Solutions manual for
Understanding NMR spectroscopy
James Keeler and Andrew J. Pell
University of Cambridge, Department of Chemistry
Version 1.0 c James Keeler and Andrew J. Pell July 2005
This solutions manual may be downloaded and printed for personal use. It may not be
copied or distributed, in part or whole, without the permission of the authors.
Preface
We hope that this solutions manual will be a useful adjunct to Understanding NMR Spectroscopy
(Wiley, 2005), and will encourage readers to work through the exercises. The old adage that
‘practice makes perfect’ certainly applies when it comes to getting to grips with the theory of
NMR.
We would be grateful if users of this manual would let us know (by EMAIL to jhk10@cam.ac.uk)
of any errors they come across. A list of corrections will be maintained on the spectroscopyNOW
website
http://www.spectroscopynow.com/nmr (follow the link ‘education’)
Cambridge, August 2005
Contents
2 Setting the scene 1
3 Energy levels and NMR spectra 4
4 The vector model 10
5 Fourier transformation and data processing 19
6 The quantum mechanics of one spin 26
7 Product operators 31
8 Two-dimensional NMR 44
9 Relaxation and the NOE 55
10 Advanced topics in two-dimensional NMR 69
11 Coherence selection: phase cycling and ﬁeld gradient pulses 85
12 How the spectrometer works 97
Chapter 2
Setting the scene
2.1
We need Eq. 2.1 on p. 6:
δ(ppm) = 106
×
υ − υref
υref
.
For the ﬁrst peak
δ(ppm) = 106
×
500.135 021 − 500.134 271
500.134 271
= 1.50 ppm .
For the second peak the shift is 7.30 ppm .
Using Eq. 2.3 on p. 9
δ(ppm) = 106
×
υ − υref
υrx
,
with υrx = 500.135 271 MHz gives the two shifts as 1.50 ppm and 7.30 ppm i.e. identical values
to three signiﬁcant ﬁgures. To all intents and purposes it is perfectly acceptable to use Eq. 2.3.
The separation of the two peaks can be converted to Hz using Eq. 2.2 on p. 7:
frequency separation in Hz = (δ1 − δ2) × υref(in MHz).
So the separation is
(7.30 − 1.50) × 400.130 000 = 2321 Hz .
The conversion to rad s−1
is made using Eq. 2.4 on p. 18
ω = 2π × υ = 2π × 2321 = 14 583 rad s−1
.
2.2
For JAB = 10 Hz & JAC = 2 Hz, the line positions are −6, −4, +4, +6 Hz. For JAB = 10 Hz &
JAC = 12 Hz, the line positions are −11, −1, +1, +11 Hz; note that compared to the ﬁrst multiplet
the two central lines swap positions. For JAB = 10 Hz & JAC = 10 Hz, the line positions are −10,
0, 0, +10 Hz; in this case, the line associated with the spin states of spins B and C being α and β,
and the line in which the spin states are β and α, lie of top of one another giving a 1:2:1 triplet.
Chapter 2: Setting the scene 2
0 10-10 0 10-10 0 10-10
JAB JAB
JAC
JAB
JAC
JAC
JAB = 10 Hz JAC = 2 Hz JAB = 10 Hz JAC = 12 Hz JAB = 10 Hz JAC = 10 Hz
Introducing a third coupling gives a doublet of doublet of doublets. The line positions are ±1.5,
±3.5, ±6.5, ±8.5 Hz. For clarity, only the spin state of the fourth spin, D, are shown by the
grey-headed arrows on the last line of the tree.
0 10-10
JAB
JAC
JAD
JAB = 10 Hz JAC = 2 Hz JAD = 5 Hz
2.3
The frequency, in Hz, is 1/period:
υ =
1
2.5 × 10−9
= 4 × 108 Hz or 400 MHz.
Converting to rad s−1
gives:
ω = 2πυ = 2.51 × 109 rad s−1
.
(a) 90◦ is one quarter of a rotation so will take 1
4 × 2.5 × 10−9 = 6.25 × 10−10 s .
(b) As 2π radians is a complete rotation, the fraction of a rotation represented by 3π/2 is
(3π/2)/(2π) = 3/4, so the time is 0.75 × 2.5 × 10−9 = 1.875 × 10−9 s .
(c) 720◦ is two complete rotations, so the time is 2 × 2.5 × 10−9 = 5.0 × 10−9 s .
Chapter 2: Setting the scene 3
To convert from angular frequency to Hz we need Eq. 2.4 on p. 18
υ =
ω
2π
=
7.85 × 104
2π
= 12 494 Hz .
The period is 1/frequency:
T =
1
υ
=
1
12 494
= 8.00 × 10−5 s .
2.4
time
x
y
x
y
x
y
φ = 3π/2
φ = 135˚
φ = 0 or 2π
(a) & (c) (b) (d)
y-comp.
x-comp.
For φ = 0 or 2π radians, the x-component is a cosine wave, and the y-component is a sine wave.
For φ = 3π/2, the y-component is minus a cosine wave, and the x-component is a sine wave.
2.5
We need the identity
sin (A + B) ≡ sin A cos B + cos A sin B.
Using this we ﬁnd:
sin (ωt + π) = sin (ωt) cos π + cos (ωt) sin π
= − sin (ωt),
where to go to the second line we have used cos π = −1 and sin π = 0. So the y-component is
indeed −r sin (ωt).
Chapter 3
Energy levels and NMR spectra
3.1
The expression for ˆHone spin is given by Eq. 3.2 on p. 31:
ˆHone spin = −γB0 ˆIz.
We need to work out the eﬀect that ˆHone spin has on ψ−1/2
:
ˆHone spinψ−1/2
= −γB0 ˆIzψ−1/2
= −γB0 −1
2 ψ−1/2
= 1
2 γB0ψ−1/2
.
To go to the second line we have used Eq. 3.3 on p. 32 i.e. that ψ−1/2
is an eigenfunction of ˆIz. The
wavefunction has been regenerated, multiplied by a constant; ψ−1/2
is therefore an eigenfunction of
ˆHone spin with eigenvalue 1
2 γB0.
3.2
The Larmor frequency, in Hz, of a nucleus with zero chemical shift is deﬁned by Eq. 3.8 on p. 35:
υ0 =
−γB0
2π
=
−6.7283 × 107 × 9.4
2π
= −1.01 × 108 Hz or −101 MHz.
To convert to rad s−1
, we need to multiply the frequency in Hz by 2π:
ω0 = 2πυ0 = 2π × −1.01 × 108
= −6.32 × 108
rad s−1
.
In the case of a non-zero chemical shift, the Larmor frequency, in Hz, is:
υ0 =
−γ(1 + 10−6δ)B0
2π
=
−6.7283 × 107 × (1 + 77 × 10−6) × 9.4
2π
= −1.01 × 108 Hz .
Chapter 3: Energy levels and NMR spectra 5
This is an identical value to three signiﬁcant ﬁgures. We need to go to considerably more ﬁgures
to see the diﬀerence between these two Larmor frequencies. To seven ﬁgures the frequencies are
1.00659 × 108 Hz and 1.00667 × 108 Hz.
3.3
We let ˆHone spin act on the wavefunction ψ+1/2
:
ˆHone spinψ+1/2
= ω0 ˆIzψ+1/2
= 1
2ω0ψ+1/2
,
where the Hamiltonian has been expressed in angular frequency units. To go to the second line,
we have used the fact that ψ+1/2
is an eigenfunction of ˆIz with eigenvalue +1
2.
In the same way,
ˆHone spinψ−1/2
= −1
2ω0ψ−1/2
.
Hence, ψ±1/2
are eigenfunctions of ˆHone spin with eigenvalues ±1
2ω0.
3.4
Following the approach in section 3.5 on p. 37, we let the Hamiltonian act on the product
wavefunction:
ˆHtwo spins, no coupl.ψα,1ψα,2 = υ0,1 ˆI1z + υ0,2 ˆI2z ψα,1ψα,2
= υ0,1 ˆI1zψα,1ψα,2 + υ0,2 ˆI2zψα,1ψα,2
= υ0,1 ˆI1zψα,1 ψα,2 + υ0,2ψα,1 ˆI2zψα,2 .
To go to the third line, we have used the fact that ˆI1z acts only on ψα,1 and not on ψα,2. Similarly,
ˆI2z acts only on ψα,2.
Using Eq. 3.11 on p. 37 i.e. that ψα,1 and ψα,2 are eigenfunctions of ˆI1z and ˆI2z, the terms in the
square brackets can be rewritten:
ˆHtwo spins, no coupl.ψα,1ψα,2 = υ0,1 ˆI1zψα,1 ψα,2 + υ0,2ψα,1 ˆI2zψα,2
= 1
2 υ0,1ψα,1ψα,2 + 1
2 υ0,2ψα,1ψα,2
= 1
2υ0,1 + 1
2υ0,2 ψα,1ψα,2.
Hence, ψα,1ψα,2 is an eigenfunction of ˆHtwo spins, no coupl. with eigenvalue 1
2υ0,1 + 1
2υ0,2.
Letting the coupling term act on the product wavefunction:
J12 ˆI1z ˆI2zψα,1ψα,2 = J12 ˆI1zψα,1 ˆI2zψα,2
= J12
1
2ψα,1
1
2ψα,2
= 1
4 J12ψα,1ψα,2.
Chapter 3: Energy levels and NMR spectra 6
ψα,1ψα,2 is indeed an eigenfunction of the coupling term, with eigenvalue 1
4 J12: this corresponds
to the energy.
ˆHtwo spins, no coupl. and the coupling term share the same eigenfunctions (a result of the fact that the
two terms commute). Since the Hamiltonian for two coupled spins can be represented as the sum
of these two terms,
ˆHtwo spins = ˆHtwo spins, no coupl. + 2π J12 ˆI1z ˆI2z,
it follows that it must also have the same eigenfunctions. Hence, ψα,1ψα,2 is an eigenfunction of
ˆHtwo spins with energy eigenvalue 1
2υ0,1 + 1
2υ0,2 + 1
4 J12, i.e. the sum of the individual eigenvalues
of ˆHtwo spins, no coupl. and J12 ˆI1z ˆI2z.
3.5
Reproducing Table 3.2 on p. 40 for υ0,1 = −100 Hz, υ0,2 = −200 Hz and J12 = 5 Hz:
number m1 m2 spin states eigenfunction eigenvalue/Hz
1 +1
2 +1
2 αα ψα,1ψα,2 +1
2υ0,1 + 1
2 υ0,2 + 1
4 J12 = −148.75
2 +1
2 −1
2 αβ ψα,1ψβ,2 +1
2υ0,1 − 1
2υ0,2 − 1
4 J12 = 48.75
3 −1
2 +1
2 βα ψβ,1ψα,2 −1
2υ0,1 + 1
2υ0,2 − 1
4 J12 = −51.25
4 −1
2 −1
2 ββ ψβ,1ψβ,2 −1
2υ0,1 − 1
2 υ0,2 + 1
4 J12 = 151.25
The set of allowed transitions is:
transition spin states frequency/Hz
1 → 2 αα → αβ E2 − E1 = 197.50
3 → 4 βα → ββ E4 − E3 = 202.50
1 → 3 αα → βα E3 − E1 = 97.50
2 → 4 αβ → ββ E4 − E2 = 102.50
80 100 120 140 160 180 200 220
frequency / Hz
24 3413 12
spin 1 flips
flipsspin 2
spin 1
spin 2
α β
α β
Chapter 3: Energy levels and NMR spectra 7
If J12 = −5 Hz, the table of energies becomes:
number m1 m2 spin states eigenfunction eigenvalue/Hz
1 +1
2 +1
2 αα ψα,1ψα,2 +1
2υ0,1 + 1
2 υ0,2 + 1
4 J12 = −151.25
2 +1
2 −1
2 αβ ψα,1ψβ,2 +1
2υ0,1 − 1
2υ0,2 − 1
4 J12 = 51.25
3 −1
2 +1
2 βα ψβ,1ψα,2 −1
2υ0,1 + 1
2υ0,2 − 1
4 J12 = −48.75
4 −1
2 −1
2 ββ ψβ,1ψβ,2 −1
2υ0,1 − 1
2 υ0,2 + 1
4 J12 = 148.75
80 100 120 140 160 180 200 220
frequency / Hz
13 1224 34
spin 1 flips
flipsspin 2
spin 1
spin 2
β α
β α
The spectrum in unchanged in appearance. However, the labels of the lines have changed; the spin
state of the passive spin for each line of the doublet has swapped over.
3.6
The allowed transitions in which spin two ﬂips are 1–2, 3–4, 5–6 and 7–8. Their frequencies are:
transition state of spin one state of spin three frequency/Hz
1–2 α α −υ0,2 − 1
2 J12 − 1
2 J23 = 193
3–4 β α −υ0,2 + 1
2 J12 − 1
2 J23 = 203
5–6 α β −υ0,2 − 1
2 J12 + 1
2 J23 = 197
7–8 β β −υ0,2 + 1
2 J12 + 1
2 J23 = 207
The multiplet is a doublet of doublets centred on minus the Larmor frequency of spin two.
There are two lines associated with spin three being in the α state, and two with this spin being in
the β state. Changing the sign of J23 swaps the labels associated with spin three, but leaves those
associated with spin one unaﬀected.
Chapter 3: Energy levels and NMR spectra 8
190 195 200 205 210 190 195 200 205 210
frequency / Hz
12 56 34 78
spin 1
spin 3α
α
β
α
α
β
β
β
56 12 78 34
spin 1
spin 3β
α
α
α
β
β
α
β
−υ0,2 −υ0,2
J12
J12 = 10 Hz J23 = 4 Hz J12 = 10 Hz J23 = -4 Hz
J23
3.7
The six zero-quantum transitions have the following frequencies:
transition initial state ﬁnal state frequency
2–3 αβα βαα −υ0,1 + υ0,2 − 1
2 J13 + 1
2 J23
6–7 αββ βαβ −υ0,1 + υ0,2 + 1
2 J13 − 1
2 J23
3–5 βαα ααβ υ0,1 − υ0,3 + 1
2 J12 − 1
2 J23
4–6 ββα αββ υ0,1 − υ0,3 − 1
2 J12 + 1
2 J23
2–5 αβα ααβ υ0,2 − υ0,3 + 1
2 J12 − 1
2 J13
4–7 ββα βαβ υ0,2 − υ0,3 − 1
2 J12 + 1
2 J13
ααα
αβα
ββα
βαα
1
2 3
4
ααβ
αββ
βββ
βαβ
5
6
7
8
Chapter 3: Energy levels and NMR spectra 9
The six transitions can be divided up into three pairs:
• 2–3 and 6–7 in which spins one and two ﬂip, and spin three is passive,
• 3–5 and 4–6 in which spins one and three ﬂip, and spin two is passive,
• 2–5 and 4–7 in which spins two and three ﬂip, and spin one is passive.
Each pair of transitions is centred at the diﬀerence in the Larmor frequencies of the two spins
which ﬂip, and is split by the diﬀerence in the couplings between the two active spins and the
passive spin.
Chapter 4
The vector model
4.1
z
x
θ
ωeff
Ω
ω1
The oﬀset of the peak is 5 ppm. This can be converted to Hz using Eq. 2.2 on p. 7:
Ω
2π
= 10−6
∆δ υref = 10−6
× 5 × 600 × 106
= 5 × 600 = 3000 Hz or 3 kHz.
From the diagram,
tan θ =
ω1
Ω
=
25 × 103 × 2π
3 × 103 × 2π
=
25
3
= 8.33,
so θ = 83 ◦ .
For a peak at the edge of the spectrum, the tilt angle is within 7◦ of that for an on-resonance pulse;
the B1 ﬁeld is therefore strong enough to give a reasonable approximation to a hard pulse over the
full shift range.
For a Larmor frequency of 900 MHz, the peak at the edge of the spectrum has an oﬀset of 4.5 kHz,
so the tilt angle is 80 ◦ . The larger oﬀset results in the same B1 ﬁeld giving a poorer approximation
to a hard pulse.
Chapter 4: The vector model 11
4.2
From Fig. 4.16 on p. 63, the y-component of the magnetization after a pulse of ﬂip angle β is
M0 sin β. The intensity of the signal will, therefore, vary as sin β, which is a maximum for β = 90◦.
(a) If β = 180◦, the magnetization is rotated onto the −z-axis. As sin 180◦ = 0, the signal
intensity is zero.
(b) If β = 270◦, the magnetization is rotated onto the y-axis. As sin 270◦ = −1, the signal will
have negative intensity of the same magnitude as for β = 90◦.
4.3
From Fig. 4.16 on p. 63, the intensity of the signal is proportional to sin β, where the value of the
ﬂip angle β is given by Eq. 4.5 on p. 62:
β = ω1tp.
The pulse lengths of 5 and 10 µs correspond to ﬂip angles below 90◦. Increasing tp further causes
β to increase past 90◦, and so the value of sin β (and hence the signal intensity) decreases. The null
at 20.5 µs corresponds to β = 180◦.
From the expression for the ﬂip angle, it follows that π = ω1t180. Therefore,
ω1 =
π
t180
=
π
20.5 × 10−6
= 1.5 × 105
rad s−1
or 2.4 × 104 Hz .
Another way to answer this question is to see that since a 180◦ pulse has a length of 20.5 µs, a
complete rotation of 360◦ takes 41.0 µs. The period of this rotation is thus 41.0 µs, so the frequency
is
1
41.0 × 10−6
= 2.4 × 104 Hz .
This frequency is ω1/2π, the RF ﬁeld strength in Hz.
The length of the 90◦ pulse is simply half that of the 180◦ pulse:
t90 = 1
2 × 20.5 = 10.25 µs.
The further null occurs at a pulse length that is twice the value of t180. This corresponds to a ﬂip
angle of 360◦, for which the magnetization is rotated back onto the z-axis.
Chapter 4: The vector model 12
4.4
x
-y
x
-yx
-y
x
-y
x
-y
x
-y
φ 2π−φ
180˚ pulse
about y
starting
position
final
position
resolved into x- and
y- components
components after
180˚ pulse
The vector has been reﬂected in the yz-plane, and has a ﬁnal phase of 2π − φ, measured anticlockwise
from the −y-axis.
4.5
x
-y
time
phase,φ
0
0
τ/2 τ τ
τ
3τ/2 2τ
2τ
180
o
(y)pulse
π
2π
π/2
3π/2
Ωτ
φ = Ωτ
φ = 2π−Ωτ
2π−Ωτ
The spin echo sequence 90◦(x)−τ−180◦(x)−τ− results in the magnetization appearing along the yaxis.
In contrast, the 90◦(x)−τ−180◦(y)−τ− sequence results in the magnetization appearing along
the −y-axis. Shifting the phase of the 180◦ pulse by 90◦ thus causes the phase of the magnetization
to shift by 180◦.
A 180◦(−x) pulse rotates the magnetization in the opposite sense to a 180◦(x) pulse, but
the net eﬀect is still to reﬂect the magnetization vectors in the xz-plane. The sequence
Chapter 4: The vector model 13
90◦(x) − τ − 180◦(−x) − τ− will, therefore, have the same eﬀect as the 90◦(x) − τ − 180◦(x) − τ−
sequence i.e. the vector appears on the y-axis at the end of the sequence.
4.6
From section 4.11 on p. 71, the criterion for the excitation of a peak to at least 90% of its theoretical
maximum is for the oﬀset to be less than 1.6 times the RF ﬁeld strength. The Larmor frequency
of 31P at B0 = 9.4 T is:
υ0 = −
γB0
2π
= −
1.08 × 108 × 9.4
2π
= −1.62 × 108
Hz or −162 MHz.
If the transmitter frequency is placed at the centre of the spectrum, the maximum oﬀset is approximately
350 ppm. In Hz, this is an oﬀset of
Ω
2π
= 350 × 162 = 5.66 × 104
Hz or 56.6 kHz.
According to our criterion, the RF ﬁeld strength must be at least 56.6/1.6 = 35.3 kHz, from which
the time for a 360◦ pulse is simply 1/(35.3 × 103) = 28.28 µs. Thus, the 90◦ pulse length is
1
4 × 28.28 = 7.07 µs .
4.7
The ﬂip angle of a pulse is given by Eq. 4.5 on p. 62:
β = ω1tp
So,
ω1 =
β
tp
.
For a 90◦ pulse, β = π/2, so the B1 ﬁeld strength in Hz is:
ω1
2π
=
(π/2)
2π tp
=
1
4 × 10 × 10−6
= 2.5 × 104 Hz or 25 kHz.
The oﬀset of 13C from 1H is 300 MHz, which is very much greater than the B1 ﬁeld strength. The
13C nuclei are therefore unaﬀected by the 1H pulses.
Chapter 4: The vector model 14
4.8
From Eq. 4.4 on p. 61,
ωeﬀ = ω2
1 + Ω2.
If we let Ω = κω1, ωeﬀ can be written
ωeﬀ = ω2
1 + κ2ω2
1 = ω1 1 + κ2. (4.1)
If tp is the length of a 90◦ pulse, we have ω1tp = π/2 and so
ω1 =
π
2tp
,
and hence substituting this into Eq. 4.1 gives
ωeﬀ =
π
2tp
1 + κ2.
Therefore the angle of rotation about the eﬀective ﬁeld, ωeﬀtp, is given by
ωeﬀtp =
π
2tp
1 + κ2 × tp
=
π
2
1 + κ2.
The null condition is when there is a complete rotation about the eﬀective ﬁeld i.e. ωeﬀtp = 2π:
2π =
π
2
1 + κ2.
Rearranging this gives
4 = 1 + κ2 i.e. κ =
√
15 or Ω =
√
15 ω1,
which is in agreement with Fig. 4.28 on p. 73.
The next null appears at ωeﬀtp = 4π i.e. two complete rotations; this corresponds to κ =
√
63 .
