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What is the difference between an impulse and a step response diagram?

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    $\begingroup$ If the impulse response is $h[n]$, $n \geq 0$, then the step response $s[m]$ at any time $m \geq 0$ is $$s[m] = h[0]+h[1]+h[2]+\cdots+h[m].$$ What other information do you want to get? The transfer function of the filter? That's easy: $$H(z) = h[0]+h[1]z^{-1} + h[2]z^{-2}+\cdots$$ $\endgroup$ Apr 16, 2013 at 20:26
  • $\begingroup$ So, because my filter was a second order filter then the step response is 2 times the impulse response! That makes also sense graphically. Thank you i understand it now. $\endgroup$
    – 20317
    Apr 16, 2013 at 20:37
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    $\begingroup$ Nope, the step response is the integral of the impulse response or the impulse response is the first derivative of the step response. That has nothing to do with filter order $\endgroup$
    – Hilmar
    Apr 16, 2013 at 20:40
  • $\begingroup$ So in the graphic representation above the fact that the step response is a "double" version is due to integration of the impulse response? $\endgroup$
    – 20317
    Apr 16, 2013 at 20:45
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    $\begingroup$ It looks "double" because every second sample of the impulse response is zero. So, when integrating it, every second sample is exactly zero added to the previous value, thus it is the same. $\endgroup$
    – lxop
    Apr 16, 2013 at 20:50

1 Answer 1

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If the impulse response is $h[n], n\geq 0$, then the step response $s[m]$ at any time $m≥0$ is $$s[m]=h[0]+h[1]+h[2]+\cdots +h[m].$$ Note that if the filter is a finite impulse response (FIR) filter so that $h[m] = 0$ for $m > M$, then the step response at $M$ is $$s[M] = h[0]+h[1]+h[2]+\cdots +h[M]$$ and remains at this value forever afterwards, that is, $s[m]=s[M]$ for all $m \geq M$.

What other information do you want to get? The transfer function of the filter? That's easy: $$H(z)=h[0]+h[1]z^{−1}+h[2]z^{−2}\cdots $$ which works out to be polynomial of degree $M$ in $z^{-1}$ for a FIR filter.

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