0
$\begingroup$

I have an algorithm which takes an input signal, $x[t]$, and (among other things) complex basebands the data and outputs the phase of the complex numbers.

I'm new to the field and the Matlab function which produces the basebanded signal is a kind of black box to me. I've tried reading the code, but it's pretty impenetrable and not commented. I also tried to input different shapes of signals, but couldn't learn much from it. My questions are:

  1. What exactly is the complex baseband of a signal? I sort of understand what "baseband" means, but where does the complex phase come from?

  2. Related, does the phase have any relation to the signal? If I input a sinusoid, what should I expect the phase output to be?

$\endgroup$
6
  • $\begingroup$ your title asks for intuition, but your first question asks for "exactly"; intuitively, that is just the RF signal mixed to 0Hz. Don't read the matlab functions, read about complex baseband! $\endgroup$ – Marcus Müller Jan 11 at 14:04
  • $\begingroup$ The phase isn't complex, the value is, and a complex value can have an angle, ="a phase". Re 2.: The phase is as much important to the signal as its amplitude. What you get out if you put in a sinusoid can only be said after you defined the frequency of both that sinusoid and your mixing oscillator and the phase of that oscillator relative to the sinusoid. $\endgroup$ – Marcus Müller Jan 11 at 14:05
  • $\begingroup$ I see. So when I input a signal, each data point corresponds to an $Ae^{i\phi}$, for an amplitude $A$ and phase $\phi$. I guess my question boils down to "how does the algorithm figure out $A$ and $\phi$ from just a number input"? $\endgroup$ – Dan Pollard Jan 11 at 14:16
  • $\begingroup$ that's given by the definition of complex baseband: you take your passband signal, remove the negative-frequency component and multiply it with $e^{i f_{\text{carrier}} t}$. $\endgroup$ – Marcus Müller Jan 11 at 14:19
  • $\begingroup$ It literally just shifts down what is around $f_\text{carrier}$ to 0 Hz, and removes what used to be at $-f_\text{carrier}$ (and would have ended up at $-2f_\text{carrier}$. $\endgroup$ – Marcus Müller Jan 11 at 14:20
0
$\begingroup$

What exactly is the complex baseband of a signal? I sort of understand what "baseband" means, but where does the complex phase come from?

A baseband signal is one where the center frequency is $\approx 0$ Hertz. There may be some small offset, but the idea is that it is no where near a passband signal whose center frequency is in the millions or billions of Hertz.

For a real signal, the frequency spectrum is symmetric. Due to the up-conversion process (converting baseband to passband so that the center frequency is no longer $0$ Hertz), we can put data on two orthogonal carriers, $\text{cos}$ and $\text{sin}$, in the passband signal:

\begin{align} x(t)&=\text{Re}\bigg(s(t)e^{j2\pi f_ct} \bigg) \\ &= x_I(t)\text{cos}\big(2\pi f_ct\big) - x_Q(t)\text{sin}\big(2\pi f_ct \big) \end{align}

where $s(t)=x_I(t) + jx_Q(t)$, is the complex baseband signal. For a nice explanation of what is probably happening in your MATLAB function on how to convert passband to baseband, check out this previous answer: https://dsp.stackexchange.com/a/43281/31316.

Related, does the phase have any relation to the signal? If I input a sinusoid, what should I expect the phase output to be?

There is a relation...it is the phase, that is all. The phase is helpful though and can help to pack more bits of information into a single symbol.

$\endgroup$
1
  • $\begingroup$ Thank you very much, this was the clearest explanation I've seen. It's starting to become a bit clearer now! $\endgroup$ – Dan Pollard Jan 11 at 15:12

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.