### Clockwork

Consider a sample of pure water, with a single NMR resonance. After we excite it, we have created a magnetization vector that rotates into the xy plane. We observe the magnetization at regular intervals. The show soon becomes dull and repetitive. Actually, after the first two observation, we can already predict all the remaining ones. Let Φ₁ be the phase observed initially, and Φ₂ the next sampled phase. We can calculate:

Δ = Φ₂ - Φ₁.

Δ remains the same for every couple of consecutive observations. For example, we can predict:

Φ₃ = Φ₂ + Δ.

Φ₄ = Φ₃ + Δ.

Because the intervals are all equal. The value of Δ must be related to the frequency of the signal, of course, but we need not to know neither the relation nor the frequency.

There's a difference between the magnetization vector and the hands of a clock: the former is subject to exponential decay. After we have measured the intensity of the magnetization twice, let I₁ and I₂ be the two values, we can calculate:

α = I₂ / I₁.

α must be < 1, because I₂ < I₁, otherwise it's not a decay. It's also α > 0, because we are considering the absolute intensities here.

If the decay is exponential, we can predict:

I₃ = α ⋅ I₂.

I₄ = α ⋅ I₃.

The value of α must be related to the relaxation time of the magnetization and the broadness of the signal, of course, but we need not to know neither the relation nor the broadness.

Phase and intensity can be observed simultaneously. If we observe the magnetization twice, we can completely predict its future. Remember this number 2. The number of observations must be at least twice the number of signals. Information about phase and intensity can be stored into a single complex value:

S₁ = I₁ cos Φ₁ + i I₁ sin Φ₁.

α and I can be similarly stored into a single complex number p, called prediction coefficient. The number of coefficients must be at least equal to the number of signals.

p = α cos Δ + i α sin Δ.

S₂ = p ⋅ S₁.

S₃ = p ⋅ S₂.

Instead of water, we may have dioxane, with different frequency and relaxation time, therefore a different prediction coefficient q:

q = α' cos Δ' + i α' sin Δ'.

Finally, let's mix water and dioxane. The spectrum will be a sum of predictable spectra, therefore itself predictable. When there are 2 signals, we need 4 observations S₁, S₁, S₃ and S₄. Using them, let's write down a system of equations:

S₃ = x ⋅ S₂ + y ⋅ S₁.

S₄ = x ⋅ S₃ + y ⋅ S₂.

There is no need to explain the above equations. They are the definitions of two new coefficients x and y. You don't demonstrate a definition. If we solve the system of equations, we also know the values of x and y. They predict all the future points. For example:

S₅ = x ⋅ S₄ + y ⋅ S₃.

This is forward linear prediction. Backward prediction is specular (Δ changes sign and α becomes its own reciprocal). We assume that 2 coefficients are enough, because the complexity of the problem is proportional to the number of our signals. It's not forbidden to define more coefficients than necessary, the prediction will still work.

x and y are related to the coefficients p and q of the isolated signals. For completeness:

x = p + q.

y = - p ⋅ q.

When you have more than 2 signals, even hundreds of them, build a polynomial an solve for its roots. The order of the polynomial is equal to the number of signals. For example:

z² - xz - y = (z - p) (z - q) = 0.

In theory, all we have seen is valid. Tomorrow we'll follow an empirical approach.

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