Download Question 7

The Rotating Frame

Let’s consider a collection of spins. There will be a few more that will align with the field than against
the field

Along each direction however the spins will be randomly distributed about the cone. As a consequence
their transverse components cancel.
z
B
y
x

Imagine adding up all the magnetic moments, the resulting sum is known as the bulk magnetization
vector, Mo, which is parallel to the field.
B
Mo
y
x

In order to describe transitions between spin states caused by an applied magnetic field we have to go to
a frame of reference, which rotates at the Larmor frequency, since the applied radiation will be on
resonance with the precession frequency of the nucleus. The alternative is to imagine the interaction
between the rotating magnetic field and the ensemble of precessing spins in the Laboratory frame, which
is static. (This would be extremely difficult to visualize)

Transition are induced by the magnetic component of a resonant radio frequency electromagnetic pulse.
In this case electric field portion of the pulse is ignored. (This form of spectroscopy is not an absorption
emission type phenomenon, but a direct interaction between magnetic fields.)

The oscillating magnetic field can be thought of as being composed of two counter rotating components.
y
y
-
+
x
x
-
+
y
-
+
x
y
x
y
x

Let’s go an axis system which rotates about the z–axis the rotation rate of the positive component of the
oscillating magnetic field.
z
z
+

B1

y
B1
+
B1
B1
x
Laboratory Frame

y
x
Rotating Frame
The bulk magnetization in the presence of a resonant r.f. pulse in the rotating frame will look like. In
this case the counter-rotating component is ignored, since it will have no effect in most cases.
z
z
+
+
Mo
Mo

y
B1

x
y
B1
The applied magnetic field will induce the magnetization to precess around it at the rate of :
B1.

x
Since the pulse is applied over a period, the angle through which it precesses is:
B1
Question 7
Given that the gyromagnetic ratios of 1H and 13C are 26,753 and 6,728 rad/(sec G), compute the pulse length
required for a 900 pulse for both nuclei, given a B1 field strength of 11.7 Gauss.
For 1H :
= Brad26,753 rad/(s G)][11.7 G]→s
For 13C :
= Brad6,728 rad/(s G)][11.7 G]→s
 After a 900 pulse the bulk magnetization is entirely in the transverse plane. Going back to the ensemble
picture, one can see that the magnetic moments are bunched together along the y-axis and that the
populations have equalized. The spins are said to have coherence.
z
B
z
+

Mo
B1

y
y'
x'
x
For many experiments it is important to accurately calibrate the pulse angle in order to be quantitative
and to avoid artifacts. In order to compensate for pulse miscalibration composite-pulse approaches are
employed for excitation and population inversion, and phase cycling is used to cancel unwanted signal.

The excitation angle is 1800, the bulk magnetization is aligned about the negative z-axis. In this case the
population have been inverted, where the excess is along the negative z-axis.
Population inversion as seen in the rotating frame for both the bulk magnetization and the
spin ensemble
z
B
z
+

y'
Mo
B1
y'
x'
x'
Question 8
Draw the bulk magnetization and spin ensemble in the rotating frame for pulse angles  and 3/2.

The axis along which the field is applied determines the phase of the pulse. In the above diagrams the
axis of rotation is about the x axis. Thus this pulse is a x pulse, where  indicates the excitation angle.
The phase of the pulse can be controlled by the instrument.
Illustrating pulse phase
z

Mo
B1

Mo
B1
y'
x'
x'
90x Pulse
B1
90y Pulse


Mo
y'
B1

y'
Mo
x'
90-x Pulse
y'
x'
90-y Pulse
The signal observe when the pulse phases are changed by 90o, are different. Depending on the receiver
coil position one will be in absorption mode whilst the other will be in dispersive mode. (This will be
explained later in more detail.)
Detection and the One pulse Experiment

The NMR signal is detected in the transverse plane, by placing a coil along either the x or y axis. The
bulk magnetization induces an electrical current the coil, which is recorded. The resulting signal is
known as the Free Induction Decay (FID). (see below) The FID is a function of time. In order to
obtain the spectrum this signal has to undergo a Fourier transform to represent it in the frequency
domain.(see below)

Note that the coil is in the laboratory frame not the rotating frame so the bulk magnetization rotates in
the transverse plane at the Larmor frequency.

Mo
y
Signal Intensity
x
Col 1 vs Col 2
Signal Intensity
Time


Frequency

In most cases the pulse is not exactly on resonance. In this case the pulse frequency determines the
rotating frame frequency. The magnetization will no precess in the x’-y’ frame at rate, which is the
difference in their frequencies, denoted below as 


y'
Mo
x'


The frequency of precession of a nucleus with respect to a standard value, is known as the chemical
shift. In solution state spectroscopy of 1H, the standard is the methyl signal of tetramethylsilane, TMS.
Since the precession frequency is dependent on the strength of the magnetic filed use in the experiment,
the relative frequency difference must be reported in units that are relative to the field. That is why the
chemical shift values are reported at parts per million of the magnetic field. This is given as the
frequency difference, , divided by the spectrometer operating frequency, o, as follows:
ppmref)/
Question 9:
The 1H resonance frequency of TMS is 1000 Hz, and that of a line in the spectrum is 2500 Hz, the
spectrometer field is 11.7 T. What is the chemical shift of the line?
Question 10:
The difference in resonance frequency of two 15N lines is 250 Hz, what is their chemical shift difference if
the spectrometer field is 11.7 T. Repeat for a spectrometer field of 7 T.

Often a nucleus is coupled to another nucleus, resulting in the magnetization to be split into two
components, which differ in their precession frequencies by the value of the coupling constant in Hz.

J
y'
J
x'

In this case one can visualize this effect by thinking that the local field experienced the nucleus is
affected by the orientation of the other nucleus. It can be thought of as adding or subtracting a small
amount from the applied field, where the amount is given by the coupling constant. So if the other
nucleus is aligned with the field it will increase the effective field and vice versa, thus perturbing the
resonance frequency.
Question 11
Two 1H nuclei are coupled with a coupling constant of 100 Hz. There chemical shifts are 1.5 and 7
ppm. Determine the resonance frequency of the four line in the spectrum in Hz. The operating
frequency of the spectrometer is 500 MHz.
1 = *o = 1.5 ppm *500 MHz = 750 Hz
2 = *o = 7 ppm * 500 MHz = 3500 Hz
Both lines are split into two differing in frequency by 100 Hz, therefore 50 Hz (J/2) is added and
subtracted from each line.
J750 Hz + 50 Hz = 800 Hz, and J750 Hz - 50 Hz = 700 Hz
J3500 Hz + 50 Hz = 3550 Hz, and J3500 Hz - 50 Hz = 3450 Hz