Chem 781
Fall Semester 2014
•
Instructor:
•
Times:Lectures: Tuesday, someThursday, 3:00 - 4:15 pm, CHM 193
Lab: some Thursdays 3:00- 4:15 pm, Individual shifts, CHM B-10
F. Holger Försterling, Room 341; Phone (414)-229-5896
e-mail: holger@csd.uwm.edu
First Meeting: Tue, Sept.2, 3:00 pm
•
Assignments:
•
Grading: The final grade will be made on the basis of the mid-term (20%), the final exam (30%)
and the average of the lab reports (50%).
•
Recommended Textbook: J. W. Akitt and B. E. Mann; "NMR and Chemistry: An Introduction to
Modern NMR Spectroscopy" Stanley Thornes, Cheltenham 2000
•
Lecture notes are available at http://marvin.chem.uwm.edu/nmr/chem781/lecturenotes.html and on
D2L at http://d2l.uwm.edu/
7 lab reports
1 Midterm (in class presentation)
1 Final Exam (take-home)
Spectroscopy: energy and
frequency
• Nuclear Magnetic Resonance spectroscopy (NMR)
involves the interaction of the material being examined with
electromagnetic radiation.
• Radiation is characterized by the wavelength λ, frequency ν,
or energy E :
λ = c/ν = hc/E
(h = Plank's constant, c = velocity of light)
• Hence the energy involved in the excitation relates to the
frequency of the radiation:
ΔE = h ν (1.1)
Basic Characteristics of NMR
change of nuclear
orientation
n.q.r
1 x 104
Figure adapted from: C. Banwell, Fundamentals of Molecular Spectroscopy,Third edition,McGraw Hill, London 1983
•
NMR employs radio frequencies (60 - 1000 MHz), it is therefore at the low energy end of
spectroscopic methods (NQR is still lower)
•
looks directly at the nucleus (change in nuclear spin state) That means we can get an “atom
by atom” view of the molecule
•
requires external magnetic field: Magnet is one of the most prominent parts of the
spectrometer
Spin Angular momentum
Classical View
Spin is the intrinsic angular momentum of particles. Often it is convenient to
visualize the spin of an elementary particle as rotation about its axis resulting in an
angular momentum J (direction of J is according to right hand rule) .
In the case of a charged particle rotation also results
in magnetism with a magnetic moment m
Depending on the sign of the charge m
is parallel or antiparallel to J
Why should spin not be viewed
as self rotation of a particle:
One has to keep in mind that this picture of a self rotation of the particle is not correct. On a basic
level spin differs from orbital angular momentum, and can not fully explained as classical rotation
•
Elementary particles like the electron are described as a point like particles and thus there is nothing
to rotate.
•
Even if regarding them as extended spheres the definition of elementary means there is no way to
define distinct points on the surface that move.
•
The experimentally maximum possible radius of the electron would require a surface speed larger
than the speed of light to achieve the observed angular momentum.
•
As opposed to molecular rotation, each particular particle has one and only one specific spin
quantum number, that is an intrinsic constant for each particle and can not be altered.
•
Spin quantum numbers an be integer or half integer as shown by experiment. One can show half
integer spin quantum numbers can not be derived quantum mechanically from rotations
•
Spin appears as necessary property of the wave function when deriving a relativistic wave function
(Dirac equation). The (non relativistic) Schroedinger equation does not contain spin, even though
spin can be added empirically to the Schroedinger equation (Pauli matrices)
Quantum Mechanics of angular
momentum:
Like molecular rotation about a free axis the value of the spin angular momentum can
only take multiples of Plancks constant determined by the spin quantum number I:
J
h
2
I ( I  1)   I ( I  1)
I can be Integer or Half-Integer and is an intrinsic constant of the particle.
Examples for particles with different spin quantum numbers:
I = 1/2
:electron (e-), proton (1H), neutron (1n), 13C, neutrino (νe)
I=1
:photon (γ), deuteron (2H) (I=1 of photon is important for selection rules)
I>1
: 7Li (3/2 ); 17O (5/2); 59Co (7/2)
I=0
: 12C, 16O
Spin of complex nuclei
• The nuclei of all elements are composed of protons (p+)
and neutrons (n), both of which have I = ½ .
• The total spin of a nucleus is a combination of spinand orbital contributions by its constituents and can be
determined by employing the shell model of the
nucleus.
• The Shell model of the nucleus uses potential of a
modified harmonic oscillator (particle in a spherical
box). Note that the potential is different from the
Coulomb potential describing electron orbits around
nucleus
Nuclear Shell model
Energy levels of the harmonic oscillator including spin- and
orbital angular momentum:
Strong spin-orbit coupling combines spin- and orbit angular momentum with the
higher total angular momentum lowest in energy
E
2(n-1)+l
2 (1d,2s)
1 (1p)
0 (1s)
))
))
))
n l
with spin-orbit coupling:
j=l±½
1 2
))) )))
1d(33 /22 )
2 0
)))
2s(½)
1 2
))) ))) )))
1d(55 /22 )
1 1
)))
1p(½)
1 1
))) )))
1p(33 /22 )
1 0
))
1s(½)
Nuclear Shell model:
Rules for filling the levels
• Protons and neutrons are initially treated independently with separate
energy levels
• A high spin pairing energy results in orbitals filled up one by one (which
is the reverse of Hund’s rule)
• filled shells (even number of p+ or n) have zero angular momentum and
don’t contribute to overall spin
• for each type of particle the unpaired spin (½) combines with its integer
orbital angular momentum (strong spin-orbit coupling), with the higher
angular momentum state considerably lower in energy.
•
p+ and n contributions combine to total spin of nucleus
Nuclear Shell model
•
protons and neutrons are initially treated independently with separate
energy levels
• A high spin pairing energy results in orbitals filled up one by one (which
is the reverse of Hund’s rule)
• filled shells (even number of p+ or n) have zero angular momentum
• p+ and n contributions combine to total spin of nucleus
↑
↑
↑↓ ↑↓
↑↓
7Li
(3 p+ and 4 n, consider only p+
↑↓
13C
(6 p+, 7 n, consider only n)
Examples of nuclear
configurations:
•
even number of both p+ and n: all spins paired
example: 12C: 6p+, 6n
•
3 p+, 4n, => p+ in 1p(3/2): I = 3/2
11B: 5p+ 6 n => p+ in 1p(3/ ) : I= 3/
2
2
13C: 6p+, 7n => n in 1p(½): I = ½
31P: 15 p+, 16n: p+ in 2s(½): I= ½
2H:
8p+, 8n
even p+ (n), odd n (p+): one upaired spin
7Li:
•
16O:
odd p+, odd n:
1 p+, 1 n, I = 1;
10B: 5 p+, 5n, I = 3;
I=0
I = n/2 (half integer)
17O:
8p+, 9n => n in 1d(5/2) state: I = 5/2
19F : 9p+, 10n : I = ½ not obvious
15N: 7 p+, 8n => 1 p+ in 1p( ) state: I = ½
½
two unpaired spins I = n (integer)
14N:
7 p+, 7 n, I = 1;
6Li: 3p+, 3n, I = 1 (!!!)
Classification by usefulness for
NMR:
• I = 0:
useless (no NMR signal)
• I = ½ : sharp lines because of spherical
symmetry of nucleus
• I > ½ : often broad lines due to quadrupole
interactions (non spherical symmetry of
nucleus can interact with non spherical
distribution of electrons)
Magnetic Moment associated with
spin
Classical electromagnetism:
• A charged particle with angular momentum has a magnetic
moment.
• Example electron in orbital:

