## The Effect Of Rf Pulses

In the rotating frame of reference, the effect of an RF pulse can be described in terms of the interaction of the stationary magnetic field vector (B1) of the pulse and the net magnetization of the sample. At equilibrium, the sample net magnetization vector lies along the axis. The RF pulse magnetic field, which is referred to as B\ to distinguish it from the Bo field, exerts a torque on the net magnetization vector that rotates it in a plane perpendicular to the B\ field vector (Fig. 6.6). For example, an RF pulse with appropriate phase to place the B1 field vector along the xX axis will rotate the net magnetization vector in a counterclockwise direction from the axis toward the — y' axis in the rotating frame. This rotation is the same as the precession under the influence of the Bo field (vo = yBo/2n), except that it is much slower because the Bi field is much weaker than Bo even for the highest power RF pulses. In the rotating frame, for an on-resonance spin, the Bo field goes away and the only field affecting the net magnetization vector is the B1 field. The net magnetization vector rotates faster if the B1 amplitude is greater. The rate of precession of the net magnetization vector around B1 can be written as v1 = yB1/2n. Be careful to recognize that v1 is not the frequency of the pulse (that would be vr), but rather the frequency of rotation of the sample net magnetization around the B1 vector during the pulse. This rotation rate, yB1/2n, is often used as a measure of the amplitude of the pulse because it is in more convenient units of hertz rather than tesla. This is analogous to the way we measure Bo field strength in megahertz ( yhBo/2n) rather than tesla.

The extent of precession of the sample net magnetization vector under the influence of the B1 field depends on the duration of the pulse. The longer we leave the pulse on, the farther the net magnetization vector rotates. A pulse that lasts just long enough to rotate the net magnetization vector by an angle of 90o is called a "90o pulse." A stronger B1 field (higher RF power during the pulse) will rotate the net magnetization faster and will lead to a shorter duration for the 90o pulse. If we are measuring pulse amplitude in units of hertz (yB1/2n), we can calculate the 90o pulse width from the amplitude:

t360 = time to rotate one cycle aroundB1 = 1/v1 = 1/(yB1 /2n)

t90 = time to rotate one-fourth cycle around B1 = 1/(4 x v1) = 1/(4yB1 /2n)

Likewise if we calibrate the pulse width to give the maximum peak height in the spectrum (90o pulse), we can calculate the pulse amplitude in hertz:

All pulse rotations are counterclockwise when viewed with the B1 vector pointing toward you.

As we look at the effect of pulses on the net magnetization vector, it is useful to keep track of the population difference between the a and 5 states at the same time. This can be done by drawing the energy diagram (two levels) and using filled circles to represent excess population and open circles to represent population deficits relative to an equal division of spins between the two energy states. For example, at equilibrium we have N/2 + 8 spins in the a state and N/2 — 8 spins in the 5 state, where 8 is a number much smaller than N (e.g., 10-5 N). We can draw an open circle (—8) in the upper energy level and a closed circle (+8) in the lower energy state (Fig. 6.7, left). Note that as we are only concerned with the population difference AP (in this case Pa - P5 = 28), we ignore the N/2 term in our circle representation.

At the end of the 90o pulse with B1 on the X axis, the net magnetization is on the — y' axis, and we have no z component. We will refer to this spin state as —Iy. Because the z component of net magnetization results from the population difference between the a and 5 states, we can say that there is no population difference at the end of a 90o pulse (Fig. 6.7). With the 90o pulse, we have effectively converted the population difference into coherence. If we record the FID right after this pulse, we would get a normal spectrum with a positive absorptive peak.

A 180o pulse, which lasts twice as long as a 90o pulse, will rotate the net magnetization from the +z axis to the — z axis (Fig. 6.7, center). We can call this spin state —Iz. There is no coherence, and we would not observe any spectrum if we collected an FID at this point. As net magnetization on the +z axis at equilibrium results from the slight excess of spins in the lower energy (a) state, rotating this to the — z axis means that we have inverted the population difference, so that now there is a slight excess of spins in the higher energy (¡) state. We have the same population difference but now it is negative: AP = Pa — P5 = —28. This is represented by a filled circle (N/2 + 8) in the higher energy (¡) state and an open circle (N/2 — 8) in the lower energy (a) state. We call this an "inversion" pulse because it inverts the populations in the a and 5 states. The really bizarre thing is that the 180o pulse does not simply move 28 spins from the a to the 5 state; instead, it moves every single spin that was in the a state to the j state and every spin that was in the j state to the a state. From the point of view of the spins, the entire world is turned upside down! The N/2 + 8 spins that were in the a state are now in the j state, and the N/2 - 8 spins that were in the j state are now all in the a state. This is truly an inversion pulse! This may seem like a trivial point, but it will be very important later on when we talk about /-coupling relationships.

