The question is: Are the following systems linear time invariant?

  1. $y(t)=x(t)$, with $x(t)=u(t)$ (unit step function)
  2. $y[n]=x[n]$, with $x[n]=\delta[n]$ (Dirac delta function)

The reason I am asking is because I don't see the relevance of the values given to the input functions. LTI is a property of the system not its inputs, so I don't understand what is meant!

  • Is it a restriction?
  • In that case how would I check for linearity? One probably can't, right?
  • $\begingroup$ Would Signal Processing be a better home for this question? $\endgroup$
    – Qmechanic
    Nov 9 '14 at 23:44
  • $\begingroup$ Studying how linear systems (time invariant or not) respond to step responses is an important concept. I suspect you are confused by the presentation or by the concept of step responses. $\endgroup$
    – David Hammen
    Nov 10 '14 at 5:36
  • $\begingroup$ @Qmechanic Thank you, I didn't know there was such a thing. $\endgroup$
    – Undertherainbow
    Nov 10 '14 at 13:10

You are absolutely right: the data you are given say nothing on their own about whether the system concerned is linear or time shift invariant.

To check that a system is linear, you need to know its behaviour for sums of inputs and multiples of inputs to check that the sum / multiples of the outputs in response to different inputs is the response when the sum of those separate inputs is the lone system input. You only have the behaviour for one input $x(t) = u(t)$. You need to be able to say that $x(t)\mapsto A\,x(t)\Rightarrow y(t)\mapsto A\,y(t)\,\forall\,A\in\mathbb{R}$ and also that if $x_1(t),\,x_2(t)$ lead to outputs $y_1(t),\,y_2(t)$, then input $x_1(t)+x_2(t)$ gives rise to output $y_1(t)+y_2(t)$ for any $x(t),\,x_1(t),\,x_2(t){}{}{}{}{}{}{}{}$ in a suitable space of functions.

To check time shift invariance, you need to know how the system behaves when we input $y(t-t_0)$ instead of $y(t)$ to check whether the output is the same, but shifted by the amount $t_0$. You only know the behaviour for one value of the delay $t_0$, that is $t_0=0$. You need to be able to say that input $x(t)$ gives rise to output $y(t)$ implies that $x(t-t_0)$ gives rise to output $Y(t-t_0)$ for any constant $t_0\in\mathbb{R}$.

So you can say nothing about the linearity or the time shift invariance of this system.

  • $\begingroup$ The nature of the input doesn't say anything about whether the system is LTI, but the y(t) = x(t) and y[n] = x[n] do. They are clearly LTI. $\endgroup$
    – Jim Clay
    Nov 10 '14 at 15:01
  • $\begingroup$ @JimClay Why are the systems clearly LTI? I think WetSavannaAnimal's answer is dead on target. $\endgroup$ Nov 10 '14 at 19:54
  • $\begingroup$ @DilipSarwate I guess it depends on what was meant by "y(t) = x(t)". I interpreted that as meaning that the system, by definition, outputs the input. This would make it LTI. If the statement just meant that we have a single observation, and in that singular case it happened to output the same thing as the input, then he/she is correct that we do not know. $\endgroup$
    – Jim Clay
    Nov 10 '14 at 20:11
  • $\begingroup$ @DilipSarwate It seems to me, though, that the "single observation" interpretation is not terribly useful because if we want to be pedantic about it we could never say that a system is LTI based on observations because there are an infinite number of possible inputs so we can't test them all. $\endgroup$
    – Jim Clay
    Nov 10 '14 at 20:12
  • 1
    $\begingroup$ @JimClay I'm guessing from the particular wording of this homework problem that it is designed to test a student's understanding of the concepts of linear and time shift invariant and to make sure that they understand, like the OP, that these are properties defined by a system's behaviour, not by lone inputs and outputs. $\endgroup$ Nov 10 '14 at 22:21

there is an SE specifically for signal processing. sorta surprized to see tags for this at the Physics SE.

what you are suspecting is precisely true. LTI is a property of the system, not of the input. both 1. and 2. are obviously linear and time-invariant.

one slight misconception. in DSP we usually use square brackets, as in "$x[n]$", to depict discrete-time signals (as an alternative to "$x_n$"). a discrete-time signal is just a sequence, but we want to leave the subscript for the same purpose we might with continuous-time signals. i.e. given a vector of continuous-time signals: $\{x_1(t), x_2(t), ... x_N(t) \}$, if we were to sample those signals, the corresponding discrete-time vector would be $\{x_1[n], x_2[n], ... x_N[n] \}$. the argument (inside of square brackets) to a discrete-time signal is always an integer.

that said, "$\delta[n]$" is the Kronecker delta, not the Dirac delta. (the latter the math guys like to tell us is not even a "function".)

$$ \delta[n] \ \triangleq \begin{cases} 1, & \text{if }n = 0 \\ 0, & \text{if }n \ne 0 \end{cases} $$

  • 1
    $\begingroup$ Why are both systems LTI? With the given information, with respect to System 1, all we know is that if the input is $u(t)$, the output is $u(t)$. It says nothing about what the output is when the input is something else, not even when the input is $2\cdot u(t)$. Similarly for System 2. $\endgroup$ Nov 10 '14 at 19:57
  • $\begingroup$ maybe i am interpreting this wrong (i gotta think about what "with" means), but for me the operative part is $$y(t) = x(t)$$. but i think you're right, Dilip, i didn't read this well. $\endgroup$ Nov 10 '14 at 21:16
  • $\begingroup$ See my comments addressed to Jim Clay on the currently accepted answer by WetSavannaAnimal, and if you feel like it, please upvote that answer. $\endgroup$ Nov 12 '14 at 12:45

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