# Does a phase rotation introduce a delay?

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 $$t$$?

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.

• Do you have a source you can recommend which goes over this material? – Isaac Gerg Apr 17 at 21:40
• Optimum Array Processing Vol 4 by Harry Van Trees covers it. Most books on beam forming will have something on narrow band processing. This is not an esoteric subject – Stanley Pawlukiewicz Apr 17 at 21:45

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:

$$Y_1(\omega) = e^{ \mp j \omega d } X_1(\omega)$$

• I can measure phase rotation in time domain. Think of a QPSK modulated signal which I rotate slightly by pi/eps. I can measure this rotation by looking at the constillation in time domian. – Isaac Gerg Apr 17 at 21:36
• so you don't have a misunderstanding of phase and delay then... :-) sorry. – Fat32 Apr 17 at 22:03