At the end of a 270° pulse, the net magnetization vector lands on the +y' axis, opposite to where it would be at the end of a 90° pulse. We call this state Iy. There is no z component, so we have no population difference. If the 90° pulse gives a normal (positive absorptive) peak in the spectrum, a 270° pulse will give an upside down peak. In this case we would call the —y' axis the "reference axis," and any magnetization vector that is on this axis at the start of the FID would give a normal (positive absorptive) peak in the spectrum. If the magnetization vector is on the opposite axis (the +y' axis) at the start of the FID, it will give an upside-down (negative absorptive) peak. Vectors on the +X or — X axes would lead to dispersive (up/down or down/up) peaks. We can choose which axis we want to use for the reference axis, and this is referred to in the pulse program (the software that drives the experiment) as the "receiver phase." NMR data processing software can always change the phase of peaks, but it cannot (or at least should not) change the relative phase of peaks, so that if the 90° pulse on the X axis gives an upside-down peak, the 270° pulse on the X axis will give a normal peak (i.e., we have changed the phase reference to the +y' axis by phase correction in software).

At the end of a 360° pulse, the net magnetization vector has made one complete rotation around the B1 vector and lands on the +z axis, exactly where it started (Fig. 6.7, right). The spin state is identical to the equilibrium state, Iz. We have the equilibrium (Boltzmann) population distribution, represented with one open circle in the upper state and one filled circle in the lower state. If we collect an FID right after the 360° pulse, we will see no spectrum.

Pulse calibration is the process of collecting a series of FIDs, each with the pulse width increased a little bit from the last one. For example, we might acquire 18 13 C spectra starting with a pulse width of 0 and increasing by 3 ^s each time (i.e., 0, 3, 6, 9, 12, etc.) (Fig. 6.8). The fifth spectrum (12 ^s pulse) gives the maximum positive peak (90° pulse), the ninth (24 ^s pulse) gives a very weak negative peak—nearly a null (180° pulse), the 13th spectrum (36 ^s pulse) gives a maximum negative peak (270° pulse), and the second null (360° pulse) occurs halfway between the 45 ^s pulse and the 48 ^s pulse. A very long relaxation delay (70 s) is necessary to make sure we are starting with the equilibrium state (Mz = +Mo) each time. It is important to understand that only one peak is shown in the spectrum, and we are repeating the experiment each time with a new pulse width value, plotting it to the right-hand side of the previous spectrum. As it is easier to pin down the null point exactly rather than the maximum spectrum, we usually find the 180° pulse width and divide by 2, or the 360° pulse width and divide by 4, to get the 90° pulse width. In this case, the 360° pulse can be interpolated as 46.5 ^s, so the 90° pulse is 11.6 ^s (46.5/4). The calibrated 90° pulse is always reported at a particular RF power setting, in this case 60 dB. Because the 90° pulse width depends on the Bi amplitude (RF power level), we need to specify that value, or the calibration will be meaningless. The B1 field strength in hertz (yB1/2n) is 1/(4 x t90) = 1/(4 x 11.6 jxs) = 1/(4 x 0.0000116 s) = 1,000,000/(4 x 11.6 s) = 1,000,000/(46.4 s) = 21,552 Hz or 21.552 kHz. This is the reciprocal of the 360° pulse: the rate of rotation of the sample net magnetization vector around the B1 vector during the pulse (v1). Note that this is only 0.0287% (100 x 0.021552 MHz/75 MHz) of the Bo field strength expressed in hertz for a 7.05-T instrument. Here we are comparing yB1/2n to yBo/2n, so we have

Figure 6.8

to use y for 13C, which gives us 75 MHz for yBo/2n on a "300-MHz" instrument. The B\ field (oscillating at 75 MHz and oriented along the X or y' axis) is very small (short vector) compared to the Bo field (static and oriented along the +z axis), but we can not get much higher than this without heating the sample and/or burning up the amplifiers.

