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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.

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  • $\begingroup$ 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
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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

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

and

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

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.

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