From the plots of Fig. 4.28 on page 68 we see that there are some offsets at which the transverse magnetization goes to zero. Recall that during the pulse the magnetization starts on +z and is rotated about the effective field; the nulls in the excitation are when the magnetization has been rotated all the way back to +z i.e. when the rotation about the effective field is through 2n radians, or some multiple of this angle. We can work out the offset at which this occurs in the following way.
The effective field is given by

To simplify things, we will express the offset as a multiple K of the RF field strength:
Ω = kω1
Show that, using this expression for Ω, ωeff is given by:

The null condition is when the rotation is 2Ï€:
null condition ωefftp = 2π,
where tp is the length of the pulse. The final thing to note is that the on-resonance flip angle is π/2; this means that
on resonance ω1tp = π/2.
Combine the last three equations to show that the null occurs when
4 = ˆšl + k2 i.e. k = ˆšl5.
The predicted null is at K = Ω/ω1 = ˆš5 i.e. = Ω = ˆšl5ω1. Does this agree with Fig. 4.28 on page 68?
Predict the value of K at which the next null will occur.
Further nulls continue to occur at larger offsets; show that at large offsets, which means K » 1, the nulls occur at K = 4n, where n is an integer.
[The nulls occur at rotation angles of 2nÏ€; for K » 1, ˆšl + K2 can be approximated.)

Transcribed Image Text:

+Ω . Weff =