# I and Q Channels

My understanding of I and Q channels is as follows (please correct me if I am wrong):

• I = In-phase, or real component
• Q = Quadrature (90° shift of real component)

Where do these two channels come from in the first place? Is one the electric field and the other the magnetic field of a EM wave? I was under the impression that these channels are only present in digital waveforms; if this is true, and if so, why?

How can this be used to find the vector of on incoming signal, and would the signal modulation make a difference (assuming you can invoke the proper filtering necessary)?

• I am not at all very knowledgeable of RF and signal processing concepts, so please forgive me if my questions, tags, etc. are not accurate. – Biff Jan 23 '13 at 23:56
• You may want to check out this site. – Aaron D. Marasco Jan 24 '13 at 2:26
• I just wanted to comment, but don't have that privilege yet. The top answer here gives a simple and clear explanation of how to obtain complex samples from real samples. Getting real samples from a real signal (analog) is fairly straightforward as long as Nyquist is heeded. – Dave C Jan 24 '13 at 21:20

The two channels exist only inside a transmitter or a receiver; the channels are physically combined in a single signal (or channel) in the physical medium (wire, coax cable, free space, etc). At the transmitter, two signals $s_I(t)$ and $s_Q(t)$ (called the I (or inphase) signal and Q (or quadrature) signal respectively) are combined into a single signal $s(t)$ that is transmitted over the physical medium in a frequency band centered at $\omega_c$ radians per second. Note that $$s(t) = s_I(t)\cos(\omega_c t) - s_Q(t)\sin(\omega_c t)$$ The receiver separates out the two signals $s_I(t)$ and $s_Q(t)$ from this by multiplying $s(t)$ by $2\cos(\omega_c t)$ and $-2\sin(\omega_c t)$ respectively, and low-pass filtering the two products. That is, \begin{align*} s_I(t) &= \text{result of low-pass filtering of}~ 2s(t)\cos(\omega_c t)\\ s_Q(t) &= \text{result of low-pass filtering of}~ -2s(t)\sin(\omega_c t) \end{align*} Note that \begin{align*} 2s(t)\cos(\omega_c t) &= 2s_I(t)\cos^2(\omega_c t) - 2s_Q(t)\sin(\omega_ct)\cos(\omega_c t)\\ &= s_I(t) + \bigr [s_I(t)\cos(2\omega_c t) - s_Q(t)\sin(2\omega_c t)\bigr]\\ -2s(t)\sin(\omega_c t) &= -2s_I(t)\cos(\omega_c t)\sin(\omega_c t) + 2s_Q(t)\sin^2(\omega_ct)\\ &= s_Q(t) + \bigr [-s_Q(t)\cos(2\omega_c t) - s_I(t)\sin(2\omega_c t)\bigr]\\ \end{align*} where the quantities in square brackets are double-frequency terms that are eliminated by the low-pass filtering.