μ = -e/(2me ) 
L
e : elementary charge

L: orbital angular momentum
me : mass of electron => minus sign because of negative charge
by defining the gyromagnetic ratio γl = e/(2me ) one obtains

μ = γl 
L
Spin magnetic momentum
Electron spin with spin angular momentum 
J:

μ = ge γl 
J = γe 
J
(1.4)
(ge 
2.0023 or γe = 1.760 859 770
1011 rad s-1 T-1 )
The ge factor can not be explained by classical electrodynamics or the
Schrodinger equation.
Nuclear spin:

μ = gN e/(2mp ) 
J = γN 
J
with
(1.5)
mp mass of proton
Naively, one would expect gN = ge and γp is expected to be a factor 1860
smaller than μB. (γp = 9.467⋅107 rad T-1s-1)
Neutrons and protons are no true elementary particles, but are composed of
quarks, and determination of gN is not straightforward:
gp = 5.7 for p+ and gn = -3.9 for neutron
γp = 2.67522⋅108 rad T-1s-1
γn = -1.8679⋅108 rad T-1s-1
Spin orientation in a magnetic field
Jz
Jz
3/2 h
h
m = +1
h
m = 1/2
h/2
h/2
J = h 3 /2
J=h 2
m=0
-h/2
-h/2
m = -1/2
-h
-h
m = -1
-3/2 h
I = 1/2
• Jz
• μz
=
=
h ms
γh ms
I=1
ms = -I, -I+1, ..., I-1, I
Precession of the magnetic
moment vector:
The magnetic quantum number ms
restricts the component of the angular
momentum parallel to the magnetic field
Jz (and therefore z), but not the
components perpendicular to the field.