For large oﬀsets, κ 1, so
√
1 + κ2 ≈ κ. The general null condition is ωeﬀtp = 2nπ, where
n = 1, 2, 3, . . . Combining these two conditions gives
2nπ =
π
2
1 + κ2 ≈
π
2
κ,
for which we ﬁnd κ = 4n.
Chapter 4: The vector model 15
4.9
In section 4.11.3 on p. 75, it was demonstrated that, on applying a hard 180◦ pulse, the range of
oﬀsets over which complete inversion is achieved is much more limited than the range over which
a 90◦ pulse gives signiﬁcant excitation. Therefore, only peaks with small oﬀsets will be inverted
completely. Peaks with large oﬀsets will not exhibit a null on the application of the 180◦ pulse.
4.10
The initial 90◦(x) pulse rotates the magnetization from the z-axis to the −y-axis; after this the
evolution in the transverse plane is as follows:
x
-y
x
-y
Ωτ 90˚(+x)delay τ
x
-y
The x-, y- and z-components after each element of the pulse sequence are:
component after ﬁrst 90◦(x) after τ after second 90◦(x)
x 0 M0 sin Ωτ M0 sin Ωτ
y −M0 −M0 cos Ωτ 0
z 0 0 −M0 cos Ωτ
The ﬁnal pulse is along the x-axis, and so leaves the x-component of the magnetization unchanged,
but rotates the y-component onto the −z-axis. The overall result of the sequence is My = 0 and
Mx = M0 sin Ωτ.
0 π/2 π 3π/2 2π
-M0
M0
Mx
Ωτ
A null occurs when Mx = 0, i.e. when Ωτ = nπ, where n = 0, 1, 2, . . .
Chapter 4: The vector model 16
4.11
The initial spin echo sequence refocuses the oﬀset, and aligns the magnetization along the y-axis.
If the ﬁnal pulse is about the y- or −y-axis, then it has no eﬀect on the magnetization as the vector
is aligned along the same axis as the B1 ﬁeld. The magnetization remains along y.
If the ﬁnal pulse is about the x-axis, then it rotates the magnetization from the y-axis to the z-axis.
Overall, the sequence returns the magnetization to its starting position.
If the ﬁnal pulse is about the −x-axis, then the magnetization is rotated from the y-axis to the
−z-axis. Overall, the magnetization has been inverted.
4.12
The initial 90◦(x) pulse rotates the magnetization from the z-axis to the −y-axis. For on-resonance
peaks, Ω = 0, so the magnetization does not precess during the delay τ. The ﬁnal 90◦(−x) then
simply undoes the rotation caused by the ﬁrst pulse. Overall, the magnetization is returned to its
starting position.
Ωτ = π/2. During the delay, the magnetization rotates to the x-axis and is therefore not aﬀected
by the ﬁnal 90◦(−x) pulse. The net result is that the magnetization appears along the x-axis.
Ωτ = π. During the delay, the magnetization rotates onto the y-axis. The ﬁnal pulse rotates
the magnetization onto the −z-axis. The equilibrium magnetization is inverted: no observable
transverse magnetization is produced.
x
-y
x
-y
90˚(-x)
90˚(-x)
90˚(-x)
delay τ
x
-y
x
-y
x
-y
delay τ
x
-y
x
-y
x
-y
delay τ
x
-y
Ω = 0
Ωτ = π/2
Ωτ = π
π/2
π
Chapter 4: The vector model 17
The overall eﬀect of the sequence is to produce x-magnetization which varies as M0 sin(Ωτ).
0 π/2 π 3π/2 2π
-M0
M0
Mx
Ωτ
To suppress a strong solvent peak, it is placed on-resonance. The delay τ is then chosen so that
Ωavτ = π/2, where Ωav is the average value of the oﬀset of the peaks we wish to excite.
4.13
The initial 90◦ pulse rotates the equilibrium magnetization to the −y-axis; from there the magnetization
precesses about the z-axis through an angle of Ωτ. The 90◦(y) pulse rotates the x-component
of the magnetization onto the −z-axis.
x
-y
x
-y
Ωτ 90˚(y)delay τ
x
-y
The y-component of the magnetization varies as −M0 cos Ωτ:
0 π/2 π 3π/2 2π
-M0
M0
My
Ωτ
The nulls are located at Ωτ = (2n + 1)π/2, where n = 0, 1, 2, . . .
To suppress the solvent peak, the transmitter frequency is placed in the middle of the peaks of
interest, and then τ is chosen so that Ωτ = π/2, where Ω is the oﬀset of the solvent. With such a
choice, the solvent will not be excited.
Chapter 4: The vector model 18
4.14
Line A is on-resonance, so its magnetization does not precess during the delay τ. The pulse
sequence is, eﬀectively, a 180◦(x) pulse, and so the magnetization is inverted.
For line B, the x-, y- and z-components of the magnetization after each element of the sequence
are:
component after ﬁrst 90◦(x) after τ after second 90◦(x)
x 0 M0 sin Ωτ M0 sin Ωτ
y −M0 −M0 cos Ωτ 0
z 0 0 −M0 cos Ωτ
The ﬁnal pulse is along the x-axis, so leaves the x-component of the magnetization unchanged.
Substituting in the values of Ω and τ we ﬁnd (note that the oﬀset of 100 Hz has to be converted to
rad s−1
):
Mx = M0 sin(2π × 100 × 5 × 10−3
) = M0 sin π = 0
Mz = −M0 cos(2π × 100 × 5 × 10−3
) = −M0 cos π = M0.
The magnetization is therefore returned to the z-axis.
The 90◦ pulse rotates the equilibrium magnetization onto the −y-axis. During the delay τ, the
vector precesses about z to give the following x- and y-components:
Mx = M0 sin Ωτ My = −M0 cos Ωτ.
For line A, oﬀset 50 Hz:
Mx = M0 sin(2π × 50 × 5 × 10−3
) = M0 sin(π/2) = M0
My = −M0 cos(2π × 50 × 5 × 10−3
) = −M0 cos(π/2) = 0.
For line B, oﬀset −50 Hz:
Mx = M0 sin(2π × −50 × 5 × 10−3
) = M0 sin(−π/2) = −M0
My = −M0 cos(2π × −50 × 5 × 10−3
) = −M0 cos(−π/2) = 0.
The two magnetization vectors rotate at the same rate in the opposite sense. After a delay of
τ = 5 ms, they are both aligned along the x-axis, but pointing in opposite directions.
Chapter 5
Fourier transformation and data
processing
5.1
One desirable feature of the dispersion lineshape is that it crosses the frequency axis at the
frequency of the transition. This allows for a more accurate measurement of the chemical shift
than might be possible for the absorption lineshape, especially in the case of broad lines.
In a spectrum containing many peaks, the following features of the dispersion lineshape make it
undesirable:
• It is broader than the absorption lineshape – the long ‘dispersive tails’ may interfere with
nearby, low intensity peaks.
• It is half the height of the absorption lineshape – the SNR is therefore reduced by half.
• The positive part of one peak may be cancelled by the negative part of an adjacent one – in
a complex spectrum, the result can be very diﬃcult to interpret.
5.2
Setting A(ω) = S0/2R, we obtain
S0
2R
=
S0R
R2 + ω2
.
Cancelling the factor of S0 from both sides and inverting the quotient, we obtain
2R =
R2 + ω2
R
.
Hence,
ω2
= 2R2
− R2
= R2
ω = ±R .
Chapter 5: Fourier transformation and data processing 20
The width of the line is therefore 2R in rad s−1
, or R/π in Hz.
5.3
D(ω) can be diﬀerentiated using the product rule:
dD(ω)
dω
=
d
dω
−ω
R2 + ω2
=
−1
R2 + ω2
+
2ω2
(R2 + ω2)2
=
−R2 − ω2 + 2ω2
(R2 + ω2)2
=
ω2 − R2
(R2 + ω2)2
.
At the turning points
dD(ω)
dω
= 0,
so,
ω2 − R2
(R2 + ω2)2
= 0.
The denominator is always non-zero, so the equation can be solved by setting the numerator to
zero:
ω2
− R2
= 0
ω = ±R .
Substituting these values into D(ω):
D(±R) = ∓
R
2R2
= ∓
1
2R
.
These values are the maximum and minimum heights in the lineshape.
There are two values of ω at which D(ω) is half its maximum positive height. At these frequencies,
D(ω) = 1/(4R). Hence,
−ω
R2 + ω2
=
1
4R
.
Inverting the quotients we obtain,
R2 + ω2
ω
= −4R,
so,
ω2
+ 4Rω + R2
= 0.
Chapter 5: Fourier transformation and data processing 21
This is a quadratic equation in ω that can be solved by applying the usual formula:
ω = 1
2 −4R ± 16R2 − 4R2 = R(−2 ±
√
3) .
Similarly, D(ω) = −1/(4R) has two solutions: ω = R(2 ±
√
3) .
The width, Wdisp, is the distance between the outer two half-maximum points, as shown in the
diagram. Its value is
Wdisp = R(2 +
√
3) − R(−2 −
√
3) = 2(2 +
√
3)R.
frequency / rad s-1
1/(2R)
1/(4R)
Wdisp
R
R(2+√3)
R(2-√3)
R(-2-√3)
R(-2+√3)
-R
For comparison, the width of the absorption mode is Wabs = 2R. Therefore, the ratio Wdisp/Wabs =
2 +
√
3 ≈ 3.7 . The dispersion lineshape is almost four times wider than the absorption lineshape.
Chapter 5: Fourier transformation and data processing 22
5.4
Sx
Sy
Sx
SySx
Sy
real imag
real imagreal imag
y
x
y
x
y
x
φ = 3π/4
φ = 2π
φ = 3π/2
(c)
(a) (b)
Sx
Sy
real imag
y
x
φ = 5π/2
(d)
5.5
A 90◦(x) pulse rotates the equilibrium magnetization onto the −y-axis. The resulting spectrum is
phased to absorption, so that magnetization along −y can be said to have a phase φ = 0.
A 90◦(y) pulse rotates the equilibrium magnetization onto the x-axis. This corresponds to a phase
shift of φ = π/2 with respect to the initial experiment.
Chapter 5: Fourier transformation and data processing 23
realreal
y
x
90˚(x)
real
y
x
90˚(-x)
real
y
x
270˚(x)
y
x
90˚(y)
(a) (b)
(a) Applying the pulse about −x rotates the magnetization vector onto y. This corresponds to a
phase shift of φ = π, therefore the spectrum will exhibit a negative absorption lineshape.
(b) A 270◦(x) pulse is equivalent to a 90◦(−x) pulse. The spectrum will be the same as in (a).
5.6
The Larmor frequency of 31P at B0 = 9.4 T is:
ω0
2π
= −
γB0
2π
= −
1.08 × 108 × 9.4
2π
= −1.62 × 108
Hz or −162 MHz.
The phase correction needed at the edge of the spectrum is given by Ωmaxtp, where Ωmax is the
maximum oﬀset. For 31P the maximum oﬀset is 350 ppm, therefore the phase correction is
2π × 162 × 350 × 20 × 10−6
= 7.1 radians .
This corresponds to 407◦ , a signiﬁcant frequency-dependent phase error.
Chapter 5: Fourier transformation and data processing 24
5.7
The intensity of the noise in the spectrum depends on both the amplitude of the noise in the
time-domain, and the acquisition time. So, recording the time-domain signal long after the NMR
signal has decayed just continues to measure noise and no signal. The resulting spectrum will
consequently have a lower SNR than it would for a shorter acquisition time.
A full discussion on how line broadening can be used to improve the SNR is given in section 5.4.2
on p. 96; the matched ﬁlter is discussed in section 5.4.3 on p. 98.
5.8
Shortening the acquisition time discards the time-domain data that contains mostly noise and little
signal. Applying a line broadening weighting function does not discard this section of the timedomain,
but reduces its amplitude relative to the earlier part of the FID. Thus, both methods reduce
the intensity of the noise in the spectrum.
5.9
Enhancing the resolution of the spectrum by the use of a weighting function that combines a rising
exponential and a Gaussian is discussed in section 5.4.4 on p. 98.
Zero ﬁlling improves the ‘deﬁnition’ of the line in the spectrum by increasing the density of data
points in the frequency domain. However, it does not improve the fundamental linewidth as no
real data is added to the time-domain.
5.10
Plots of the sine bell weighting functions are given in Fig. 5.21 on p. 102.
A sine bell that is phase-shifted by 45◦ initially increases over time, therefore partly cancelling
the decay of the FID; the linewidth of the spectrum will therefore be decreased. The subsequent
decay of the sine bell attenuates the noise at the end of the time-domain. The overall eﬀect will be
to enhance the resolution, assuming that the original FID has decayed close to zero by the end of
the acquisition time.
The sine bell with a phase shift of 90◦ is purely a decaying function, which will broaden the lines
in just the same way as a decaying exponential does.
Chapter 5: Fourier transformation and data processing 25
5.11
The peak due to TMS is likely to be a sharp line. Hence, the corresponding time-domain signal
decays slowly, and is therefore more likely to be truncated. The other lines in the spectrum will
usually be broader than TMS, so their time-domain signals decay more rapidly and are less likely
to be truncated.
Truncation artefacts (‘sinc wiggles’) can be suppressed by applying a decaying weighting function.
This will decrease the resolution, and may reduce the SNR.
Chapter 6
The quantum mechanics of one spin
6.1
ˆIzψβ = −1
2ψβ Dirac notation: ˆIz |β = −1
2 |β
ψβ ψα dτ Dirac notation: β|α
ψβ ψβ dτ Dirac notation: β|β
ψ ˆQψ dτ Dirac notation: ψ| ˆQ|ψ
(a) α|α = 1
(b) α|β = 0 or β|α = 0
(c) ˆIz |α = 1
2|α
(d) |ψ = cα|α + cβ |β .
6.2
The expectation value of ˆIy is given by:
Iy =
ψ|ˆIy|ψ
ψ|ψ
.
If |ψ is normalized, ψ|ψ = 1, so the expectation value is given by
Iy = ψ|ˆIy|ψ .
Substituting in |ψ = cα|α + cβ |β , we obtain
Iy = cα α| + cβ β| ˆIy cα |α + cβ |β
= cα cα α|ˆIy|α + cαcβ α|ˆIy|β + cβ cα β|ˆIy|α + cβ cβ β|ˆIy|β
= 1
2 i cαcα α|β − 1
2i cαcβ α|α + 1
2i cβ cα β|β − 1
2 i cβ cβ β|α
= 1
2 i cβ cα − 1
2 i cαcβ .
Chapter 6: The quantum mechanics of one spin 27
To go to the third line, we have used Eq. 6.11 on p. 115,
ˆIy |α = 1
2i |β ˆIy |β = −1
2i |α ,
and to go to the last line, we have used the fact that |α and |β are orthonormal (Eq. 6.5 and Eq. 6.6
on p. 112).
Iy can be interpreted as the average value of the y-component of angular momentum when
measured for a large number of spins, each of which has the same wavefunction |ψ .
6.3
The matrix representation of ˆIx is
Ix =
⎛
⎜⎜⎜⎜⎜⎜⎜⎝
α|ˆIx|α α|ˆIx|β
β|ˆIx|α β|ˆIx|β
⎞
⎟⎟⎟⎟⎟⎟⎟⎠
=
⎛
⎜⎜⎜⎜⎜⎜⎜⎝
1
2 α|β 1
2 α|α
1
2 β|β 1
2 β|α
⎞
⎟⎟⎟⎟⎟⎟⎟⎠
=
⎛
⎜⎜⎜⎜⎜⎜⎜⎝
0 1
2
1
2 0
⎞
⎟⎟⎟⎟⎟⎟⎟⎠ .
To go to the second line, we have used Eq. 6.10 on p. 115,
ˆIx|α = 1
2|β ˆIx|β = 1
2|α ,
and to go to the last line we have used the fact that |α and |β are orthonormal (Eq. 6.5 and Eq. 6.6
on p. 112).
Similarly,
Iy =
⎛
⎜⎜⎜⎜⎜⎜⎜⎝
α|ˆIy|α α|ˆIy|β
β|ˆIy|α β|ˆIy|β
⎞
⎟⎟⎟⎟⎟⎟⎟⎠
=
⎛
⎜⎜⎜⎜⎜⎜⎜⎝
1
2i α|β −1
2i α|α
1
2i β|β −1
2i β|α
⎞
⎟⎟⎟⎟⎟⎟⎟⎠
=
⎛
⎜⎜⎜⎜⎜⎜⎜⎝
0 −1
2i
1
2i 0
⎞
⎟⎟⎟⎟⎟⎟⎟⎠ .
Chapter 6: The quantum mechanics of one spin 28
6.4
Starting with the expression for Iy , and substituting in cα = rα exp(i φα) and cβ = rβ exp(i φβ) we
ﬁnd:
Iy = 1
2i cβ cα − 1
2 i cαcβ
= 1
2i rαrβ exp(−i φβ) exp(i φα) − rαrβ exp(−i φα) exp(i φβ)
= 1
2i rαrβ exp −i (φβ − φα) − exp i (φβ − φα)
= 1
2i rαrβ exp i (φβ − φα) − exp −i (φβ − φα) ,
where to go to the last line we have multiplied top and bottom by i.
Applying the identity
exp(iθ) − exp(−iθ) ≡ 2i sin θ
to the above expression gives
Iy = rαrβ sin(φβ − φα).
The bulk y-magnetization is then given by
My = γ Ix
(1)
+ γ Ix
(2)
+ . . .
= γr(1)
α r(1)
β sin(φ(1)
β − φ(1)
α ) + γr(2)
α r(2)
β sin(φ(2)
β − φ(2)
α ) + . . .
= γNrαrβ sin(φβ − φα).
At equilibrium, the phases φ are randomly distributed, and so sin(φβ − φα) is randomly distributed
between ±1. As a result, the equilibrium y-magnetization is zero.
6.5
Starting from Eq. 6.31 on p. 124 and premultiplying by β|, we obtain:
dcα(t)
dt
|α +
dcβ (t)
dt
|β = −1
2i Ωcα(t)|α + 1
2i Ωcβ (t)|β
β|
dcα (t)
dt
|α + β|
dcβ (t)
dt
|β = β| −1
2i Ωcα(t) |α + β| 1
2i Ωcβ (t) |β .
The derivatives of cα and cβ , and the quantities in square brackets, are numbers, so the above
expression can be rearranged to give
dcα (t)
dt
β|α +
dcβ (t)
dt
β|β = −1
2i Ωcα(t) β|α + 1
2i Ωcβ (t) β|β
dcβ (t)
dt
= 1
2i Ωcβ (t).
Chapter 6: The quantum mechanics of one spin 29
To go to the last line, we have used the orthonormality property of |α and |β .
Substituting Eq. 6.58 into the left-hand side of Eq. 6.57 gives:
dcβ (t)
dt
=
d
dt
cβ (0) exp 1
2i Ωt
= 1
2 i Ωcβ (0) exp 1
2i Ωt
= 1
2 i Ωcβ (t).
Eq. 6.58 is indeed the solution.
6.6
The expectation value of ˆIy is
Iy = 1
2i cβ cα − 1
2i cα cβ .
Substituting in the expressions for how cα and cβ vary under free evolution (Eq. 6.34 on p. 125)
gives:
Iy (t) = 1
2i cβ (0) exp −1
2i Ωt cα(0) exp −1
2i Ωt − 1
2 i cα (0) exp 1
2i Ωt cβ (0) exp 1
2i Ωt
= 1
2i cβ (0)cα (0) exp (−i Ωt) − 1
2 i cα(0)cβ (0) exp (i Ωt)
= 1
2i cβ (0)cα (0) [cos(Ωt) − i sin(Ωt)] − 1
2i cα(0)cβ (0) [cos(Ωt) + i sin(Ωt)]
= cos(Ωt) 1
2i cβ (0)cα (0) − 1
2 i cα(0)cβ (0) + sin(Ωt) 1
2cα(0)cβ (0) + 1
2cβ (0)cα (0)
= cos(Ωt) Iy (0) + sin(Ωt) Ix (0).
To go to the third line, the identities
exp(iθ) ≡ cos θ + i sin θ exp(−iθ) ≡ cos θ − i sin θ
were used, and to go to the last line, the expressions for Ix and Iy in terms of cα and cβ were
used (Eqs 6.12 and 6.13 on p. 115).
This result is summarized in the diagram below. The grey arrow shows the initial position, and the
black arrow shows the position after time t.
<Ix>
<Ix>(0)
<Iy>(0)
<Iy>
Ωt
Chapter 6: The quantum mechanics of one spin 30
6.7
The matrix representation of the density operator is given by:
ρ =
⎛
⎜⎜⎜⎜⎜⎜⎜⎝
α|ˆρ|α α|ˆρ|β
β|ˆρ|α β|ˆρ|β
⎞
⎟⎟⎟⎟⎟⎟⎟⎠ ≡
⎛
⎜⎜⎜⎜⎜⎜⎜⎝
ρ11 ρ12
ρ21 ρ22
⎞
⎟⎟⎟⎟⎟⎟⎟⎠ .
We can now calculate the ρ11 element (for clarity, the overbars indicating the ensemble averaging
have been omitted until the last line):
ρ11 = α|ˆρ|α
= α|ψ ψ|α
= α| cα |α + cβ |β cα α| + cβ β| |α
= cα α|α + cβ α|β cα α|α + cβ β|α
= cαcα.
To go to the second line, the deﬁnition of ˆρ was inserted, and on the third line |ψ was expressed
as a superposition of |α and |β .
The other elements can be calculated in a similar way to give:
ρ12 = cαcβ ρ21 = cβ cα ρ22 = cβ cβ .
Hence,
ρ =
⎛
⎜⎜⎜⎜⎜⎜⎜⎝
cαcα cαcβ
cβ cα cβ cβ
⎞
⎟⎟⎟⎟⎟⎟⎟⎠ .
