---
title: "How a spin echo recovers a signal that seemed lost"
description: "An uneven field makes the signal fade faster than T2. See how a 180° pulse realigns the phases, how T2* differs from T2, and why motion still costs signal."
canonical_url: "https://sandboxphysics.com/stories/mri-spatial-encoding/spin-echo-and-t2/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/mri-spatial-encoding/spin-echo-and-t2/"
last_modified: "2026-10-29"
---

[MRI spatial encoding](https://sandboxphysics.com/stories/mri-spatial-encoding/)

L2 · Chapter 2

# How a spin echo recovers a signal that seemed lost

Sandbox Physics · 2026-10-29

After a radio-frequency pulse the coil signal often dies much faster than $T_2$ predicts, because the field is not exactly the same throughout the sample. Surprisingly, that part of the signal is not really lost: a 180° pulse makes it reappear a little later as a spin echo. This chapter derives when the echo appears, separates the decay that can be recovered from the decay that cannot, and shows why the echo weakens when molecules move between the pulses. Chapter 7 turns that weakness into diffusion imaging.

## Before you begin

Precession, the flip angle and T2 decay from Chapter 1, and the idea of phase: the angle a precessing arrow has turned through. You only need to average a group of turning arrows.

- Explain why an uneven static field makes the signal decay with T2\*, and write the relation between T2\*, T2 and T2′.
- Derive that the phases realign at twice the pulse time after a 180° pulse, and compute the echo signal.
- Explain why motion keeps the echo from recovering fully.

## Why the signal fades faster than T2

Suppose the field in a small region of the sample exceeds $B_0$ by $\Delta B$. In the frame turning at $\omega_0$ the magnetization there drifts round at $\gamma\Delta B$ and after a time $t$ has gained a phase $\varphi=\gamma\Delta B\,t$. The coil measures the vector sum of the transverse magnetization of all regions, and the more the phases spread, the smaller the sum.

When Bloch defined $T_2$ he already put “small and irregular inhomogeneities” of the static field $H_0$ together with the action of neighbouring nuclei, as one effective field irregularity. Today the two are kept apart: the irreversible decay caused by fluctuating fields is still called $T_2$, the extra decay from the static field spread is $T_2'$, and the total decay we observe is $T_2^*$.

If the static frequency offsets follow a Lorentzian distribution, their average is exactly an exponential and the two decay rates add. The nuclear induction experiment is set up this way, which is why its fitted effective decay cannot tell the two apart.

$$
S(t)=S_0\,e^{-t/T_2}\,\bigl\langle e^{-i\gamma\Delta B\,t}\bigr\rangle
$$

The angle brackets average over all static field offsets in the sample.

$$
\frac{1}{T_2^*}=\frac{1}{T_2}+\frac{1}{T_2'}
$$

Holds when the offsets are Lorentzian with half width $1/(2\pi T_2')$; for other distributions the average is not strictly exponential.

Source: F. Bloch · J. Hrabe

## A 180° pulse turns the phases round

At $t=\tau$ apply a 180° pulse. It flips every transverse magnetization about a transverse axis, so every phase changes sign: a spin $\varphi$ ahead is now $\varphi$ behind. After the pulse each region keeps turning at its old rate, because its field has not changed.

At $t=2\tau$ every region therefore has phase $-\gamma\Delta B\tau+\gamma\Delta B\tau=0$. All arrows line up again and the signal reappears: the spin echo, with echo time $T_E=2\tau$. The static unevenness cancels completely and only the $T_2$ decay remains.

Hrabe and co-workers sketch the history in their introduction: Hahn noticed as early as 1950 that self-diffusion weakens spin echoes, and Carr and Purcell then used many echoes to reduce the effect. The ISMRM diffusion problem set treats the two same-sign gradients on either side of the 180° pulse as one dephasing vector $q$ that grows during the first pulse and returns to zero during the second, which is the phase flip in another form.

$$
\varphi(2\tau)=-\gamma\Delta B\,\tau+\gamma\Delta B\,\tau=0
$$

The first term is the phase just after the flip, the second the phase gained in the second half.

$$
S(T_E)=S_0\,e^{-T_E/T_2}
$$

The top of the echo decays with $T_2$ alone; away from the top the static unevenness matters again.

Source: J. Hrabe · F. Szczepankiewicz and J. Hutter · F. Bloch

## What the echo cannot recover: motion

The derivation relies on each region keeping its field. A water molecule that sits in different places during the two halves feels different fields, and its phases do not cancel. Many molecules each keep a random leftover phase, and the echo weakens.

Hrabe and co-workers point out that uneven magnetic susceptibility creates local gradients that act like unknown weak diffusion gradients. A spin echo reduces this weighting, and the more echoes, the smaller the b-value these local gradients contribute. Chapter 7 turns the weakness round: it applies a known, strong pair of gradients on purpose and uses the lost echo to measure how far molecules travel.

Source: J. Hrabe · F. Szczepankiewicz and J. Hutter

## Worked example: the same echo time read two ways

Problem: a tissue has $T_2=80\,\mathrm{ms}$ and a static field spread with $T_2'=40\,\mathrm{ms}$ (both values assumed for the calculation). Read the signal 50 ms after excitation in two ways: by simply waiting, and with a 180° pulse at 25 ms.

First the total decay: $1/T_2^*=1/80+1/40$, so $T_2^*=26.7\,\mathrm{ms}$. Waiting leaves $e^{-50/26.7}=0.153$ of the signal; the spin echo leaves $e^{-50/80}=0.535$, 3.49 times as much.

How small is this unevenness? The Lorentzian half width is $1/(2\pi T_2')=3.98\,\mathrm{Hz}$, which corresponds to $\Delta B=3.98/42.576\times10^6\,\mathrm T=0.0935\,\mu\mathrm T$, or 0.031 ppm of 3 T. A field difference of a few parts in a hundred million wipes out most of the signal within tens of milliseconds.

