How does complex data of an FFT and IQ modulation differ?

If I have an arbitrary time domain signal (e.g. a simple cosine) i can sample it with an ADC and get a time discrete representation. I can perform an FFT on the samples and get a complex output. The phase of the signal is:

$$\phi=\arctan\frac\Im\Re$$

Using an IQ down modulation of the same signal gives complex data, too, with the phase being:

$$\phi=\arctan\frac QI$$

Is there a difference between these two and when is one preferred over the other?

EDIT

Thanks for the input!

@Dilip Sarwate: Here is what I did: set $\phi=60°$ (see your example)

The results are exactly what I expected. I can see the frequency of 1 Hz in the spectrum (plot 3) and the initial phase of $60°$ in the phase spectrum (plot 4). I get the same result when I use a non-complex function, e.g. $x(t)=cos(2\pi t+\phi)$.

I read everywhere that phase information is lost if I don't use IQ down-modulation, but that does not seem to be the case, as I can see the phase information in the plots.

Or am I missing something here?

• Please edit your question to add new information. What you added is NOT an answer. – Peter K. Sep 15 '17 at 10:55

Since you are interested in cosines, take, for example, the signal $x(t) = \exp(j(2\pi t + \theta))$ which is a complex sinusoid of period $1$ and sample it $16$ times per second to get $16$ samples $x[n]$, $n = 0, 1, \dots, 15$, where $x[n] = x\left(\frac{n}{16}\right) = \exp\left(j\left(\pi \frac{n}{8} + \theta\right)\right), n=0,1,\ldots, 15.$ Then, take the 16-point FFT of $\mathbf{x} =\big(x, x, \ldots, x\big)$ which will give you $\mathbf{X} =\big(X, X, \ldots, X\big)$. Report back to us what $\mathbf{x}$ and $\mathbf X$ are and whether you notice any differences between them.