Quadrature Signals Notes

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Quadrature Signals

Two signals are in quadrature if:

They have the same frequency
They differ in phase by exactly 90 degrees

Example:

$\begin{aligned} I (t) & = \cos (2 ω t) \\ Q (t) & = \sin (2 ω t) \end{aligned}$

These are quadrature signals.

The signals are orthogonal, i.e., they do not interfere with eachother which, for example, allows two independent signals to share the same carrier frequency - very useful!

Sine and cosine are othogonal because:

$\int_{0}^{T} \cos (ω t) \sin (ω t) d t = 0$

I think of the orthogonality of the signals in the same way I was taught orthogonality for vectors (see linear algrebra notes on orthogonality and indpendence).

In linear algebra world, assuming the normal cartesian basis vectors, if I have a vector, $[a, b]$ , I can decompose it into basis vectors like this - $a [1, 0] + b [0, 1]$ - and I can only decompose it this way. There is no other combination of basis vectors that can yield the vector $[a, b]$ . It's like I could mix two "signals", the vectors $[0, 1]$ and $[1, 0]$ in various quantities into a third signal, and then recover the original two signals and their "amplitudes" from that third, mixed up signal.

Same with the signal $x (t) = a \cos (ω t) + b \sin (ω t)$ . If I just have the signal $x (t)$ , in the same way as I did above, I can decompose it into the basis functions $\cos (ω t)$ and $\sin (ω t)$ . I.e. $x (t)$ is some mixed up jumble of two sinusoidal signals and I can recover the exact mix of the two sinusoids because those sinusoids are orthogonal or quadrature signals.

How?

To get $a$ , do the following...

We have:

$x (t) = a \cos (ω t) + b \sin (ω t) $$

Multiplying both sides by $\cos (ω t)$ gives:

$\begin{aligned} x (t) \cos (ω t) & = [a \cos (ω t) + b \sin (ω t)] \cos (ω t) \\ = a \cos (ω t) \cos (ω t) + b \sin (ω t) \cos (ω t) \\ = a \cos^{2} (ω t) + b \sin (ω t) \cos (ω t) \end{aligned}$

Thus, we have

$x (t) \cos (ω t) = a \cos^{2} (ω t) + b \sin (ω t) \cos (ω t)$

Integrate that result over one period.

$\begin{aligned} = \int_{0}^{T} a \underset{integrates to T / 2}{\underset{⏟}{\cos^{2} (ω t)}} + b \overset{integrates to 0}{\overset{⏞}{\sin (ω t) \cos (ω t)}} d t \\ = a \frac{T}{2} \end{aligned}$

Which means that

$\int_{0}^{T} x (t) \cos (ω t) = a \frac{T}{2}$

Thus we have a forumla that results in $a \frac{T}{2}$ . Multiple by $\frac{2}{T}$ to remove the half-period to ge just $a$ . This gives us the result

$a = \frac{2}{T} \int_{0}^{T} x (t) \cos (ω t) d t$

This completely an unambiguously extracts the "amount", $a$ of the basis vector $\cos (ω t)$ contained in our signal $x (t)$ .

To find $b$ the same method is repeated, but with sine, to get:

$b = \frac{2}{T} \int_{0}^{T} x (t) \sin (ω t) d t$

IQ (Quadrature) Demodulator

This [All About Circuits article] is a good ref(https://www.allaboutcircuits.com/textbook/radio-frequency-analysis-design/radio-frequency-demodulation/understanding-i-q-signals-and-quadrature-modulation/).

A signal composed of two quadrature waveforms can be "decoded" by an IQ demodulator to extract the in phase component and the quad component.

If I have a singal I

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Quadrature Signals Notes

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Quadrature Signals

IQ (Quadrature) Demodulator