| Line 26: | Line 26: | ||
Then, <math>h(t) = e^t[x(t)]</math> | Then, <math>h(t) = e^t[x(t)]</math> | ||
| − | Now, if x(t) is bounded ( | + | Now, if x(t) is bounded (Why? - refer SYSTEM 1), then h(t) is bounded too. |
Hence, h(t) is stable. | Hence, h(t) is stable. | ||
Revision as of 14:20, 30 June 2008
SYSTEM 1 - $ h[n] = \delta[n+1] + \delta[n-1] $
A - For an LTI system to be memoryless, the output value of 'n' should only depend on the input value of 'n'. But, in this case, the output value of 'n' depends on the past and future values of 'n'. As a result, the system is NOT memoryless or in other words, has memory.
B - For an LTI system to be causal, the output should NOT depend on the future input values. But, in this case, the output does depend on the future input values and as a result, is not causal.
C - Stable (Why?)
$ \delta[n+1] $ - Bounded (Stable)
$ \delta[n-1] $ - Bounded (Stable)
Therefore, h[n] = Bounded + Bounded = Bounded (Stable)
SYSTEM 2 -
$ h(t) = e^t[u(t-2) - u(t-5)] $
A - For an LTI system to be memoryless, the output value of 'n' should only depend on the input value of 'n'. But, in this case, the output value of 'n' depends on the past value of 'n'. As a result, the system is NOT memoryless or in other words, has memory.
B - For an LTI system to be causal, the output should NOT depend on the future input values. In this case, the output only depends on the past input values and as a result, is causal.
C - Let $ x(t) = u(t-2) - u(t-5) $
Then, $ h(t) = e^t[x(t)] $
Now, if x(t) is bounded (Why? - refer SYSTEM 1), then h(t) is bounded too.
Hence, h(t) is stable.