Chapter 7
Product operators
7.1
• exp(−i θˆIx)ˆIy exp(i θˆIx) represents a rotation of ˆIy about x through angle θ. From Fig. 7.4 (a)
on p. 152, ˆIy is transformed into ˆIz. Hence,
ˆIy
θˆIx
−−→ cos θ ˆIy + sin θ ˆIz.
This is consistent with the identity on line one of Table 7.1 on p. 146.
• exp(−i θ ˆSy) ˆSz exp(i θ ˆSy). From (b) of Fig. 7.4 on p. 152, ˆSz is transformed into ˆSx by a
rotation about y:
ˆSz
θ ˆSy
−−→ cos θ ˆSz + sin θ ˆSx.
• exp(−i θˆIx)ˆIx exp(i θˆIx). Rotating ˆIx about the x-axis has no eﬀect:
ˆIx
θˆIx
−−→ ˆIx.
• exp(−i θˆIz)(−ˆIy) exp(i θˆIz). Fig. 7.4 (c) shows the eﬀect of a rotation about z on −ˆIy: the
result is a transformation to ˆIx. Hence,
−ˆIy
θˆIz
−−→ − cos θ ˆIy + sin θ ˆIx.
• exp(−i (θ/2)ˆIy)ˆIx exp(i (θ/2)ˆIy). This represents the rotation of ˆIx about y through angle θ/2.
From Fig. 7.4 (b), ˆIx is transformed to −ˆIz. Hence,
ˆIx
(θ/2)ˆIy
−−−−−→ cos(θ/2) ˆIx − sin(θ/2) ˆIz.
• exp(i θˆIz)(−ˆIz) exp(−i θˆIz). Careful inspection of the arguments of the exponentials reveals
that this represents a z-rotation through angle −θ i.e. the rotation is in a clockwise sense. In
this case, it does not matter as −ˆIz is unaﬀected by a rotation about the z-axis:
−ˆIz
−θˆIz
−−−→ −ˆIz.
Chapter 7: Product operators 32
7.2
The 90◦(x) pulse rotates the equilibrium magnetization (represented by ˆIz) onto the −y-axis:
ˆIz
(π/2)ˆIx
−−−−−→ cos(π/2) ˆIz − sin(π/2) ˆIy
= −ˆIy.
This transverse term evolves under the oﬀset during the delay τ to give
−ˆIy
ΩτˆIz
−−−→ − cos(Ωτ) ˆIy + sin(Ωτ) ˆIx,
where (c) of Fig. 7.4 on p. 152 has been used.
The 180◦(y) pulse does not aﬀect the ˆIy term, but inverts the ˆIx term:
− cos(Ωτ) ˆIy + sin(Ωτ) ˆIx
πˆIy
−−→ − cos(Ωτ) ˆIy + cos π sin(Ωτ) ˆIx − sin π sin(Ωτ) ˆIz
= − cos(Ωτ) ˆIy − sin(Ωτ) ˆIx.
Now we consider the evolution during the second delay. Taking each term separately, we obtain
− cos(Ωτ) ˆIy
ΩτˆIz
−−−→ − cos(Ωτ) cos(Ωτ) ˆIy + sin(Ωτ) cos(Ωτ) ˆIx,
− sin(Ωτ) ˆIx
ΩτˆIz
−−−→ − cos(Ωτ) sin(Ωτ) ˆIx − sin(Ωτ) sin(Ωτ) ˆIy.
Combining these terms gives the ﬁnal result as
− cos2
(Ωτ) + sin2
(Ωτ) ˆIy = −ˆIy,
where the terms in ˆIx cancel, and the identity cos2 θ + sin2
θ ≡ 1 has been used. At the end of the
sequence, the magnetization has been refocused onto the −y-axis, irrespective of the oﬀset.
7.3
ˆIy
(π/2)ˆIy
−−−−−→ ˆIy
ˆIy
−(π/2)ˆIy
−−−−−−→ ˆIy
ˆSy
π ˆSy
−−→ ˆSy.
In all three cases, the pulse is applied about the same axis along which the magnetization is aligned,
therefore the magnetization is unaﬀected.
Chapter 7: Product operators 33
In the following cases, we refer to Fig. 7.4 on p. 152 to determine how the operator is transformed
by the rotation.
ˆIx
−πˆIy
−−−→ cos(−π)ˆIx − sin(−π)ˆIz
= −ˆIx.
In this case the magnetization is simply inverted.
The diﬀerence between the next two examples is the sense of the 90◦ rotation.
ˆIz
(π/2)ˆIy
−−−−−→ cos(π/2) ˆIz + sin(π/2) ˆIx
= ˆIx.
ˆIz
−(π/2)ˆIy
−−−−−−→ cos(−π/2) ˆIz + sin(−π/2) ˆIx
= −ˆIx.
The next two are simply inversions:
ˆSz
π ˆSy
−−→ cos π ˆSz + sin π ˆSx
= − ˆSz.
ˆIz
−πˆIy
−−−→ cos(−π) ˆIz + sin(−π) ˆIx
= −ˆIz.
7.4
The 90◦(x) pulse rotates the equilibrium magnetization ˆIz to −ˆIy. Free evolution is a rotation about
z, so the state of the system after the delay τ is
− cos(Ωτ) ˆIy + sin(Ωτ) ˆIx.
The 90◦(y) pulse does not aﬀect the ˆIy term, but rotates ˆIx to −ˆIz. The ﬁnal result is
− cos(Ωτ) ˆIy − sin(Ωτ) ˆIz.
The pulse sequence has therefore produced transverse magnetization along y, whose amplitude
varies as − cos(Ωτ). This becomes zero if cos(Ωτ) = 0. Hence, there is a null at Ωτ = π/2, which
corresponds to an oﬀset of Ω = π/(2τ) in rad s−1
, or 1/(4τ) in Hz.
There is a maximum in the excitation when cos(Ωτ) = ±1. This occurs at oﬀsets satisfying
Ωτ = nπ where n = 0, 1, 2, . . . i.e. Ω = (nπ)/τ or n/(2τ) in Hz.
Chapter 7: Product operators 34
7.5
Figure 7.6 (b) on p. 156 shows that, as a result of evolution of the scalar coupling, the in-phase
term −ˆI1y is partly transformed into the anti-phase term 2ˆI1x ˆI2z; the angle of rotation is πJ12τ. This
is represented as:
−ˆI1y
2πJ12τˆI1z ˆI2z
−−−−−−−−−→ − cos(πJ12τ) ˆI1y + sin(πJ12τ) 2ˆI1x ˆI2z.
Using the same ﬁgure, we see that −2ˆI1x ˆI2z is partly transformed to −ˆI1y:
−2ˆI1x ˆI2z
2πJ12τˆI1z ˆI2z
−−−−−−−−−→ − cos(πJ12τ) 2ˆI1x ˆI2z − sin(πJ12τ) ˆI1y.
Similarly,
ˆSx
2πJIS (τ/2)ˆIz ˆSz
−−−−−−−−−−−→ cos(πJIS τ/2) ˆSx + sin(πJIS τ/2) 2ˆIz ˆSy.
ˆI2y
2πJ12τˆI1z ˆI2z
−−−−−−−−−→ cos(πJ12τ) ˆI2y − sin(πJ12τ) 2ˆI1z ˆI2x.
2ˆI1z ˆI2y
2πJ12τˆI1z ˆI2z
−−−−−−−−−→ cos(πJ12τ) 2ˆI1z ˆI2y − sin(πJ12τ) ˆI2x.
ˆI2z
2πJ12τˆI1z ˆI2z
−−−−−−−−−→ ˆI2z.
In the last example we see that z-magnetization is not aﬀected by evolution under coupling simply
because the Hamiltonian for coupling only contains ˆIz operators.
7.6
The evolution is determined by the Hamiltonian given in Eq. 7.14 on p. 154:
ˆHtwo spins = Ω1 ˆI1z + Ω2 ˆI2z + 2πJ12 ˆI1z ˆI2z.
We will now work out the eﬀect in turn of the three terms in the Hamiltonian. The ﬁrst is a rotation
about z:
ˆI1y
Ω1tˆI1z
−−−−→ cos(Ω1t) ˆI1y − sin(Ω1t) ˆI1x.
The second term, Ω2 ˆI2z, does not need to be considered as spin-two operators have no eﬀect on
spin-one operators. Finally, we consider the eﬀect of evolution under scalar coupling:
cos(Ω1t) ˆI1y − sin(Ω1t) ˆI1x
2πJ12tˆI1z ˆI2z
−−−−−−−−→
cos(πJ12t) cos(Ω1t) ˆI1y
y-magnetization
− sin(πJ12t) cos(Ω1t) 2ˆI1x ˆI2z
− cos(πJ12t) sin(Ω1t) ˆI1x
x-magnetization
− sin(πJ12t) sin(Ω1t) 2ˆI1y ˆI2z.
Chapter 7: Product operators 35
The NMR signal is given by:
S (t) = Mx + iMy
= − cos(πJ12t) sin(Ω1t) + i cos(πJ12t) cos(Ω1t)
= i cos(πJ12t) [cos(Ω1t) + i sin(Ω1t)]
= i cos(πJ12t) exp(i Ω1t)
= 1
2 i exp(i πJ12t) + exp(−i πJ12t) exp(i Ω1t)
= 1
2 i exp (i[Ω1 + πJ12]t) + 1
2i exp (i[Ω1 − πJ12]t) .
To go to the fourth line, we have used the identity cos θ + i sin θ ≡ exp(i θ), and to go to the
ﬁfth line, we have used cos θ ≡ 1
2[exp(i θ) + exp(−i θ)]. Finally, to go to the sixth line we have
multiplied out the square brackets. Fourier transformation of this signal gives a positive line at
Ω1 + πJ12, and a second positive line at Ω1 − πJ12 i.e. an in-phase doublet on spin one. The
factor of i corresponds to a phase shift of 90◦, so the imaginary part of the spectrum contains the
absorption mode lineshape.
ω
real
imaginary
Ω1-πJ12
2πJ12
Ω1+πJ12
A similar line of argument gives the observable signal arising from 2ˆI1y ˆI2z as
S (t) = 1
2i exp (i[Ω1 + πJ12]t) − 1
2i exp (i[Ω1 − πJ12]t) .
The corresponding spectrum is an anti-phase doublet on spin one. Again, the factor of i means
that the absorption mode lines will appear in the imaginary part of the spectrum.
ω
real
imaginary
Ω1-πJ12
2πJ12
Ω1+πJ12
Chapter 7: Product operators 36
7.7
ˆI1y represents in-phase magnetization on spin one, aligned along the y-axis. The resulting spectrum
will be an in-phase doublet centred on the oﬀset of spin one, both peaks of which are in the
absorption mode.
ˆI2x represents in-phase magnetization on spin two. However, it is aligned along the x-axis, so has
a phase of 3π/2 relative to the y-axis. The spectrum therefore comprises an in-phase doublet that
is centred on the oﬀset of spin two, with both peaks in the dispersion mode.
2ˆI1y ˆI2z represents magnetization on spin one that is anti-phase with respect to spin two, and aligned
along y. The spectrum is therefore an anti-phase doublet in the absorption mode.
2ˆI1z ˆI2x represents anti-phase magnetization on spin two. It is aligned along x, so the lineshape
will be dispersive. Therefore, the spectrum is an anti-phase spin-two doublet with the dispersion
lineshape.
Ω1
I1y
2I1yI2z
ω
Ω2
I2x
2I1zI2x
7.8
In-phase magnetization ˆI1x is rotated in the xz-plane towards −ˆI1z by the application of the y-pulse
of duration tp.
ˆI1x
ω1tp ˆI1y
−−−−−→ cos(ω1tp) ˆI1x − sin(ω1tp) ˆI1z
A 180◦ pulse about y applied only to spin two changes the sign of the anti-phase magnetization on
spin one.
2ˆI1x ˆI2z
−πˆI2y
−−−−→ cos(−π) 2ˆI1x ˆI2z + sin(−π) 2ˆI1x ˆI2x
= −2ˆI1x ˆI2z
Chapter 7: Product operators 37
In-phase magnetization on spin one is allowed to evolve under coupling for time t, thus generating
anti-phase magnetization on the same spin.
−ˆI1x
2πJ12tˆI1z ˆI2z
−−−−−−−−→ − cos(πJ12t) ˆI1x − sin(πJ12t) 2ˆI1y ˆI2z
Letting each term act sequentially, we obtain
2ˆI1x ˆI2z
(π/2)ˆI1y
−−−−−−→ −2ˆI1z ˆI2z
(π/2)ˆI2y
−−−−−−→ −2ˆI1z ˆI2x.
Note that the spin-one operators do not act on spin-two operators and vice versa. The net result is
that the non-selective 90◦(y) pulse has caused a coherence transfer from spin one to spin two.
Transverse, in-phase magnetization on the S spin evolves under oﬀset for time t. The oﬀset term
for the I spin has no eﬀect on the ˆSx.
ˆSx
ΩItˆIz
−−−−→ ˆSx
ΩS t ˆSz
−−−−→ cos(ΩS t) ˆSx + sin(ΩS t) ˆSy
Anti-phase magnetization on spin two evolves under coupling to generate in-phase magnetization
on the same spin.
−2ˆI1z ˆI2y
2πJ12tˆI1z ˆI2z
−−−−−−−−→ − cos(πJ12t) 2ˆI1z ˆI2y + sin(πJ12t) ˆI2x
7.9
The Hamiltonian for free evolution is given by Eq. 7.14 on p. 154:
ˆHtwo spins = Ω1 ˆI1z + Ω2 ˆI2z + 2πJ12 ˆI1z ˆI2z.
The spin echo refocuses the evolution due to oﬀset, so we only need to consider the evolution of
2ˆI1x ˆI2z under coupling, which gives
2ˆI1x ˆI2z
2πJ12τˆI1z ˆI2z
−−−−−−−−−→ cos(πJ12τ) 2ˆI1x ˆI2z + sin(πJ12τ) ˆI1y.
The π pulse about the x-axis acts on both spins, leaving ˆI1x unaﬀected, but inverting ˆI2z and ˆI1y:
cos(πJ12τ) 2ˆI1x ˆI2z + sin(πJ12τ) ˆI1y
π(ˆI1x+ˆI2x)
−−−−−−−→ − cos(πJ12τ) 2ˆI1x ˆI2z − sin(πJ12τ) ˆI1y.
Finally, evolution under coupling during the second delay gives
− cos(πJ12τ) 2ˆI1x ˆI2z − sin(πJ12τ) ˆI1y
2πJ12τˆI1z ˆI2z
−−−−−−−−−→
− cos2
(πJ12τ) ˆI1x ˆI2z − sin(πJ12τ) cos(πJ12τ) ˆI1y − cos(πJ12τ) sin(πJ12τ) ˆI1y + sin2
(πJ12τ) 2ˆI1x ˆI2z
= − cos2
(πJ12τ) − sin2
(πJ12τ) 2ˆI1x ˆI2z − [2 cos(πJ12τ) sin(πJ12τ)] ˆI1y
= − cos(2πJ12τ) 2ˆI1x ˆI2z − sin(2πJ12τ) ˆI1y.
Chapter 7: Product operators 38
To go to the last line, we have used the identities cos2 θ − sin2
θ ≡ cos 2θ and 2 cos θ sin θ ≡ sin 2θ.
By a similar method we can show:
2ˆI1y ˆI2z
τ−πx−τ
−−−−−→ cos(2πJ12τ) 2ˆI1y ˆI2z − sin(2πJ12τ) ˆI1x.
The eﬀect of the τ − πy − τ spin echo on spin-one and spin-two terms is shown in the table below:
ﬁnal state
initial state × cos (2πJ12τ) × sin (2πJ12τ)
ˆI1x −ˆI1x −2ˆI1y ˆI2z
ˆI1y ˆI1y −2ˆI1x ˆI2z
2ˆI1x ˆI2z 2ˆI1x ˆI2z ˆI1y
2ˆI1y ˆI2z −2ˆI1y ˆI2z ˆI1x
ˆI2x −ˆI2x −2ˆI1z ˆI2y
ˆI2y ˆI2y −2ˆI1z ˆI2x
2ˆI1z ˆI2x 2ˆI1z ˆI2x ˆI2y
2ˆI1z ˆI2y −2ˆI1z ˆI2y ˆI2x
The results for the in- and anti-phase operators on spin two can be obtained from those for spin
one simply by swapping the labels 1 and 2.
Likewise for the τ − πx − τ spin echo:
ﬁnal state
initial state × cos (2πJ12τ) × sin (2πJ12τ)
ˆI2x ˆI2x ˆI1z ˆI2y
ˆI2y −ˆI2y 2ˆI1z ˆI2x
2ˆI1z ˆI2x −2ˆI1z ˆI2x −ˆI2y
2ˆI1z ˆI2y 2ˆI1z ˆI2y −ˆI2x
7.10
A spin echo in a homonuclear two-spin system is equivalent to:
(a) evolution of the coupling for time 2τ,
(b) a 180◦(x) pulse.
Chapter 7: Product operators 39
Applying this to the ﬁrst example, we obtain
ˆI2y
τ−πx−τ
−−−−−→ − cos(2πJ12τ) ˆI2y + sin(2πJ12τ) 2ˆI1z ˆI2x.
For complete transformation to 2ˆI1z ˆI2x, we need sin(2πJ12τ) = 1 and cos(2πJ12τ) = 0. These
occur when 2πJ12τ = π/2, i.e. τ = 1/(4J12).
ˆI1x
τ−πx−τ
−−−−−→ cos(2πJ12τ)ˆI1x + sin(2πJ12τ)2ˆI1y ˆI2z.
Setting 2πJ12τ = π/4 gives cos(2πJ12τ) = sin(2πJ12τ) = 1/
√
2. The required delay is therefore
τ = 1/(8J12).
To achieve conversion to −ˆI1x, we need cos(2πJ12τ) = −1 and sin(2πJ12τ) = 0 i.e. τ = 1/(2J12).
2ˆI1z ˆI2x
τ−πx−τ
−−−−−→ − cos(2πJ12τ)2ˆI1z ˆI2x − sin(2πJ12τ)ˆI2y.
Setting the delay to τ = 1/(4J12) gives complete conversion to in-phase magnetization.
7.11
The pulse sequence is given in Fig. 7.14 on p. 168:
I
S
ττ
The 180◦(x) pulse is applied to only the S spin, so the evolution of the oﬀset of the S spin will be
refocused. We need to consider the evolution of the coupling. Starting with ˆSx, the state of the
system after the ﬁrst delay is
cos(πJ12τ) ˆSx + sin(πJ12τ) 2ˆIz ˆSy.
The 180◦(x) pulse is applied only to the S spin, and so does not aﬀect ˆIz or ˆSx. However, the term
in ˆSy changes sign to give:
cos(πJ12τ) ˆSx − sin(πJ12τ) 2ˆIz ˆSy.
Evolution of the coupling during the second delay gives
cos2
(πJ12τ) + sin2
(πJ12τ) ˆSx + [sin(πJ12τ) cos(πJ12τ) − cos(πJ12τ) sin(πJ12τ)] 2ˆIz ˆSy = ˆSx,
where the anti-phase terms cancel, and the identity cos2 θ + sin2
θ ≡ 1 has been used. The evolution
of the coupling has therefore been refocused.
Repeating the calculation for the anti-phase term, we see that 2ˆIz ˆSx is unaﬀected by the spin echo
sequence. Again, the coupling is refocused.
Chapter 7: Product operators 40
Both operators are unchanged, which is the same eﬀect that a 180◦(x) pulse to the S spin would
have:
ˆSx
π ˆSx
−−−→ ˆSx
2ˆIz ˆSx
π ˆSx
−−−→ 2ˆIz ˆSx.
Likewise, the operators ˆIx and 2ˆIx ˆSz will have their evolution under coupling refocused. However,
as the 180◦(x) pulse is not applied to the I spin, the oﬀset will not be refocused, but will evolve for
the duration of the spin echo (time 2τ).
7.12
The pulse sequence for the INEPT experiment is reproduced below from Fig. 7.15 on p. 172:
τ1 τ1 τ2 τ2
y
A C
B
I
S
At the end of period A it was shown in section 7.10.2 on p. 172 that the state of the spin system is
kI cos(2πJIS τ1) ˆIy − kI sin(2πJIS τ1) ˆIx ˆSz.
The purpose of the two 90◦ pulses in period B is to transfer the anti-phase magnetization (the
second term) from the I spin to the S spin. This requires the pulse acting on the I spin to cause the
transformation ˆIx → ˆIz, which requires a rotation about the y-axis.
If the initial 90◦ pulse is about the −x-axis, it rotates the equilibrium kI ˆIz to kI ˆIy. At the end of the
spin echo in period A, the system is in the following state:
−kI cos(2πJIS τ1) ˆIy + kI sin(2πJIS τ1) 2ˆIx ˆSz.
As before, the ˆIy term is not aﬀected by the 90◦(y) pulse on the I spin, and can be discarded. The
two pulses aﬀect the ant-phase term as follows:
kI sin(2πJIS τ1) 2ˆIx ˆSz
(π/2)ˆIy
−−−−−→ −kI sin(2πJIS τ1) 2ˆIz ˆSz
(π/2) ˆSx
−−−−−→ kI sin(2πJIS τ1) 2ˆIz ˆSy.
This term evolves under coupling during the spin echo in C to give:
kI cos(2πJIS τ2) sin(2πJIS τ1) 2ˆIz ˆSy − kI sin(2πJIS τ2) sin(2πJIS τ1) ˆSx,
the observable term of which is the one in ˆSx.
The 90◦(x) pulse acting on the S spin during B also rotates equilibrium kS ˆSz to −kI ˆSy, which
evolves during the spin echo in C to give:
−kS cos(2πJIS τ2) ˆSy + kS sin(2πJIS τ2) 2ˆIz ˆSx.
Chapter 7: Product operators 41
This also has an observable term in ˆSy. Hence, the two observable terms are combined to give:
−kS cos(2πJIS τ2) ˆSy − kI sin(2πJIS τ2) sin(2πJIS τ1) ˆSx.
