I was posed the following homework problem:

2.10 The following input-output pairs have been observed during the operation of a time-invariant system: \begin{align} x_1(n)&=\{\underset{\uparrow}{1}, 0, 2\}\overset{\mathcal T}{\longleftrightarrow} y_1(n)=\{\underset{\uparrow}{0}, 1, 2\}\\ x_2(n)&=\{\underset{\uparrow}{0}, 0, 3\}\overset{\mathcal T}{\longleftrightarrow} y_2(n)=\{\underset{\uparrow}{0}, 1, 0, 2\}\\ x_3(n)&=\{\underset{\uparrow}{0}, 0, 0,1\}\overset{\mathcal T}{\longleftrightarrow} y_3(n)=\{1,\underset{\uparrow}{2}, 1\}\\ \end{align} Can you draw any conclusions regarding the linearity of the system ? What is the impulse response of the system ?

I thought that I had a decent grasp on linearity and time-invariance, but I am having trouble relating them to a specific problem. I know that linearity means that the system satisfies the superposition principle - and that the weighted sum of the input signals is equal to the weighted sum to each output signal.

How do I apply that to this problem? I have the solution below, but I am unsure how they get there:


The system is nonlinear. This is evident form observation of the pairs $$x_3(n)\leftrightarrow y_3 (n)\text{ and }x_2(n)\leftrightarrow y_2(n).$$ If the system were linear, $y_2(n)$ would be of the form $$y_2(n)=\{3, 6, 3\}$$ because the system is time-invariant. However, this is not the case.

  • $\begingroup$ What is the meaning of the vertical arrow pointing up? $\endgroup$
    – MBaz
    Commented Aug 30, 2016 at 0:09
  • $\begingroup$ It means that is the point where n = 0. Essentially giving you an idea of where the graph starts/ends on the discrete plot. $\endgroup$
    – Gary
    Commented Aug 30, 2016 at 0:45
  • $\begingroup$ which way does time go? is n=1 left or right from the vertical arrow? I gotta say, this is a quirky notation. $\endgroup$
    – Hilmar
    Commented Aug 30, 2016 at 17:24

2 Answers 2


If you assume that the system is time invariant, then a shift to the right of $x_2$, say $$x^s_2 = \{\underset{\uparrow}{0}, 0,0, 3\}$$ should give you the shifted version of $y_2(n)$, in other words: $$y^s_2(n)=\{\underset{\uparrow}{0}, 0, 1, 0, 2\}\,.$$ But $x^s_2 = 3 x_3$ (a linear factor), and $y^s_2$ is not a multiple of $y_3$. Linearity is not preserved.

One example of non-linearity suffices to claim the system cannot be both time-invariant and linear.


This looks like a very contrived question, but the idea is simple: if the system were linear, then, since $x_2[n]=3x_3[n+1]$, then $y_2[n]$ should be $3y_3[n+1]$. The impulse response can be obtained by time-shifting $x_3$ so that the $1$ is at $n=0$, and then time-shift $y_3$ the same amount.


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