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<math>= \delta[n-1] + \delta[n-2] + \dots + \delta[n-(k+1)] + \dots = \sum_{k=0}^{\infty} \delta[n-(k+1)] = \sum_{k=1}^{\infty} \delta[n-k] = u[n-1]</math> | <math>= \delta[n-1] + \delta[n-2] + \dots + \delta[n-(k+1)] + \dots = \sum_{k=0}^{\infty} \delta[n-(k+1)] = \sum_{k=1}^{\infty} \delta[n-k] = u[n-1]</math> | ||
| − | Thus, the signal that produces output y[n] = u[n-1] is the input <math>x[n] = \sum_{k=0}^{\infty} \frac{x_k[n]}{(k+1)^2}</math> | + | Thus, the signal that produces output y[n] = u[n-1] is the input <math>x[n] = \sum_{k=0}^{\infty} \frac{x_k[n]}{(k+1)^2} = \sum_{k=0}^{\infty} \frac{ \delta[n-k]}{(k+1)^2}</math> |
Latest revision as of 11:29, 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].
$ u[n-1] = \delta[n-1] + \delta[n-2] + \dots = \sum_{k=1}^{\infty} \delta[n-k] $
Let's try the following first:
$ f(\sum_{k=0}^{\infty} x_k) = \sum_{k=0}^{\infty} f(x_k) = \delta[n-1] + 4\delta[n-2] + \dots + (k+1)^2 \delta[n-(k+1)] + \dots = \sum_{k=0}^{\infty} (k+1)^2 \delta[n-(k+1)] $
(Note: The first equivalence is true because f (the system) is assumed linear.) We're pretty close, but there's the problem with the pesty constants out in front of the delta terms, and we need to get rid of those. Since f is assumed to be linear, we can simply multiply each term by some constant (namely, $ (k+1)^{-2} $) to get rid of this:
$ f(\sum_{k=0}^{\infty} \frac{x_k}{(k+1)^2}) = \sum_{k=0}^{\infty} f(\frac{x_k}{(k+1)^2}) = \sum_{k=0}^{\infty} \frac{1}{(k+1)^2}f(x_k) $ (Because the system f is linear)
$ = \frac{\delta[n-1]}{1} + \frac{4 \delta[n-2]}{4} + \dots + \frac{(k+1)^2 \delta[n-(k+1)]}{(k+1)^2} + \dots $
$ = \delta[n-1] + \delta[n-2] + \dots + \delta[n-(k+1)] + \dots = \sum_{k=0}^{\infty} \delta[n-(k+1)] = \sum_{k=1}^{\infty} \delta[n-k] = u[n-1] $
Thus, the signal that produces output y[n] = u[n-1] is the input $ x[n] = \sum_{k=0}^{\infty} \frac{x_k[n]}{(k+1)^2} = \sum_{k=0}^{\infty} \frac{ \delta[n-k]}{(k+1)^2} $
