0
$\begingroup$

Trying to wrap my mind around the concepts of this one...

Consider the following impulse response $h[n]$ for a linear, time-invariant system: $$ h[n] =\left\{\underline{1} , -2, 2, -1\right\} $$ where n=0 starts at 1 for h[n]

  • (a) Is the system causal? Why or why not?
  • (b) Is the system stable? Why or why not?
  • (c) Is the system FIR or IIR? Give a reason for your answer.
  • (d) Does the system have memory? why or why not?

Based on my understanding I've concluded...

  • (a): ?
  • (b): Yes the system is stable because the summation of |h[n]| is less than infinity.
  • (c): The system is an FIR (Finite Impulse System) because the duration of h[x] is not infinite. It has a finite number of values.
  • (d): This system is not memoryless, therefore has memory. This is because in order to be memoryless (to my understanding) an impulse response should not have values at any times other than n=0. h[n] has multiple values

Updated the question to reflect how the values relate to time. The underlined 1 in h[n] denotes the origin n=0

Improve this question
$\endgroup$
4
  • $\begingroup$ This is homework-style kind of question, the question won't be well-received unless you provide what you have tried or tell the community where you are stuck. $\endgroup$
    – Gilles
    Commented Feb 13, 2018 at 6:36
  • $\begingroup$ @OlliNiemitalo h(n) is the system's response to the input. where n is a unit in time. h is the impulse response of the system $\endgroup$
    – crazyCoder
    Commented Feb 13, 2018 at 8:05
  • $\begingroup$ All your conclusions are correct, but (a) can't be answered, as mentioned by Olli, because we don't know where in time to place those values given for $h[n]$. $\endgroup$
    – Matt L.
    Commented Feb 13, 2018 at 8:44
  • $\begingroup$ @OlliNiemitalo yes it is. I updated the question to reflect that $\endgroup$
    – crazyCoder
    Commented Feb 13, 2018 at 17:13

1 Answer 1

Reset to default
3
$\begingroup$

As soon as you've figured out how the given values for $h[n]$ relate to the time index $n$ (which is unclear in the current problem formulation), it should be easy to decide whether the given system is causal or not by considering the input-output relation of a discrete-time LTI system:

$$y[n]=\sum_{k=-\infty}^{\infty}h[k]x[n-k]\tag{1}$$

where $x[n]$ is the input sequence, and $y[n]$ is the output sequence.

For the system to be causal, the output at time $n$ must only depend on the current value or on past values of the input $x[n]$. What does that imply for the impulse response $h[n]$?

Improve this answer
$\endgroup$
4
  • $\begingroup$ Updated the question to reflect how the values relate to time. The underlined 1 in h[n] denotes the origin n=0 $\endgroup$
    – crazyCoder
    Commented Feb 13, 2018 at 16:52
  • $\begingroup$ @crazyCoder: OK, so now you should be able to answer (a), right? $\endgroup$
    – Matt L.
    Commented Feb 13, 2018 at 18:31
  • $\begingroup$ No, I think I'm missing something conceptually. But I'm thinking that the system IS causal because for h[0] = 1, h[1] = -2, etc there is never an instance where h[n]=value where n < value. so for every instance n only depends on current and past values? $\endgroup$
    – crazyCoder
    Commented Feb 13, 2018 at 21:59
  • 1
    $\begingroup$ @crazyCoder: If you look at Eq. (1) in my answer you see that only the current and past values of $x[n]$ are used to compute $y[n]$ if and only if $h[n]=0$ for $n<0$. So that's simply the condition for causality. $\endgroup$
    – Matt L.
    Commented Feb 13, 2018 at 22:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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