The Basics of MRI

Joseph P. Hornak

Chapter 4

NMR SPECTROSCOPY

The Time Domain NMR Signal

As transverse magnetization rotates about the Z axis, it will induce a current in a coil of wire located around the X axis.

Plotting current as a function of time gives a sine wave.

This wave will of course decay with time constant T2* due to dephasing of the spin packets. This signal is called a free induction decay (FID).

We will see in Chapter 5 how the FID is converted into a frequency domain spectrum.

The +/- Frequency Convention

Transverse magnetization vectors rotating faster than the rotating frame of reference are said to be rotating at a positive frequency relatve to the rotating frame (+ν).

Vectors rotating slower than the rotating frame are said to be rotating at a negative frequency relative to the rotating frame (-ν).

The 90-FID Sequence

A set of RF pulses applied to a sample to produce a specific form of NMR signal is called a pulse sequence. In the 90-FID pulse sequence, net magnetization is rotated down into the X'Y' plane with a 90o pulse.

The net magnetization vector begins to precess about the +Z axis.

The magnitude of the vector also decays with time.

A timing diagram is a multiple axis plot of some aspect of a pulse sequence versus time. A timing diagram for a 90-FID pulse sequence has a plot of RF energy versus time and another for signal versus time. [1] [2]

When this sequence is repeated, for example when signal-to-noise improvement is needed, the amplitude of the signal after being Fourier transformed (S) will depend on T1 and the time between repetitions, called the repetition time (TR), of the sequence. In the signal equation below, k is a proportionality constant and ρ is the density of spins in the sample.

S = k ρ ( 1 - e-TR/T1 )

The Spin-Echo Sequence

Another commonly used pulse sequence is the spin-echo pulse sequence. [1] [2] Here a 90° pulse is first applied to the spin system. The 90° pulse rotates the magnetization down into the X'Y' plane.

The transverse magnetization begins to dephase.

At some point in time after the 90° pulse, a 180° pulse is applied. This pulse rotates the magnetization by 180° about the X' axis.

The 180° pulse causes the magnetization to at least partially rephase and to produce a signal called an echo.

A timing diagram shows the relative positions of the two radio frequency pulses and signal.

The signal equation for a repeated spin-echo sequence as a function of the repetition time, TR, and the echo time (TE) defined as the time between the 90° pulse and the maximum amplitude in the echo is

S = k ρ ( 1 - e-TR/T1 ) e-TE/T2

This equation is only valid when TR >> TE.

The Inversion-Recovery Sequence

An inversion-recovery (IR) pulse sequence can also be used to record an NMR spectrum. In this sequence, a 180° pulse is first applied. This rotates the net magnetization down to the -Z axis.

The magnetization undergoes spin-lattice relaxation and returns toward its equilibrium position along the +Z axis.

Before it reaches equilibrium, a 90° pulse is applied which rotates the longitudinal magnetization into the XY plane. In this example, the 90° pulse is applied shortly after the 180° pulse.

Once magnetization is present in the XY plane it rotates about the Z axis and dephases giving a FID.

Once again, the timing diagram shows the relative positions of the two radio frequency pulses and the signal.

The signal as a function of TI when the sequence is not repeated is

S = k ρ ( 1 - 2e-TI/T1 )

It should be noted at this time that the zero crossing of this function occurs for TI = T1 ln2.

When an inversion-recovery sequence is repeated every TR seconds, for signal averaging or imaging purposes, the signal equation becomes

S = k ρ ( 1 - 2e-TI/T1 + e-TR/T1) .

To prove this equation, start with the Bloch equations introduced in Chapter 3. The signal in a pulse sequence is proportional to the amount of longitudinal magnetization rotated in to the xy-plane. For this reason, we only need to examine the component which describes in differential form the time (t) evolution of the Mz magnetization to equilibrium Mzo with time constant T1 after perturbation by an RF pulse.

Grouping like terms and setting up the integrals between the time limits of 0 and TI, and the Mz limits of -Mzo(1-e-(TR-TI)/T1) and Mz we have

Integration yields

and evaluating between the limits yields the following equations.


Which after rearrangement gives the IR equation.

Chemical Shift

When an atom is placed in a magnetic field, its electrons circulate about the direction of the applied magnetic field. This circulation causes a small magnetic field at the nucleus which opposes the externally applied field. [3]

The magnetic field at the nucleus (the effective field) is therefore generally less than the applied field by a fraction σ.

B = Bo (1-σ)

The electron density around each nucleus in a molecule varies according to the types of nuclei and bonds in the molecule. The opposing field and therefore the effective field at each nucleus will vary. This is called the chemical shift phenomenon.

Consider the methanol molecule. The resonance frequency of two types of nuclei in this example differ. This difference will depend on the strength of the magnetic field, Bo, used to perform the NMR spectroscopy. The greater the value of Bo, the greater the frequency difference. This relationship could make it difficult to compare NMR spectra taken on spectrometers operating at different field strengths. The term chemical shift was developed to avoid this problem.

The chemical shift of a nucleus is the difference between the resonance frequency of the nucleus and a standard, relative to the standard. This quantity is reported in ppm and given the symbol delta, δ.

δ = (ν - νREF)x106 / νREF

In NMR spectroscopy, this standard is often tetramethylsilane, abbreviated TMS. In the body there is no TMS, but there are two primary hydrogen containing substances, water and fat. The chemical shift difference between these two types of hydrogens is approximately 3.5 ppm.

References

  1. T.C. Farrar, An Introduction To Pulse NMR Spectroscopy. Farragut Press, Chicago, 1987.
  2. T.C. Farrar, E.D. Becker, Pulse And Fourier Transform NMR. Academic Press, New York, 1971.
  3. R.K. Harris, Nuclear Magnetic Resonance Spectroscopy. Pitman, London, 1983.

