$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?

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.