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The unit step can be written as an infinite sum of shifted delta functions. Since the system is linear, a sum of shifted delta functions as input will yield a sum of shifted delta functions as output. In this system the fact that the desired step function output begins at n = 1 implies that the first delta function in the input should begin at n = 0 because the system always shifts the input by 1 unit. The only remaining task is to counteract the <math>(k+1)^2</math> term inherent in the system, so that the magnitude of each delta function in the output is 1. Therefore, the desired input is as follows: | The unit step can be written as an infinite sum of shifted delta functions. Since the system is linear, a sum of shifted delta functions as input will yield a sum of shifted delta functions as output. In this system the fact that the desired step function output begins at n = 1 implies that the first delta function in the input should begin at n = 0 because the system always shifts the input by 1 unit. The only remaining task is to counteract the <math>(k+1)^2</math> term inherent in the system, so that the magnitude of each delta function in the output is 1. Therefore, the desired input is as follows: | ||
| − | <math>X[n]=\sum_{k=0}^{\infty}\frac{1}{(k+1)^2}\delta | + | <math>X[n]=\sum_{k=0}^{\infty}\frac{1}{(k+1)^2}\delta[n-k]</math> |
Revision as of 08:23, 11 September 2008
a)Time-Invariant?
It is almost trivial to show that the system is time-invariant because all it does is time shift and magnitude scale the input (there is no frequency scaling). If the input itself is shifted, this same shift will appear in the output. The same result could be obtained by shifting the output instead of the input. This is, by definition, time-invariance.
b)Input to Output
$ X[n]= \; ? \longrightarrow Y[n]=u[n-1] $
The unit step can be written as an infinite sum of shifted delta functions. Since the system is linear, a sum of shifted delta functions as input will yield a sum of shifted delta functions as output. In this system the fact that the desired step function output begins at n = 1 implies that the first delta function in the input should begin at n = 0 because the system always shifts the input by 1 unit. The only remaining task is to counteract the $ (k+1)^2 $ term inherent in the system, so that the magnitude of each delta function in the output is 1. Therefore, the desired input is as follows:
$ X[n]=\sum_{k=0}^{\infty}\frac{1}{(k+1)^2}\delta[n-k] $
