(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> | + | : <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) \\ | + | : <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> | ||
| + | |||
| + | : <math> \Rightarrow h[n] = \mathfrak{F}^{-1}(H(\omega)) = u[n] - \frac{1}{2}\ </math> | ||
| + | |||
| + | 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] \\ | ||
| + | &= \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} </math> | ||
| + | : (this seems unreasonable!) | ||
| + | |||
| + | Note: The original eq. <math> y[n] - y[n-1] = x[n]\ </math> can be expressed as: | ||
| + | |||
| + | : <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>. | ||
| + | |||
| + | 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?