The ﬁrst term is unaﬀected by changing the phase of the I spin 90◦ pulse from x to −x, whereas
the second term changes sign.
7.13
• By deﬁnition, ˆI+ has coherence order +1.
• ˆIz is unaﬀected by a z-rotation, so has coherence order zero.
• ˆI− has coherence order −1, again by deﬁnition.
• Using the deﬁnitions of ˆI1+ and ˆI1− (Eq. 7.28 on p. 178) as applied to spin one:
ˆI1+ ≡ ˆI1x + i ˆI1y
ˆI1− ≡ ˆI1x − i ˆI1y,
we can write ˆI1x as:
ˆI1x ≡ 1
2
ˆI1+ + ˆI1− .
Therefore, ˆI1x is an equal mixture of coherence orders +1 and −1.
• Similarly, ˆI2y can be written as
ˆI2y ≡ 1
2i
ˆI2+ − ˆI2− .
Hence, 2ˆI1z ˆI2y can be written as
2ˆI1z ˆI2y ≡ 2 × 1
2i
ˆI1z ˆI2+ − ˆI2− ,
which is an equal mixture of coherence orders +1 and −1, found by summing the coherence
orders of spins one and two (spin one has coherence order zero).
• Since both ˆI1z and ˆI2z have coherence order zero, so does 2ˆI1z ˆI2z.
• 2ˆI1+ ˆI2− has coherence order zero since the coherence order of spin one is +1 and that of
spin two is −1.
• 2ˆI1x ˆI2y can be written as:
2ˆI1x ˆI2y ≡ 2 × 1
2
ˆI1+ + ˆI1− × 1
2i
ˆI2+ − ˆI2−
≡ 1
2i
ˆI1+ ˆI2+ − ˆI1− ˆI2− − ˆI1+ ˆI2− + ˆI1− ˆI2+ .
2ˆI1x ˆI2y is therefore an equal mixture of coherence orders +2 and −2, double-quantum
coherence, and coherence order 0, zero-quantum coherence.
Chapter 7: Product operators 42
7.14
Using the deﬁnitions of ˆIi± given by Eq. 7.28 on p. 178, we can write 2ˆI1x ˆI2y as:
2ˆI1x ˆI2y ≡ 2 × 1
2
ˆI1+ + ˆI1− × 1
2i
ˆI2+ − ˆI2−
≡ 1
2i
ˆI1+ ˆI2+ − ˆI1− ˆI2−
double-quantum part
+ 1
2i
ˆI1− ˆI2+ − ˆI1+ ˆI2−
zero-quantum part
.
The other relationships in the table can be veriﬁed in the same way.
7.15
The ﬁrst 90◦(x) pulse rotates the equilibrium ˆI1z to −ˆI1y. During the spin echo sequence, the oﬀset
is refocused, but the coupling evolves throughout. The state of the spin system at the end of the
spin echo is
cos(2πJ12τ) ˆI1y − sin(2πJ12τ) 2ˆI1x ˆI2z.
The ﬁnal pulse acts to give
cos(2πJ12τ) ˆI1z + sin(2πJ12τ) 2ˆI1x ˆI2y.
Using the deﬁnitions of ˆDQy and ˆZQy given in the last table of section 7.12.1 on p. 178, we see
that we can rewrite the second term as
1
2 sin(2πJ12τ) ˆDQy − ˆZQy ,
which is a mixture of double- and zero-quantum coherence.
The amplitude of this multiple quantum term is a maximum when sin(2πJ12τ) = 1, which occurs
when τ = 1/(4J12).
Starting with equilibrium magnetization on spin two, ˆI2z, the terms present after the ﬁnal pulse are
cos(2πJ12τ) ˆI2z + sin(2πJ12τ) 2ˆI1y ˆI2x;
we have taken the terms from the previous calculation and swapped the labels 1 and 2. Again, from
the deﬁnitions of ˆDQy and ˆZQy in section 7.12.1 on p. 178, we can write the multiple quantum
term as
1
2 sin(2πJ12τ) ˆDQy + ˆZQy .
Therefore, adding this term to the one originating from ˆI1z, we obtain;
1
2 sin(2πJ12τ) ˆDQy − ˆZQy + 1
2 sin(2πJ12τ) ˆDQy + ˆZQy = sin(2πJ12τ) ˆDQy,
which is pure double-quantum coherence. It is a rather unusual feature of this sequence that, in a
two-spin system, it generates pure double-quantum coherence.
Chapter 7: Product operators 43
7.16
From the table on p. 180, ˆZQx is equal to 2ˆI1x ˆI2x + 2ˆI1y ˆI2y . Zero-quantum coherence between
spins one and two does not evolve under the coupling between these two spins, so we need only
consider the evolution under oﬀset. Considering ﬁrst the 2ˆI1x ˆI2x term:
2ˆI1x ˆI2x
Ω1tˆI1z+Ω2tˆI2z
−−−−−−−−−−→ 2 cos(Ω1t) ˆI1x + sin(Ω1t) ˆI1y cos(Ω2t) ˆI2x + sin(Ω2t) ˆI2y .
We will now look at the 2ˆI1y ˆI2y term;
2ˆI1y ˆI2y
Ω1t ˆI1z+Ω2t ˆI2z
−−−−−−−−−−−→ 2 cos(Ω1t) ˆI1y − sin(Ω1t)ˆI1x cos(Ω2t) ˆI2y − sin(Ω2t) ˆI2x .
Collecting these terms together, we obtain:
[cos(Ω1t) cos(Ω2t) + sin(Ω1t) sin(Ω2t)] (2ˆI1x ˆI2x + 2ˆI1y ˆI2y)
+ [sin(Ω1t) cos(Ω2t) − cos(Ω1t) sin(Ω2t)] (2ˆI1y ˆI2x − 2ˆI1x ˆI2y).
Using the identities:
cos(A − B) = cos A cos B + sin A sin B
sin(A − B) = sin A cos B − cos A sin B,
and the deﬁnitions of ˆZQx and ˆZQy:
ˆZQx ≡ (2ˆI1x ˆI2x + 2ˆI1y ˆI2y) ˆZQy ≡ (2ˆI1y ˆI2x − 2ˆI1x ˆI2y),
we obtain
cos ([Ω1 − Ω2]t) ˆZQx + sin ([Ω1 − Ω2]t) ˆZQy.
Chapter 8
Two-dimensional NMR
8.1
In each example, the preparation period is highlighted with a grey box, and the mixing period with
a grey box with a dashed border.
t2t1 t2t1
t2
t1
τ τ
τ1 τ1
y
τ2 τ2 t2
t1
I
S
τ τ
t2
t1
I
S
τ
t2
t1
I
S
t2
t1
τmix
τ1 τ1 τ2 τ2
t2
t1
I
S
COSY
HSQC
HMQC HMBC
HETCOR
DQF COSY
DQ spectroscopy TOCSY
Chapter 8: Two-dimensional NMR 45
8.2
ω2
1
2
3
4
5
6
t1
ω2
t1
4
5
61
2 3
1, 2 and 3 are cross-sections of the damped cosine wave, whose amplitude provides the modulation
in t1. The period is the same for each wave, and the amplitude increases as we approach the centre
of the peak in ω2.
4, 5 and 6 are cross-sections through the ω2 dimension. The amplitude and sign of the peak is
modulated by a damped cosine wave in t1.
8.3
The COSY pulse sequence is given in Fig. 8.8 on p. 195.
t2t1
Starting with equilibrium magnetization on spin two, the state of the system at t2 = 0 can be
determined from terms [1]–[4] on p. 195 by swapping the spin labels 1 and 2. The result is:
− cos (πJ12t1) cos (Ω2t1) ˆI2z [1]
− sin (πJ12t1) cos (Ω2t1) 2ˆI1y ˆI2x [2]
+ cos (πJ12t1) sin (Ω2t1) ˆI2x [3]
− sin (πJ12t1) sin (Ω2t1) 2ˆI1y ˆI2z. [4]
The observable terms are [3] and [4]. The operator in term [3] is ˆI2x, which will give rise to a
doublet on spin two in the ω2 dimension. It is modulated in t1 by sin(Ω2t1) i.e. at the oﬀset of spin
two. Thus, [3] produces a diagonal-peak multiplet.
Chapter 8: Two-dimensional NMR 46
The operator in term [4] is 2ˆI1y ˆI2z; this gives rise to an anti-phase doublet centred at the oﬀset of
spin one in the ω2 dimension. It is also modulated in t1 by sin(Ω2t1). Therefore, it produces a
cross-peak multiplet.
It was shown in section 7.5.2 on p. 158 that the evolution of 2ˆI1y ˆI2z during t2 gives rise to the
following time domain signal:
1
2i exp(i[Ω1 + πJ12]t2) − 1
2i exp(i[Ω1 − πJ12]t2).
Imposing an exponential decay on this signal and Fourier transforming, we obtain the following
spectrum
1
2i [A2(Ω1 + πJ12) + iD2(Ω1 + πJ12)] − 1
2i [A2(Ω1 − πJ12) + iD2(Ω1 − πJ12)] .
To ensure that the absorption mode lineshape appears in the real part of the spectrum, we multiply
the expression above by a −90◦ phase correction factor i.e. by exp(−i π/2). Noting that
exp(−i π/2) ≡ −i, we obtain:
1
2 [A2(Ω1 + πJ12) + iD2(Ω1 + πJ12)] − 1
2 [A2(Ω1 − πJ12) + iD2(Ω1 − πJ12)] .
Clearly this is an anti-phase doublet on spin one.
The t1 modulation of term [4] has the form − sin (πJ12t1) sin (Ω2t1). Applying the identity
sin A sin B ≡ 1
2 [cos(A − B) − cos(A + B)] ,
gives
1
2 [cos(Ω2 + πJ12)t1 − cos(Ω2 − πJ12)t1] .
Imposing an exponential decay and taking the cosine Fourier transform yields the spectrum
1
2 [A1(Ω2 + πJ12) − A1(Ω2 − πJ12)] .
This is clearly an anti-phase doublet on spin two.
Multiplying the ω1 and ω2 spectra together, and taking the real part, gives the following four lines
which form the cross-peak multiplet. Note that they form an anti-phase square array.
+ 1
4 A1(Ω2 + πJ12)A2(Ω1 + πJ12) − 1
4 A1(Ω2 + πJ12)A2(Ω1 − πJ12)
− 1
4 A1(Ω2 − πJ12)A2(Ω1 + πJ12) + 1
4 A1(Ω2 − πJ12)A2(Ω1 − πJ12).
The operator in the diagonal peak term [3] is ˆI2x. Evolution of this operator during t2 gives the
following time domain signal:
1
2 exp(i[Ω2 + πJ12]t2) + 1
2 exp(i[Ω2 − πJ12]t2).
Imposing an exponential decay to this, and Fourier transforming gives the spectrum
1
2 [A2(Ω2 + πJ12) + iD2(Ω2 + πJ12)] + 1
2 [A2(Ω2 − πJ12) + iD2(Ω2 − πJ12)] .
This is an in-phase doublet on spin two.
The t1 modulation is:
cos(πJ12t1) sin(Ω2t1) ≡ 1
2 [sin(Ω2 + πJ12)t1 + sin(Ω2 − πJ12)t1] ,
Chapter 8: Two-dimensional NMR 47
where we have used the identity
sin A sin B ≡ 1
2 [sin(A + B) + sin(A − B)] .
Assuming an exponential decay and applying a sine Fourier transform gives the spectrum:
1
2 [A1(Ω2 + πJ12) + A1(Ω2 − πJ12)] .
This is an in-phase doublet on spin two.
Multiplying together the ω1 and ω2 parts of the spectrum and taking the real part yields the
following four components of the diagonal-peak multiplet. Note that they all have the same sign.
+ 1
4 A1(Ω2 + πJ12)A2(Ω2 + πJ12) + 1
4 A1(Ω2 + πJ12)A2(Ω2 − πJ12)
+ 1
4 A1(Ω2 − πJ12)A2(Ω2 + πJ12) + 1
4 A1(Ω2 − πJ12)A2(Ω2 − πJ12).
8.4
The DQF COSY pulse sequence is given in Fig. 8.15 on p. 205.
t2t1
Starting with equilibrium magnetization on spin two, ˆI2z, the state of the spin system after the
second pulse is exactly the same as for the COSY experiment at t2 = 0 as calculated in Exercise
8.3. Of the four terms present, the only one that contains double-quantum coherence is [2]:
− sin (πJ12t1) cos (Ω2t1) 2ˆI1y ˆI2x.
In section 7.12.1 on p. 178, it was shown that 2ˆI1y ˆI2x is a mixture of double- and zero-quantum
coherence. The double-quantum operator ˆDQy, and the zero-quantum operator ˆZQy, are deﬁned
as:
ˆDQy ≡ 2ˆI1x ˆI2y + 2ˆI1y ˆI2x
ˆZQy ≡ 2ˆI1y ˆI2x − 2ˆI1x ˆI2y.
Hence,
2ˆI1y ˆI2x = 1
2
ˆDQy + ˆZQy .
The double-quantum part that is retained is therefore:
−1
2 sin (πJ12t1) cos (Ω2t1) ˆDQy = −1
2 sin (πJ12t1) cos (Ω2t1) 2ˆI1x ˆI2y + 2ˆI1y ˆI2x .
The third 90◦ pulse acts to give:
−1
2 sin (πJ12t1) cos (Ω2t1) 2ˆI1x ˆI2z + 2ˆI1z ˆI2x .
2ˆI1x ˆI2z and 2ˆI1z ˆI2x represent anti-phase magnetization on spins one and two, respectively. Both
are modulated in t1 at Ω2, so the ﬁrst term therefore gives the cross-peak multiplet, and the second
the diagonal-peak multiplet.
Chapter 8: Two-dimensional NMR 48
Expanding the t1 modulation, we obtain
−1
2 sin (πJ12t1) cos (Ω2t1) ≡ −1
4 [sin(Ω2 + πJ12)t1 − sin(Ω2 − πJ12)t1] ,
which is an anti-phase doublet on spin two. Hence, both the cross- and diagonal-peak multiplets
are anti-phase in both dimensions. Furthermore, both terms have the same t1 modulation, and both
appear along the x-axis at the start of acquisition, so the spectrum can be phased so that all the
peaks appear in the double absorption mode.
8.5
The pulse sequence is given in Fig. 8.19 on p. 209.
t2
t1
τ τ
The ﬁrst 90◦ pulse rotates equilibrium ˆI1z to −ˆI1y, which then evolves under coupling during the
spin echo (the oﬀset is refocused) to give
cos(2πJ12τ) ˆI1y − sin(2πJ12τ) 2ˆI1x ˆI2z.
This is rotated by the second 90◦ pulse to give
cos(2πJ12τ) ˆI1z + sin(2πJ12τ) 2ˆI1x ˆI2y.
We select just zero-quantum coherence at this point. From the table on p. 180, the zero-quantum
part of 2ˆI1x ˆI2y is −1
2
ˆZQy, so at the start of t1 we have:
−1
2 sin(2πJ12τ) ˆZQy.
This evolves during t1 according to the rules in section 7.12.3 on p. 180:
−1
2 sin(2πJ12τ) ˆZQy
Ω1t1 ˆI1z+Ω2t1 ˆI2z
−−−−−−−−−−−−→ − 1
2 cos ([Ω1 − Ω2]t1) sin(2πJ12τ) ˆZQy
+ 1
2 sin ([Ω1 − Ω2]t1) sin(2πJ12τ) ˆZQx,
where
ˆZQx ≡ 2ˆI1x ˆI2x + 2ˆI1y ˆI2y
ˆZQy ≡ 2ˆI1y ˆI2x − 2ˆI1x ˆI2y.
Note that the zero-quantum coherence between spins one and two does not evolve due to the
coupling between these two spins.
The ﬁnal pulse rotates the zero-quantum terms to give
− 1
2 sin(2πJ12τ) cos ([Ω1 − Ω2]t1) 2ˆI1z ˆI2x − 2ˆI1x ˆI2z
+ 1
2 sin(2πJ12τ) sin ([Ω1 − Ω2]t1) 2ˆI1x ˆI2x + 2ˆI1z ˆI2z ,
the observable terms of which are:
1
2 sin(2πJ12τ) cos ([Ω1 − Ω2]t1) 2ˆI1x ˆI2z − 2ˆI1z ˆI2x .
The spectrum has the same form as the double-quantum spectrum shown in Fig. 8.20 on p. 210
with the following diﬀerences:
Chapter 8: Two-dimensional NMR 49
• In ω2 the anti-phase doublet on spin two, which arises from the 2ˆI1z ˆI2x term, appears with
the opposite sign.
• The frequency of the peaks in ω1 is (Ω1 − Ω2) i.e. the zero-quantum frequency.
The information that can be gained from this spectrum is the same as for the double-quantum
spectrum.
8.6
From section 8.8 on p. 214, the terms present after the ﬁrst spin echo are
cos(2πJIS τ1)ˆIy − sin(2πJIS τ1)2ˆIx ˆSz.
The subsequent 90◦ pulses are required to transfer the anti-phase magnetization (the second term)
to the S spin, so that it can evolve under the oﬀset of the S spin during t1. This requires the I spin
pulse to rotate ˆIx to ˆIz, which is only possible if the pulse is about y.
Applying the I spin pulse about −y gives:
− sin(2πJIS τ1) 2ˆIx ˆSz
(−π/2)ˆIy
−−−−−−→ − sin(2πJIS τ1) 2ˆIz ˆSz
(π/2) ˆSx
−−−−−→ sin(2πJIS τ1) 2ˆIz ˆSy.
The 2ˆIz ˆSy term, present at the start of t1, simply changes sign when the I spin pulse is changed in
phase from +y to −y.
8.7
The pulse sequence is given in Fig. 8.23(a).
τ1 τ1
y
y
A C
B D
t2
t1
I
S
The state of the spin system after the spin echo (A) is, from section 8.8 on p. 214:
cos(2πJIS τ1) ˆIy − sin(2πJIS τ1) 2ˆIx ˆSz.
The pulses during period B have the following eﬀect on the anti-phase term:
− sin(2πJIS τ1)2ˆIx ˆSz
(π/2)(ˆIy+ ˆSy)
−−−−−−−−−→ sin(2πJIS τ1) 2ˆIz ˆSx.
Chapter 8: Two-dimensional NMR 50
Period C is a spin echo, during which the coupling is refocused, but the oﬀset of the S spin evolves
for time t1. At the end of this period, the terms are:
− cos(ΩS t1) sin(2πJIS τ1) 2ˆIz ˆSx − sin(ΩS t1) sin(2πJIS τ1) 2ˆIz ˆSy.
The ﬁnal two pulses (period D) produce the following state at t2 = 0:
cos(ΩS t1) sin(2πJIS τ1) 2ˆIy ˆSx + sin(ΩS t1) sin(2πJIS τ1) 2ˆIy ˆSz.
The observable signal is due to the 2ˆIy ˆSz term, and is now modulated in t1 according to sin(ΩS t1).
So, shifting the phase of the ﬁrst 90◦ pulse to the S spin from x to y does indeed alter the modulation
in t1 from cosine to sine.
8.8
The pulse sequence is given in Fig. 8.25 (a) on p. 218. We will now modify it so that the ﬁrst 90◦
S spin pulse is about −x.
τ
A
τ
EC
F
B D
t2
t1
I
S
-x
As argued in section 8.9 on p. 217, the oﬀset of the I spin is refocused over the whole of period F.
The ﬁrst pulse creates −ˆIy, which evolves during period A under coupling to give
− cos(πJIS τ) ˆIy + sin(πJIS τ) 2ˆIx ˆSz.
Taking just the second term (the ﬁrst does not produce any useful peaks), and applying to it the
ﬁrst S spin pulse (with phase −x) gives:
sin(πJIS τ) 2ˆIx ˆSy,
which is of opposite sign to the corresponding term in section 8.9 on p. 217. This sign change
propagates throughout the rest of the calculation so that the observable term
sin2
(πJIS τ) cos(ΩS t1)ˆIy,
also has the opposite sign. The same result is produced on changing the phase of the second 90◦ S
spin pulse to −x.
I spins that are not coupled to S spins do not give rise to anti-phase magnetization, and so are
not aﬀected by the S spin pulses. This I spin magnetization is therefore unaﬀected by altering the
phase of the ﬁrst S spin pulse. So, recording two spectra, the ﬁrst with the ﬁrst S spin pulse about
x, and the second with it about −x, and then subtracting one from the other will retain the wanted
signal and eliminate the unwanted signal.
Chapter 8: Two-dimensional NMR 51
8.9
It was shown in section 8.9 on p. 217 that the observable term at the start of acquisition is
− sin2
(πJIS τ) cos(ΩS t1) ˆIy.
The amplitude of the signal is given by sin2
(πJIS τ), which has a maximum value of 1. This occurs
when the argument of the sine is an odd multiple of π/2 i.e. when πJIS τ = nπ/2, n = 1, 3, 5, . . .
Hence, τ = n/(2JIS ), n = 1, 3, 5, . . .
sin2
(πJIS τ) = 0 when πJIS τ = nπ/2, n = 0, 2, 4, . . . i.e. is an even multiple of π/2. Hence the
amplitude will be zero when τ = n/(2JIS ), n = 0, 2, 4, . . .
8.10
The HSQC pulse sequence, without decoupling during acquisition, is shown in Fig. 8.23 (b) on
p. 215.
τ1 τ1
y
A C
B D
t2
t1
I
S
At the start of acquisition, the observable terms are:
− cos(2πJIS τ2) sin(2πJIS τ1) cos(ΩS t1) 2ˆIy ˆSz
+ sin(2πJIS τ2) sin(2πJIS τ1) cos(ΩS t1) ˆIx.
The modiﬁcations for detecting long-range correlation are essentially the same as those discussed
for the HMQC experiment in section 8.9 on p. 217. They are:
• Increase the length of the delay τ1 so that sin(2πJIS τ1) is signiﬁcant for typical values of the
long-range coupling constants.
