1. CLARIDGE CHAPTER 2.2; 3.2

2. The Vector Model of NMR
A nucleus is placed in a static magnetic field (B0), nuclei with spin will precess at their given Larmor frequency.
The field is applied along the z-axis of an x,y,z coordinate system.
The motion of the nucleus is represented by a vector moving on the surface of a cone.

3. The Vector Model of NMR
There are many nuclei in one NMR sample, in the vector model the many spins are represented by a single vector which is the bulk magnetization or average.
The small population difference between the ground and the excited states causes the bulk magnetization to be pointed to +z axis.

4. B1 Field and Vector Diagrams
The B1 or Radio-frequency field (RF) is the pulse that excites the nuclei.
The RF field is an alternating voltage applied across the end of probe coil; this induces an alternating magnetic field on the sample; it is applied perpendicular to the static field (B0) which is aligned along the z-axis. This means that the RF pulse is in the x-y plane.
Vector diagram of the oscillating magnetization due to the RF field (half of a cycle):
Two magnetization vectors of constant amplitude rotating in opposite directions:
These two diagrams represent the same thing. Using two rotating vectors is useful for analyzing how the RF field interacts with the sample.

5. The Rotating Frame
The magnetization is rotating around the static field and the RF field is oscillating, to understand what is happening to the sample magnetization as a result of the RF field is confusing.
The purpose of the rotating frame is to simplify the view of what is happening by having the coordinate axes rotating at a chosen speed- the Larmor frequency.

6. The Rotating Frame
In the rotating frame, then each nuclear magnetic moment appears static.
There is no apparent precession- the effect of the external field has apparently disappeared. The net magnetization is still along the z-axis.
One of the two components of the oscillating RF field is now static in the x-y plane.
The other is now rotating at twice the Larmor precession frequency and thus does not effect the experiment, so it is neglected.

7. Radio-Frequency Pulse
The magnetization and the B0 field are static, aligned along the z-axis in the rotating frame; the B1 field is static in the rotating frame, aligned along an axis of our choosing.
The B1 field is applied at a right angle to the sample magnetization and thus exerts a force on the sample magnetization.
Since the nuclei have angular momentum, the B1 field is a torque about the B1 field vector.
The length of time required to align the magnetization along the y-axis exactly, on the first cycle, is a 90˚ pulse (or /2 expressing the pulse in radians).

8. Phase of the Pulse

9. Radio-Frequency Pulse
What is happening to the magnetization in the laboratory frame?
Remember our detector of magnetization is the same coil in the probe as the RF field, so our detector is in the x-y plane.
After the pulse the magnetization is detectable, generating a radio frequency signal
The frequency of that signal depends upon how fast the magnetization is traveling (the frequency difference relative to the Larmor frequency).

10. Chemical shift in the Rotating Frame

11. Spin-Spin Coupling in Rotating Frame

12. B1 Field Strength tan ( = DB/B1 gDB = Dn
How do we determine how large the B1 field should be?
We chose a rotating frame based upon the bulk magnetization, the Larmor frequency.
To successfully rotate all signals exactly 90˚, the pulse have sufficient power.
tan ( = DB/B1
gDB = Dn
where tan () must be small in order to be sure all frequencies are rotated equivalently
Dn is the difference from the center frequency (on resonance), the chosen rotating frame speed, and the edge of the spectrum in Hz

13. B1 Field Strength tan ( = DB/B1 gDB = Dn
If we want even excitation tan () to be small, B1 >> DB
pw90 (in seconds) = 1/(4)*(field strength in Hz) = 1/(4*B1)
Thus, pw90 needs to be small- ~ms

14. Choosing a Reference Frequency
The signals detected by the receiver need to be compared to a reference frequency.
Put the reference frequency in the center?
Logical, but positive and negative frequencies are not distinguishable.

15. Choosing a Reference Frequency
2) Put the reference frequency at the edge of the spectrum?
All frequencies will be positive relative to the reference so frequencies are distinguishable, but:
a) This wastes computer disk space or memory as twice the actual spectral width must be digitized
b) This adds noise to the spectrum.
c) More off-resonance effects.

16. Quadrature Detection
What if we put the reference frequency in the center, but found a way to detect the difference between positive and negative signals?
There would be no loss of signal-to-noise, less off-resonance effects, less computer memory
How do we accomplish a reference at the center of the spectrum while distinguishing between positive and negative signals?
Detect in quadrature. Two separate detectors on two axes; the two detectors have the same reference frequency, but the phases are different by 90˚.

17. Hardware Setup for Quadrature Detection
Problems with quadrature detection:
The two detectors must have amplitudes that are exactly equal, or the subtraction will not work.
2) The phases must differ by exactly 90˚.
These problems lead to quad images that are reflections of peaks about the center of the spectrum. These peaks should show a different phase than real peaks and be affected by where the center of the spectrum is set.

18. Phase Cycling The purpose of phase cycling is to reduce quad images
CYCLOPS =
CYCLically Ordered Phase Sequence

19. Phase Corrections
Phase correction of a frequency spectrum is standard for all NMR spectra acquired.
Phase errors result from:
1) The reference frequency is not adjusted so that the pure sine and cosine components are detected which translates to pure real and pure imaginary or pure absorptive and pure dispersive lines. What we get instead is a linear combination of the two:
Real = Acos() + Dsin()
Imag = Asin() - Dcos()
where A is absorptive and D is the dispersive component and  is the phase error.
2) Off-resonance effects. As the pulse becomes further and further away from the resonance a phase error develops

20. 3) The receiver cannot be turned on instantly after a pulse (or during the pulse either). The electronics to need time to recover from the pulse (on the order of 10 s).
4) Receiver electronics cannot transmit all frequencies with equal phase.
All of the phase errors are put together in this equation:
() =  + 
 and  are adjustable parameters (rp and lp).
Also, note that with low signal-to-noise and poor digitization or acquisition times, what appears to be a phase error is not really a phase error (this is particularly a problem for 13C spectra).

21. 90˚ 1H pulse
A 90˚ pulse is the pulse length that results in a maximum and positive signal, a 180˚ is when the signal is null…
The length of the 90˚ pulse is primarily dependent upon 4 factors:
1) Probe- the type of probe (inverse, broadband, etc.), the quality, the age, etc.
2) Tuning- a poorly tuned probe will have a long 90˚ pulse length.
3) Power of the pulse- tpwr
4) Solvent
To measure, setup array of different values of pulse width:

22. Field Strength (Hz) = 1/(4*pulse width90)
Low Power Pulses
Some pulse sequences require pulses that are not at maximum or near maximum power.
Such pulses are used for many purposes including decoupling, small frequency range saturation, and spin-locking.
Why they are sometimes used:
1) Lower heat to the probe.
2) Sometimes only smaller frequency ranges are required
Field Strength (Hz) = 1/(4*pulse width90)
If you want to selective excite a particular resonance, you can apply a long pulse

23. Power of Pulses Transmitter power = tpwr is db scale
db is a log scale- 10 less db = 10X less power
~6 less db = Voltage/2
Voltage/2 leads to pw90*2
If pw90 is twice as long, then field strength of pulse is ½ as wide
pw90 ~6.5 us, at tpwr = 58
Thus, at tpwr = 46, pw90 ~26 us

24. Measure 90 pulse