Latest revision as of 08:35, 9 March 2011

Issue with LTI system defined by Difference Equation

I do not understand this difference equation, so could someone please explain it...

The Fourier Transform gives one answer, but it seems as though there are really more answers. One in paticular, c=0, appears to give better results for the step response...please see attached for explanation of what I am talking about...

$ 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?

once again --mireille.boutin.1, Fri, 26 Oct 2007 14:30:38

Dividing by zero is not recommended.

Please clarify --ross.a.howard.1, Fri, 26 Oct 2007 18:16:00

Can you please clarify where I divided by zero? I still do not know what to do with the arbitrary constant that comes up in the solution...

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