• Acquire immediately after the ﬁnal transfer pulses D, thus avoiding loss of signal due to
relaxation during the ﬁnal spin echo E, as in sequence (b) of Fig. 8.23.
• Acquire without broadband decoupling, as the wanted term is anti-phase with respect to JIS .
Chapter 8: Two-dimensional NMR 52
8.11
The diagonal peak is
A1→1 cos(πJIS t1) cos(Ω1t1)ˆI1y.
It was shown in section 7.5.1 on p. 157 that evolution of ˆI1y during t2 gives the following timedomain
signal:
1
2i exp(i[Ω1 + πJ12]t2) + 1
2i exp(i[Ω1 − πJ12]t2).
Imposing an exponential decay and Fourier transforming yields the following spectrum:
1
2i [A2(Ω1 + πJ12) + iD2(Ω1 + πJ12)] + 1
2i [A2(Ω1 − πJ12) + iD2(Ω1 − πJ12)] .
Applying a −90◦ phase correction and taking the real part, we obtain an in-phase doublet on spin
one:
1
2 A2(Ω1 + πJ12) + 1
2 A2(Ω1 − πJ12).
The modulation with respect to t1 is A1→1 cos(πJIS t1) cos(Ω1t1), which can be expanded using the
identity
cos A cos B ≡ 1
2 [cos(A + B) + cos(A − B)] ,
to give
1
2 A1→1 [cos(Ω1 + πJIS )t2 + cos(Ω1 − πJIS )t2] .
Imposing an exponential decay, and then taking the cosine Fourier transform gives:
1
2 A1→1 [A1(Ω1 + πJ12) + A1(Ω1 − πJ12)] ,
which is an in-phase doublet in ω1.
Multiplying the spectra in the ω1 and ω2 dimensions together gives the following four peaks for
the diagonal-peak multiplet:
+ 1
4 A1→1A1(Ω1 + πJ12)A2(Ω1 + πJ12) + 1
4 A1→1A1(Ω1 + πJ12)A2(Ω1 − πJ12)
+ 1
4 A1→1A1(Ω1 − πJ12)A2(Ω1 + πJ12) + 1
4 A1→1A1(Ω1 − πJ12)A2(Ω1 − πJ12).
All the peaks are positive and in the absorption mode.
The cross peak term
A1→2 cos(πJIS t1) cos(Ω1t1)ˆI2y
has the same modulation in t1 as the diagonal peak, and in t2 the operator is ˆI2y, rather than ˆI1y, so
in ω2 the doublet appears at Ω2. We can simply write down the four peaks which contribute to the
cross-peak multiplet as:
+ 1
4 A1→2A1(Ω1 + πJ12)A2(Ω2 + πJ12) + 1
4 A1→2A1(Ω1 + πJ12)A2(Ω2 − πJ12)
+ 1
4 A1→2A1(Ω1 − πJ12)A2(Ω2 + πJ12) + 1
4 A1→2A1(Ω1 − πJ12)A2(Ω2 − πJ12).
Again, these are in the absorption mode, and are all positive.
Chapter 8: Two-dimensional NMR 53
8.12
The phase-twist lineshape is
S (ω1, ω2) = [A1(ΩA)A2(ΩB) − D1(ΩA)D2(ΩB)]
real
+i [A1(ΩA)D2(ΩB) + D1(ΩA)A2(ΩB)]
imaginary
.
The plot shows the imaginary part.
8.13
The observable signal, acquired with broadband decoupling, is
sin(2πJIS τ2) cos(ΩS t1) sin(2πJIS τ1) ˆIx.
(a) Applying the SHR method to the HSQC sequence requires the acquisition of two timedomain
signals: one with cos(ΩS t1) modulation in t1, the second with sin(ΩS t1) modulation
in t1. It was shown in section 8.13.1 on p. 232 that the modulation can be changed from
cosine to sine by shifting the phase of the ﬁrst 90◦ S spin pulse by 90◦.
(b) For TPPI, each time t1 is incremented, the phase of the ﬁrst 90◦ pulse on the S spin must be
incremented by 90◦.
Chapter 8: Two-dimensional NMR 54
8.14
In order to obtain a sine modulated data set from
cos ([Ω1 + Ω2]t1 + 2φ),
we need to set 2φ = −π/2 i.e. φ = −π/4. To show this explicitly, we expand the argument of the
cosine using the identity
cos (A − B) ≡ cos A cos B + sin A sin B,
hence
cos ([Ω1 + Ω2]t1 − π/2) ≡ cos ([Ω1 + Ω2]t1) cos (π/2) + sin ([Ω1 + Ω2]t1) sin (π/2)
≡ sin ([Ω1 + Ω2]t1),
where we have used cos (π/2) = 0 and sin (π/2) = 1. So, shifting the phase by −π/4 alters the
modulation from cosine to sine. Thus, to implement TPPI, each time we increment t1 the phases
of the pulses preceding t1 are incremented by −45◦.
Chapter 9
Relaxation and the NOE
9.1
The equilibrium populations of the α and β levels are given by Eq. 9.6 on p. 264:
n0
α = 1
2 N exp(−Eα/kBT) n0
β = 1
2 N exp(−Eβ/kBT),
where
Eα = −1
2 γB0 Eβ = +1
2 γB0.
Evaluating the energies yields:
Eα = −1
2 × 1.055 × 10−34
× 2.675 × 108
× 9.4 = −1.326 × 10−25
J,
Eβ = +1.326 × 10−25
J.
Hence, at 298 K, the populations are:
n0
α = 1
2 × 1013
× exp(1.326 × 10−25
/(1.381 × 10−23
× 298))
= 5.00016 × 1012 ,
n0
β = 1
2 × 1013
× exp(−1.326 × 10−25
/(1.381 × 10−23
× 298))
= 4.99984 × 1012 .
On account of the very small energy gap, these populations are very similar, although as expected
n0
α > n0
β.
The energy of the system is given by
E = nαEα + nβEβ
= 1
2 γB0 nβ − nα .
Initially, nα = nβ, so Einitial = 0. At equilibrium,
Eequ. = 1.326 × 10−25
× (4.99984 × 1012
− 5.00016 × 1012
)
= −4.243 × 10−17
J.
The total change in energy is therefore
∆E = Eequ. − Einitial = −4.243 × 10−17 J .
Chapter 9: Relaxation and the NOE 56
The thermal energy of N molecules is of the order
NkBT = 1013
× 1.381 × 10−23
× 298 = 4.115 × 10−8 J ,
which is nine orders of magnitude greater than the value of ∆E calculated above. This reinforces
the point that the energy of interaction between the spins and the magnetic ﬁeld is minuscule
compared to the thermal energy.
9.2
The reduced spectral density function is given by Eq. 9.4 on p. 262
j(ω) =
2τc
1 + ω2τ2
c
.
For a ﬁxed frequency ω, the maximum value of j(ω) occurs at a value of τc given by
d j(ω)
dτc
= 0.
Using the product rule, we obtain:
d
dτc
j(ω) =
2
1 + ω2τ2
c
−
4ω2τ2
c
1 + ω2τ2
c
2
=
2 + 2ω2τ2
c − 4ω2τ2
c
1 + ω2τ2
c
2
=
2 1 − ω2τ2
c
1 + ω2τ2
c
2
.
The denominator is always non-zero, so the above expression can be solved by setting the numerator
to zero:
2 1 − ω2
τ2
c = 0
τc =
1
ω
Since the rate constant for longitudinal relaxation depends on j(ω0), the above result indicates that
this rate constant has its maximum value when τc = 1/ω0.
Chapter 9: Relaxation and the NOE 57
9.3
At equilibrium, the lower state (α) must have a greater population than the upper state (β), as predicted
by the Boltzmann distribution (assuming that the gyromagnetic ratio is positive). Suppose
we start with equal populations of the α and β states. The only way in which the population of
the α state can increase relative to that of the β state is for the rate of transitions from β to α to
exceed the rate from α to β. As the populations are equal, this implies that the rate constant for
the transition from β to α must be greater than that for the transition from α to β.
9.4
In the inversion–recovery experiment, the peak height S (τ) is given by
S (τ) = S (0) 2 exp(−Rzτ) − 1 ,
where S (0) is the peak height at time zero. Rearranging this, we get:
ln
S (τ) + S (0)
2S (0)
= −Rzτ,
from which we can see that a plot of ln[(S (τ) + S (0))/(2S (0))] against τ will be a straight line of
gradient −Rz = −1/T1.
τ / s 0.0 0.1 0.5 0.9 1.3 1.7 2.1 2.9
S (τ) −129.7 −93.4 7.6 62.6 93.4 109.5 118.9 126.4
ln[(S (τ) + S (0))/(2S (0))] 0.000 −0.151 −0.754 −1.353 −1.968 −2.554 −3.179 −4.370
0.0 0.5 1.0 1.5 2.0 2.5 3.0
-5
-4
-3
-2
-1
0
τ / s
ln[(S(τ)+S(0))/2S(0)]
The gradient is −1.508 s−1, so Rz = 1.508 s−1 and T1 = 0.663 s .
Chapter 9: Relaxation and the NOE 58
9.5
In section 9.5.2 on p. 271, it was shown that an estimate for T1 is given by τnull/ ln 2. The values
of T1 are therefore:
τnull / s 0.5 0.6 0.8
T1 / s 0.72 0.87 1.15
The fact that the solvent was still inverted after a delay of 1.5 s shows that it has a T1 value that is
greater than 1.5/ ln 2 = 2.16 s i.e. the solvent relaxes at a slower rate than the other spins.
9.6
The z-magnetization relaxes according to Eq. 9.15 on p. 269:
Mz(t) = Mz(0) − M0
z exp(−Rzt) + M0
z .
Setting Mz(0) = 0 and t = τ, we obtain
Mz(τ) = M0
z 1 − exp(−Rzτ) .
0
0
1
τ
Mz(τ)/Mz
0
The peak height S (τ) is proportional to the z-magnetization present just before the 90◦ pulse. Thus,
S (τ) can be written as
S (τ) = c 1 − exp(−Rzτ) .
Letting τ → ∞, S∞ = c; this will be the height of the peak in a simple 90◦–acquire experiment.
Substituting this into the above equation gives
S (τ) = S∞ 1 − exp(−Rzτ) .
Rearranging this yields:
S (τ) = S∞ 1 − exp(−Rzτ)
S (τ)
S∞
= 1 − exp(−Rzτ)
S∞ − S (τ)
S∞
= exp(−Rzτ)
ln
S∞ − S (τ)
S∞
= −Rzτ,
Chapter 9: Relaxation and the NOE 59
where we have taken the natural logarithm to go to the last line. Hence, a plot of
ln[(S∞ − S (τ))/S∞] against τ gives a straight line of gradient −Rz.
9.7
Assuming that the rate is proportional to the deviation from the equilibrium population, we can
write the rate of change of the population of level 1 (using the labelling in Fig. 9.17 on p. 273) as
dn1
dt
= −W(2,α)
1 n1 − n0
1 − W(1,α)
1 n1 − n0
1 − W2 n1 − n0
1
loss from level 1
+W(2,α)
1 n2 − n0
2
gain from level 2
+W(1,α)
1 n3 − n0
3
gain from level 3
+W2 n4 − n0
4
gain from level 4
.
Similarly, the rates of change of the populations of the other levels are:
dn2
dt
= −W(2,α)
1 n2 − n0
2 − W0 n2 − n0
2 − W
(1,β)
1 n2 − n0
2
loss from level 2
+W(2,α)
1 n1 − n0
1
gain from level 1
+W0 n3 − n0
3
gain from level 3
+W
(1,β)
1 n4 − n0
4
gain from level 4
.
dn3
dt
= −W(1,α)
1 n3 − n0
3 − W0 n3 − n0
3 − W
(2,β)
1 n3 − n0
3
loss from level 3
+W(1,α)
1 n1 − n0
1
gain from level 1
+W0 n2 − n0
2
gain from level 2
+W(2,β)
1 n4 − n0
4
gain from level 4
.
dn4
dt
= −W2 n4 − n0
4 − W
(1,β)
1 n4 − n0
4 − W
(2,β)
1 n4 − n0
4
loss from level 4
+W2 n1 − n0
1
gain from level 1
+W
(1,β)
1 n2 − n0
2
gain from level 2
+W
(2,β)
1 n3 − n0
3
gain from level 3
.
9.8
(a) The expression for b is (from section 9.6.3 on p. 277)
b =
µ0γ2
H
4πr3
=
4π × 10−7 × (2.675 × 108)2 × 1.055 × 10−34
4π × (1.8 × 10−10)3
= 1.294 × 105
s−1
.
Hence, b2 = 1.675 × 1010 s−2 .
Chapter 9: Relaxation and the NOE 60
(b) The expressions for the transition rate constants are given in section 9.6.3 on p. 277:
W(1)
1 = 3
40b2 j(ω0,1) W(2)
1 = 3
40 b2 j(ω0,2)
W2 = 3
10b2 j(ω0,1 + ω0,2) W0 = 1
20b2 j(ω0,1 − ω0,2).
In the fast motion limit, j(ω) = 2τc for all frequencies ω, so the rate constants have the
following numerical values:
W(1)
1 = 3
20 b2τc = 3
20 × 1.675 × 1010 × 20 × 10−12 = 0.0503 s−1 ,
W(2)
1 = 3
20 b2τc = 3
20 × 1.675 × 1010 × 20 × 10−12 = 0.0503 s−1 ,
W2 = 3
5 b2τc = 3
5 × 1.675 × 1010 × 20 × 10−12 = 0.201 s−1 ,
W0 = 1
10 b2τc = 1
10 × 1.675 × 1010 × 20 × 10−12 = 0.0335 s−1 .
From Eq. 9.19 on p. 277:
R(1)
z = 2W(1)
1 + W2 + W0 = (2 × 0.0503) + 0.201 + 0.0335 = 0.335 s−1 ,
R(2)
z = 2W(2)
1 + W2 + W0 = (2 × 0.0503) + 0.201 + 0.0335 = 0.335 s−1 ,
σ12 = W2 − W0 = 0.201 − 0.0335 = 0.168 s−1 .
(c) Substituting j(ω) = 2τc for all values of ω in Eq. 9.20 on p. 278, we obtain:
R(1)
z = b2 3
20 j(ω0,1) + 3
10 j(ω0,1 + ω0,2) + 1
20 j(ω0,1 − ω0,2)
= b2
τc
= 1.675 × 1010
× 20 × 10−12
= 0.335 s−1 .
Similarly, R(2)
z = 0.335 s−1 , and σ12 = 0.168 s−1 .
(d) The value of R(1)
xy can be calculated from the expression in section 9.8.3 on p. 295:
R(1)
xy = b2 1
10 j(0) + 3
20 j(ω0,2) + 3
40 j(ω0,1) + 3
20 j(ω0,1 + ω0,2) + 1
40 j(ω0,1 − ω0,2)
= b2
τc
= 1.675 × 1010
× 20 × 10−12
= 0.335 s−1 .
To go to the second line, we set j(ω) = 2τc. Similarly, R(2)
xy = 0.335 s−1 .
(e) As expected in the fast motion limit, the rate constants for the self-relaxation of both
longitudinal and transverse magnetization have the same value. The rate constant for the
cross-relaxation of longitudinal magnetization has half the value of the self-relaxation rate
constant and is positive, again as expected.
(f) The Larmor frequency is:
ω0 = 2π × 500 × 106
= 3.140 × 109
rad s−1
.
Chapter 9: Relaxation and the NOE 61
From the expression for the reduced spectral density,
j(ω) =
2τc
1 + ω2τ2
c
,
we can calculate the values of j(ω0), j(2ω0) and j(0):
j(ω0) =
2τc
1 + ω2
0τ2
c
=
2 × 500 × 10−12
1 + (3.140 × 109 × 500 × 10−12)2
= 2.88 × 10−10
s,
j(2ω0) = 9.20 × 10−11
s,
j(0) = 1.00 × 10−9
s.
The values of R(1)
z , R(2)
z and σ12 can be calculated by substituting ω0,1 = ω0,2 = ω0 into
Eq. 9.20 on p. 278, giving R(1)
z = R(2)
z = 2.025 s−1 , and σ12 = −0.375 s−1 . Similarly, from
section 9.8.3 on p. 295, R(1)
xy = R(2)
xy = 3.41 s−1 .
(g) As ω0τc = 1.6, we are now outside the fast motion limit, and beyond the zero-crossing point
where σ12 = 0. As a result, σ12 is negative and the rate constant for transverse relaxation
exceeds that for longitudinal relaxation. We are not very far beyond ω0τc = 1, so the rate of
longitudinal relaxation is signiﬁcantly faster than for τc = 20 ps.
9.9
For a 13C–1H pair, the value of b is:
b =
µ0γCγH
4πr3
=
4π × 10−7 × 6.728 × 107 × 2.675 × 108 × 1.055 × 10−34
4π × (1.1 × 10−10)3
= 1.427 × 105
s−1
.
Hence, b2 = 2.035 × 1010 s−2.
In the fast motion limit (τc = 20 ps), the values of the rate constants can be calculated from those
in the previous question by multiplying by the ratio of the b2 values. Note that we can only do this
because j(ω) is independent of τc in this limit. So,
R(1)
z
13
C–1
H =
⎛
⎜⎜⎜⎜⎜⎝
b2
C–H
b2
H–H
⎞
⎟⎟⎟⎟⎟⎠ R(1)
z
1
H–1
H
=
2.035 × 1010
1.675 × 1010
× 0.335
= 0.407 s−1 .
Similarly, R(2)
z = 0.407 s−1 , σ12 = 0.204 s−1 , and R(1)
xy = R(2)
xy = 0.407 s−1 . All these values
are greater than for the 1H–1H pair due to the smaller separation between the 13C and 1H. γC is
a quarter the value of γH, so for the same distance we would expect the relaxation to be sixteen
times slower. However, the rate constant goes as 1/r6, which changes by a factor of 19.2 on going
from r = 1.8 Å to r = 1.1 Å.
Chapter 9: Relaxation and the NOE 62
9.10
The necessary equations are given in section 9.10.2 on p. 304. At B0 = 4.7 T, c2 is given by:
c2
= γB0 σ − σ⊥
2
= 6.728 × 107
× 4.7 × 100 × 10−6 2
= 1.00 × 109 s−2 ,
where we have used the gyromagnetic ratio of 13C. In the fast motion limit, j(ω) = 2τc for all
values of ω, so the rate constants are:
Rz = c2 1
15 j(ω0)
= c2 2
15 τc
= 1.00 × 109
× 2
15 × 20 × 10−12
= 0.00267 s−1 ,
Rxy = c2 2
45 j(0) + 1
30 j(ω0)
= c2 7
45τc
= 1.00 × 109
× 7
45 × 20 × 10−12
= 0.00311 s−1 .
At B0 = 11.74 T, the rate constants are greater by a factor of (11.74/4, 7)2:
c2
=
11.742
4.72
× 1.00 × 109
= 6.24 × 109 s−2 ,
Rz =
11.742
4.72
× 0.00267 = 0.0167 s−1 ,
Rxy =
11.742
4.72
× 0.00311 = 0.0194 s−1 .
The values of the CSA relaxation rate constants at B0 = 11.74 T are an order of magnitude smaller
than those for dipolar relaxation of 13C due to an attached 1H. However, as the CSA contribution
goes as B2
0 it will become more signiﬁcant at higher ﬁelds.
9.11
The formulae are as for the previous question. For B0 = 4.7 T,
c2
= γB0 σ − σ⊥
2
= 2.675 × 108
× 4.7 × 10 × 10−6 2
= 1.581 × 108 s−2 .
Chapter 9: Relaxation and the NOE 63
Hence,
Rz = c2 1
15 j(ω0)
= c2 2
15τc
= 1.581 × 108
× 2
15 × 20 × 10−12
= 0.00042 s−1 ,
Rxy = c2 2
45 j(0) + 1
30 j(ω0)
= c2 7
45τc
= 1.581 × 108
× 7
45 × 20 × 10−12
= 0.00049 s−1 .
At B0 = 11.74 T, the values are greater by a factor of (11.74/4.7)2: c2 = 9.864 × 108
s−1 ,
Rz = 0.00263 s−1 , and Rxy = 0.00307 s−1 .
At B0 = 23.5 T, c2 = 3.953 × 109
s−1 , Rz = 0.01054 s−1 , and Rxy = 0.01230 s−1 .
Even at a ﬁeld of B0 = 23.5 T, the rate constants are still an order of magnitude smaller than the
dipolar relaxation rate constants at B0 = 11.74 T.
9.12
We are going to apply the initial rate limit, in which we assume that, on the right hand side of
Eq. 9.21 on p. 279,
dI1z
dt
= −R(1)
z I1z − I0
1z − σ12 I2z − I0
2z ,
I1z and I2z have their initial values:
dI1z
dt init
= −R(1)
z I1z(0) − I0
1z − σ12 I2z(0) − I0
2z
= −R(1)
z I0
1z − I0
1z − σ12 0 − I0
2z
= σ12I0
2z.
Integrating this, we obtain:
dI1z(t) = σ12I0
2z dt
I1z(t) = σ12I0
2zt + const.
We know that at time t = 0, I1z(0) = I0
1z, so the constant of integration is I0
1z. At t = τ:
I1z(τ) = σ12I0
2zτ + I0
1z.
Chapter 9: Relaxation and the NOE 64
Now we will look at the z-magnetization on spin two in the initial rate limit. Starting from
dI2z
dt
= −R(2)
z I2z − I0
2z − σ12 I1z − I0
1z ,
we obtain:
dI2z
dt init
= −R(2)
z I2z(0) − I0
2z − σ12 I1z(0) − I0
1z
= −R(2)
z 0 − I0
2z − σ12 I0
1z − I0
1z
= R(2)
z I0
2z.
Integrating this, and noting that I2z(0) = 0, we get, at t = τ:
I0
2z(τ) = R(2)
z I0
2zτ.
