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 Aug 30 '16 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 Aug 30 '16 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 Aug 30 '16 at 17:24

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.


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.