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

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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 $$s[n]=u[n]*h[n]=\sum_{k=-\infty}^\infty h[k]u[n-k]=\sum_{k=-\infty}^n h[k]$$

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

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

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