 0  B 0
or
0 
B0
2
B0
 =  B0
dm
m
Energy of spin states:
Em = - ℏB0 ms
E
mS= - ½ ()
E
The transition energy
between the two levels
ΔE = E+½ - E -½ is
E  B0
  0
mS= +½
B0
 h 0
Resonance condition for electromagnetic radiation (photons): hνRF = hν0
Selection Rule:
Δms = ± 1 due to spin 1 of photon
()
Magnetization of an ensemble of spins
• Only z-magnetization generated
• The relative excess in the lower
energy level in then given by
N N   N 
N ( P   P  ) B0



N
N
N
2 kT
B0
ms = - 1/2
• For an ensemble of spin ½ nuclei only
the excess spins in the level ms = + ½
(lower energy) will contribute to the
overall magnetization:
2 2
0
B0 N 1
 21
2 2 N 
M z  Nmz 
  N 
B0
2kT 2
4kT
Mz
ms = + 1/2
General case (any spin):
0

2 2 I ( I  1)
M z  N 
B0
3kT
Curie Law
In a magnetic field in equilibrium, a constant
magnetization of nuclear spins parallel to B0 is present
(longitudinal magnetization)
Observable magnetization Mz0 will depend on:
• magnetic field B0: we need as strong magnetic field as possible
• number of spins N, so concentration is important
• square of gyromagnetic ratio γ: one factor due to Boltzmann
distribution, one due to dependence of magnetic moment on γ
• inverse temperature T
Generating x,y magnetization:
The effect of RF radiation on macroscopic magnetization can be best understood
in a classical vector picture. The coil inside the probe generates an oscillating
radio frequency field along x` axis:


y
B1,x
x
B1,x(t) = 2 B1 cos(ωRFt)
• If ωRF ≈ ω0 the additional B1 field exerts a torque on Mz and
will cause net magnetization M to precess ⊥ to both 0 and 1(t)
(nutation).
z
z
y
x
y
x
A maximum of transverse magnetization is observed for θ = π/2 (90̊)
Rotating reference Frame
The complex motion of the magnetization vector can be
simplified when the observer is rotating with the B1 field at the
frequency ωRF.
Effect of RF pulse in rotating reference
frame
That results in the following transformations:
Laboratory frame
Rotating frame
1(t) = B 1,x cos(ωt) + B 1,y s in(ωt)
B
ω0
B0
eff
B


eff = B
00 + B
11
Rotating frame on resonance
B 1,x
Ω = ω0 - ω
B 0 - ω RF/γ
B1,x
Ω=0
0

B
eff
eff = B11
ωeff = ω1 = γB1
z
M0
For a B1 field along the rotating x axis one
obtains:
Mz = M0 Iz cos(ω1τP)
Mx = 0
My = - M0 Iy sin(ω1τp)
M
If ω1AτP = π/2 (90̊ pulse), then after pulse
Mz = 0, Mx = 0 and My = -M0
For ω1 A τP = π (180̊ pulse)
Mz = -M0 Iz, Mx = My = 0 after the pulse.
x
B1
y
Off resonance effects
• If ω is close, but not exactly on resonance with ω0, there will
be precession about the z axis with the frequency Ω = ω0 - ωRF
during the pulse.
• If the condition ω1 ≫ Ω is fulfilled, then this precession will
be small. Thus ω1 = 2πν1 = B1 is a measure of the bandwidth
of excitation:
• The stronger B1, the shorter the duration of the pulse τP, and
the wider the excitation range for which ω1 >> Ω.
A typical example: In a proton spectrum at ν0 = 300 MHz, the typical chemical shift
range is 10 ppm, which corresponds to Ωmax = 2π ∙ 3kHz. For ω1 = 10 ∙Ω = 2π ∙ 30kHz,
we obtain a duration for a 90⁰ rotation of τP(π/2) = 1/4 ∙ (2π)/ω1 = 8.3 μs which is a
typical value.
With Ωmax/(2π) = 3 kHz, we obtain an angle of 9̊ of precession during the pulse for
signals at the edge of spectrum.
Consequences of off resonance effects:
• Finite excitation range. Problems can arise in particular
in 13C spectra (or other hetero nuclei), where the
chemical shift range is much larger, or when multiple
pulses have to be applied (and errors will accumulate).
• Phase errors arising from rotation about z- axis during
the excitation pulse.
• Application of very weak pulses with low value of B1
and long tP(90 ) will allow for selective excitation
(water suppression)
Relation to microscopic quantum
mechanical picture:
ms = - 1/2
Pulse
1tP = /2
ms = + 1/2
application of radio frequency has two effects: change in population and
alignment of the exited spins:
In the case of B1τP = π/2 the result is:
• Equal population of á and â spin state: Mz = 0
• Excess magnetization becomes aligned along y axis: My =Mz0
Free precession
If the rotating radio-frequency field B1 is turned off, the
magnetization is precessing freely about the main field 0 with the
lamor frequency ω0 = B0.