The real and imaginary channels of the NMR receiver can be considered to record the X and y components of the net magnetization vector in the rotating frame: Mx and My. Thus, if we use a 90° excitation pulse on the X axis, the net magnetization vector will rotate to the —/ axis and then undergo chemical shift evolution in a ccw direction (for peaks in the downfield half of the spectral window) or in a cw direction (for peaks in the upfield half). For example, for a positive rotating-frame frequency, the net magnetization vector will move from the — y' axis to the +X, +y, —X axis and back to the — y' axis, as it rotates. The X component (Mx) will start at zero, then increase to a positive maximum as the vector passes the X axis, then decrease to zero as the vector crosses the +y axis, and pass to a negative maximum as it crosses the —X axis. This is shown in Figure 6.9, along with the y component (My), which starts at a negative maximum, decreases to zero, increases to a positive maximum, and decreases to zero again in the same time period. If we see these two waveforms, can we deduce the position of the net magnetization vector at the beginning of the FID (the peak phase), as well as its direction of rotation (the sign of the peak frequency)? First consider that we have only the real part of the FID (the MX component): What can we say about the sample net magnetization? It could be starting on the — y' axis and moving ccw (positive frequency), or it could be starting on the +y axis and moving cw (negative

Figure 6.9

frequency). In both cases, we would see the same Mx waveform. So if we use the —y' axis as our reference axis, we could draw the spectrum as a positive absorptive peak with frequency ("chemical shift") equal to +Av, on the left-hand side of the spectral window, or as a negative absorptive peak with frequency equal to -Av, on the right-hand side of the spectral window. As both are equally likely given the information available, the Fourier transform includes both peaks with equal intensity in the spectrum (Fig. 6.9). If instead we only have the imaginary part of the FID (My), we could say that the net magnetization definitely starts on the —y' axis (positive absorptive peak), but we do not know whether it moves ccw or cw, because both directions would give the same results in the My trace: starting at a negative maximum, decreasing to zero, building to a positive maximum, and so forth. So the Fourier transform would give a spectrum with positive peaks of equal intensity at + Av and — Av. If we add these two spectra together, we get an idea of how quadrature detection works: Combining what we know from the real part of the FID with what we know from the imaginary part of the FID gives us a spectrum with a single positive peak at frequency +Av, and the "quadrature image" peak at - Av is canceled out. This gives us a sense of how the mathematics of the complex (i.e., real and imaginary combined) Fourier transform works.

Another term for quadrature detection is "phase-sensitive detection," and we can see how the complex Fourier transform is truly sensitive to the phase of the NMR signal: It can not only distinguish positive frequencies (ccw rotation) from negative frequencies (cw rotation), but also tell on which axis the net magnetization started at the beginning of the FID (corresponding to the phase of the NMR peak). In Figure 6.9 we saw how the ghost image, or quadrature image, at —Av in the spectrum was canceled out in the two spectra obtained from Mx and My. Sometimes this cancellation is not perfect and we get a small image, either positive or negative on the opposite side of the spectral window from our peak. This is especially likely for a very intense peak in the spectrum. The most likely cause of this artifact is that the two receiver channels, real and imaginary, have unequal gain or amplification so that the negative peak and the positive peak at - Av are not equal in intensity. Consider, for example, that the imaginary channel has a gain 10% higher than the real channel. We would then see a signal for Mx equal to sin(2nAvt) and a signal for My equal to — 1.1cos(2n Av t), ignoring the decay of the FID. The center spectrum with two positive peaks in Figure 6.9 would be 10% more intense than the upper one with one positive and one negative peak, and in the sum the true peak at + Av would have an intensity of 2.1 whereas the quadrature artifact at — Av would have an intensity of +0.1.

We could spend a lot of time adjusting and balancing the two receiver channels of the spectrometer to get exactly the same gain, but there is a much simpler way to eliminate the "quad" artifacts. We can acquire a four-scan spectrum, but with each scan we advance the phase of the pulse, starting with a pulse on X (scan 1), moving ccw by 90° to a pulse on y (scan 2), then ccw again to a pulse on —X (scan 3), and finally to a pulse on — y (scan 4). The table below shows the signals that we would observe for Mx and My for each of these pulses. You should draw the three axes (x/, y' and z) and verify for yourself that the two components will vary as shown as the net magnetization vector rotates ccw in the x'-y' plane after the excitation pulse.

Scan Pulse phase Vector starts on

Scan Pulse phase Vector starts on

 l +x —y' sin(2nAvt) — l.l cos(2nAvt)
0 0