What is the width of the function in the answer to question 3 in Hz and in ppm when this signal is greater than 90% of its maximum value? Why is this result significant in NMR and MRI?

This sinc function represents the B1 field as a function of frequency being sent into the sample. Only magnetization vectors in the sample with resonant frequencies that are also found in the sinc function will experience a rotation. The NMR signal is related to the amount of Z magnetization (MZ) rotated into the XY plane (MXY). For a rotation angle q, the signal is

 

Sinθ=MXY/MZ.

For MXY to be greater than or equal to 0.9, θ must be between 64.15 and 115.85 degrees. The rotation angle is proportional to B1 through the rotation equation

 

θ=2π γ B1 τ.

B1 is proportional to the magnitude (sqrt(RE2 + IM2)) of the sinc function. Adopting a normalized B1 (B1=1 for a 116 degree rotation), B1=0.553 for a 64.15 degree rotation. Define ν as the frequency in the laboratory frame of reference. The RE and IM parts of the function are:

 

RE(B1) = (sin(2π (ν-νo) t))/(2π (ν-νo) t)
  IM(B1) = -(sin2(π(ν-νo) t))/(π (ν-νo) t)
 
Taking the magnitude of this function we obtain

 

MAG(B1) = sqrt[ (sin2(2π(ν-νo) t))/(2π (ν-νo) t)2 + (4 sin4(π (ν-νo) t))/(2π (ν-νo)t)2 ]

 0.553 = sqrt[sin2(2π (ν-νo)t) + 4 sin4(π (ν-νo)t)] /(2π (ν-νo)t)2

Solving this equation numerically we get the range of frequencies, Δν = 22.6 kHz, or 359 ppm.

This result is significant because it tells us that an 359 ppm wide spectrum would easily be rotated by approximately 64 to 116 degrees (giving 90% of the possible signal) with a 50 microsecond wide pulse.