Chem 781 Fall Semester 2014

• Instructor: F. Holger Försterling, Room 341; Phone (414)-229-5896e-mail: holger@csd.uwm.edu

• Times: Lectures: Tuesday, someThursday, 3:00 - 4:15 pm, CHM 193Lab: some Thursdays 3:00- 4:15 pm, Individual shifts, CHM B-10

First Meeting: Tue, Sept.2, 3:00 pm

• Assignments: 7 lab reports1 Midterm (in class presentation)

1 Final Exam (take-home)

• 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/

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

n .q .r

1 x 1 0 4

F ig u re ad ap ted fro m : C . B an w ell, F u n d am en ta ls o f M o lecu la r S p ec tro sco p y,T h ird ed itio n ,M cG raw H ill, L o n d o n 1 9 8 3

ch an g e o f n u c lea r o rien ta tio n

• 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 momentumClassical 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:

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

Jh

I I I I 2

1 1

( ) ( )

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 spin- and 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

E2(n-1)+l n l with spin-orbit coupling: j=l±½

1 2 ))) ))) 1d(33/22)

2 (1d,2s) )) 2 0 ))) 2s(½)

1 2 ))) ))) ))) 1d(55/22)

1 1 ))) 1p(½) 1 (1p) ))

1 1 ))) ))) 1p(33/22)

0 (1s) )) 1 0 )) 1s(½)

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

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 I = 0

example: 12C: 6p+, 6n 16O: 8p+, 8n

• even p+ (n), odd n (p+): one upaired spin I = n/2 (half integer)

7Li: 3 p+, 4n, => p+ in 1p(3/2): I = 3/2 17O: 8p+, 9n => n in 1d(5/2) state: I = 5/2

11B: 5p+ 6 n => p+ in 1p(3/2) : I= 3/2 19F : 9p+, 10n : I = ½ not obvious

13C: 6p+, 7n => n in 1p(½): I = ½ 15N: 7 p+, 8n => 1 p+ in 1p(½) state: I = ½31P: 15 p+, 16n: p+ in 2s(½): I= ½

• odd p+, odd n: two unpaired spins I = n (integer)

2H: 1 p+, 1 n, I = 1; 14N: 7 p+, 7 n, I = 1;10B: 5 p+, 5n, I = 3; 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 momentumElectron spin with spin angular momentum J:

μ = ge γl J = γe J (1.4)(ge 2.0023 or γe = 1.760 859 7701011 rad s-1 T-1)

Nuclear spin:μ = gN e/(2mp ) J = γN J (1.5)with mp mass of proton

The ge factor can not be explained by classical electrodynamics or the Schrodinger equation.

Naively, one would expect gN = ge and γp is expected to be a factor 1860 smaller than μB. (γp = 9.467 10⋅ 7 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 10⋅ 8 rad T-1s-1 γn = -1.8679 10⋅ 8 rad T-1s-1

Spin orientation in a magnetic fieldJ z

h

h /2

-h /2

-h

J = h 3 /2

J z

h

h /2

-h /2

-h

J = h 2

3 /2 h

-3 /2 h

I = 1 /2 I = 1

m = + 1

m = 0

m = -1

m = 1 /2

m = -1 /2

• Jz = h ms ms = -I, -I+1, ..., I-1, I

• μz = γh ms

Precession of the magnetic moment vector:

B 0

= B 0 d

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 0 0

0

2

B orB

Energy of spin states:

B 0

E

m S = - ½

m S = + ½

E

( )

( )

E B h 0 0 0

Em = - gℏB0 ms

The transition energy between the two levels ΔE = E+½ - E -½ is

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

m = + 1 /2s

m = - 1 /2s

B 0

M z

N

N

N N

N

N P P

N

B

kT

( ) 0

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:

M N

B N

kTN

N

kTBz z

0 0 2 22 2

0

12

2

1

2 4

General case (any spin):

M N

I I

kTBz

0 2 20

1

3

( ) 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:

x

y

B 1 ,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

x

y

z

x

y

A maximum of transverse magnetization is observed for θ = π/2 (90X)

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 Rotating frame on resonanceB 1(t) = B1,x cos(ωt) + B1,y sin(ωt) B1,x B1,x

ω0 Ω = ω0 - ω Ω = 0B0 B0 - ωRF/γ 0B eff eff = B00 + B11 Beffeff = B11

ωeff = ω1 = γB1

yx

z

B 1

M 0

M

For a B1 field along the rotating x axis one obtains:

Mz = M0 Iz cos(ω1τP)Mx = 0My = - M0 Iy sin(ω1τp)

If ω1AτP = π/2 (90X pulse), then after pulseMz = 0, Mx = 0 and My = -M0

For ω1 A τP = π (180X pulse) Mz = -M0 Iz, Mx = My = 0 after the pulse.

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 = gB1 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 90 rotation of ⁰ 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 9F 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:

m = + 1 /2s

m = - 1 /2s

P u lse

1 P = /2

application of radio frequency has two effects: change in population and alignment of the exited spins:

In the case of gB1τP = 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 = gB0.