The height of the peak due to spin one is proportional to I1z, and the height of that due to spin two
is proportional to I2z. Furthermore, both spins are of the same type, so I0
1z = I0
2z. The peak heights
for the irradiated, reference and diﬀerence spectra are:
spectrum S1(τ) S2(τ)
irradiated: (a) c (σ12τ + 1) cR(2)
z τ
reference: (b) c c
NOE diﬀerence: (a) − (b) c σ12τ c R(2)
z τ − 1
Note that |σ12| τ 1 and R(2)
z τ 1 in the initial rate limit.
Ω1 Ω2
(a)
(b)
(c) = (a) - (b)
The NOE enhancement is given by:
η =
peak height in irradiated spectrum − peak height in reference spectrum
peak height in reference spectrum
=
c (σ12τ + 1) − c
c
= σ12τ.
Chapter 9: Relaxation and the NOE 65
9.13
The NOE diﬀerence spectrum is convenient as it only shows the target resonance, and the resonances
which are receiving an NOE enhancement.
9.14
(a) The observation that the NOE enhancement depends only upon the cross-relaxation rate
constant is a property of the initial rate limit i.e. the assumption that the target peak is still
fully inverted after the delay τ. We are eﬀectively ignoring self relaxation during this delay.
(b) At longer times, the inverted spin begins to relax back to equilibrium. This reduces the
z-magnetization on that spin and so slows the growth of the NOE: hence the dependence
on the self-relaxation rate constant of that spin. The spin receiving the enhancement can
also relax, resulting in the NOE enhancement being lost: hence the dependence on its self
relaxation rate constant.
(c) Spin one is held saturated throughout the experiment, so its relaxation is of no importance.
Cross relaxation gives the rate of transfer of magnetization from spin one to spin two, while
self relaxation of spin two leads to a loss of this transferred magnetization. Therefore, there
is competition between these two processes, which is reﬂected in the observation that the
enhancement depends upon the ratio of the rate constants for cross and self relaxation.
9.15
In the initial rate limit, the enhancement in a transient NOE experiment depends only upon the
cross-relaxation rate constant for the transfer of magnetization between the inverted spin and the
spin receiving the enhancement. In this example, σAB and σBC will be approximately equal, so
when HB is inverted, the enhancement of HA and HC will be the same.
On inverting HA, the enhancement at HB still depends only on σAB, so will be the same as for HA
and HC when HB is irradiated. HC is too far from HA to receive an enhancement.
In a steady state experiment, the enhancement depends upon the ratio of the cross-relaxation rate
constant to the self-relaxation rate constant of the spin receiving the enhancement. RA
z and RC
z are
equal to each other, so saturation of HB will give equal enhancements on HA and HC.
Irradiation of HA gives a smaller enhancement on HB as the self relaxation of this spin is faster
than for HA or HC. This is because HB has two nearby protons which relax it, whereas HA and HC
only have one nearby proton.
Chapter 9: Relaxation and the NOE 66
9.16
The NOESY pulse sequence is given in Fig. 9.24 on p. 287.
t2
t1 τ
We will start with equilibrium magnetization on spin one, and assume that spins one and two are
not coupled. If the phase of the ﬁrst 90◦ pulse is −x, it rotates equilibrium ˆI1z to ˆI1y. This evolves
under the oﬀset during t1 to give:
ˆI1y
Ω1t1 ˆI1z+Ω2t1 ˆI2z
−−−−−−−−−−−−→ cos(Ω1t1)ˆI1y − sin(Ω1t1)ˆI1x.
The second 90◦ pulse acts on the above terms to give:
cos(Ω1t1)ˆI1y − sin(Ω1t1)ˆI1x
(π/2)(ˆI1x+ˆI2x)
−−−−−−−−−−−→ cos(Ω1t1)ˆI1z − sin(Ω1t1)ˆI1x.
There are also similar terms due to spin two. We select only longitudinal terms after this pulse, so
at τ = 0, the z-magnetization on each spin is:
I1z = cos(Ω1t1) I0
1z and I2z = cos(Ω2t1) I0
2z.
The Solomon equations are (from Eq. 9.26 on p. 287):
dI1z(t)
dt
= −Rz I1z(t) − I0
z − σ I2z(t) − I0
z
dI2z(t)
dt
= −σ I1z(t) − I0
z − Rz I2z(t) − I0
z ,
where we have assumed that I0
1z = I0
2z = I0
z . Using the initial rate approximation with the following
initial conditions:
I1z(0) = cos(Ω1t1) I0
z and I2z(0) = cos(Ω2t1) I0
z ,
we obtain:
dI1z(t)
dt init
= −Rz [cos(Ω1t1) − 1] I0
z − σ [cos(Ω2t1) − 1] I0
z
dI2z(t)
dt init
= −σ [cos(Ω1t1) − 1] I0
z − Rz [cos(Ω2t1) − 1] I0
z .
Integrating these, and using the initial conditions to determine the values of the constants of
integration, we obtain:
I1z(τ)
I0
z
= cos(Ω1t1) (1 − Rzτ)
diagonal peak
− cos(Ω2t1)στ
cross peak
+ (Rz + σ) τ
axial peak
,
I2z(τ)
I0
z
= cos(Ω2t1) (1 − Rzτ)
diagonal peak
− cos(Ω1t1)στ
cross peak
+ (Rz + σ) τ
axial peak
.
Chapter 9: Relaxation and the NOE 67
Comparing these with Eq. 9.28 and Eq. 9.29 on p. 288, we see that the terms which give the
diagonal and cross peaks have changed sign, while the axial peak terms have not. The axial peaks
can therefore be suppressed by diﬀerence spectroscopy: we record two spectra with the phase of
the ﬁrst pulse set to +x and −x in turn, then we subtract one spectrum from the other. The cross
and diagonal peaks reinforce, and the axial peaks cancel.
9.17
The Larmor frequency in rad s−1
is
ω0 = 2π × 500 × 106
= 3.142 × 109
rad s−1
.
The value of ω0τc is 0.03, which is much less than 1. Therefore, we are working in the fast motion
limit, where j(ω) = 2τc for all frequencies. The rate constant for longitudinal relaxation is given
by Eq. 9.31 on p. 293:
Rz = γ2
B2
loc j(ω0),
where Rz = 1/T1. Substituting in this, and using the fast motion limit expression for j(ω0), we
obtain:
1
T1
= 2γ2
τcB2
loc
B2
loc =
1
2T1γ2τc
=
1
2 × 1 × (2.675 × 108)2 × 10 × 10−12
= 6.99 × 10−7 T2
.
This corresponds to a root mean square ﬁeld of 8.4 × 10−4 T, which is 10−4 times smaller than
B0. The local ﬁelds are indeed very weak.
9.18
τ τ
Any eﬀects of inhomogeneous broadening are refocused by the spin echo, so the amplitude of the
transverse magnetization present at the start of acquisition depends only upon Rxy and the time 2τ.
The peak height is therefore given by:
S (τ) = S0 exp(−2Rxyτ)
S (τ)
S0
= exp(−2Rxyτ).
Chapter 9: Relaxation and the NOE 68
Taking logarithms of both sides gives us
ln
S (τ)
S0
= −2Rxyτ,
so a plot of ln(S (τ)/S0) against τ is a straight line of gradient −2Rxy.
τ / s 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
S (τ) 65 39.4 23.9 14.5 8.8 5.34 3.24 1.96
ln(S (τ)/S0) 0 −0.501 −1.001 −1.500 −2.000 −2.499 −2.999 −3.501
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
τ / s
ln[S(τ)/S(0)]
The gradient is −5.00 s−1, giving Rxy = 2.50 s−1 , or T2 = 0.4 s.
Chapter 10
Advanced topics in two-dimensional
NMR
10.1
-10 -5 0 5 10 -10 -5 0 5 10
J23
J12
J12 = 2 Hz J23 = 6 Hz J12 = 6 Hz J23 = 6 Hz
α αβ β
α α β βspin 3
spin 1
J23 J12
α β/α β
α α/β βspin 3
spin 1
I2y
2I1zI2y
4I1zI2yI3z
2I2yI3z
Hz
Ω2/2π Ω2/2π
(a) (b)
(a) Assuming that the oﬀset of spin two is 0 Hz, the line positions are −4, −2, 2 and 4 Hz.
(b) Assuming that the oﬀset of spin two is 0 Hz, the line positions are −6, 0, 0 and 6 Hz; we
have a doublet of doublets, with the central two lines falling on top of one another, giving a
triplet.
Chapter 10: Advanced topics in two-dimensional NMR 70
10.2
We do not need to consider the 1–3 coupling as this does not aﬀect the evolution of a spin-two
operator. First, let us consider the evolution due to the 1–2 coupling:
ˆI2y
2πJ12tˆI1z ˆI2z
−−−−−−−−−→ cos (πJ12t) ˆI2y − sin (πJ12t) 2ˆI1z ˆI2x.
We will now consider the eﬀect of the 2–3 coupling separately on each of the terms on the right.
For the term in ˆI2y the evolution is straightforward:
cos (πJ12t) ˆI2y
2πJ23tˆI2z ˆI3z
−−−−−−−−−→ cos (πJ23t) cos (πJ12t) ˆI2y − sin (πJ23t) cos (πJ12t) 2ˆI2x ˆI3z.
For the − sin (πJ12t) 2ˆI1z ˆI2x term, the factor − sin (πJ12t) 2ˆI1z is unaﬀected by the evolution of the
2–3 coupling: writing this factor as A we have
A ˆI2x
2πJ23tˆI2z ˆI3z
−−−−−−−−−→ A cos (πJ23t) ˆI2x + A sin (πJ23t) 2ˆI2y ˆI3z.
Reinserting the factor A gives
− cos (πJ23t) sin (πJ12) 2ˆI1z ˆI2x − sin (πJ23t) sin (πJ12t) 4ˆI1z ˆI2y ˆI3z.
The overall result of the evolution of ˆI2y under coupling is summarized in the table:
term
dependence
on J12
dependence
on J23
axis description
ˆI2y cos (πJ12t) cos (πJ23t) y in-phase
−2ˆI1z ˆI2x sin (πJ12t) cos (πJ23t) −x anti-phase with respect to J12
−2ˆI2x ˆI3z cos (πJ12t) sin (πJ23t) −x anti-phase with respect to J23
−4ˆI1z ˆI2y ˆI3z sin (πJ12t) sin (πJ23t) −y
doubly anti-phase with
respect to J12 and J23
As expected, going anti-phase with respect to the coupling between spins i and j introduces a factor
sin (πJijt), whereas remaining in-phase with respect to this coupling introduces a factor cos (πJijt).
The in-phase term is along y, singly anti-phase terms are along −x, and the doubly anti-phase term
is along −y i.e. they follow around in the usual sequence x → y → −x → −y.
The corresponding tree diagram is
I2y
I2y
I2y
-2I1zI2x
-2I1zI2x -4I1zI2yI3z-2I2xI3z
J12
J23 J23
Chapter 10: Advanced topics in two-dimensional NMR 71
10.3
The trick to getting the signs right is just to think about the usual way in which y evolves into −x
and then into −y:
-I2x
-I2x -2I1zI2y-4I1zI2xI3z
2I2yI3z
2I2yI3z
2I2yI3z
J23
J12 J12
The term 4ˆI1z ˆI2x ˆI3z arises from splitting ﬁrst to the left, giving the coeﬃcient cos (πJ23t), and second
to the right, giving the coeﬃcient sin (πJ12t). Note also that there is a minus sign introduced.
So the overall factor multiplying 4ˆI1z ˆI2x ˆI3z is − cos (πJ23t) sin (πJ12t).
10.4
Term [1] is
cos (πJ13t1) cos (πJ12t1) sin (Ω1t1) ˆI1x.
First, let us consider the modulation in t1. We use the identity
sin A cos B ≡ 1
2 [sin (A + B) + sin (A − B)]
to combine the terms cos (πJ12t1) sin (Ω1t1) to give
1
2 cos (πJ13t1) [sin (Ω1t1 + πJ12t1) + sin (Ω1t1 − πJ12t1)] .
Next we multiply out the square brace:
1
2 cos (πJ13t1) sin (Ω1t1 + πJ12t1) + 1
2 cos (πJ13t1) sin (Ω1t1 − πJ12t1). (10.1)
Now we combine the two terms cos (πJ13t1) sin (Ω1t1 + πJ12t1) to give
1
2 sin (Ω1t1 + πJ12t1 + πJ13t1) + 1
2 sin (Ω1t1 + πJ12t1 − πJ13t1).
Doing the same for the two terms cos (πJ13t1) sin (Ω1t1 − πJ12t1) gives
1
2 sin (Ω1t1 − πJ12t1 + πJ13t1) + 1
2 sin (Ω1t1 − πJ12t1 − πJ13t1).
So overall Eq. 10.1 expands to four terms
1
4 sin (Ω1t1 + πJ12t1 + πJ13t1) + sin (Ω1t1 + πJ12t1 − πJ13t1)
+ sin (Ω1t1 − πJ12t1 + πJ13t1) + sin (Ω1t1 − πJ12t1 − πJ13t1) .
Chapter 10: Advanced topics in two-dimensional NMR 72
Therefore, what we have in ω1 is a completely in-phase doublet of doublets on spin one.
In ω2 the operator is ˆI1x, which also gives rise to an in-phase doublet of doublets on spin one.
‘Multiplying’ these two multiplets together in the manner of Fig. 10.6 on p. 325 gives rise to a
two-dimensional multiplet consisting of sixteen lines, all with the same sign; this is in contrast to
the cross-peak multiplet, which consists of four anti-phase square arrays.
Note, too, that the magnetization which gives rise to the diagonal peak is along x in t2 and is sine
modulated in t1. This is the complete opposite of the cross peak, which is along y in t2 and cosine
modulated. Thus, as in the COSY of the two-spin system, the diagonal and cross peaks are 90◦
out of phase with one another in both dimensions.
The reason why the splittings due to J12 and J13 are in-phase in the ω1 dimension is that the
modulation with respect to these couplings takes the form of a cosine: cos (πJijt1).
Chapter 10: Advanced topics in two-dimensional NMR 73
10.5
(a)
(b)
(c)
(d)
(e)
(f)
For (f) there are only two anti-phase square arrays.
Chapter 10: Advanced topics in two-dimensional NMR 74
(a) (b)
(c) (d)
ω1
ω2
Ω1
Ω2
In each case, the region plotted is ±10 Hz from the centre of the cross-peak multiplet; for clarity,
only one anti-phase square array is shown. The linewidth is 0.5 Hz in each dimension.
Chapter 10: Advanced topics in two-dimensional NMR 75
10.6
ω1
ω2
(a) (b) (c)
(d) (e) (f)
In each case ± 10 Hz is plotted from the centre of the cross-peak multiplet. Note that in (c), where
J12 = J23, the column of peaks down the centre of the cross peak no longer cancel one another
out, as four of the peaks are missing from the reduced multiplet.
In the series (a) to (c), J23 is increasing, thus increasing the ω2 separation of the two anti-phase
square arrays. In the series (d) to (f), J13 is decreasing, thus decreasing the ω1 separation of the
two anti-phase square arrays.
10.7
It is not usually possible to measure a value for the active coupling constant since this appears
as an anti-phase splitting. If the positive and negative peaks overlap signiﬁcantly, the separation
between the maxima and minima of the anti-phase peaks is no longer equal to the value of the
active coupling constant.
See section 10.3.3 on p. 332 for a description of how, under some circumstances, the values of
passive couplings may be determined from reduced multiplets.
Chapter 10: Advanced topics in two-dimensional NMR 76
10.8
ˆI1α ˆI2− ˆI3α: observable magnetization corresponding to the line of the spin-two multiplet which is
associated with spin one and spin three both being in the α state.
ˆI1α ˆI2− ˆI3−: double-quantum coherence, with p = −2, between spins two and three. More specifically,
this operator is associated with one of the lines of the double-quantum ‘doublet’ – see
section 3.7.3 on p. 44. This term is not observable.
ˆI1β ˆI2β ˆI3β: the population of the βββ energy level. This term is not observable.
ˆI1α ˆI2β ˆI3+: single-quantum coherence, with p = +1, corresponding to the line of the spin-three
multiplet which is associated with spin one being in the α state and spin two being in the β state.
Although it is single quantum, this term is not observable as only coherence order −1 is observable.
As described in section 10.4.2 on p. 336, free evolution simply gives a phase factor, with the
frequency depending on the oﬀset of the spin in question and on the spin states of the passive
spins. If the passive spin is in the α state, a term −πJ is contributed to the frequency, whereas if
it is in the β state, a term +πJ is contributed. The overall sense of the phase factor depends on
whether the operator is ˆI+ or ˆI−.
ˆI1α ˆI2− ˆI3α −→ exp (i[Ω2 − πJ12 − πJ23]t1) ˆI1α ˆI2− ˆI3α
ˆI1α ˆI2β ˆI3+ −→ exp (−i[Ω3 − πJ13 + πJ23]t1) ˆI1α ˆI2β ˆI3+
10.9
During t1 the term ˆI1+ ˆI2β ˆI3α acquires a phase factor:
exp (−i[Ω1 + πJ12 − πJ13]t1) ˆI1+ ˆI2β ˆI3α.
The small ﬂip angle pulse causes the following transfers to observable operators on spin two (the
coeﬃcients come from Eq. 10.7 on p. 338)
ˆI1+ ˆI2β ˆI3α −→ +1
2iθ +1
2iθ (1) ˆI1α ˆI2− ˆI3α
ˆI1+ ˆI2β ˆI3α −→ −1
2iθ +1
2iθ (1) ˆI1β ˆI2− ˆI3α
ˆI1+ ˆI2β ˆI3α −→ +1
2iθ +1
2iθ 1
4θ2 ˆI1α ˆI2− ˆI3β
ˆI1+ ˆI2β ˆI3α −→ −1
2iθ +1
2iθ 1
4θ2 ˆI1β ˆI2− ˆI3β.
For a small ﬂip angle, we discard the third and fourth terms as these go as θ4. This leaves
ˆI1+ ˆI2β ˆI3α −→ −1
4θ2 ˆI1α ˆI2− ˆI3α ˆI1+ ˆI2β ˆI3α −→ +1
4θ2 ˆI1β ˆI2− ˆI3α. (10.2)
These two transfers can be found in the table on p. 340.
Chapter 10: Advanced topics in two-dimensional NMR 77
Next we consider the behaviour of the term ˆI1− ˆI2β ˆI3α. For this term the sense of the phase
modulation is opposite to that of ˆI1+ ˆI2β ˆI3α:
exp (+i[Ω1 + πJ12 − πJ13]t1) ˆI1− ˆI2β ˆI3α.
For this term, the transfer ˆI1− −→ ˆI1α has associated with it a factor of (−1
2iθ), which is the opposite
sign to that for the transfer ˆI1+ −→ ˆI1α. So, the cross-peak components arising from ˆI1+ ˆI2β ˆI3α and
ˆI1− ˆI2β ˆI3α have opposite signs.
10.10
Starting with ˆI1+ ˆI2β ˆI3α the ﬁrst small ﬂip angle pulse creates four possible population terms, which
are the ones of interest in ZCOSY, in which spin one is in the α state:
ˆI1+ ˆI2β ˆI3α −→ +1
2iθ (1) (1) ˆI1α ˆI2β ˆI3α
ˆI1+ ˆI2β ˆI3α −→ +1
2iθ 1
4θ2
(1) ˆI1α ˆI2α ˆI3α
ˆI1+ ˆI2β ˆI3α −→ +1
2iθ (1) 1
4θ2 ˆI1α ˆI2β ˆI3β
ˆI1+ ˆI2β ˆI3α −→ +1
2iθ 1
4θ2 1
4θ2 ˆI1α ˆI2α ˆI3β.
Of these four terms, only the ﬁrst will be signiﬁcant for the case of a small ﬂip angle.
There are four additional transfers from ˆI1+ ˆI2β ˆI3α to operators in which spin one is in the β state,
but as before only one of these is signiﬁcant in the small ﬂip angle case:
ˆI1+ ˆI2β ˆI3α −→ −1
2iθ (1) (1) ˆI1β ˆI2β ˆI3α
So, we have just two population terms at this stage:
+1
2iθ ˆI1α ˆI2β ˆI3α and −1
2iθ ˆI1β ˆI2β ˆI3α.
From all that we have done so far we can see that, for small ﬂip angles, the signiﬁcant contributions
that these terms will make to the 1–2 cross peak arise from the transfer ˆI2β −→ ˆI2−, with both of
the other operators remaining the same:
+1
2iθ ˆI1α ˆI2β ˆI3α −→ +1
2iθ +1
2iθ ˆI1α ˆI2− ˆI3α
−1
2iθ ˆI1β ˆI2β ˆI3α −→ +1
2iθ −1
2iθ ˆI1β ˆI2− ˆI3α.
So the overall transfers from ˆI1+ ˆI2β ˆI3α caused by the two small ﬂip angles pulses are
ˆI1+ ˆI2β ˆI3α −→ −1
4θ2 ˆI1α ˆI2− ˆI3α
ˆI1+ ˆI2β ˆI3α −→ +1
4θ2 ˆI1β ˆI2− ˆI3α.
These are exactly the same as found for small ﬂip angle COSY in the previous exercise (see
Eq. 10.2 on the previous page).
Chapter 10: Advanced topics in two-dimensional NMR 78
10.11
From section 10.8.1 on p. 352, we found that at the end of the constant time T the following
operators are present:
cos (Ω1t1) cos (πJ12T) ˆI1y − sin (Ω1t1) cos (πJ12T) ˆI1x
− cos (Ω1t1) sin (πJ12T) 2ˆI1x ˆI2z − sin (Ω1t1) sin (πJ12T) 2ˆI1y ˆI2z.
