(New page: : <math> y[n] - y[n-1] = x[n]| </math> : <math> \Rightarrow H(\omega) = \frac {1}{1 - e^{-j \omega}} </math> From table in book: : <math> \begin{align} u[n] &\overset {\mathfrak{F}}{\lon...)
 
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: <math> y[n] - y[n-1] = x[n]| </math>
 
: <math> y[n] - y[n-1] = x[n]| </math>
: <math> \Rightarrow H(\omega) = \frac {1}{1 - e^{-j \omega}} </math>
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: <math> \Rightarrow H(\omega) = \frac{1}{1 - e^{-j \omega}} </math>
  
 
From table in book:
 
From table in book:
  
: <math> \begin{align} u[n] &\overset {\mathfrak{F}}{\longleftrightarrow} \frac {1}{1 - e^{-j \omega}} + \sum_{k=-\infty}^{\infty} \pi \delta (\omega - 2 \pi k) \\
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: <math> \begin{align} u[n] &\overset {\mathfrak{F}}{\longleftrightarrow} \frac{1}{1 - e^{-j \omega}} + \sum_{k=-\infty}^{\infty} \pi \delta (\omega - 2 \pi k) \\
 
1 &\overset {\mathfrak{F}}{\longleftrightarrow} 2 \pi \sum_{l=-\infty}^{\infty} \delta (\omega - 2 \pi l) \end{align} </math>
 
1 &\overset {\mathfrak{F}}{\longleftrightarrow} 2 \pi \sum_{l=-\infty}^{\infty} \delta (\omega - 2 \pi l) \end{align} </math>
 +
 +
: <math> \Rightarrow H(\omega) = \frac{1}{1 - e^{-j \omega}} + \sum_{k=-\infty}^{\infty} \pi \delta (\omega - 2 \pi k) - \frac{1}{2} \cdot 2 \pi \sum_{l=-\infty}^{\infty} \delta (\omega - 2 \pi l) </math>
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 +
: <math> \Rightarrow h[n] = \mathfrak{F}^{-1}(H(\omega)) = u[n] - \frac{1}{2}\ </math>
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 +
Find step response:
 +
 +
: <math> \begin{align} y[n] &= x[n] * h[n] = u[n] * h[n] = \sum_{k=-\infty}^{\infty} u[k]h[n-k] \\
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&= \sum_{k=-\infty}^{\infty} u[k] \left ( u[n-k] - \frac {1}{2} \right ) = \sum_{k=-\infty}^{\infty} u[k]u[n-k] - \frac{1}{2} \sum_{k=-\infty}^{\infty} u[k] \\
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&= \sum_{k=0}^{\infty} u[n-k] - \frac{1}{2} \sum_{k=0}^{\infty} 1 = -\infty \end{align} </math>
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: (this seems unreasonable!)
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 +
Note: The original eq. <math> y[n] - y[n-1] = x[n]\ </math> can be expressed as:
 +
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: <math> h[n] - h[n-1] = \delta [n]\ </math>
 +
 +
By observation, <math> h[n] = u[n] + c\ </math> for any constant c.  And further: the step response is divergent for <math> c \ne 0\ </math>.
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 +
So, what is c equal to?

Revision as of 11:56, 12 December 2008

$ y[n] - y[n-1] = x[n]| $
$ \Rightarrow H(\omega) = \frac{1}{1 - e^{-j \omega}} $

From table in book:

$ \begin{align} u[n] &\overset {\mathfrak{F}}{\longleftrightarrow} \frac{1}{1 - e^{-j \omega}} + \sum_{k=-\infty}^{\infty} \pi \delta (\omega - 2 \pi k) \\ 1 &\overset {\mathfrak{F}}{\longleftrightarrow} 2 \pi \sum_{l=-\infty}^{\infty} \delta (\omega - 2 \pi l) \end{align} $
$ \Rightarrow H(\omega) = \frac{1}{1 - e^{-j \omega}} + \sum_{k=-\infty}^{\infty} \pi \delta (\omega - 2 \pi k) - \frac{1}{2} \cdot 2 \pi \sum_{l=-\infty}^{\infty} \delta (\omega - 2 \pi l) $
$ \Rightarrow h[n] = \mathfrak{F}^{-1}(H(\omega)) = u[n] - \frac{1}{2}\ $

Find step response:

$ \begin{align} y[n] &= x[n] * h[n] = u[n] * h[n] = \sum_{k=-\infty}^{\infty} u[k]h[n-k] \\ &= \sum_{k=-\infty}^{\infty} u[k] \left ( u[n-k] - \frac {1}{2} \right ) = \sum_{k=-\infty}^{\infty} u[k]u[n-k] - \frac{1}{2} \sum_{k=-\infty}^{\infty} u[k] \\ &= \sum_{k=0}^{\infty} u[n-k] - \frac{1}{2} \sum_{k=0}^{\infty} 1 = -\infty \end{align} $
(this seems unreasonable!)

Note: The original eq. $ y[n] - y[n-1] = x[n]\ $ can be expressed as:

$ h[n] - h[n-1] = \delta [n]\ $

By observation, $ h[n] = u[n] + c\ $ for any constant c. And further: the step response is divergent for $ c \ne 0\ $.

So, what is c equal to?

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