M y   M0 I y
and
Mx  0
for  = 0
In the more general case of a B1 field not exactly on resonance
with ω0 (ωRF≈ω0) free precession occurs in the rotating frame
with the offset frequency Ω = ω0-ωRF:


M y   M 0 I y cos(t )


Mx 
M 0 Ix sin(t)
Relaxation of signal
The experimentally observed time domain signal is decaying
exponentially because the rotating x,y magnetization results from
NON equilibrium state of sample.
Three distinct processes:
• dephasing of spins due to differences in lamor frequency due
to inhomogeneity of magnetic field
• restoration of equilibrium magnetization through exchange of
energy with surroundings (spin-lattice relaxation)
• random dephasing of spins (Spin-spin relaxation)
Signal decay through
inhomogeneity
z
a
b
a
b
c
b
a, b
c ,d
y
y
y
c
c
x
x
d
x
B0
t= 0
t> 0
In real sample the field is not perfectly homogeneous. Spins in different parts of
sample will experience a different field and have different ω0
Spin-lattice relaxation
• The population of energy levels will return to
thermal equilibrium
• This process reestablishes z-magnetization
with a time constant T1
• All processes contributing to spin-lattice
relaxation will also contribute to a reduction of
transverse magnetization ( T2, see below) =>
buildup of z magnetization on expense of x,y
magnetization
Relaxation of x,y magnetization
Even in perfectly homogeneous B0 field x,y magnetization will
decay:
• Cause is interaction of nuclei with lattice AND each other.
1.
2.
Random transitions to reestablish equilibrium z magnetization
with the time constant T1
Spin spin interactions (chemical exchange, spin diffusion,
dependence of ω0 on orientation) can result in dephasing of
spins without change in population energy levels. Result is
random (statistical) change in phase => irreversible (Entropy)
• The relaxation of x,y magnetisation is described by time
constant T2 (transversal relaxation)
• It is always T1 ≥ T2
Difference between inhomogeneity
broadening and transverse relaxation
Dephasing from field
inhomogeneity is reversible
Statistical variation of the
speed of individual runners
will cause the echo signal to
be non perfect - Transversal
(x,y) relaxation
Total decay of x,y magnetization
The experimentally observed decay of the NMR signal
is a combined effect of inhomogeneity broadening and
transversal relaxation with time constant T2*
•
1/T2* = γΔB0/2 + 1/T2 where ΔB0 is the maximum
spread of field values across the sample.
• The line width is given by the decay rate of the signal
Δν1/2 = 2/(πT2* )
Bloch equations with precession
and relaxation
Considering free precession and relaxation, the three components
of the magnetization present after a 90̊ pulse along x can be
described as follows:

M

M

M
t
z
y
x




T
M 0 I z [1  e 1 ]

