Given a system, it is possible to determine whether the system is LTI given the response to input to the system, other than the unit impulse?

Specifically, my input to the system is

$$x(t)=\begin{cases} 0 & t<0, t>0.5 \\ 1 & 0 \le t \le 0.5 \end{cases}$$

The output is given by

$$y(t)=\begin{cases} 0 & t<0, t>2 \\ t & 0 \le t \le 1 \\ 2-t & 1 \le t \le 2 \end{cases}$$

I do know that, if the system is LTI, then the differential of the step response would give the impulse response. However, I'm not exactly sure how to use this information.

While I'd appreciate a specific response to my question, I'd be more interested in a general answer that characterizes the nature of input for which whether or not the system is LTI can be determined, and if it can be determine, and algorithm to do so.

  • \$\begingroup\$ The output spec doesn't make any sense. Are those 'x's supposed to be 't's? \$\endgroup\$
    – Dave Tweed
    Mar 24, 2013 at 15:34
  • \$\begingroup\$ @DaveTweed - say I have a system that takes in the input $$x(t)=\begin{cases} 0 & t<0, t>1 \\ 1 & 0 \le t \le 0.5 \end{cases}$$ but produces the same output as above. In addition, I am given the information that the system is LTI. Would it then be possible to determine the impulse response of the system uniquely, or will there be a range of possible impulse responses based on the limited data we have? \$\endgroup\$ Mar 25, 2013 at 1:56
  • \$\begingroup\$ I assume that last 0.5 is supposed to be 1. And yes, since you now know the response to an input step function, you can derive the impulse response from that. \$\endgroup\$
    – Dave Tweed
    Mar 25, 2013 at 4:32
  • \$\begingroup\$ Yes, sorry for that mistake again. \$\endgroup\$ Mar 25, 2013 at 4:39

2 Answers 2


With just the kind of information you have, you can't prove the system is LTI. (As Dave points out in his answer, you can definitively prove this particular system is not LTI)

Specifically, for a system with input \$x(t)\$ and output \$y(t)\$, for the system to be called time-invariant, you need to know that if you time-shift the input by \$t'\$ so that you have \$x(t-t')\$, then the output is \$y(t-t')\$. In your case you only know the output for the case \$t' = 0\$, and you haven't been told that the input-output relationship will be maintained if the input signal is advanced or delayed in time.

Generally, time-invariance is something you have to be told about a system, or you have to assume it based on the physical nature of the system. Because if it's possible for the system to not be time-invariant, then any observations you make on it don't prove the system behavior won't change at some later time.


In general, given an input signal and an output signal, you might be able to definitively state that the system is not LTI, but you'll never be able to definitively state that it is, only that it might be.

The general technique is to use the information in the input signal to see whether the output signal can be composed from the input features. Specifically, a rectangular pulse is the superposition of two step functions, separated in time.

In your specific example, the response to the first step function at t=0 is that the output starts rising linearly, also at t=0. The second (negative) input step function occurs at t=0.5, and if the system is LTI, we would expect the output to also change what it's doing at that time. In fact, we would expect the output to level off at whatever value it had reached at that time, because the LTI response to the second step should be a negative-going linear ramp, which, when added to the original response, should cancel out.

However, this is not the output signal that we get, so this system is definitely not LTI.


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