# How can i derive step response in terms of impulse response from the convolution sum?

If it is discrete LTI system, How can i derive step response in terms of impulse response from the convolution sum?

In an LTI system, any linear operation on inputs, is directly imposed on the outputs, that is, if an LTI system responses to $$\delta[n]$$ as $$h[n]$$, then it responses to $$u[n]=\sum_{k=-\infty}^{n}\delta[k]$$ as $$s[n]=\sum_{k=-\infty}^{n}h[k]$$.

This is also clear from the convolution since $$s[n]=u[n]*h[n]=\sum_{k=-\infty}^\infty h[k]u[n-k]=\sum_{k=-\infty}^n h[k]$$

Given a LTI System, $$h[n] = F(\delta[n])$$

Since $$u[n]= \sum_{k=0}^{ \infty} \delta [n-k],$$

Step response:

$$y[n]=F(u[n])=F(\sum_{k=0}^{ \infty} \delta [n-k])$$

Then, according to Linearity: $$f(a \times x + b) = a \times f(x) + f(b)$$ $$y[n] =\sum_{k=0}^{ \infty} F(\delta [n-k]) = \sum_{k=0}^{ \infty} h[n-k].$$

• Rewriting the final summation as in Mostafa Ayaz’s answer results in a more informative sum than what you have written above. Sep 24 '20 at 15:34