 M 0 I y co s( t ) e

M 0 I x s in (  t) e
-
t

T
*
2
t
T
*
2
Bloch Equations of free precession in a magnetic field after a 90̊x pulse
Mz(t=0) = 0
Mx(t=0) = 0
My(t=0)= -Mo Iy
Consequences of signal decay
by T2* for experiment:
• Maximizing the homogeneity will improve the resolution of
the experiment (longer T2* gives smaller 1/2 Sensitivity will
also be improved (why ? )
• Acquisition time must be long enough for signal to decay to
zero in order to avoid artifacts. This is often not possible in
2D experiments. Special filter functions have to be applied
before processing
• Extending acquisition time beyond 3 T2 will not accumulate
any more signal
Consequences of finite recovery time T1
• If many scans are to be accumulated, one will have to wait for
z-magnetization to recover between scans. Often it is more
advantageous to excite with shorter (<90⁰) pulses and use
shorter repetition time
• For quantitative determinations same z-magnetization needs to
be present before each pulse for each species. If multiple
scans are added, repetition time needs to be 5T1(max) or more
for accurate results.
• After inserting sample into magnet it will take a few seconds
for magnetization to build up. Normally no problem. But
some low ã nuclei (103Rh) or isolated spins (CHCl3 in
degassed, deuterated solvent) can have T1 of several minutes.
Further Reading:
General Textbooks:
• J.W. Akitt, NMR and Chemistry, fourth ed. Chapter 1
• Andrew Derome, Modern NMR techniques for Chemistry Research, Chapter 4.1 4.3.3 and 4.4-4.4.3
• Harald Günther, NMR Spectroscopy, second ed. Chapter 1, Chapter 7, pp221-233
• Edwin Becker, High Resolution NMR: Theory and Chemical Applications, Third
Ed; Chapter 2.1 - 2.9
• Jeremy Sanders, Brian Hunter; Modern NMR Spectroscopy; second Ed. ; Ch. 1.11.2.5 and 1.2.7
• Frank van de Ven; Multidimensional NMR in Liquids; Ch. 1.1 - 1.4
Nuclear Shell Model:
• G. Friedlander et al.; Nuclear and RadioChemistry, 3rd Ed.; Ch. 10.D
• http://en.wikipedia.org/wiki/Shell_model
• http://www.hep.phys.soton.ac.uk/hepwww/staff/D.Ross/phys3002/shell.pdf
Major components of the NMR
spectrometer
• Magnet: generates static B0 field (see scheme). Also contains
sample transport, spin assembly and shim coils.
• Probe: transmits radio frequency field and receives NMR signal.
Typically solution probe contains two coils, one for 1H/19F and 2H
(lock), and one broadband tunable from 31P downwards. Also
standard on modern probes are gradient coils to apply a linear field
gradient for controlled periods of time. Also specialty probes like
solid state probes available.
• Console: RF-electronics, generate and control frequencies, pulse
control, amplification and digitization of signal, acquisition control
• Computer (workstation): user interface, post acquisition
processing
Schematics of Magnet
Components of the NMR Spectrometer
Probe Head
Spectrometer – Observe Channel
• Preamplifier: The signal is amplified such that ADC is fully utilized,
but not overloaded (see below).
• Mixer:Transmitter frequency ωRF is subtracted from signal. That is
equivalent to observing the signal in the rotating frame: Only the offset
oscillation with respect to the carrier (Ω) is digitized and analyzed,
which is only in the order of Hz or kHz rather the initial MHz signal.
The principle is identical to a radio where subtraction of the MHz
carrier signal leaves the audio signal only.
Analog-digital converter (ADC:)
The analog audio signal (voltage) is converted into discrete data
points, each expressed by a number according to its intensity.
One obtains TD points (TD = time domain data points).
The process of digitization has implications for both sampling
speed and intensity of signals.
Sampling rate and discrete data points
• The sampling rate is determined by maximum frequency to be digitized. In
general, a sine function needs least 2 data points per period.
