> Y(t)

Then the two outputs would not be equal. The coefficients would be different.


Part B

With only inputing

x[n] = u[n]

Then the coefficient of the output is just the square of the time time delay value, which this time delay would be "1".

So u^2 is just (1)^2 = 1

So the output of the signal is:

y[n] = (u * 1)[n - 1] = u[n - 1]

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Ryne Rayburn