E.g. when having 2 signals $x(t)$ and $y(t)$ and transforming each one with adding their spectra, the operation is linear, as the result would be the same as the transformation of $(x + y)(t)$.

Even when looking at the sampling series, each of the infinite elements adds up linearly.

But how is aliasing a non-linear operation and how is it proven mathematically?


2 Answers 2


I think you mean "images", not "aliases". They become aliases if there is foldover from resampling.

It's because you are not adding two signals, $x(t)$ and $\operatorname{III}(t)$, you are multiplying them that these images appear.

$$\begin{align} x_\text{s}(t) & \triangleq x(t) \cdot \operatorname{III}(t/T) \\ &= x(t) \cdot \sum\limits_{n=-\infty}^{+\infty} T \delta(t-nT) \\ &= T \sum\limits_{n=-\infty}^{+\infty} x(t) \cdot \delta(t-nT) \\ &= T \sum\limits_{n=-\infty}^{+\infty} x(nT) \cdot \delta(t-nT) \\ &= T \sum\limits_{n=-\infty}^{+\infty} x[n] \cdot \delta(t-nT) \\ \end{align}$$

where $x[n] \triangleq x(nT)$ and

$$ \operatorname{III}(u) \triangleq \sum\limits_{n=-\infty}^{+\infty} \delta(u-n) $$

$$\begin{align} \operatorname{III}(t/T) &= \sum\limits_{n=-\infty}^{+\infty} \delta\left(\tfrac{t}{T}-n\right) \\ &= \sum\limits_{n=-\infty}^{+\infty} \delta\left(\tfrac{t-nT}{T}\right) \\ &= \sum\limits_{n=-\infty}^{+\infty} T \delta(t-nT) \\ &= \sum\limits_{k=-\infty}^{+\infty} e^{j k \frac{2 \pi}{T} t} \\ \end{align}$$

The last line is from doing fourier series. Now, if you use the shifting property of the Fourier Transform, then the Fourier Transform of $x_\text{s}(t)$ is

$$\begin{align} X_\text{s}(f) & \triangleq \mathscr{F}\{x_\text{s}(t)\} \\ &= \mathscr{F}\{x(t) \operatorname{III}(t/T) \} \\ &= \mathscr{F}\left\{x(t) \sum\limits_{k=-\infty}^{+\infty} e^{j k \frac{2 \pi}{T} t} \right\} \\ &= \mathscr{F}\left\{\sum\limits_{k=-\infty}^{+\infty} x(t) \, e^{j k \frac{2 \pi}{T} t} \right\} \\ &= \sum\limits_{k=-\infty}^{+\infty} \mathscr{F}\left\{ x(t) \, e^{j k \frac{2 \pi}{T} t} \right\} \\ &= \sum\limits_{k=-\infty}^{+\infty} X\left(f-\tfrac{k}{T}\right) \\ \end{align}$$

where $ X(f) \triangleq \mathscr{F}\{ x(t) \} $ .

That non-linear process of multiplication generates frequency components than did not previously exist in $x(t)$. Those new components are simply shifted versions of $X(f)$ and are called "images".

  • 1
    $\begingroup$ From the definition of linearity of a system, the process of aliasing under sampling operator seems to be a linear one... I suppose the term aliasing distortion somewhat creates the misleading understanding that the process is nonlinear. Sampling is a linear (but time varying) operation and aliasing is just a specific case of sampling in which there is spectral overlap. But the time domain exposition is obviously linear: $ y(t) = T\{x(t)\} = (\sum {\delta(t-kT)})x(t) $ OR I'm missing something here. Btw: Aliasing cannot be inverted but should all linear transform be (I'm not sure)? $\endgroup$
    – Fat32
    Apr 27, 2017 at 12:15


  • Can an LTI system generate components in some frequency $\omega_0$ if the input signal $x(n)$ was such that $X(e^{j\omega_0})=0$?
  • Does aliasing do such thing?

The answers to these questions are straightforward and, combined, they answer the original question.

  • $\begingroup$ I know that sampling can be depicted as a succession of linear building blocks, but I fail to remember the source and page. That must be the key, right? $\endgroup$
    – Starhowl
    Apr 26, 2017 at 18:12
  • 3
    $\begingroup$ @Tendero: it is perfectly possible for a linear system to generate new frequencies, if it is time-variant. It is only an LTI system that can't. $\endgroup$
    – MBaz
    Apr 26, 2017 at 18:43
  • $\begingroup$ @MBaz Thanks for pointing that out - I've edited the answer. $\endgroup$
    – Tendero
    Apr 28, 2017 at 2:52

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