It looks good! Consider the structure in the example solution:
             0 1 2 3 4 5 6
           a 0 0 0 0 0 0 ...
           b 1 0 1 0 1 0 ...
    [a SU b] 1 0 1 0 1 0 ...
   X[a SU b] 0 1 0 1 0 1 ...
          Xa 0 0 0 0 0 0 ...
   [Xa SU b] 1 0 1 0 1 0 ...
Note that [a SU b] becomes equivalent to "b" if "a" is always false. Thus, the two formulas X[a SU b] and [Xa SU b] just become equivalent to Xb and b, respectively.
Your first structure generates the following
             0 1 2 3 4 5 6
           a 0 1 0 0 0 0 ...
           b 0 0 1 1 1 1 ...
    [a SU b] 0 1 1 1 1 1 ...
   X[a SU b] 1 1 1 1 1 1 ...
          Xa 1 0 0 0 0 0 ...
           b 0 0 1 1 1 1 ...
   [Xa SU b] 0 0 1 1 1 1 ...
So, here X[a SU b] holds, and [Xa SU b] is false. 
Finally, your second structure:
             0 1 2 3 4 5 6
           a 0 0 0 0 0 0 ...
           b 0 1 1 1 1 1 ...
    [a SU b] 0 1 1 1 1 1 ...
   X[a SU b] 1 1 1 1 1 1 ...
          Xa 0 0 0 0 0 0 ...
           b 0 1 1 1 1 1 ...
   [Xa SU b] 0 1 1 1 1 1 ...
Again, here X[a SU b] holds, and [Xa SU b] is false.