23 votes 23 votes Design a synchronous counter to go through the following states:$$1, 4, 2, 3, 1, 4, 2, 3, 1, 4 \dots $$ Digital Logic gate1998 digital-logic normal descriptive synchronous-asynchronous-circuits + – Kathleen asked Sep 26, 2014 Kathleen 5.1k views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments Abir Mazumder commented Sep 25, 2020 reply Follow Share We need 2 FF counter , 3 FF system has redundancy . PARDON MY DRAWING :p 1 votes 1 votes jatinmittal199510 commented Apr 12, 2021 reply Follow Share @Abir Mazumder, correct solution. Moreover, the function that we will use to map to the desired state number will be $2Q_1 + Q_0 + 1$ 0 votes 0 votes svas7246 commented Aug 15, 2022 reply Follow Share This counter has various way to derive it for the states it depends on the flipflop you choose to use and when you actually draw the k maps which dont care’s do you use all these will come in to play 0 votes 0 votes Please log in or register to add a comment.
Best answer 28 votes 28 votes Sequence given is as $ 1,4,2,3,1\ldots$ From the given sequence of states we can design the state table and Suppose we are using T-FF for sequential circuit of counter.$$\small \begin{array}{|ccc|ccc|ccc|}\hline \rlap{\textbf{Present State}}& & & \rlap{\textbf{Next State}} & & & \rlap{\textbf{FF Inputs}} & & & \\\hline \;\; A\;\; &\;\; B\;\; &\;\; C \;\; & A^{+} & B^{+} & C^{+} & T_{A} & T_{B} & T_{C} \\\hline 0 & 0 & 0 & x& x & x & x & x & x \\\hline 0 & 0 & 1 & 1&0 &0 &1 &0 & 1 \\\hline 0 & 1 & 0 & 0&1 &1 &0 &0 & 1 \\\hline 0 & 1 & 1 & 0&0 &1 &0 &1 &0 \\\hline 1 & 0 & 0 & 0&1&0 &1 &1 & 0 \\\hline 1 & 0 & 1 & x& x & x & x & x & x \\\hline 1 & 1 & 0 & x& x & x & x & x & x \\\hline 1 & 1 & 1 & x& x & x & x & x & x \\\hline \end{array}$$From the above table, we will find the equation of $T_A$, $T_B$ and $T_C$ Praveen Saini answered Jan 13, 2016 • edited Jul 16, 2019 by Lakshman Bhaiya Praveen Saini comment Share Follow See all 22 Comments See all 22 22 Comments reply Show 19 previous comments jatinmittal199510 commented Apr 12, 2021 reply Follow Share Can be done using $2$ FF's. Function that we will use to map to the desired state number will be $2Q_1 + Q_0 + 1$ 0 votes 0 votes JAINchiNMay commented Jun 17, 2022 reply Follow Share In the k-map for Tb why you have not considered 4sized subcube i.e. A+BC 2 votes 2 votes Abhrajyoti00 commented Jan 6, 2023 reply Follow Share @JAINchiNMay Yes, it should be A+BC. 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes Using D-FlipFlops An Bn Cn An+1 Bn+1 Cn+1 0 0 0 X X X 0 0 1 1 0 0 0 1 0 0 1 1 0 1 1 0 0 1 1 0 0 0 1 0 1 0 1 X X X 1 1 0 X X X 1 1 1 X X X After minimizing An+1, Bn+1, Cn+1 using K-Map An+1 = Bn’C Bn+1 = Cn’ Cn+1 = B mayankso answered Dec 8, 2020 mayankso comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Generally synchronous counter are designed by using D- flip flop and asynchronous counter are designed by using T- flip flop. So here counter designed by using D flip-flop ravi rajak 3 answered Oct 14, 2017 ravi rajak 3 comment Share Follow See 1 comment See all 1 1 comment reply rajinder singh commented Aug 24, 2018 reply Follow Share D1=not(Q0) only 2 votes 2 votes Please log in or register to add a comment.