Now one pair of spins, one 4 Hz faster than the other. At 25 ms they differ by $2\pi\times4\times0.025=0.628\,\mathrm{rad}$, or 36°. The 180° pulse turns the difference into −36°, the faster spin keeps catching up, and at 50 ms the difference is exactly zero again.

Source: J. Hrabe · F. Bloch

## Try it in the Lab

### [See static broadening shorten the signal](https://sandboxphysics.com/labs/nmr-free-induction-spectrometer/)

1. With the “Reference pulse” preset, note how fast the voltage envelope decays and how wide the spectral line is.
2. Switch to the “Broader field” preset. Predict: wider static broadening, shorter $T_2^*$, wider line. Check against the record.
3. Think: this instrument applies one pulse and fits one exponential. Why can it not measure $T_2$ on its own, and what is missing from the sequence to separate the two?

### [Which dephasing one inversion can undo](https://sandboxphysics.com/labs/spin-echo-noise/)

1. Choose the “No inversions” preset and read the model refocused coherence. Then choose “Exact static refocusing”: the offsets stay fixed and one inversion sits at the midpoint. Predict the coherence before you read it.
2. Choose “Hahn echo”: the offsets have the same size but drift slowly during the sequence. Why is the coherence lower than with static offsets? Then try “Four-pulse train” and explain why more inversions recover more.
3. Which preset matches the static unevenness behind $T_2'$ in this chapter, and which is closer to the fluctuating fields behind $T_2$?

### [What the echo cannot recover is the phase left by motion](https://sandboxphysics.com/labs/mri-diffusion-encoding/)

1. Choose the “Free water, a clinical b-value” preset and watch the colours of the molecules: the first gradient pulse spreads them, and after the 180° pulse the second gradient pulse gathers them back, but not completely.
2. Switch to the “Pulses close together” preset. How does the echo ratio change? Explain it by the molecules moving less between the two pulses.

## Exercises with explained answers

1. A 180° pulse is applied at 30 ms. When does the echo appear, and how much phase has a spin with a 10 Hz offset gained at 30 ms?

   Show the answer

   The echo comes at $T_E=60\,\mathrm{ms}$. At 30 ms the phase is $2\pi\times10\times0.03=1.885\,\mathrm{rad}$, or 108°. The flip makes it −108°, and it is back to zero at 60 ms.

2. Waiting gives $T_2^*=20\,\mathrm{ms}$ and a spin echo gives $T_2=100\,\mathrm{ms}$. What $T_2'$ does the static unevenness correspond to?

   Show the answer

   From $1/T_2'=1/T_2^*-1/T_2=1/20-1/100$ we get $T_2'=25\,\mathrm{ms}$, assuming the static offsets are close to Lorentzian.

3. Why does a 180° pulse cancel static unevenness but not the $T_2$ decay?

   Show the answer

   A static offset is the same before and after the pulse, so the second half exactly undoes the flipped phase. $T_2$ comes from fields that fluctuate in time; the phases gained in the two halves are unrelated, and flipping cannot make them cancel.

[Previous: How proton magnetization precesses, tips over and recovers](https://sandboxphysics.com/stories/mri-spatial-encoding/precession-and-relaxation/) [Next: How a gradient turns position into frequency](https://sandboxphysics.com/stories/mri-spatial-encoding/gradients-make-position-into-frequency/)

## Sources and what we recomputed

1. [F. Bloch, Nuclear induction, Physical Review 70, 460 (1946), doi:10.1103/PhysRev.70.460](https://doi.org/10.1103/PhysRev.70.460)

   Read in full from a scanned copy. Eq. 11 (precession of the polarization), Eq. 32 (longitudinal relaxation time), Eqs. 33–35 (transverse relaxation time from an effective field irregularity, with the estimate of about 1e-4 s), Eq. 37 and Eqs. 38a–c (the equations with both relaxation times), Eqs. 40–42 (the rotating-frame variables). Bloch writes in Gaussian units with H for the field; the course restates the equations in SI units with B.

2. [J. Hrabe, G. Kaur and D. N. Guilfoyle, Principles and limitations of NMR diffusion measurements, Journal of Medical Physics 32(1), 34–42 (2007), doi:10.4103/0971-6203.31148](https://pmc.ncbi.nlm.nih.gov/articles/PMC3003887/)

   Read in full (PubMed Central). Introduction: Hahn’s 1950 observation, the Stejskal–Tanner pulsed gradients and Eqs. 1–3. Eqs. 4–10 (the binomial random walk, D from Eq. 7, Einstein’s relation in Eqs. 8 and 10). Eq. 11 (Larmor frequency, with 2π × 42.576 MHz/T for the proton). Eqs. 12–14 (signal of the lattice walk, the b-value of Eq. 13 and the discrete Stejskal–Tanner form) and Eqs. 15–17 (the continuous limit). The pitfalls section: imaging gradients, local gradients, dependence on the diffusion time and tissue compartments. The 1950 and 1965 papers it cites were not opened.

3. [F. Szczepankiewicz and J. Hutter, Diffusion MRI acquisition basics, problems and answers, ISMRM 2019 educational session](https://www.ismrm.org/dmff-uploads/2s.pdf)

   Read in full. Problem A1 and its answer (Eqs. 2–4): two 40 mT/m, 10 ms pulses with onsets 50 ms apart give 0.53 ms/µm², and three axes together triple b. Problem A2 and its answer (Eqs. 1 and 5–7): b as the time integral of q squared and its closed form for rectangular pulses. Part B: the echo-time budget of a spin-echo diffusion sequence. The value of γ used there is 2.68 × 10⁸ per second per tesla.