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I am given the impulse response
$$h[n]=2\left (\frac{1}{2} \right )^n\sigma [n]-\delta [n]$$
where $\sigma$ is the unit step function and $\delta$ is the unit impulse function. I know that
$$a[n]=\sum_{k=-\infty}^{\infty}h[k]\sigma [n-k]=\sum_{k=-\infty}^{n}h[k]$$
Thus, my step response should be
$$a[n]=\sum_{k=-\infty}^{n} \left( 2 \left (\frac{1}{2} \right )^k \sigma [k]-\delta [k] \right)$$
How do I calculate this?
Taking the $\mathcal Z$ transform of the impulse response,
$$H (z) = \dfrac{2}{1 - \frac 12 z^{-1}} - 1$$
Hence, the $\mathcal Z$ transform of the step response is
$$S (z) := \dfrac{H (z)}{1 - z^{-1}} = \cdots = \dfrac{3}{1 - z^{-1}} - \dfrac{2}{1 - \frac 12 z^{-1}}$$
Taking the inverse $\mathcal Z$ transform, we obtain the step response
$$s (n) = \begin{cases} 3 - 2 \left( \frac 12 \right)^n & \text{if } n \geq 0\\ 0 & \text{if } n < 0\end{cases}$$
