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Why the impulse response of this system:

\$y(n)=\sum_{|k|\leq3}^{}x(n-k)\$

is:

\$h(n)=\sum_{|k|\leq3}^{}\delta(n-k)\$

and the impulse response of this one:

\$y(n)=\sum_{k=0}^{+\infty}x(n-k)\$

is:

\$h(n)=u(n)\$ ?

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  • \$\begingroup\$ For a unit impulse input, \$y(n)=1\$ only when a value of \$k\$ exists such that \$(n-k)=0\$. For the first system, this is only possible for \$n\$ values:\$ [-3,\:-2,\:-1\:, 0,\: 1\:, \:2,\: 3]\$, since \$k\$ is limited to the range \$-3\le k\le3\$. For the second system \$k\$ is limited to the range \$0\le k\le \infty\$, so \$y(n)=1\$ for the entire positive (and zero) range of \$n\$ values; this is defined as the unit step, \$u(n)\$ \$\endgroup\$ – Chu Apr 27 '17 at 13:01
  • \$\begingroup\$ @Chu You should check your math here. \$\endgroup\$ – Enric Blanco Apr 27 '17 at 13:33
  • \$\begingroup\$ @Enric Blanco, that won't be the first time; where have I gone wrong? \$\endgroup\$ – Chu Apr 27 '17 at 13:37
  • \$\begingroup\$ @Chu Sorry, I thought I was asking another comment from you - which seems to have disappeared?? \$\endgroup\$ – Enric Blanco Apr 27 '17 at 13:46
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    \$\begingroup\$ @Enric Blanco, Yes, I made a mistake in saying system 1 is not moving average; it clearly is. Sorry for confusion. \$\endgroup\$ – Chu Apr 27 '17 at 14:01
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First system is a kind of moving average filter (3 samples before + current sample + 3 samples after) with finite memory (7 samples). The impulse response reflects this - it's also finite in time.

Second system is an integrator - the impulse response is a step. It has infinite memory (note that the sum goes from \$k=0\$ to \$+\infty\$), that's why the output is latched to 1 well past the time when you excited it with an unity impulse - an infinite time response.

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