M I M xy yM 0 0an d fo r = 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 I

M Ix x

y yM t

M

0

0

co s( )

s in ( t)

Relaxation of signalThe 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

x

y

x

y

x

y

b

ca ,

,b

c d

a

b

c

d

ab

c

t = 0 t > 0B 0

z

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. Random transitions to reestablish equilibrium z magnetization with the time constant T1

2. 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 90X pulse along x can be described as follows:

M I

M I

M I x

z z

y y

x

M

M t

M

0

02

0

1[ ]

co s( )

t

T *

e

e

s in ( t) e

-t

T1

-t

T2*

Bloch Equations of free precession in a magnetic field after a 90+x pulseMz(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 Dn1/2 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. 1.1-

1.2.5 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

D W

v

00

0123

54

-1-2-3-4-5

• 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 With modern spectrometers only a problem in 2D NMR

Relationship between sampling rate and frequency

• The required Spacing between data points (dwell time) is given by the maximum frequency nmax which is half the spectral width in Hz (SWH) set on the spectrometer

• The time required to accumulate TD data points is then

DW = 2π/(2Ωmax) = 1/(2SWH) (2.1)DW = 2π/(2Ωmax) = 1/(2SWH) (2.1)

AQ = TDDW = TD/(2SWH) (2.2)AQ = TDDW = TD/(2SWH) (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

D W

v

0

0123

54

-1-2-3-4-5

5

-5

Signals too small or too big will not be properly digitized:

• 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 c⋅ H2O) =(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.

Consequences of dynamic range

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 distin-guished by one detector alone.

y

x

M y

t M x

t

c o s ( t)

sin( t)

Solution: Use of two phase shifted detectors

• Use two detectors which differ in phase by 90X and collect both cosine and sine components of signal separately:

receiver 1 (cosine component) receiver 2 (sine component)

Mx,y = - M0 Iy cos(Ωt) + M0 Ix sin(Ωt)

d e tec to r 2

d e tec to r 1

N M R S ig n a l

A D C 1

A D C 2

m em o ry A

(rea l d a ta )

m em o ry B

(im ag in a ry d a ta )

R efe ren ces ig n a l

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 90F.

• 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 t i t d t

S t S t i t d

( ) ( )[co s( ) s in ( )]

( ) ( )[co s( ) s in ( )]

an d

Common fourier pairs

• Sine or cosine / well defined frequency spike

• Exponential decay / lorenzian

• Square function / sinc

t0

- 11

t 0

- 11

t

1 /e

T 2

*

0

1T 2

*

0 p t 0p

1p

1-

0

0

0

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

• Fourier transform yields a complex frequency signal S(ω) = R(ω)+i I(ω):

• 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)

S t R t I t e t i tt

T( ) ( ) ( ) [co s( ) s in ( )]*

2

f(t) [ ( ) ( )][ s( )

R iI co t i s in ( t)]d0

Complex NMR Signal: Absorbtion and dispersion spectrum

AT

T

( )( ) ( )

*

*

1

12

2

2 2

D

T

( )( * ) ( )

1

2

2 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

21 2

2

2 1/ * / *

To r

T

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

x r f

y r f

y re c

y re c

• 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 x- and 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

First Order Phase errors (linear)

= p 1p

d e aq

1( ) = d e+ p

• 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

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

0

FTD C

M 0

Imperfect baseline, will cause artefact at zero frequency after FT:

solution: remove artefact by turning phase of pulse by 180X 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 ) rec. 2 (Ix ) R(t) I(t) receiver phase

1 x -cos +sin + rec.1 + rec.2 x

2 y +sin +cos - rec. 2 + rec. 1 y

CYCLOPS phase cycle

• Quad suppression is combined with DC-offset correction to a 4 step phase cycle

scan P1 rec. 1 ( Iy ) rec. 2 (Ix ) R(t) I(t) receiver phase

1 x -cos +sin + 1 +2 x

2 y +sin +cos - 2 + 1 y

3 -x +cos -sin - 1 - 2 -x

4 -y -sin -cos + 2 - 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

• Signal to noise ratio will therefore grow only as NS/ = i.e. to

achieve twice the signal / noise one needs four times as many scans.

N S

N S N S

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

b

c

B 0

z

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.

• 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

How are gradients used in high resolution NMR

Gradient Echo:The effect of two consecutive gradiends on x,ymagnetization will cancel out.

1H

G ra d ien t +G z

-G z

m agnetization dephased- no signal

m agnetization rephased- gradient echo

• Gradients can be used to selec- tively dephase x,y magnetization while not effecting z-magnetization.

• A signal is also observed for signals where subsequent gradients cancel out, while every other signal is dephased (short T2*).

Examples of Gradients

G zG z

G z

1H

13C

p=1 p=1 p=1 p = -1

p = 0

p = 1

p = -1

p = 0

a b

c

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

d1

p1=P

de

dw =1

2·swh

times ns

aq = dw td =. td2 swh

td data pointssignal intensity given by rg

B set by pl11

Experiment time = ns (d1+aq+p1+de)

RF set byo1p or o1

Implementation of one-pulse experi-ment 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