A system that perfoms Fourier Transform operation - is it an LTI system?

Question

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?

Answer 1

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.

Answer 2

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).