《数字电》英文版 chapter4-2 Flip-flops

Flip-flops Chapter4 Transformation between different flip-flops

Flip-flops Chapter4 Transformation between different flip-flops

a Add the certain combinational logic circuit to the input of the original flip flop New Original excitation Converting circuit input flip-flop New flip-flop

◼ Add the certain combinational logic circuit to the input of the original flip￾flop Original flip-flop Converting circuit Q’ Q New flip-flop New excitation input

Simultaneous characteristic equation X Q Converting Original circuit flip-flop converted flip-flop

Simultaneous characteristic equation Original flip-flop Converting circuit Q’ Q converted flip-flop X, Q Y

a y=F X, Q a x be the input variable set of the new flip-flop ay be the input variable set of the original flip-flop Q be the present-state of the original flip-flop X, Q Converting Original circuit flip-flop Q converted flip-flop

◼ Yi=Fi (X,Q)  X be the input variable set of the new flip-flop  Y be the input variable set of the original flip-flop  Q be the present-state of the original flip-flop Original flip-flop Converting circuit Q’ Q converted flip-flop X, Q Y

Convert JK to D,T Rs JK>>D Q 口J=f1(DQ),K=f2(DQ) 口Qm+1=D=DQ+DQ C Q Qn+l=JQ,+K'Q K 口J=D,K=D ■JK>T 口J=f1(TQ),K=f2(TQ) C a Qn+=TQ+TQ Q 口Qn+1=JQ+kQ K 口J=T,K=T

Convert JK to D, T, RS ◼ JK >> D  J=f1(D,Q), K=f2(D,Q)  Qn+1=D=DQ’+D’Q  Qn+1=JQ’+K’Q  J=D, K=D’ ◼ JK >> T  J=f1(T,Q), K=f2(T,Q)  Qn+1=TQ’+T’Q  Qn+1=JQ’+K’Q  J=T, K=T JK Q’ C Q D JK Q’ C Q T

Convert j to D,T Rs ■JK>RS O J=fi(R, S, Q), K=f2(R, S, Q S 口Qm+1=S+RQ(RS=0) + Q ES(Q+Q)+R'Q =SQ+R'Q+SQ(R+R) -SQ+RQ+SRQ =SQ+RQ 口J=s,k=Q

Convert JK to D, T, RS ◼ JK >> RS  J=f1(R,S,Q), K=f2(R,S,Q)  Qn+1=S+R’Q (RS=0) =S(Q’+Q)+R’Q =SQ’+R’Q+SQ(R+R’) =SQ’+R’Q+SR’Q =SQ’+R’Q  J=S, K=Q JK Q’ C Q RS

Convert d tO JK T Rs ■D>>JK Q 口D=f1(KQ) 口Qm+1=JQ+KQ Qn+=D 口D=JQ+kQ ■D>RS 口D=f1(RSQ a Qn+=S+RQ 口Qn+1=D D=S+R'Q K J

Convert D to JK, T, RS ◼ D >> JK  D=f1(J,K,Q)  Qn+1=JQ’+K’Q  Qn+1=D  D=JQ’+K’Q ◼ D >> RS  D=f1(R,S,Q)  Qn+1=S+R’Q  Qn+1=D  D=S+R’Q D Q Q’ CK J D Q Q’ C S R

Simultaneous excitation table Qn+1Qn TD 00d00d00 00 110dd001

Simultaneous excitation table 0 0 1 1 d 1 0 0 0 0 1 d Qn+1Qn R S J K T D 0 1 0 1 0 d 1 d d 1 d 0 0 0 1 1 0 1 1 0

T>> RS Qn+1Qn S 00d0 0 TEf(Q, R, s) 110d 0 RS TESQ'+RQ Q0001110 1 d d

◼ T >> RS 0 0 1 1 d 1 0 0 0 0 1 d Qn+1Qn R S J K T D 0 1 0 1 0 d 1 d d 1 d 0 0 0 1 1 0 1 1 0 00 01 11 10 0 1 0 1 2 3 6 7 4 5 RS Q 1 1 d d T=SQ’+RQ ◼ T =f(Q,R,S)

T>>D 口T=f(QD) Qn+1Qn DQ+DQ 00 0 10 0

◼ T >>D 0 1 0 1 0 1 2 3 D Q 1 1 T=DQ’+D’Q  T =f(Q,D) 0 0 1 1 d 1 0 0 0 0 1 d Qn+1Qn R S J K D T 0 1 0 1 0 d 1 d d 1 d 0 0 0 1 1 0 1 1 0