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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.) | ||
| − | + | We are given some signal <math>x_k=\delta[n-k]</math> and a system <math>f(x_k) = f(\delta[n-k])) = (k+1)^2 \delta[n-(k+1)]</math>. To show f is time-invariant, we must prove the following statement: | |
| − | <math>f( | + | <math>S_{k_0}(f(x_k)) = f(S_{k_0}(x_k)) \forall k_0 \text{ and }\forall k \in\mathbb{N}\cup{0}</math> |
| − | + | <math>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)]</math> | |
| − | <math> | + | <math>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)]</math> |
| − | + | Pick <math>k=k_0=1</math>: | |
| + | |||
| + | <math>f(S_{k_0}(x_k)) = (3)^2 \delta[n-(3)] \neq (2)^2 \delta[n-(3)] = S_{k_0}(f(x_k))</math> | ||
| + | |||
| + | Since <math> \exists k_0, k \ s.t. \ S_{k_0}(f(x_k)) \neq f(S_{k_0}(x_k)) </math> (e.g., if <math>k=k_0=1</math>), f (ie, the "system") is '''time variant'''. | ||
==Part B== | ==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... | |
Revision as of 06:03, 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...