• If the frequency becomes larger than the required sampling rate (i.e. if there are
signals outside the selected window) the signal will still be detected, but
digitized with the wrong frequency and would appear as a “folded” peak inside
the window
v  With modern spectrometers only a problem in 2D NMR
DW
0
5
4
3
2
1
0
-1
-2
-3
-4
-5
Relationship between sampling rate
and frequency
•
The required Spacing between data points (dwell time) is given by the maximum
frequency max which is half the spectral width in Hz (SWH) set on the spectrometer
DW = 2π/(2Ωmax) = 1/(2SWH)
•
(2.1)
The time required to accumulate TD data points is then
AQ = TDDW = TD/(2SWH)
(2.2)
A typical 1D 1H NMR spectrum has 215 = 32,768 data points with a
spectral width of 20 ppm: At 300 MHz that is SWH = 6000 Hz.
One obtains AQ = 2.7 s.
Dynamic Range of receiver
Signals too small or too big will not be properly digitized:
v
5
0
-5
5
4
3
2
1
0
-1
-2
-3
-4
-5
DW
Consequences of dynamic range
• only a limited number of bits will be available for the receiver: There will
be maximum and minimum value that can be digitized. A 16 bit ADC can
take values from -32767 to 32767 (±[215 -1]).
• Any signal larger than the maximum number will be cut off at the top.
•
If the signal is too small it will not be digitized at all since any number
smaller than “1" will be represented as “0“.
• optimization of the signal intensity can be performed automatically, on the
Bruker DPX300 the command is rga
Example:1mM sample in H2O Mz /Mz(H2O) = c/(2⋅cH2O)
=(1⋅10-3 mol l-1 ) /(2⋅55 mol l-1) ≈ 1⋅10-5 = 1:100,000
• Problems occur if big signals exist next to small signals
• Deuterated solvents or solvent suppression techniques are required to
properly detect signal in dilute samples.
Quadrature detection: Problem of
sign of rotation
• After subtraction of carrier
frequency νRF the sign of rotation
Ω can be positive (faster than νRF)
or negative (slower than νRF).
• One detector measures only the
projection of the signal on one
axis (x OR y in the rotating
frame).
• The direction of the rotation
(clockwise or counter clockwise
rotation, positive or negative
frequency) can not be distinguished by one detector alone.
My
cos (t)
Mx
t
y
x
t
sin(t)
Solution: Use of two phase shifted
detectors
•
Use two detectors which differ in phase by 90̊ and collect both cosine and sine
components of signal separately:
Mx,y =
receiver 1 (cosine component)
- M0 Iy cos(Ωt)
+
detector 2
ADC 2
receiver 2 (sine component)
M0 Imemory
x sin(Ωt)
B
(imaginary data)
Reference
signal
NMR Signal
memory A
detector 1
ADC 1
(real data)
In practice, simultaneous recording of x- and y- component with two detectors is
realized by splitting the NMR signal and subtracting one time the carrier
frequency, the other time the carrier signal phase shifted by 90̊.
• Other possible solution: Set the offset frequency on edge of spectrum:
Impractical for general use, and one would waste range by acquiring half of the
window without signal.
• Advantage of two detectors: The carrier frequency can be set in center of
spectrum, and the AD converter can run with lower frequency since
νmax = 2π Ωmax = ½ SWH
• Problem of two detectors: two pieces of hardware are used, so subtraction
artefacts arising from the combination of the two data can occur (Quadratur
images)
• Sequential acquision of x- and y- datapoints is another possibility of sampling
both components using only one detector, but sampling speed needs to be twice
as fast
Fourier transform and lineshape
•
The time dependent signal has to be converted into a function of frequency. This
can be achieved by a process called fourier transformation (“extraction of
frequencies contained in oscillation”)
•
The process works both ways, and the time dependent function S(ω) and the
frequency dependent function S(t) form a Fourier pair.
S ( ) 

