# Intuitive meaning of complex time domain signal representation (quadrature sampling)

My question is about I Q samples as recorded from an SDR receiver (e.g. RTL-SDR).

My understanding is that we can call these I Q samples a complex time domain signal and these can have both non-zero real and imaginary components (the question I linked above is talking about zero imaginary part in the time domain).

Is there an intuitive explanation of what is the meaning of this type of complex time domain signal?

I can think of taking a simple case of an input sine wave of a frequency below the sampling frequency which would make the IQ signal's phasor rotate with the frequency equal to the difference between the input sine wave's frequency and the sampling frequency thus allowing to unambiguously reconstruct frequencies up to (but not including) the sampling frequency from the IQ samples.

But I'm struggling to extend this special case to an input composed of several sinusoids of different frequencies.

• Are you familiar with Euler’s identities that express sines and cosines in terms of exponentials such as $e^{j\omega t}$ and do you have an intuitive understanding of that exponential expression, and what “positive” and “negative” frequencies are? If not this was all answered in other posts here we can link you to as I think starting with that will really help you and is already answered. Dec 25, 2019 at 23:26
• @Dan Boschen, I'm familiar with the identities, but don't have an intuitive understanding of what a negative frequency would mean in the time domain. The phasor would rotate in the opposite direction I suppose, but I don't know what it means in the time domain.
– axk
Dec 26, 2019 at 8:26
• The "rotating phasor" is in the time domain (the constant magnitude phasor rotating at a constant rate is just a single impulse in the frequency domain). Look at this answer by @endolith where he shows the spinning phasor, and the sine and cosine components: dsp.stackexchange.com/questions/431/…. So a positive frequency is cosine + j sine and a negative frequency is cosine - j sine Dec 26, 2019 at 12:55
• And here are other related posts that may further help: dsp.stackexchange.com/questions/52826/… and dsp.stackexchange.com/questions/31355/… Dec 26, 2019 at 13:02
• Yes that makes sense to say. Also note that we can represent that signal at baseband (carrier =0) as long as we use the complex signal representation meaning we need I and Q (or magnitude and phase) for each sample to represent it. For linear processes it doesn’t matter then what carrier the signal is at. Multiplying by $e^{j\omega_c t}$ just shifts it to a new frequency. Did any other the answers below meet your question? If so best to select the one that did to close this out. Dec 26, 2019 at 18:42

$$r(t)=I(t)\cos(2\pi f_ct+\phi)-Q(t)\sin(2\pi f_ct+\phi)\tag{1}$$
where $$I(t)$$ and $$Q(t)$$ are the in-phase and quadrature components, respectively. After demodulation you're left with the two signals $$I(t)$$ and $$Q(t)$$, and if you like you can combine them into a complex-valued baseband signal $$x(t)=I(t)+jQ(t)$$.