The third of these is rotated by the second 90◦ pulse to a mixture of double- and zero-quantum
coherence:
− cos (Ω1t1) sin (πJ12T) 2ˆI1x ˆI2z
(π/2)(ˆI1x+ˆI2x)
−−−−−−−−−−→ + cos (Ω1t1) sin (πJ12T) 2ˆI1x ˆI2y.
Following section 7.12.1 on p. 178, the pure double quantum part of 2ˆI1x ˆI2y is 1
2(2ˆI1x ˆI2y +2ˆI1y ˆI2x),
so the double quantum term between the ﬁnal two pulses is
1
2 cos (Ω1t1) sin (πJ12T) (2ˆI1x ˆI2y + 2ˆI1y ˆI2x).
The ﬁnal 90◦ pulse makes this observable:
1
2 cos (Ω1t1) sin (πJ12T) (2ˆI1x ˆI2y + 2ˆI1y ˆI2x)
(π/2)(ˆI1x+ˆI2x)
−−−−−−−−−−→
1
2 cos (Ω1t1) sin (πJ12T) (2ˆI1x ˆI2z + 2ˆI1z ˆI2x).
The term cos (Ω1t1) sin (πJ12T) 2ˆI1x ˆI2z gives rise to a diagonal peak centred at {Ω1, Ω1}, as it is
modulated in t1 at Ω1 and appears on spin one in t2. There is a single modulating frequency of Ω1
in ω1 i.e. no splitting due to couplings, as expected. In ω2 the multiplet is in anti-phase.
The term cos (Ω1t1) sin (πJ12T) 2ˆI1z ˆI2x gives rise to a cross peak centred at {Ω1, Ω2}, as it is
modulated in t1 at Ω1 and appears on spin two in t2. Like the diagonal peak, it is in anti-phase in
ω2. Furthermore, note that the terms which give rise to both the diagonal and cross peak appear
along x, so they will have the same lineshape in ω2: this contrasts with the simple constant time
COSY experiment.
The intensity of both the diagonal and cross peaks goes as sin (πJ12T): again, this contrasts with
the simple constant time COSY, where the two kinds of peaks have a diﬀerent dependence on T.
The advantage of double-quantum ﬁltration is that it results in both diagonal and cross peaks
having the same lineshape in ω2, as well as in ω1.
Chapter 10: Advanced topics in two-dimensional NMR 79
10.12
Following the same kind of analysis as in section 10.8.1 on p. 352, we ﬁrst let −ˆI1y evolve under
the coupling for time T: a ‘tree’ is perhaps useful here:
-I1y
-I1y
-I1y
2I1xI2z
2I1xI2z 4I1yI2zI3z2I1xI3z
J12
J13 J13
Using this, we can simply read oﬀ the four terms which arise as a result of the evolution of the
coupling
− cos (πJ13T) cos (πJ12T) ˆI1y + sin (πJ13T) cos (πJ12T) 2ˆI1x ˆI3z
+ cos (πJ13T) sin (πJ12T) 2ˆI1x ˆI2z + sin (πJ13T) sin (πJ12T) 4ˆI1y ˆI2z ˆI3z.
The 180◦ pulse in the constant time period simply ﬂips the sign of any y or z operators:
+ cos (πJ13T) cos (πJ12T) ˆI1y − sin (πJ13T) cos (πJ12T) 2ˆI1x ˆI3z
− cos (πJ13T) sin (πJ12T) 2ˆI1x ˆI2z − sin (πJ13T) sin (πJ12T) 4ˆI1y ˆI2z ˆI3z.
Now we have to let each of these terms evolve under the oﬀset of spin one for time t1. The result
will be all of the above terms, multiplied by cos (Ω1t1):
cos (Ω1t1) + cos (πJ13T) cos (πJ12T) ˆI1y − sin (πJ13T) cos (πJ12T) 2ˆI1x ˆI3z
− cos (πJ13T) sin (πJ12T) 2ˆI1x ˆI2z − sin (πJ13T) sin (πJ12T) 4ˆI1y ˆI2z ˆI3z ,
and a related set of terms multiplied by sin (Ω1t1):
sin (Ω1t1) − cos (πJ13T) cos (πJ12T) ˆI1x − sin (πJ13T) cos (πJ12T) 2ˆI1y ˆI3z
− cos (πJ13T) sin (πJ12T) 2ˆI1y ˆI2z + sin (πJ13T) sin (πJ12T) 4ˆI1x ˆI2z ˆI3z .
After the ﬁnal 90◦ pulse the ﬁrst set of terms become
cos (Ω1t1) + cos (πJ13T) cos (πJ12T) ˆI1z + sin (πJ13T) cos (πJ12T) 2ˆI1x ˆI3y
+ cos (πJ13T) sin (πJ12T) 2ˆI1x ˆI2y − sin (πJ13T) sin (πJ12T) 4ˆI1z ˆI2y ˆI3y ,
none of which are observable.
The second set of terms, those multiplied by sin (Ω1t1), become
sin (Ω1t1) − cos (πJ13T) cos (πJ12T) ˆI1x + sin (πJ13T) cos (πJ12T) 2ˆI1z ˆI3y
+ cos (πJ13T) sin (πJ12T) 2ˆI1z ˆI2y + sin (πJ13T) sin (πJ12T) 4ˆI1x ˆI2y ˆI3y .
Chapter 10: Advanced topics in two-dimensional NMR 80
The term in ˆI1x is the diagonal peak: in ω2 it will appear as the in-phase doublet of doublets of
spin one, and as the t1 modulation is simply sin (Ω1t1), there will be a single frequency in ω1 i.e.
no splitting due to couplings, as expected.
The term in 2ˆI1z ˆI2y is the 1–2 cross peak: in ω2 it will appear as the doublet of doublets of spin
two, anti-phase with respect to the 1–2 coupling, but in-phase with respect to the 2–3 coupling. In
ω1 there is a single modulating frequency of Ω1, just as for the diagonal peak.
The cross- and diagonal-peak terms have the same modulation in t1, and so will have the same
lineshape is this dimension. However, in t2 the magnetization which gives rise to the diagonal
peak appears along x, whereas that which gives rise to the cross peak appears along y. So, as for
the two-spin case, in ω2 the cross and diagonal peaks will have diﬀerent lineshapes.
The intensities of the two type of peaks show a diﬀerent dependence on the couplings:
spin-one diagonal peak : cos (πJ13T) cos (πJ12T) 1–2 cross peak : cos (πJ13T) sin (πJ12T).
As for the two-spin case, the cross-peak goes as sin (πJactiveT), whereas the diagonal peak goes as
cos (πJactiveT): here the active coupling is J12. The two kinds of peaks have a common dependence
on the passive coupling J13, going as the cosine: cos (πJ13T). In words, to give rise to the 1–2
cross peak, the magnetization needs to be anti-phase with respect to the 1–2 coupling, and in-phase
with respect to the 1–3 coupling, hence the sine dependence on J12 and the cosine dependence on
J13.
For the cross peak to have a maximum intensity πJ12T must be an odd multiple of π/2, whereas
πJ13T must be an even multiple of π/2. It might be diﬃcult to satisfy this requirement exactly.
This analysis reveals the main problem with constant time experiments, which is the complex
dependence of the cross-peak intensity on the couplings in the system, and the value of the ﬁxed
delay T.
10.13
Following section 8.8 on p. 214, we found for a two-spin system the following S spin operator
after the ﬁrst S spin 90◦ pulse:
− sin (2πJIS τ1) 2ˆIz ˆSy.
We need to adapt this for the more complex spin system we are dealing with here. Firstly, the S
spin has to become the spin S1, and the coupling becomes that between I and S1, JIS1 :
− sin (2πJIS1 τ1) 2ˆIz ˆS1y.
If τ1 = 1/(4JIS1 ), then the sine term goes to 1 and so we just have −2ˆIz ˆS1y at the start of t1.
Just as before, we now allow the homonuclear coupling, which in this case is between S1 and S2,
to evolve for the whole time T, giving
− cos (πJ12T) 2ˆIz ˆS1y + sin (πJ12T) 4ˆIz ˆS1x ˆS2z,
where J12 is the coupling between the two S spins. Note the generation of anti-phase magnetization
with respect to this coupling. We also need to take account of the S spin 180◦ pulse which inverts
the operators ˆS1y and ˆS1z to give
+ cos (πJ12T) 2ˆIz ˆS1y − sin (πJ12T) 4ˆIz ˆS1x ˆS2z.
Chapter 10: Advanced topics in two-dimensional NMR 81
We now allow the S spin oﬀset terms to act for time t1; only the oﬀset of S1 has an eﬀect, giving
cos (ΩS1 t1) cos (πJ12T) 2ˆIz ˆS1y − cos (ΩS1 t1) sin (πJ12T) 4ˆIz ˆS1x ˆS2z
− sin (ΩS1 t1) cos (πJ12T) 2ˆIz ˆS1x − sin (ΩS1 t1) sin (πJ12T) 4ˆIz ˆS1y ˆS2z.
Finally, we need to take account of the I spin 180◦ pulse, which inverts all of the terms, as they all
contain ˆIz:
− cos (ΩS1 t1) cos (πJ12T) 2ˆIz ˆS1y + cos (ΩS1 t1) sin (πJ12T) 4ˆIz ˆS1x ˆS2z
+ sin (ΩS1 t1) cos (πJ12T) 2ˆIz ˆS1x + sin (ΩS1 t1) sin (πJ12T) 4ˆIz ˆS1y ˆS2z.
Note that we do not need to worry about the evolution of the heteronuclear coupling as this is
refocused by the 180◦ pulses in periods A and B.
Next comes the 90◦ pulses to the I and S spins: these have the following eﬀect on the operators
(the trigonometric terms have been left out):
2ˆIz ˆS1y −→ −2ˆIy ˆS1z 4ˆIz ˆS1x ˆS2z −→ 4ˆIy ˆS1x ˆS2y 2ˆIz ˆS1x −→ −2ˆIy ˆS1x 4ˆIz ˆS1y ˆS2z −→ 4ˆIy ˆS1z ˆS2y.
Of these terms, only the ﬁrst becomes observable on the I spin. We can see that the feature of this
term is that it has remained in-phase with respect to the S1–S2 coupling, and is cosine modulated
in t1.
After these two 90◦ pulses the observable term on the I spin is
cos (ΩS1 t1) cos (πJ12T) 2ˆIy ˆS1z.
After the following spin echo, assuming τ1 = 1/(4JIS1 ), this term simply becomes in-phase along
−x:
− cos (ΩS1 t1) cos (πJ12T) ˆIx.
We then observe this term with broadband S spin decoupling, giving a single peak at {ΩS1, ΩI}.
As a result of using the constant time procedure, there is no splitting in ω1 due to the coupling
between the S spins.
The intensity of the peak depends on cos (πJ12T); for a maximum, πJ12T must be a multiple of π,
i.e. πJ12T = nπ or T = n/J12 n = 1, 2, . . .. This condition corresponds to the magnetization
being in-phase with respect to the coupling between the S spins at the end of the constant time T.
In the case that the S spins are 13C in a globally labelled sample, the couplings we need to worry
about are the one-bond 13C–13C couplings, simply because these are the largest. Such couplings
do not vary very much with structure, so it should be possible to ﬁnd a value of T which is a
reasonable compromise for all the carbons in the system.
If there are further S spins coupled to S1, then we can see that the intensity of the cross peak will
go as cos (πJ12T) cos (πJ13T) . . .. Again, if the couplings do not cover too wide a range, we can
ﬁnd a value of T which will give reasonable intensity for all cross peaks.
Chapter 10: Advanced topics in two-dimensional NMR 82
10.14
We just use the idea that the selective 180◦ pulse ‘drags’ the curly line (the coherence) from the
energy level shared by the pulse and the coherence, to the energy level at the ‘other end’ of the
180◦ pulse.
1αα
2αβ
4ββ
3βα
1αα
2αβ
4ββ
3βα
1αα
2αβ
4ββ
3βα
1αα
2αβ
4ββ
3βα
1αα
2αβ
4ββ
3βα
1αα
2αβ
4ββ
3βα
180˚ to 1-3 180˚ to 3-4
180˚ to 1-3 180˚ to 1-2
result
The same idea is used below. Note that the selective 180◦ pulse and the coherence must share an
energy level for anything to happen.
1αα
2αβ
4ββ
3βα
1αα
2αβ
4ββ
3βα
1αα
2αβ
4ββ
3βα
1αα
2αβ
4ββ
3βα
1αα
2αβ
4ββ
3βα
1αα
2αβ
4ββ
3βα
180˚ to 3-4 180˚ to 2-4
180˚ to 2-4180˚ to 1-2
result
(a)
(b)
Transfer (a) can also be achieved by pulses to 3–4 and then 1–3; similarly, transfer (b) can also be
achieved by pulses to 1–2 and 1–3.
Chapter 10: Advanced topics in two-dimensional NMR 83
10.15
After the 90◦(y) pulse to the I spin and the ﬁrst 90◦ pulse to the S spin, and assuming that
τ = 1/(4JIS ), we have already worked out that the state of the system is −2ˆIz ˆSy (see section 8.8 on
p. 214). It is then just a question of following the evolution of this term under the inﬂuence of the
S spin oﬀset and the I–S coupling.
The ﬁnal stage is to use the trigonometric identities (given in the appendix). For example, the term
ˆSx is multiplied by the trigonometric term cos (ΩS t1) sin (πJIS t1). Applying the identity
cos A sin B ≡ 1
2 [sin (A + B) − sin (A − B)]
gives
cos (ΩS t1) sin (πJIS t1) ≡ 1
2 [sin (ΩS t1 + πJIS t1) − sin (ΩS t1 − πJIS t1)] .
This is indeed 1
2(s+ − s−), as stated.
We now follow through the fate of the term ˆSx for the rest of the sequence (Fig. 10.33 on p. 362).
The I spin 90◦ pulse at the start of period A has no eﬀect, and there then follows a spin echo
of total duration 1/(2JIS ) during which the in-phase term is completely transferred to anti-phase,
giving 2ˆIz ˆSy. We need to take account of the two 180◦ pulses which invert both ˆIz and ˆSy, leaving
the term overall unaﬀected. The 90◦(y) pulse to the I spin transforms this term to 2ˆIx ˆSy; this brings
us to the end of period A.
The 90◦ pulse to the S spin which starts period B rotates the operator to 2ˆIx ˆSz, and this anti-phase
term evolves completely into in-phase during the subsequent spin echo, giving ˆIy. We need to take
account of the two 180◦ pulses in the spin echo, which invert this term to give −ˆIy. This term is
unaﬀected by the ﬁnal 90◦ pulse to the S spin, so the observable term arising from ˆSx is
−1
2(s+ − s−) ˆIy.
This term can be found on the ﬁrst line of the table on p. 363.
10.16
The combinations S3 and S4 are:
S3 = 1
2 [(a) + (d)] S4 = 1
2 [(a) − (d)] ,
where (a) and (d) are given in the table on p. 363:
observable operator at t2 = 0
expt φI φS ˆIx ˆIy 2ˆIx ˆSz 2ˆIy ˆSz
(a) y y (−c+ − c−) (−s+ + s−) (c+ − c−) (s+ + s−)
(b) −y y (−c+ − c−) (s+ − s−) (−c+ + c−) (s+ + s−)
(c) y −y (−c+ − c−) (−s+ + s−) (−c+ + c−) (−s+ − s−)
(d) −y −y (−c+ − c−) (s+ − s−) (c+ − c−) (−s+ − s−)
Chapter 10: Advanced topics in two-dimensional NMR 84
Forming these combinations we have
S3 = 1
2 [(a) + (d)]
= (−c+ − c−) ˆIx
E
+ (c+ − c−) 2ˆIx ˆSz
F
S4 = 1
2 [(a) − (d)]
= (−s+ + s−) ˆIy
G
+ (s+ + s−) 2ˆIy ˆSz.
H
As before, we have a clean separation of x- and y-magnetization. If the two combinations are
processed separately, and a 90◦ phase correction applied to one combination in both dimensions,
we will have two spectra in which all peaks are in the absorption mode.
Term E is in-phase in ω2 and also in-phase in ω1, so all four peaks of the multiplet have the same
sign, which is this case is negative. The multiplet is the same as from term A given in Eq. 10.12 on
p. 363. Term F is anti-phase in each dimension, so gives rise to an anti-phase square array. Note,
however, that the overall sign is opposite to that of term B given in Eq. 10.12.
Term G is in-phase in ω2 and anti-phase in ω1, and is again opposite in overall sign to term C in
Eq. 10.13 on p. 363. Finally, term H is anti-phase in ω2, but in-phase in ω1: it is identical to term
D in Eq. 10.13.
The multiplets from the four terms, along with the way they combine to give S3 and S4, are shown
in the diagram below, which should be compared to Fig. 10.34 on p. 364.
E F S3 S3 + S4
S3 - S4G H S4
+ =
=+
We see from the digram that by combining the spectra S3 and S4, either as (S3 + S4) or (S3 − S4),
we are left with just one line of the multiplet, either top left, or bottom right.
10.17
Aside from the extra complication of the pulse sequence and data processing, probably the only
signiﬁcant diﬃculty is that the peak does not appear at {ΩS , ΩI}, but oﬀset from this by 1
2 JIS in
each dimension. Account needs to be take of this when comparing TROSY type spectra with other
spectra.
Chapter 11
Coherence selection: phase cycling
and ﬁeld gradient pulses
11.1
ˆIi− ≡ ˆIix − iˆIiy
φˆIiz
−−−→ cos φ ˆIix + sin φ ˆIiy − i cos φ ˆIiy − sin φ ˆIix
= cos φ ˆIix − i ˆIiy + i sin φ ˆIix − i ˆIiy
= (cos φ + i sin φ) ˆIix − i ˆIiy
= exp (iφ) ˆIi−.
Assigning coherence orders
ˆI1+ ˆI2− : p1 = 1 p2 = −1 p = p1 + p2 = 0
2ˆI1+ ˆI2+ ˆI3z : p1 = 1 p2 = 1 p3 = 0 p = p1 + p2 + p3 = 2
ˆI1x ≡ 1
2
ˆI1+ + ˆI1− : p = ±1
ˆI2y ≡ 1
2i
ˆI2+ − ˆI2− : p = ±1
2ˆI1z ˆI2y ≡ 2 × 1
2i
ˆI1z ˆI2+ − ˆI2− : p1 = 0 p2 = ±1 p = ±1
2ˆI1x ˆI2x + 2ˆI1y ˆI2y ≡ 21
2
1
2
ˆI1+ + ˆI1− ˆI2+ + ˆI2− + 2 1
2i
1
2i
ˆI1+ − ˆI1− ˆI2+ − ˆI2−
≡ 1
2
ˆI1+ ˆI2+ + ˆI1+ ˆI2− + ˆI1− ˆI2+ + ˆI1− ˆI2− − ˆI1+ ˆI2+ + ˆI1+ ˆI2− + ˆI1− ˆI2+ − ˆI1− ˆI2−
≡ ˆI1+ ˆI2− + ˆI1− ˆI2+
hence p1 = 1 p2 = −1 or p1 = −1 p2 = 1 p = 0
Chapter 11: Coherence selection: phase cycling and ﬁeld gradient pulses 86
Heteronuclear spin system
ˆIx ≡ 1
2
ˆI+ + ˆI− : pI = ±1
ˆSy ≡ 1
2i
ˆS+ − ˆS− : pS = ±1
2ˆIx ˆSz ≡ 2 × 1
2
ˆI+ + ˆI− ˆSz : pI = ±1 pS = 0
2ˆIx ˆSy ≡ 21
2
1
2i
ˆI+ + ˆI− ˆS+ − ˆS− : pI = ±1 pS = ±1
Following section 11.1.2 on p. 372, free evolution results in these operators acquiring a phase
exp −i Ω(p1+p2+...)
t ,
where Ω(p1+p2+...) = p1Ω1 + p2Ω2 + . . .. The table gives this phase term for each operator:
operator p1 or pI p2 or pS p3 Ω(p1+p2+...) phase term
ˆI1+ +1 Ω1 exp (−i Ω1t)
ˆI2− −1 −Ω2 exp (i Ω2t)
ˆI1+ ˆI2+ +1 +1 Ω1 + Ω2 exp (−i [Ω1 + Ω2]t)
ˆI+ ˆS− +1 −1 ΩI − ΩS exp (−i [ΩI − ΩS ]t)
ˆI1− ˆI2− ˆI3− −1 −1 −1 Ω1 + Ω2 + Ω3 exp (i [Ω1 + Ω2 + Ω3]t).
11.2
t1
t2
+1
-1
+2
+3
-2
-3
0
(a) TQF COSY (c) ZCOSY
(d) HSQC
(b) zero-quantum spectroscopy
t1
t2
τ τ
+1
-1
+2
-2
0
+1
-1
+2
-2
0
t2t1
θ θ
I
S
+1
-1
0
+1
-1
0
pI
pS
τ τ τ τ
y
t2
t1
Chapter 11: Coherence selection: phase cycling and ﬁeld gradient pulses 87
Note that in HSQC, sequence (d), we have pS = ±1 and pI = 0 during t1 i.e. S spin single-quantum
coherence, and that during t2 we have pI = −1 and pS = 0, as these are the coherence orders for
observable signals on the I spin.
τ τ
I
S
t2
t1
+1
-1
0
+1
-1
0
pI
pS
HMQC
(a) As described in section 11.3 on p. 377, the P-type spectrum has the same sign of p in t1 and
t2: this is the solid line in the CTP. The resulting spectrum will be phase modulated in t1,
and so is frequency discriminated.
(b) The N-type spectrum has the opposite sign of p in t1 and t2: this is the dashed line in the
CTP; like the P-type spectrum, the N-type spectrum is frequency discriminated.
(c) To be able to give absorption mode lineshapes we need to retain symmetrical pathways in t1
i.e. pS = ±1. Thus we need to select both the solid and dashed CTP. The resulting spectrum
is not frequency discriminated, but discrimination can be achieved using the SHR or TPPI
methods (section 8.13 on p. 231).
