jagsee1972
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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.
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.