 S (t )[cos( t )  i sin( t )]dt

and

S (t ) 
 S ( )[cos( t )  i sin( t )]d

Common fourier pairs
t
•
t
0
Sine or cosine / well
defined frequency spike
Exponential decay /
lorenzian
t

1

t
- t1
0
t
1
T2*
1/e
*
t
T2
0
•
1
0
t
0
•
- t1

0
Square function / sinc
0
tp
t
-1 0
tp
1
tp

Fourier Transform and quadrature
detection
•
•
NMR signal consists of TWO time dependent signals: (cosine from detector 1 and
sine from detector 2)
Mathematically these are real and imaginary components of a complex function
S ( t )  R( t )  I ( t )  e
•

t
T2*
[cos( t )  i sin( t )]
Fourier transform yields a complex frequency signal S(ω) = R(ω)+i I(ω):

f(t)   [ R( )  iI ( )][co s( t )  isin( t)]d
0
•
In the absence of phase errors the real component will give a pure absorption
spectrum and the imaginary component will give rise to a pure dispersion spectrum:
R(ω) = A(ω)
and
I(ω) = D(ω)
Complex NMR Signal: Absorbtion
and dispersion spectrum
1
T2*
A( ) 
(
1 2
2
)

(



)
T2*
 
D( ) 
(
1 2
)  (   )2
T*
2
The absorption signal is a Lorentz
function. The maximum of the signal is
at ω = Ω, and the width at half height is
  1/ 2 
2
T2*
or
  1/ 2
1
 T2*
Consequence of complex fourier
trasform:
•
Normally one is only interested in the absorption signal (real signal, cosine
component).
•
The dispersion part (imaginary, sine component) can be discarded but is important
for phase correction (below).
•
The number of points in final spectrum (SI) is half of that in original time domain
signal (TD). TD points means TD/2 cosine (“real”) and TD/2 sine (“imaginary”)
points, therefore SI = TD/2.
•
In practice, one normally sets SI = TD, equivalent to doubling the number of time
domain data points by adding zeros at the end of the signal.
•
For computational reasons, SI always needs to be a power of 2.
Window (apodization) functions:
•
•
•
•
Exponential: Multiplying the FID with an exponentially decaying function will
enhance the first high signal to noise part of the signal compared to the end of
signal resulting in improvement of S/N at cost of resolution
– decay constant of exponential has to be matched to decay of signal (parameter
LB)
– default: for 1H LB = 0.3, 13C LB = 3
– LB < 0 gives growing exponential: results in improved resolution, but
tremendous cost of S/N
Gaussian: resolution enhancement with less cost of S/N (function goes towards 0 at
end)
LB < 0, 1> GB > 0 gives shift of maximum of gaussian
Sine, squared sine: Brings a not fully decayed signal smoothly to zero at the end.
Important for short sampling times (2D NMR)
Phase Errors:
•
Assumption so far: Magnetization fully aligned with y-axis at beginning of
acquisition, resulting is pure sine- and cosine functions in time domain and pure
absorption after fourier transform
•
However in reality the signal most likely is not pure absorption, but contains
dispersion contribution, because the initial magnetization is not perfectly aligned
with the -y axis.
•
The error normally can be expressed by two terms, one constant for all signals (zero
order) and one dependent on the position of the signal in the spectrum (first order)
Zero Order phase errors
• The reference frequency in the phase
sensitive detector is not exactly in
phase with the transmitter
• This is equivalent to the x- and y-axes
of the receiver in the rotating reference
frame not perfectly aligned with the xand y axes of the transmitter. The
reason is that the signals for the
transmitter pulse travels a different
pathway in the spectrometer as the
reference signal.
• In practice it is more convenient to
correct for that error by manipulating
the x and y (sine and cos) components
of the spectrum afterwards than trying
to perfect the spectrometer
yrec

yrf
yrec
xrf
First Order Phase errors (linear)
•
proportional to frequency offset of
signal
• The excitation pulse is of finite length,
and for and for a nucleus with ν0 ≠νRF
(Ω ≠ 0) some precession will take place
during τP.
• In addition, the pulse will not fall off to
zero power instantaneously but will
require a dead time after the pulse (pre
acquisition delay).
• The begin of data acquisition will
therefore not at t = 0, but at t = τP + DE,
and nuclei with different ν0 will no
longer be aligned, but dephased
proportional to their offset frequency
tp
= p1
de
aq
1() =de+ tp
Phase correction
•
If one combines zero and first order effects the total phase error can be described as
φ = φ0 + φ1 (Ω) = φ0 + (n/SI ) φ1 (n = specific data point, SI total number of points)
•
Phase error means that the two components detected during quadrature detection
are not pure sine and cosine, but mixed, and therefore the final real and imaginary
spectra will be not pure absorption and dispersion, but also mixed phase.
•
Pure absorption and dispersion spectra can be obtained by a linear combination of
real and imaginary data:
A(ω) = R(ω) cos φ + I(ω) sin φ and D(ω) = I(ω) cos φ - R(ω) sin φ
•
This correction can be obtained interactively or automatically by the command apk
on the Bruker spectrometer, as long both real and imaginary data are available.
Artefact suppression by phase
cycling- Constant DC-offset
M0
DC
FT
0

