Impulse response of a LTI TD system

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:

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

• 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, $u(n)$ – Chu Apr 27 '17 at 13:01
• @Chu You should check your math here. – Enric Blanco Apr 27 '17 at 13:33
• @Enric Blanco, that won't be the first time; where have I gone wrong? – Chu Apr 27 '17 at 13:37
• @Chu Sorry, I thought I was asking another comment from you - which seems to have disappeared?? – Enric Blanco Apr 27 '17 at 13:46
• @Enric Blanco, Yes, I made a mistake in saying system 1 is not moving average; it clearly is. Sorry for confusion. – Chu Apr 27 '17 at 14:01

1 Answer

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.