Why the quadratic formula works - RF spoiling: why not just use random phases?

The goal of spoiling¶

A spoiled gradient echo (SPGR / FLASH) sequence is meant to behave as if all transverse magnetization disappeared before the next RF pulse. The steady-state signal then follows the Ernst equation:

S_\text{Ernst}(\alpha) = M_0 \sin\alpha \,\frac{1 - E_1}{1 - E_1\cos\alpha}, \qquad E_1 = e^{-\mathrm{TR}/T_1}.

(1)

With $\mathrm{TR} \ll T_2$ , the transverse magnetization has not decayed by the next pulse, so something has to destroy it.

Gradient spoiling alone is not enough¶

A spoiler gradient at the end of each TR dephases the voxel. The isochromat at position $\theta\in[0,2\pi)$ picks up an extra phase $\theta$ every TR. The net transverse magnetization at the end of the TR is then zero, but the individual isochromats still carry their transverse components. Because the gradient is the same every TR, later RF pulses refocus these components into echoes (stimulated echoes and higher-order pathways) that land exactly at the next readout. The steady state then depends on $T_2$ and is far from (1), as shown by the green curves on the next pages.

RF spoiling: change the RF phase every TR¶

If the phase of each RF pulse is changed, and the receiver is demodulated with the same phase, the refocused pathways add up with different phases and cancel. The standard choice (Zur et al., 1991) is a quadratic phase schedule:

\phi_n = \phi_0\,\frac{n(n+1)}{2} \quad\Longleftrightarrow\quad \Delta\phi_n = \phi_n - \phi_{n-1} = n\,\phi_0 .

(2)

The phase increment grows linearly, so the second difference is a constant $\phi_0$ (often $117^\circ$ or $50^\circ$ ).

Why the quadratic schedule gives a steady state¶

Work in the receiver frame, demodulated by $\phi_n$ at TR $n$ . Let $\mathbf m_n(\theta)$ be the magnetization of isochromat $\theta$ just after pulse $n$ . Going from pulse $n$ to pulse $n+1$ , the isochromat relaxes and precesses by $\theta$ . The frame then rotates by $-\Delta\phi_{n+1}$ , and the RF pulse (now always about $x$ ) is applied. The only thing that depends on $n$ is the effective precession angle

\theta_\text{eff} = \theta - \Delta\phi_{n+1} = \theta - (n+1)\,\phi_0 .

(3)

Now relabel the isochromats and follow $\mathbf u_n(\vartheta) = \mathbf m_n(\vartheta + n\phi_0)$ . Substituting gives

\mathbf u_{n+1}(\vartheta) = \mathcal F\!\left[\mathbf u_n(\vartheta + \phi_0);\ \vartheta\right],

(4)

where $\mathcal F$ (relax, precess by $\vartheta$ , rotate by $\alpha$ ) does not depend on $n$ at all. Relaxation makes this map contractive, so $\mathbf u_n \to \mathbf u^\star$ , a fixed pattern. In the original labels,

\mathbf m_n(\theta) = \mathbf u^\star(\theta - n\phi_0):

(5)

the magnetization pattern across the voxel is the same every TR, just shifted by $\phi_0$ along the dephasing axis. The voxel signal is the average over $\theta\in[0,2\pi)$ , and a shift doesn’t change that average:

s_n = \frac{1}{2\pi}\int_0^{2\pi} m_{xy,n}(\theta)\,d\theta = \text{constant}.

(6)

That is a true steady state. The value of the constant depends on $\phi_0$ , $T_1$ , $T_2$ and $\alpha$ . For well-chosen $\phi_0$ it is close to (1), but not exactly equal to it, because some refocused pathways still add up coherently.

Why random phases do not¶

With random phases, $\Delta\phi_{n+1}$ is a new random number every TR. No relabelling of isochromats makes the map independent of $n$ , so there is no fixed point, and the voxel signal is a random variable at every TR.

In the echo-pathway picture, each refocused pathway reaching the readout carries a phase that is a sum of several different random phases. Averaged over many TRs these contributions cancel, which is why random spoiling is unbiased: the mean signal equals the Ernst signal. At any single TR, though, they don’t cancel, so the signal jitters around the Ernst value in magnitude and phase. The longer $T_2$ is compared with $\mathrm{TR}$ , the more pathways survive and the larger the jitter.

The next page shows this with a handful of spins.