11.3
By inspecting Fig. 11.5 on p. 380 we can determine the form of the signal from detectors A and B
using simple trigonometry. For example in (b) it is clear that the component along A is − sin (Ωt)
whereas that along B is cos (Ωt). The table gives these components and the required combinations
for all four cases:
A B combination result
(a) cos (Ωt) sin (Ωt) A + i B cos (Ωt) + i sin (Ωt) = exp (i Ωt)
(b) − sin (Ωt) cos (Ωt) B − i A cos (Ωt) − i [− sin (Ωt)] = exp (i Ωt)
(c) − cos (Ωt) − sin (Ωt) −A − i B −[− cos (Ωt)] − i [− sin (Ωt)] = exp (i Ωt)
(d) sin (Ωt) − cos (Ωt) −B + i A −[− cos (Ωt)] + i sin (Ωt) = exp (i Ωt)
Each combination gives modulation of the form exp (i Ωt), which will all give the same lineshape
on Fourier transformation.
Following the approach of Fig. 11.6 on p. 382, for the case where the pulse goes [x, y, −x, −y] and
the receiver phase goes [−180◦, −270◦, 0◦, −90◦] we have
Chapter 11: Coherence selection: phase cycling and ﬁeld gradient pulses 88
x x x x
y y y y
90˚(-y)90˚(x)
-180˚ -270˚ 0˚ -90˚
90˚(-x)90˚(y)
rx phase
A 90◦(x) pulse places the magnetization along −y and then precession through an angle Ωt rotates
the vector towards +x. Similarly, a 90◦(y) pulse places the magnetization along +x, and then
precession rotates the vector towards +y.
The receiver phase is measured clockwise from 3 o’clock, and is indicated by the bullet •. We
see that in each diagram there is a constant angle between the position of the magnetization and
the receiver phase. As a result, each combination of pulse and receiver phase will give the same
lineshape, and so all four spectra will add up.
11.4
For ∆p = −1 the phase shift experienced by the pathway when the pulse is shifted in phase by ∆φ
is −∆p × ∆φ = −(−1)∆φ = ∆φ. Similarly for ∆p = 0 the phase shift is −0 × ∆φ = 0, and for
∆p = 5 the phase shift is −5 × ∆φ = −5∆φ. The table gives the phase shifts for each of these three
pathways:
pulse phase ∆p = −1 ∆p = 0 ∆p = 5
step ∆φ ∆φ 0 −5 ∆φ equiv(−5 ∆φ)
1 0◦ 0◦ 0◦ 0◦ 0◦
2 90◦ 90◦ 0◦ −450◦ 270◦
3 180◦ 180◦ 0◦ −900◦ 180◦
4 270◦ 270◦ 0◦ −1350◦ 90◦
These phases can be represented in the manner of Fig. 11.8 on p. 386:
Chapter 11: Coherence selection: phase cycling and ﬁeld gradient pulses 89
(1) (3)(2) (4)
signal phases for ∆p = -1, receiver phases to select ∆p = -3
(1) (3)(2) (4)
signal phases for ∆p = 0, receiver phases to select ∆p = -3
For ∆p = −1, steps (1) and (3) have the signal and receiver in alignment, whereas in steps (2) and
(4) the signal and the receiver are opposed. As a result steps (1) and (3) will cancel steps (2) and
(4).
For ∆p = 0, steps (1) and (3) will cancel as the signal and the receiver are aligned in one and
opposed in the other. Similarly, steps (2) and (4) will cancel as in step (2) the signal is 90◦ ahead
of the receiver, whereas in step (4) it is 90◦ behind i.e. there is an overall shift of 180◦.
For ∆p = 5 the signal phase shifts are exactly the same as those for ∆p = −3, so both pathways
are selected. This is of course exactly what is expected for a four-step cycle since −3 + 2 × 4 = +5
i.e. ∆p = −3 and ∆p = 5 are separated by a multiple of four.
11.5
The second pulse has ∆p = −2, so if the pulse phase goes [0◦, 90◦, 180◦, 270◦] the receiver
phase shifts must be [0◦, 180◦, 0◦, 180◦]. The ﬁrst pulse has ∆p = +1, so if the pulse phase goes
[0◦, 90◦, 180◦, 270◦] the receiver phase shifts must be [0◦, 270◦, 180◦, 90◦].
In the ﬁrst four steps, ∆φ2 therefore goes [0◦, 90◦, 180◦, 270◦], ∆φ1 remains ﬁxed, and the receiver
goes [0◦, 180◦, 0◦, 180◦].
In the second group of four steps, ∆φ2 does the same, but ∆φ1 is now 90◦, and this results in
an extra 270◦ which must be added to the receiver phase shifts from the ﬁrst group of four. The
required receiver phase shifts are therefore
[0◦+270◦, 180◦+270◦, 0◦+270◦, 180◦+270◦] ≡ [270◦, 90◦, 270◦, 90◦].
In the third group of four steps ∆φ1 is 180◦, and this results in an extra 180◦ which must be added
to the receiver phase shifts from the ﬁrst group of four. Finally, for the fourth group of four steps
∆φ1 is 270◦, and 90◦ must be added to the receiver phase shifts. The complete sixteen-step cycle
is therefore
Chapter 11: Coherence selection: phase cycling and ﬁeld gradient pulses 90
step ∆φ1 ∆φ2 φrx
1 0◦ 0◦ 0◦
2 0◦ 90◦ 180◦
3 0◦ 180◦ 0◦
4 0◦ 270◦ 180◦
5 90◦ 0◦ 270◦
6 90◦ 90◦ 90◦
7 90◦ 180◦ 270◦
8 90◦ 270◦ 90◦
step ∆φ1 ∆φ3 φrx
9 180◦ 0◦ 180◦
10 180◦ 90◦ 0◦
11 180◦ 180◦ 180◦
12 180◦ 270◦ 0◦
13 270◦ 0◦ 90◦
14 270◦ 90◦ 270◦
15 270◦ 180◦ 90◦
16 270◦ 270◦ 270◦
Selection of ∆p = −1 and then ∆p = +3
The ﬁrst pulse has ∆p = −1, so if the pulse phase goes [0◦, 90◦, 180◦, 270◦] the receiver phase
shifts must be [0◦, 90◦, 180◦, 270◦]. The second pulse has ∆p = +3, so if the pulse phase goes
[0◦, 90◦, 180◦, 270◦] the receiver phase shifts must be [0◦, 90◦, 180◦, 270◦]. For these four-step
cycles the receiver phases needed to select ∆p = −1 and +3 are, of course, the same.
The sixteen-step cycle is:
step ∆φ1 ∆φ2 φrx
1 0◦ 0◦ 0◦
2 90◦ 0◦ 90◦
3 180◦ 0◦ 180◦
4 270◦ 0◦ 270◦
5 0◦ 90◦ 90◦
6 90◦ 90◦ 180◦
7 180◦ 90◦ 270◦
8 270◦ 90◦ 0◦
step ∆φ1 ∆φ3 φrx
9 0◦ 180◦ 180◦
10 90◦ 180◦ 270◦
11 180◦ 180◦ 0◦
12 270◦ 180◦ 90◦
13 0◦ 270◦ 270◦
14 90◦ 270◦ 0◦
15 180◦ 270◦ 90◦
16 270◦ 270◦ 180◦
11.6
For ∆p = −2 the phase shift experienced by the pathway when the pulse is shifted in phase by ∆φ
is −∆p × ∆φ = −(−2)∆φ = 2∆φ. So, as the pulse goes [0◦, 120◦, 240◦] the pathway experiences
phase shifts of [0◦, 240◦, 480◦] which are equivalent to [0◦, 240◦, 120◦]. So, to select ∆p = −2,
we would use the cycle:
pulse: [0◦, 120◦, 240◦] receiver: [0◦, 240◦, 120◦].
Chapter 11: Coherence selection: phase cycling and ﬁeld gradient pulses 91
On modern spectrometers, the receiver phase can be shifted by arbitrary amounts, not just multiples
of 90◦.
The selectivity of this three-step sequence can be represented in the manner described on p. 386:
−5 (−4) (−3) −2 (−1) (0) 1 (2) (3) 4
Here the boldface numbers are the values of ∆p which are selected, and the numbers in brackets
are the values which are rejected; these selected values are separated by three, as we are dealing
with a three-step cycle.
The CTP for N-type COSY is:
+1
-1
0
t1
t2
The second pulse has ∆p = −2, so we can use the three-step cycle described above to select this.
As the ﬁrst pulse can only generate p = ±1, this three step cycle is suﬃcient to select the overall
pathway we require. To be speciﬁc, p = −1 present during t1 would only lead to observable
coherence via the pathway ∆p = 0 on the second pulse, which is blocked by this three-step cycle.
For P-type COSY (Fig. 11.4 (b) on p. 378), ∆p = 0 on the second pulse. This is selected by the
following three-step cycle of the second pulse:
pulse: [0◦, 120◦, 240◦] receiver: [0◦, 0◦, 0◦].
Such a cycle would be suﬃcient to select the wanted pathway as it would reject the ∆p = −2
pathway on the second pulse.
11.7
t1
t2
+1
-1
+2
+3
-2
-3
0
φ1 φ2 φ3 φrx
Grouping together the ﬁrst two pulses means that they are required to achieve the transformation
∆p = ±3. Concentrating for the moment on the pathways with ∆p = −3, shifting the phase of the
ﬁrst two pulses by ∆φ will result in a phase shift of −∆p × ∆φ = −(−3)∆φ = 3∆φ.
If the pulse goes through the phases [0◦, 60◦, 120◦, 180◦, 240◦, 300◦] then the phase acquired by
the pathway with ∆p = −3 is [0◦, 180◦, 360◦, 540◦, 720◦, 900◦]. Reducing these to the range 0◦
to 360◦ gives [0◦, 180◦, 0◦, 180◦, 0◦, 180◦]. So the phase cycle needed is
φ1 and φ2: [0◦, 60◦, 120◦, 180◦, 240◦, 300◦] receiver: [0◦, 180◦, 0◦, 180◦, 0◦, 180◦].
Chapter 11: Coherence selection: phase cycling and ﬁeld gradient pulses 92
This six-step phase cycle also selects ∆p = +3.
Since p = ±3 has been selected prior to the last pulse, and as the ﬁrst pulse can only generate
p = ±1, no further phase cycling is needed (with the possible exception of axial peak suppression,
see section 11.7 on p. 391).
Other pathways selected by this six-step cycle include ∆p = +3 + 6 = +9 and ∆p = −3 − 6 = −9.
These involve such high orders of coherence that we can safely ignore them.
The ﬁnal pulse has ∆p = −4 and ∆p = +2; as these are separated by 6, they will both be selected
by a six-step cycle. The phase experienced by the pathway with ∆p = −4 will be 4 ∆φ so as
the pulse goes [0◦, 60◦, 120◦, 180◦, 240◦, 300◦] then the phase acquired by the pathway will be
[0◦, 240◦, 480◦, 720◦, 960◦, 1200◦]. Reducing these to the range 0◦ to 360◦ gives the following
cycle:
φ3: [0◦, 60◦, 120◦, 180◦, 240◦, 300◦] receiver: [0◦, 240◦, 120◦, 0◦, 240◦, 120◦].
11.8
+1
-1
+2
-2
0
φ2φ1 φ3 φrx
t2t1 τ
The ﬁrst two pulses achieve the transformation ∆p = 0, so a four-step cycle will be:
φ1 and φ2: [0◦, 90◦, 180◦, 270◦] receiver: [0◦, 0◦, 0◦, 0◦].
Axial peak suppression (section 11.7 on p. 391) involves shifting the phase of the ﬁrst pulse
[0◦, 180◦] and similarly for the receiver. Combining these two cycles gives eight steps:
step 1 2 3 4 5 6 7 8
φ1 0◦ 90◦ 180◦ 270◦ 180◦ 270◦ 0◦ 90◦
φ2 0◦ 90◦ 180◦ 270◦ 0◦ 90◦ 180◦ 270◦
φrx 0◦ 0◦ 0◦ 0◦ 180◦ 180◦ 180◦ 180◦
The CTP for N-type NOESY is
+1
-1
+2
-2
0
φ2φ1 φ3 φrx
t2t1 τ
Chapter 11: Coherence selection: phase cycling and ﬁeld gradient pulses 93
We need to select ∆p = −1 on the last pulse. A suitable four-step cycle is [0◦, 90◦, 180◦, 270◦] for
φ3 and [0◦, 90◦, 180◦, 270◦] for the receiver.
We also need to select ∆p = +1 on the ﬁrst pulse. The four-step cycle [0◦, 90◦, 180◦, 270◦] for φ1
and [0◦, 270◦, 180◦, 90◦] for the receiver achieves this selection.
The complete sixteen-step cycle is
step ∆φ1 ∆φ3 φrx
1 0◦ 0◦ 0◦
2 90◦ 0◦ 270◦
3 180◦ 0◦ 180◦
4 270◦ 0◦ 90◦
5 0◦ 90◦ 90◦
6 90◦ 90◦ 0◦
7 180◦ 90◦ 270◦
8 270◦ 90◦ 180◦
step ∆φ1 ∆φ3 φrx
9 0◦ 180◦ 180◦
10 90◦ 180◦ 90◦
11 180◦ 180◦ 0◦
12 270◦ 180◦ 270◦
13 0◦ 270◦ 270◦
14 90◦ 270◦ 180◦
15 180◦ 270◦ 90◦
16 270◦ 270◦ 0◦
It is not necessary to add explicit axial peak suppression to this cycle as we are selecting ∆p = +1
on the ﬁrst pulse, and so all of the peaks we see in the spectrum must derive from the ﬁrst pulse.
11.9
RF
G τ1
G1
G2
τ2
+1
+2
0
The spatially dependent phase is given by Eq. 11.8 on p. 400:
φ(z) = −p × γ G z t.
Hence the phases due to the two gradient pulses are
φ1 = −(1) × γG1zτ1 and φ2 = −(2) × γG2zτ2.
The refocusing condition is that the total phase, φ1 + φ2, is zero:
φ1 + φ2 = −γG1zτ1 − 2γG2zτ2 = 0.
The factors of z and γ cancel to give, after some rearrangement:
G2τ2
G1τ1
= −
1
2
Chapter 11: Coherence selection: phase cycling and ﬁeld gradient pulses 94
(a) If the gradients have the same length, then G2/G1 = −1
2 , i.e. the second gradient needs to
be half the strength of the ﬁrst, and applied in the opposite sense.
(b) If the gradients have the same absolute strength, they still have to be applied in the opposite
sense i.e. G1 = −G2. Inserting this gives the refocusing condition as (G2τ2)/(−G2τ1) = −1
2,
which means that τ2 = 1
2τ1 .
11.10
G
G1
G2
τ2τ1
I
S
+1
-1
0
+1
-1
0
pI
pS
In the heteronuclear case we use Eq. 11.9 on p. 402 to ﬁnd the spatially dependent phase:
φ(z) = − (pIγI + pS γS ) G z t.
During the ﬁrst gradient pI = 0 and pS = −1, whereas during the second pI = −1 and pS = 0. So
the spatially dependent phases are
φ1 = γS G1zτ1 and φ2 = γIG2zτ2.
The refocusing condition is φ1 + φ2 = 0, which in this case rearranges to
G1τ1
G2τ2
= −
γI
γS
.
If I is 1H and S is 15N, then γI/γS = 10/(−1) so the refocusing condition becomes
G1τ1
G2τ2
= 10.
If the gradients have the same duration, τ1 = τ2 then G1 = 10G2 .
Note that ratio of the gyromagnetic ratios of 1H and 15N is in fact 9.86 : −1.
Chapter 11: Coherence selection: phase cycling and ﬁeld gradient pulses 95
11.11
t1
t2
+1
-1
+2
-2
0
G
G1 G2 G3
τ1 τ2 τ3
(a) P-type DQF COSY
t1
t2
+1
-1
0
G
G1 G2
τ1 τ2
(c) N-type COSY
G
G1 G2 G3
τ1 τ2 τ3
t1
t2
+1
-1
+2
+3
-2
-3
0
(b) N-type TQF COSY
(d) double-quantum spectroscopy
G
G1 G1 G2 G3
τ1 τ1 τ2 τ3
t1
t2
τ τ
+1
-1
+2
-2
0
(e) N-type HSQC
G
G1 G2 G3
τ1 τ2 τ3
I
S
+1
-1
0
+1
-1
0
pI
pS
τ τ τ τ
y
t2
t1
(a) P-type DQF COSY We have chosen p = 2 in the interval between the last two pulses, but
it would have been just as acceptable to choose p = −2. The pathway will give a P-type
spectrum as p = −1 is present during t1. The refocusing condition is
G1τ1 − 2G2τ2 + G3τ3 = 0.
If the gradients are all the same length, then one choice is for the strengths to be in the ratio
G1 : G2 : G3 = 1 : 1 : 1.
(b) N-type TQF COSY We have chosen p = 3 in the interval between the last two pulses, but
it would have been just as acceptable to choose p = −3. The pathway will give an N-type
spectrum as p = +1 is present during t1. The refocusing condition is
−G1τ1 − 3G2τ2 + G3τ3 = 0.
If the gradients are all the same length, then one choice is for the strengths to be in the ratio
G1 : G2 : G3 = 1 : 1 : 4.
Chapter 11: Coherence selection: phase cycling and ﬁeld gradient pulses 96
(c) N-type COSY The pathway will give a N-type spectrum as p = +1 is present during t1.
The refocusing condition is
−G1τ1 + G2τ2 = 0.
If the gradients are all the same length, then the strengths must be in the ratio
G1 : G2 = 1 : 1
(d) Double-quantum spectroscopy The two gradients G1 serve to ‘clean up’ the 180◦ pulse
in the spin echo (see section 11.12.3 on p. 406). Double-quantum coherence is dephased by
G2 and then rephased by G3; to control phase errors due to the underlying evolution of the
oﬀsets, both gradients are placed within spin echoes (see section 11.12.5 on p. 407). We will
need to record separate P- and N-type spectra, and then recombine them in order to obtain
an absorption mode spectrum (see section 11.12.2 on p. 405); the N-type pathway is given
by the solid line, and the P-type by the dashed line.
The refocusing condition for the N-type pathway is
2G2τ2 + G3τ3 = 0.
If the gradients are the same length, then the strengths must be in the ratio
G2 : G3 = 1 : −2.
The refocusing condition for the P-type pathway is
−2G2τ2 + G3τ3 = 0.
If the gradients are the same length, then the strengths must be in the ratio
G2 : G3 = 1 : 2.
(e) N-type HSQC G1 is a ‘purge’ gradient (see section 11.12.6 on p. 408). S spin magnetization
is dephased by G2 and rephased after transfer to I by G3. The refocusing condition
is
−γS G2τ2 + γIG3τ3 = 0.
If the gradients are both the same length, then the strengths must be in the ratio
G2 : G3 = γI : γS .
For the case where the I spin is 1H and the S spin is 13C, γI : γS = 4, and so the refocusing
condition is
G2 : G3 = 4 : 1.
Chapter 12
How the spectrometer works
12.1
The magnetic ﬁeld strength can be computed from the given Larmor frequency, f0, and gyromagnetic
ratio using 2π f0 = γB0. Hence
B0 =
2π f0
γ
=
2π × 180 × 106
1.08 × 108
= 10.47 T.
A homogeneity of one part in 108 means that the magnetic ﬁeld varies by ∆B = 10−8 × 10.47 =
1.047 × 10−7 T. This translates to a variation in frequency, ∆f, of
∆f =
γ∆B
2π
=
1.08 × 108 × 1.047 × 10−7
2π
= 1.8 Hz.
This is signiﬁcantly less than the expected linewidth of 25 Hz, so the magnet is useable.
The calculation is much simpler if we realise that a homogeneity of 1 part in 108 means that the
Larmor frequency will vary by 10−8 times its nominal value i.e.
∆f = 10−8
× 180 × 106
= 1.8 Hz.
12.2
For a 180◦ (or π) pulse, π = ω1t180, so
ω1 =
π
t180
=
π
24.8 × 10−6
= 1.27 × 105
rad s−1
.
Therefore ω1/(2π) = 20.2 kHz.
The same result can be found more simply by noting that a 360◦ pulse takes 2 × 24.8 = 49.6 µs;
this is the period of the rotation about the RF ﬁeld, so the frequency is just the reciprocal of this:
ω1/(2π) = 1/(49.6 × 10−6) = 20.2 kHz.
Using Eq. 12.2 on p. 434 we have
attenuation = 20 log
2
20.2
= −20.1 dB .
Chapter 12: How the spectrometer works 98
12.3
To go from a 90◦ pulse width of 20 µs to 7.5 µs, the RF ﬁeld has to be increased by a factor of
20/7.5 = 2.67, since the pulse width is inversely proportional to the RF ﬁeld strength. As the RF
ﬁeld strength is proportional to the square root of the power, the power would need to increase by
a factor of (2.67)2 = 7.11, so the transmitter power would be 7.11 × 100 = 711 W .
This is a very large increase and, unless the probe is designed to take this much power, there would
be a signiﬁcant risk of probe arcing.
12.4
The output of a two-bit ADC is two binary digits which are capable of representing the numbers
00, 01, 10 and 11 i.e. just four levels.
00
01
10
11
Note how the data points, because they are constrained to correspond to one of the four levels, are
not a particularly good representation of the smooth curve.
Having a larger number of bits means that there are more possible output levels, and hence the
digital representation of the signal will be more precise. As a result, the digitization sidebands are
reduced.
12.5
15 ppm at 800 MHz is 15×800 = 12 000 Hz. The range of frequencies, assuming that the receiver
reference frequency is placed in the middle, is thus −6 000 Hz to +6 000 Hz. From section 12.5.2
on p. 436, the sampling interval, ∆, is given by
∆ =
1
2 × fmax
=
1
2 × 6 000
= 83.3 µs .