# Impulse response of an accumulator

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#### jagsee1972

##### Newbie level 5
Hello All,
I am a bit puzzled on what may be a simple question (only simple if you know).
To determine the impulse response for a system the input signal is replaced by the delta function.

e.g. the impulse response for the following function (system) is:

$y[n] = \sum_{k = -\infty}^\infty..x[k]$

$h[n] = \sum_{k = -\infty}^\infty\delta[k]$

The impulse response for the accumulator is the discrete step function u[n].

However I have seen this version of the accumulator which has its impulse response bieng u[n].

$y[n] = \sum_{k = -\infty}^\infty..x[n-k]$

$h[n] = \sum_{k = -\infty}^\infty\delta[n-k]$

I agree with the impulse function but can not see how it equals the discrete unit step function: u[n].

for example if n=-1, then this clearly should equal 0, however one of the deltas in the whole summation will equal 0 i.e. $\delta[-1-(-1)]$ this will give an amplitude of 1 when n=-1. Which can not be true if the function is equal to u[n].

I would really appreciate some advice on where I am going wrong.

Woops, Looks like a made a mistake. The lower limit has to be 0, which makes perfect sense.

Hi jagsee1972,

there is something missing in

$y[n] = \sum_{k = -\infty}^\infty..x[k]$

Note that the right-side member is independent of n (i.e. y would have the same value for all times) and in addition the sum can be not convergent.
Probably you intended to write

$y[n] = \sum_{k = -\infty}^{\infty}h[n-k]x[k]$