//Example 8.10 clc clear close b=[0 1 0 1 0 1 0 1];//given truth tble for the fictitious flip flop a = [0 0 1 1 0 0 1 1]; qn = [0 0 0 0 1 1 1 1]; for i=1:8 if(a(i)==0 & b(i)==0) then qn1(i) = 0; end; if (a(i)==1 & b(i)== 1) then qn1(i) = 1 ; end; if (a(i)==1 & b(i)== 0) then qn1(i) = bitcmp(qn(i),1) ; end; if (a(i)==0 & b(i)== 1) then qn1(i) = (qn(i)) ; end; end; for i = 1 : 8 // printin the truth table Y(i,1)=qn(i); Y(i,2)=a(i); Y(i,3)=b(i); Y(i,4)=qn1(i); end disp('The given truth table is :'); disp(' Qn A B Qn+1'); disp(Y); disp('The transitions are shown below'); c1=0; c2=0; c3=0; c4=0; for j=1:8 //checking all possible cases to make a transition table if(qn(j)==0) then if(qn1(j) == 0) then if(c1==0) then disp('transition from 0 ----> 0'); disp(' A B '); c1=1; end; disp(Y(j,2:3)); end; end; end; for j=1:8 if(qn(j)==0) then if(qn1(j) == 1) then if(c2==0) then disp('transition from 0 ----> 1'); disp(' A B '); c2=1; end; disp(Y(j,2:3)); end; end; end; for j=1:8 if(qn(j)==1) then if(qn1(j) == 0) then if(c3==0) then disp('transition from 1 ----> 0'); disp(' A B '); c3=1; end; disp (Y(j,2:3)); end; end; end; for j=1:8 if(qn(j)==1) then if(qn1(j) == 1) then if(c4==0) then disp('transition from 1 ----> 1'); disp(' A B '); c4=1; end; disp(Y(j,2:3)); end; end; end;