Why the impulse response of this system:




and the impulse response of this one:



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

  • \$\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
  • 1
    \$\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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.