3
$\begingroup$

If a system takes input as the time domain signal and outputs the frequency domain signal, is such a system an LTI system? For if the input time domain signal can be represented as a linear combination of sinusoids, then since the operation is transformation to the frequency domain, the outputs will also be the same Linear Combinations of the individual outputs (frequency domain representation of the individual sinusoids).. right? so we can say it's Linear? Since the system performs the same operation independent of time, it would be Time-Invariant too right?

$\endgroup$
  • 2
    $\begingroup$ the concept of time-invariance is a bit funky when the input is a function of time (like $x(t)$) but the output is a function of frequency (like $X(f)$). offsetting the input by a known time offset does not offset the output (a function of $f$) by the same time offset. but the Fourier Transform is most certainly a linear operation. it's the convolution operation (that has a function of time, $x(t)$, going in and another function of time, $y(t)$, coming out) that is LTI. $\endgroup$ – robert bristow-johnson Sep 30 '17 at 20:16
  • $\begingroup$ Makes sense that the output is in a different domain and how we cant quite speak of time-invariance. So could it imply that such a system cannot be represented by an impulse response? $\endgroup$ – Disciple Sep 30 '17 at 20:33
  • $\begingroup$ Abstractly, the Fourier transform operation takes a function as an input and produces another function as its output. "Time-invariance" should really be interpreted as "shift-invariance". Then the issue of "different domains" doesn't arise. There is no time. No frequency. Just a function, transformed into another function. $\endgroup$ – Atul Ingle Sep 30 '17 at 21:11
8
$\begingroup$

The Fourier transform operator $\mathscr{F}$ is a linear one; i.e., $$\mathscr{F}\{x(t)\}=X(f) ~,~ \mathscr{F}\{y(t)\}=Y(f) \implies \mathscr{F}\{\alpha x(t) + \beta y(t) \} = \alpha X(f) + \beta Y(f)$$ And therefore the system that implements it will be linear.

However time invariance, which is tested by the method

$$\mathcal{T}\{x(t)\}=y(t) \implies \mathcal{T}\{x(t-d)\}=y(t-d)$$

does not apply to the Fourier Transform operator, because the input and output functions do not belong to the same domain (namely time here)

However, the physical system which produces Fourier transform of the given signal $x(t)$ can be associated with a time-dependent Fourier transform which computes the function

$$X(\omega,t)= \mathcal{TF} \{ x(t) \} = \int_{-\infty}^{\infty} x(t+\tau) w(\tau) e^{-j \omega \tau} d\tau $$

Then it can be shown that this operator is time invariant in the sense that,

$$ \mathcal{TF} \{ x(t) \} = X(\omega,t) \implies \mathcal{TF} \{ x(t-d) \} = X(\omega,t-d)$$

Lets elaborate on this. Now lets define STFT as $$X(\omega,t) = STFT \{ x(t) \} = \int_{-\infty}^{\infty} x(t+\tau) w(\tau) e^{- \omega \tau} d\tau $$ It's easy to show that $$X_2(\omega,t) = STFT \{ x(t-d) \} = \int_{-\infty}^{\infty} x(t-d+\tau) w(\tau) e^{- \omega \tau} d\tau = X(\omega,t-d) $$ Hence it's a shif-invariant operator.

EDIT: Below shown is a trivial redefinition of the above STFT, which does not confirm to the standard notation, hence may not apply. I don't delete it for now however. Shift-invariance is shown for the above definition only.

Now a similar form discussed in the provided link is obtained by a change of variable $t+\tau = \tau'$ on the STFT definition integral:

$$ X(\omega,t) = \int_{-\infty}^{\infty} x(\tau') w(\tau'-t) e^{-j \omega (\tau'-t)} d \tau' $$ and replacing $\tau'$ with $\tau$ yields the equivalent form of STFT as $$X(\omega,t) = STFT \{x(t)\}= e^{j \omega t} \int_{-\infty}^{\infty} x(\tau) w(\tau-t) e^{-j \omega \tau'} d \tau $$

Now lets show its time invariance by the input $x_2(t) = x(t-d)$ : $$ X_2(\omega,t) = STFT \{ x_2(t) \} = STFT \{ x(t-d) \} = e^{j \omega t} \int x(\tau-d) w(\tau-t) e^{-j \omega \tau} d\tau $$ Make a change of variabes inside the integral operator as $\tau-d = \tau'$ yields

$$ X_2(\omega,t) = STFT \{ x(t-d) \} = e^{j \omega t} \int x(\tau') w(\tau'-(t-d)) e^{-j \omega (\tau'+d)} d\tau' $$ which simplifies by replacing $\tau'$ with $\tau$ and extracting $e^{-j\omega d}$ outside of the integral

$$ X_2(\omega,t) = STFT \{ x(t-d) \} = e^{j \omega t} e^{-j \omega d} \int x(\tau) w(\tau-(t-d)) e^{-j \omega \tau} d\tau $$

$$ X_2(\omega,t) = STFT \{ x(t-d) \} = e^{j \omega (t-d)} \int x(\tau) w(\tau-(t-d)) e^{-j \omega \tau} d\tau = X(\omega,t-d)$$

Which proves that this form is also shift-invariant on the $t$ variable.

$\endgroup$
  • $\begingroup$ i'm gonna poke at this a little Fat. $\endgroup$ – robert bristow-johnson Sep 30 '17 at 20:20
  • $\begingroup$ @robertbristow-johnson the time-invariance part ? $\endgroup$ – Fat32 Sep 30 '17 at 20:22
  • $\begingroup$ yeah. for my own benefit i changed $\omega$ to $f$. but i don't think the last equation is exactly correct. and your STFT is not quite correct either. $\endgroup$ – robert bristow-johnson Sep 30 '17 at 20:28
  • $\begingroup$ @robertbristow-johnson its ok with $f$ too. Also my STFT is derived from the discrete STFT version and may not exactly associate with the standard form. I didin't mean it either. I have just provided a model for the system so I belive the last statement is correct. But its association with the real system and/or the stadard STFT may be questioned. $\endgroup$ – Fat32 Sep 30 '17 at 20:31
  • $\begingroup$ @robertbristow-johnson your last edit makes the notation not the one I intend. I will take it back to mine ok ? $\endgroup$ – Fat32 Sep 30 '17 at 20:33
2
$\begingroup$

A Fourier Transform of a finite width window, whose position is related to a point in time, is an LTI system with respect to the window position (which is in the time domain).

$\endgroup$
  • $\begingroup$ hot, can you explain (maybe with an equation or two) how the Fourier Transform is time-invariant? $\endgroup$ – robert bristow-johnson Sep 30 '17 at 23:00

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.