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==Part A== | ==Part A== | ||
We are given the following: | We are given the following: | ||
| − | <math>X_k[n]=\delta[n-k] \rightarrow \text{ system } \rightarrow Y_k[n]=(k+1)2 \delta[n-(k+1)] \ (k \in \mathbb{Z}, k \geq 0)</math> | + | <math>X_k[n]=\delta[n-k] \rightarrow \text{ system } \rightarrow Y_k[n]=(k+1)^2 \delta[n-(k+1)] \ (k \in \mathbb{Z}, k \geq 0)</math> |
Translate this into math: (See [[HW2-D_Brian_Thomas_ECE301Fall2008mboutin#Definition_of_Time_Invariance|this]] for symbology.) | Translate this into math: (See [[HW2-D_Brian_Thomas_ECE301Fall2008mboutin#Definition_of_Time_Invariance|this]] for symbology.) | ||
Revision as of 06:04, 11 September 2008
Part A
We are given the following: $ X_k[n]=\delta[n-k] \rightarrow \text{ system } \rightarrow Y_k[n]=(k+1)^2 \delta[n-(k+1)] \ (k \in \mathbb{Z}, k \geq 0) $
Translate this into math: (See this for symbology.)
We are given some signal $ x_k=\delta[n-k] $ and a system $ f(x_k) = f(\delta[n-k])) = (k+1)^2 \delta[n-(k+1)] $. To show f is time-invariant, we must prove the following statement:
$ S_{k_0}(f(x_k)) = f(S_{k_0}(x_k)) \forall k_0 \text{ and }\forall k \in\mathbb{N}\cup{0} $
$ f(S_{k_0}(x_k)) = f(S_{k_0}(\delta[n-k])) = f(\delta[n-(k+k_0)]) = (k+k_0+1)^2 \delta[n-(k+k_0+1)] $
$ S_{k_0}(f(x_k)) = S_{k_0}(f(\delta[n-k])) = S_{k_0}((k+1)^2 \delta[n-(k+1)]) = (k+1)^2 \delta[n-(k+k_0+1)] $
Pick $ k=k_0=1 $:
$ f(S_{k_0}(x_k)) = (3)^2 \delta[n-(3)] \neq (2)^2 \delta[n-(3)] = S_{k_0}(f(x_k)) $
Since $ \exists k_0, k \ s.t. \ S_{k_0}(f(x_k)) \neq f(S_{k_0}(x_k)) $ (e.g., if $ k=k_0=1 $), f (ie, the "system") is time variant.
Part B
We are told to assume f is linear. We would like to find input x[n] s.t. we get output y[n] = u[n-1]. To be continued...
