I and Q Channels

Question

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)?

Answer 1

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.

Answer 2

I think it's worth adding another answer from the perspective of orthonormal basis functions ($\sin$/$\cos$). Any time you encode bits onto a carrier by phase shifts, you can do that using two "basis functions”— $\sin$ and $\cos$, basically. $\cos(x) = \sin(x + \tfrac{\pi}2)$. So a $\sin$ is just a $\cos$ with a phase shift. These are like unit vectors from geometry/physics free body diagrams; e.g. $\hat{x}$ and $\hat{y}$.

Just as you can create any vector in two dimensions by various linear combinations of $\hat{x}$ and $\hat{y}$, you can create any phase shift by linear combinations of $\sin$ and $\cos$. They aren't orthogonal unit vectors; they're orthonormal basis functions.

So when a receiver gets the phase-shifted signal, it splits it and sends it on two paths. One is correlated with a $\cos$, the other is correlated with a $\sin$, which lets the receiver determine how much $\cos$ is in the signal on the one path and how much $\sin$ is on the signal on the other path. This correlation is like taking a dot product for vectors. And the output is a scalar constant value--a number.

In phase, and quadrature. I values, and $Q$ values. $\cos$ is "in phase" (to itself). $\sin$ is quadrature—$90^{o}$ from the $\cos$.