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?


1 Answer 1


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

$$s (n) = \begin{cases} 3 - 2 \left( \frac 12 \right)^n & \text{if } n \geq 0\\ 0 & \text{if } n < 0\end{cases}$$


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