You are using a spin-echo pulse sequence and the adipose tissue sample in question number two. If the minimum TE value you can obtain is 20 ms, how much more signal could you obtain with a 90-FID sequence?

S(TR,TE) = kρ(1-e-TR/T1)e-TE/T2

where:

    TR = repetition time
    TE = echo time
    S(TR) = signal at time TR from a 90-FID
    ρ = spin density
    k = proportionality constant
    T1 = spin-spin relaxation time
    T2 = spin-echo relaxation time
Taking the ratio of the observed signal (S(TR,TE) from above) to the signal of a 90o FID, gives:

S(TR) / S(TR,TE) = kρ(1-eTR/T1) / kρ(1-e-TR/T1)e-TE/T2
S(TR) / S(TR,TE) = eTE/T2

Given: TE = 20 ms

S(TR) / S(TR,TE) = e20 ms/53 ms = 1.46
S(TR) / S(TR,TE) = e20 ms/94 ms = 1.24

Therefore, under these conditions, the 90o FID sequence would provide 1.24 to 1.46 times the signal of a spin-echo sequence.