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I'm studying MRI and I couldn't understand this assumption(?)

enter image description here

The sentence 'The signal from one channel is no more or less "real" than that from the other channel.' Does not make sense to me. (+ and the one before that, too)

all the equations are using that expression and its actually working. Why treating real numbers like imaginary numbers doesn't cause any issues?

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    $\begingroup$ That's called "complex baseband", and works because the projection onto inphase (I) cosine components and onto quadrature (Q) sine components is orthogonal and hence behaves like $\mathbb R^2$, or like $\mathbb C$. $\endgroup$ Jan 26 at 10:33
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    $\begingroup$ @ConstantineA.B. this picky AlexTP would say that it's not $\mathbb{R}^2$, it's $\mathbb{C}$ :) $\endgroup$
    – AlexTP
    Jan 26 at 14:55
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    $\begingroup$ Yeah, I think the picky AlexTP would be right to point that out. But let's ask the laid back @AlexTP whether he would want to explain "isomorphic" to the asker? $\endgroup$ Jan 27 at 0:07
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    $\begingroup$ @ConstantineA.B. ha ha AlexTP is a lazy lad. $\endgroup$
    – AlexTP
    Jan 27 at 11:18

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The MRI system is taking two sets of measurements: the in-phase and quadrature measurements. Both measurements are real. There is no way for a physical system to measure $i5 V$, we measure $5 V$.

The fact that we've identified places in the system that are in-phase and places in the system that are quadrature means that our mathematical model of the system has these two signal streams.

Once the $5 V$ measurement sample has been taken, we end up with a $5$ or $101_2$ or $0x05$ or $5.0f$ inside the computer.

Now we can decide that, if the sample came from the in-phase signal stream, it's just $5$. Or, if the sample came from the quadrature signal stream, it's $i5$.

After that, we can manipulate the mathematical model of our system as we wish.

There are many places in signal processing and elsewhere where this mathematical "trick" is useful.

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    $\begingroup$ You haven't seen my imaginary voltmeter. $\endgroup$ Jan 26 at 15:48
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    $\begingroup$ @robertbristow-johnson Have you seen your imaginary voltmeter or is it merely a figment of your imagination? That is, does the adjective imaginary qualify voltmeter or does it qualify volt in which case the name might be better written as (imaginary volt)meter? $\endgroup$ Jan 26 at 16:39
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    $\begingroup$ That's for me to know (or imagine) and you to find out. $\endgroup$ Jan 26 at 17:08
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For historical reasons, the two primary types of complex numbers have names that are completely inaccurate. One type of number is referred to as "real" even though there's nothing particularly real about them, and one type is referred to as "imaginary" even though there's nothing particularly imaginary about them.

A better pair of words for the two kinds of numbers may be "vertical" and "horizontal." If we use those names, the passage becomes:

The MR signal can be represented as a vector with vertical (Ve) and horizontal (Ho) components recorded from the I and Q channels respectively. An equivalent/alternative representation of the signal is as a complex number

Signal (Ve, Ho) = Ve + h Ho

where h² = -1, the horizontal unit.

As you can see, we're not treating anything real as being imaginary. We're simply using complex numbers as a tool for studying the signal.

Unfortunately, the words that people actually use for the two kinds of numbers are "real" and "imaginary." You'll just have to remember that the word "real" means something completely different from "real" and the word "imaginary" means something completely different from "imaginary."

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  • $\begingroup$ Before "imaginary" became established, the term "impossible part" (of a polynomial root) was in use -- at least in "On Governors" from 1864-ish, that's the terminology that James Clerk Maxwell used. I found it an interesting insight into how our mathematical forebears viewed complex numbers. $\endgroup$
    – TimWescott
    Jan 28 at 17:09
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In many situations, such as with these MRI calculations, we want to measure a gain and a phase shift. However, in many situations, phase shifts are difficult to measure directly. Instead, we measure something related and calculate the phase shift from it.

