Suppose I have 2 complex time series, $x_1$ and $x_2$.
If I measure the delay between $x_1$ and $x_2$, I get $t$.
If I make $x_2 = x_2 e^{-j 2 \pi 10}$, and measure the delay between $x_1$ and $x_2$, do I still get $t$?
Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. It only takes a minute to sign up.
Sign up to join this communitySuppose I have 2 complex time series, $x_1$ and $x_2$.
If I measure the delay between $x_1$ and $x_2$, I get $t$.
If I make $x_2 = x_2 e^{-j 2 \pi 10}$, and measure the delay between $x_1$ and $x_2$, do I still get $t$?
if your signal is a tone, phase rotation and time delay do the same thing.
In general, a rotation of a complex signal isn’t equivalent to a time delay.
A good portion of the beamforming literature uses the “narrow band” assumption and if the travel time of a wavefront over an array is small compared the inverse bandwidth of the envelope of the tone, beam steering (delay) is well approximated by phase multiplication.
I would assume a discrete-time context (though your notation suggests continuous-time).
Then you have a misunderstanding of the relation between the "delay" and "phase". Delay is measured in time-domain whereas the phase rotation is defined in frequency-domain.
So if there was a delay $\pm d$ between $x_1[n]$ and $x_2[n]$, then $y_1[n]$ (defined as the inverse DTFT of $Y_1(\omega)$ defined below) will have zero delay wrt $x_2[n]$ in time:
$$ Y_1(\omega) = e^{ \mp j \omega d } X_1(\omega)$$