Problems

  1. From the 1H NMR perspective, the human body is composed primarily of fat hydrogens (-CH2-) and water hydrogens (H2O). The resonance frequency difference between the NMR signal from these two types of hydrogens is approximately 220 Hz on a 1.5 Tesla imager. What is the chemical shift difference?
  2. The hydrogen T1, T2 and spin density values for common brain tissues are listed in the following table.

    TissueT1 (s) T2 (ms) ρ*
    CSF0.8 - 20 110 - 2000 70-230
    White0.76 - 1.08 61-100 70-90
    Gray1.09 - 2.15 61 - 109 85 - 125
    Meninges0.5 - 2.2 50 - 165 5 - 44
    Muscle0.95 - 1.82 20 - 67 45 - 90
    Adipose0.2 - 0.75 53 - 94 50 - 100
    *Based on ρ = 111 for 12mM aqueous NiCl2
     
    At what TI value is the signal from fat approximately equal to zero in an inversion recovery sequence?

  3. When using a 90-FID pulse sequence and a sample containing all the tissues in question number two, what TR value would guarantee at least 98% of the signal from all the tissues?
  4. You are using a spin-echo pulse sequence and the adipose tissue sample in question number two. If the minimum TE value you can obtain is 20 ms, how much more signal could you obtain with a 90-FID sequence?
  5. From the 1H NMR perspective, the human body is composed primarily of fat hydrogens (-CH2-) with a chemical shift of ~ 1 PPM, and water hydrogens (H2O) with a chemical shift of ~ 4.5 PPM. What is the resonance frequency difference between the NMR signal from these two types of hydrogens?
  6. The hydrogen T1, T2 and spin density values for common brain tissues are listed in the following table.

    TissueT1 (s) T2 (ms) ρ*
    CSF0.8 - 20 110 - 2000 70-230
    White0.76 - 1.08 61-100 70-90
    Gray1.09 - 2.15 61 - 109 85 - 125
    Meninges0.5 - 2.2 50 - 165 5 - 44
    Muscle0.95 - 1.82 20 - 67 45 - 90
    Adipose0.2 - 0.75 53 - 94 50 - 100
    *Based on ρ = 111 for 12mM aqueous NiCl2

    At what TR and TE values is the signal from white matter approximately equal to that from gray matter in a spin-echo pulse sequence?
  7. When using an inversion recovery pulse sequence and a sample containing all the tissues in question number two, what TI value greater than 2 ms would guarantee at least 90% of the signal from all the tissues?
  8. You are using a 90-FID pulse sequence and the brain tissue sample in question number two. For adipose tissue, how much less signal will be present when TR=100 ms then when TR=500 ms?

Answers

  1. δ = (ν - νref) x 106 / νref
    νref = γBo
    δ = (ν - νref) x 106 / γBo
    where:
      νref = resonance frequency
      γ = gymagnetic ratio for the nuclei in question
      Bo = magnetic field strength
      ν = resonance frequency of second component
      δ = chemical shift difference
    Given:
      Bo = 1.5 T
      ν - νref = 220 Hz
      γBo = 42.58 MHz/T * 1.5 T = 63.87 MHz
    δ NT>= 220 Hz / 63.87 MHz * 106 = 3.444 = chemical shift difference
     
  2. S(TI) = kρ(1 - 2 exp(-TI/T1>)) = 0
    TI(s=0) = T1 * ln( 2 )
    where:
    Given:
    TI(s=0) = 0.2 s * ln( 2 ) = 0.14 s
    TI(s=0) = 0.75 s * ln( 2 ) = 0.52 s
    TI(s=0) would vary from 0.14 to 0.52 seconds
     
  3. S(TR) = kρ(1-e-TR/T1)
    where:
      TR = repetition time
      S(TR) = signal at time TR
      ρ = spin density
      k = proportionality constant
      T1 = spin-spin relaxation time
    Taking the ratio of the observed signal (S(TR) from above) to the signal at infinitely long TR, gives:
    S(TR) / S(TR= ∞) = kρ(1-e-TR/T1) / kρ(1-e-∞/T1)
    S(TR) / S(TR= ∞) = kρ(1-e-TR/T1) / kρ(1)
    Setting S(TR) / S(TR= ∞) = 0.98 we have
    0.98 = 1-e-TR/T1
    ln( 1 - 0.98) = -TR/T1
    TR = -T1 * ln( 0.02 )
    Using the longest T1 (20 s) to determine TR would ensure 98% signal or better for all of the listed tissues.
    TR = -20 s * ln( 0.02 ) = 78.24 s
     
  4. S(TR,TE) = kρ(1-e-TR/T1)e-TE/T2
    where:
      TR = repetition time
      TE = echo time
      S(TR) = signal at time TR from a 90-FID
      ρ = spin density
      k = proportionality constant
      T1 = spin-spin relaxation time
      T2 = spin-echo relaxation time
    Taking the ratio of the observed signal (S(TR,TE) from above) to the signal of a 90o FID, gives:
    S(TR) / S(TR,TE) = kρ(1-eTR/T1) / kρ(1-e-TR/T1)e-TE/T2
    S(TR) / S(TR,TE) = eTE/T2
    Given: TE = 20 ms
    S(TR) / S(TR,TE) = e20 ms/53 ms = 1.46
    S(TR) / S(TR,TE) = e20 ms/94 ms = 1.24
    Therefore, under these conditions, the 90o FID sequence would provide 1.24 to 1.46 times the signal of a spin-echo sequence.
     
Go to the: [ next chapter | chapter top | previous chapter | contents | cover ]

Copyright © 1996-2020 J.P. Hornak.
All Rights Reserved.