If we consider the mapping $\mathcal{H} : x[n]\mapsto y[n]$ and define the following output signal $y_{1}[n]:=\mathcal{H}\{x[n]\}:=x^{2}[n]$, then one can easily verify that such system is non-linear for it rejects the superposition property. But I am having a hard time applying the superposition property to $y_{2}[n]:=\mathcal{H}\{x[n]\}:=x[n^{2}]$ because its confuses with me with $y_{1}[n]$. I hope someone can offer any assitance I would be very much grateful.

  • $\begingroup$ Showing whether $\mathcal{H}\{x[n^2]\}$ is linear (or not) follows the exact same steps of showing that $\mathcal{H}\{x^2[n]\}$ is non-linear...? On which step did you have the trouble? $\endgroup$
    – Fat32
    Feb 19 '21 at 20:45
  • $\begingroup$ The first step, how do I substitute $a_{1}x_{1}+a_{2}x_{2}$ in place of $x[n^{2}]$ because its supposedly not same as replacing it in place of $x^{2}[n]$. @Fat32 $\endgroup$ Feb 19 '21 at 22:06
  • $\begingroup$ but it's not the first step. You shall first define y1 and y2 from x1 and x2. $\endgroup$
    – Fat32
    Feb 19 '21 at 22:15
  • $\begingroup$ Indeed, we define $x_{1}[n]\mapsto y_{1}[n]$ and $x_{2}[n]\mapsto y_{2}[n]$ by the means of the system $\mathcal{H}$ for we shall prove that $\mathcal{H}\{a_{1}x_{1}+a_{2}x_{2}\}=a_{1}\mathcal{H}\{x_{1}\}+a_{2}\mathcal{H}\{x_{2}\}$. @Fat32 $\endgroup$ Feb 19 '21 at 22:19

Given the system I/O definition:

$$y[n] = \mathcal{H}\{x[n]\} = x[n^2] \tag{1} $$

you can easily show that it's a linear (but time-varying) system.

Following the standard procedure

let $$y_1[n] = \mathcal{H}\{x_1[n]\} = x_1[n^2] \tag{2.1}$$ and $$y_2[n] = \mathcal{H}\{x_2[n]\} = x_2[n^2] \tag{2.2}$$

then define

$$x_3[n] = a x_1[n] + b x_2[n] \tag{3}$$


$$ \begin{align} y_3[n] &= \mathcal{H}\{x_3[n]\} \tag{4}\\\\ &= x_3[n^2] \tag{5}\\\\ &= a x_1[n^2] + b x_2[n^2]\tag{6}\\\\ &= a y_1[n] + b y_2[n]\tag{7}\\ \end{align} $$

where Eqs(4)-(5) follow Eq.(1), Eq.(6) follows Eq.(3), and Eq.(7) follows Eq.(2).

Equation (7) indicates that the system is linear.


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.