Since i am not sure if i have understood IQ samples correctly, i want to ask the following question:

If i look at inphase and quadrature in an IQ diagramme, the length of the arrow describes the amplitude of the signal, while the angle between the arrow and the x-axis describes the phase shift. That however would mean, that these samples could still be transmitted on different frequencys, maintaining the same amplitude and phase shift.

Is this right or am i misunderstanding something?

  • 2
    $\begingroup$ If your IQ samples refer to the baseband complex signal, then yes, this is independent of the carrier frequency, because the up- and downconversion to passband cancel each other. $\endgroup$ Dec 14, 2017 at 14:49
  • $\begingroup$ That's why we call it equivalent baseband. It's really equivalent to the bandwidth around the carrier. But in baseband. No matter where the carrier originally was. $\endgroup$ Dec 14, 2017 at 19:12

1 Answer 1


I'm not sure I understand your question, but here are some thoughts.

Say you have a passband (that is, with a carrier) quadrature signal: $$s(t) = x(t)\cos(2\pi f_c t) + y(t)\sin(2\pi f_c t).$$ Here $x(t)$ and $y(t)$ have bandwidth $B<<f_c$. If you sample this signal, the specific samples you get will obviously depend on $f_c$.

One almost never does that; for many reasons, it's better to remove the carrier. So, you can obtain the low-pass equivalent (complex envelope) signal $$s_{BB}(t) = x(t) + j y(t),$$ and sample that instead. Then, the samples will not depend on $f_c$.

  • $\begingroup$ Thanks, i think i've understood this now. Just to clarify: if i would want to transmit the carrier-independent samples i would multiply the samples with the sine/cosine carrier signal and would combine those, like in the first equation in your answer, right? $\endgroup$
    – lalu
    Dec 15, 2017 at 9:26
  • $\begingroup$ That is correct, except that you need to convert samples (a discrete-time signal) to $x(t)$ and $y(t)$, which are continuous-time. The process to do that is called pulse-shaping. $\endgroup$
    – MBaz
    Dec 15, 2017 at 15:36

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