Imperfect baseline, will cause artefact at zero frequency after FT:
solution: remove artefact by turning phase of pulse by 180̊ and subtracting spectra:
1st scan: pulse (π/2)x
-M0Iycos(Ωt) + M0 Ix sin(Ωt) + DC;
add to memory
2nd scan: pulse (π/2)-x
+M0 Iy cos(Ωt) - M0 Ix sin(Ωt) + DC;
subtract from memory
total
-2 M0 [ Iy cos(Ωt) + Ix sin(Ωt) ]
Phase cycling adds desired signal and subtracts unwanted signal.
Artefact suppression - Imbalance of
the two receivers
•
Two physically different pieces of hardware are used for quadrature detection.
Imbalance will cause an image peak symmetric to the center of spectrum
•
Solution: Cycle phase of pulse x/y and interchange detector 1 and 2:
Two Phase sensitive Detectors
Two Memory Blocks
scan P1 rec. 1 ( Iy )
1
x -cos
2
y +sin
rec. 2 (Ix )
+sin
+cos
R(t)
+ rec.1
- rec. 2
I(t)
+ rec.2
+ rec. 1
receiver phase
x
y
CYCLOPS phase cycle
•
Quad suppression is combined with DC-offset correction to a 4 step phase cycle
scan
1
2
3
4
•
•
•
•
P1
x
y
-x
-y
rec. 1 ( Iy )
-cos
+sin
+cos
-sin
rec. 2 (Ix )
+sin
+cos
-sin
-cos
R(t)
+1
- 2
-1
+2
I(t) receiver phase
+2
x
+1
y
-2
-x
-1
-y
The different combinations of adding/substracting the receiver outputs to real and
imaginary data are equivalent to having the receiver along different axes.
Short notation:
P1
x y -x -y or in Bruker pulse programs:
PH1 0 1 2 3
receiver
x y -x -y
PH31 0 1 2 3
Effect of phase cycling on spectrum
A one scan spectrum shows artifacts
arising from DC-offset and
quadrature detector imbalance.
A four step phase cycle eliminates
those artefacts.
Signal accumulation
•
Phase cycle subtracts artefacts, adds up signal:
•
NS scans: signal will grow by factor NS, artifacts will subtract for NS = n 4 (for
CYCLOPS cycle)
•
however: random noise will not cancel out (opposed to popular belief) but grow
by NS
•
Signal to noise ratio will therefore grow only as NS/ NS =
NS i.e. to
achieve twice the signal / noise one needs four times as many scans.
Pulsed Field Gradients
•
Disadvantage of phase cycling: Several scans are required to remove undesired
signals. => Problem of subtraction artefacts, extra time spent in case of strong
signals
•
If the undesired signal is larger than the desired signal the receiver gain (rg) has still
to be adjusted using the undesired signal rather the desired one.
•
Modern spectrometers allow removal of artefacts/unwanted signals within one scan
using pulsed field gradients
The advantage over phase cycling is that it is all done in one scan, no subtraction
artifacts occur and in case of good signal to noise experiment time is kept to
minimum
•
How do Gradients work
• gradients coils are built into probe in addition to shim coils
• The effect is linear z-gradient similar to a bad z-shim, but usually much stronger.
• Also pulsed field gradients can be turned on for only a few ms at a time during the
experiment
z
b
c
B0
Effect of field gradient applied during acquisition: With a z-gradient
applied, the frequency will depend on the position along the z-axis in your
sample (principle of imaging). If the gradient is strong enough, the signal is
effectively eliminated from the spectrum.
How are gradients used in high
resolution NMR
1
H
magnetization dephased
- no signal
Gradient
• Gradients can be used to selectively dephase x,y magnetization
while not effecting zmagnetization.
magnetization rephased
- gradient echo
+Gz
-Gz
• A signal is also observed for
signals where subsequent
gradients cancel out, while every
other signal is dephased (short
T2*).
Gradient Echo:
The effect of two consecutive gradiends on x,y
magnetization will cancel out.
Examples of Gradients
a
p=1
b
p=1
p=1
p = -1
Gz
Gz
c
1
H
p=0
p = -1
p=1
p=0
13
C
Gz
a) Reverse dephasing by applying two opposing gradients b) Reverse
dephasing by applying a 180o pulse between two gradients c) Select specific
signal (C-13 bound protons) by applying matched gradients.
Summary of the one-pulse
experiment
signal intensity
given by rg
 RF set by
o1p or o1
td data points
dw =
1
2·swh
B1 set by pl1
d1
de
aq = dw. td =
p1=tP
times ns
Experiment time = ns (d1+aq+p1+de)
td
2 swh
Implementation of one-pulse experiment on Bruker spectrometers:
(; denotes comments):
;zg
1 ze
;Clear Memory
2 d1
;relaxation delay, wait for z magnetization to build up
p1 ph1 ;apply rf pulse with relative phase according to phase program ph1
go=2 ph31
;Wait delay de, turn on receiver, monitor signal according to phase
program ph31
jump back to “2" for ns times to repeat more scans
wr #0 ;write spectrum to disk
exit
ph1=0 2 2 0 1 3 3 1
;CYCLOPS phase cycling of p1
ph31=0 2 2 0 1 3 3 1
;and the receiver
;pl1 : f1 channel - power level for pulse (default)
;p1 : f1 channel - high power pulse
;d1 : relaxation delay; 1-5 * T1