This is the all question.
I thought that because Tcq>Th we will only count Tcq. If we need to know the minimum clock period, we should calculate the duration from the beginning to the output of DFF1. It makes TpdX+Tsu+Tcq+TpdA+TpdX+Tsu+Tcq = 51ns. Is that right?
I didn't get how the minimum clock period will occur when there are 2 flip flops. Because second flip flop depends on output of the first flip flop. Will we count Tcq twice?