Euler's Formula provides an effective way to do it: $e^{i\theta}=\cos\theta + i\sin\theta$. If one has a sinusoid of a constant frequency, we can write $\theta=\omega t + \phi$ where $\omega$ is the angular frequency and $\phi$ is a phase shift. So we can write $e^{i(\omega t + \theta)}=\cos(\omega t + \phi) + i\sin(\omega t + \phi)$

One can think of the real and imaginary parts of the above equation as just two "real" axes. And that's perfectly legitimate. However, if we think of them as real and imaginary parts of a complex number, then we can apply Euler's Formula to convert measurements in those two axes into an exponential form. This turns out to be very convenient because many transforms we care about multiply two signals, and $e^{xy}=e^xe^y$. This makes many operations we want to do much easier.

The formula can also be used in reverse to prove nice things. It shows that, given any sinusoidal signal, you can always think of it as a complex number just by taking the exponential of the signal. This can then be used to prove out that the "Real" and "Imaginary" measurements (which are really just measurements taken 90 degrees out of phase) completely capture phase and gain information. There's no reason not to think of it as a complex number.

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  • $\begingroup$ " $e^{xy}=e^xe^y" missing an addition there. "It shows that, given any sinusoidal signal, you can always think of it as a complex number just by taking the exponential of the signal." The complex number isn't the exponential of the signal, the signal is the exponential of the complex number. $\endgroup$ Jan 28 at 4:46
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There's different semantics at play, and they're being mixed. This is not uncommon in signal processing, or in the STEM in general. You can either get perplexed and bothered, or you can just sit back and enjoy the absurdity*.

A "real number" is the thing you were taught in grade school, it has values like $\pi$ or $\frac 1 3$ or $42$.

A real quantity -- in this context -- is some physical measurement, like the number of cups of milk in your morning pancake recipe, or the field strength of an RF signal from an MRI.

"Real" and "real" in this context mean two different things.

An "imaginary number" is complicated -- hopefully you already know, but it's numbers that are real multiples of $\sqrt{-1}$. They can't be represented as numbers on the real number line, but they have mathematical meaning. An imaginary number is, arguably more abstract than a "real number" -- but you can't touch $\pi$ with your fingers**, so it's no more real than $\sqrt{-1}$ in the physical sense.

Just as a note, the terms "inphase" and "quadrature" (the author's I and Q) mean yet another thing. When you're analyzing a narrowband signal around some sinusoidal carrier, then the inphase part is what you get when you demodulate the signal with $\cos \omega t$, and the quadrature part is what you get when you demodulate the signal with $-\sin \omega t$. The inphase and quadrature channel are each real (in the "real number") sense, and they each derive from some real quantity (i.e., varying field strength). In spite of this dual "realness", if you treat the inphase and quadrature parts as the real and imaginary part of a complex number, then all the arithmetic you do on them is just as correct as if you carried all of the $\cos \omega t$ and $\sin \omega t$ terms throughout your calculations -- it's just a lot more compact.


* If you're ever at a party with a civil engineer and a circuit designer, ask them what "permeability" means. Or ask an army officer and a physicist what "force" means. Etc.

** Opportunity for puns accidental, but welcomed.

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  • $\begingroup$ Levitating pie to my mouth without touching it sounds like telekinesis! $\endgroup$
    – Peter K.
    Jan 27 at 16:53
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In wave mechanics, one often encounters solutions of the form $f(t) = A\cos(\omega t -\phi)$. If we have a constant $\omega$, then this gives a two-parameter family of functions. If we take $u = A\cos(\phi)$ and $v=-A\sin(\phi)$, then $f(t)=A(\cos(\omega t)(\cos(\phi)-\sin(\omega t)\sin(\phi)) = u\cos(\omega t)+v\sin(\omega t)$

Given two functions in this family $f_1(t) = A_1\cos(\omega t -\phi_1)$ and $f_2(t) = A_2\cos(\omega t -\phi_2)$, we have

$$(f_1+f_2)(t)= A_1\cos(\omega t -\phi_1)+ A_2\cos(\omega t -\phi_2)=$$ $$u_1\cos(\omega t)+v_1\sin(\omega t)+u_2\cos(\omega t)+v_2\sin(\omega t)=$$ $$(u_1+u_2)\cos(\omega t)+(v_1+v_2)\sin(\omega t)$$

That is, the parameters of a sum are the sums of the parameters, i.e. this two-parameter family is isomorphic to a two-dimensional vector space, with $\cos(\omega t)$ and $\sin(\omega t)$ acting as basis vectors. Since the complex numbers are also isomorphic to a two-dimensional vector space over the reals, we can represent each function as a complex number.

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