# Are IQ-samples independent from the carrier frequency?

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?

• 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. – Maximilian Matthé Dec 14 '17 at 14:49
• 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. – Marcus Müller Dec 14 '17 at 19:12

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 $f_c$.
• 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. – MBaz Dec 15 '17 at 15:36