Mod 4 Up Down Counter g2z2r
This document was ed by and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this report form. Report 3i3n4
Overview 26281t
& View Mod 4 Up Down Counter as PDF for free.
More details 6y5l6z
- Words: 1,152
- Pages: 14
SC 5155
Chapter 7
Slide 1
Modulo–4 Up–Down Counter This is a counter with input. If X = 0, the device counts up: If X = 1, the device counts down:
0, 1, 2, 3, 0, 1, 2, 3, etc. 0, 3, 2, 1, 0, 3, 2, 1, etc.
Step 1a: Derive the state diagram and state table for the circuit.
Note two transitions between the state pairs: one is up and one is down.
Slide 1 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 2
Step 1b: Derive the State Table Present State 0 1 2 3
Next State X=0 X=1 1 3 2 0 3 1 0 2
This is just a restatement of the state diagram. Note: Two columns for the “Next State”.
Slide 2 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 3
Step 2: Count the States and Determine the Flip–Flop Count Count the States There are four states for any modulo–4 counter. N = 4 The states are simple: 0, 1, 2, and 3. Calculate the Number of Flip–Flops Required Let P be the number of flip–flops. Solve 2P-1 < N 2P. So 2P-1 < 4 2P and P = 2. We need two flip–flops.
Slide 3 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 4
Step 3: Assign a unique P-bit binary number (state vector) to each state. Here P = 2, so we are asg two–bit binary numbers. Vector is denoted by the binary number Y1Y0.
State 0 1 2 3
2-bit Vector Y1 Y0 0 0 0 1 1 0 1 1
Each state has a unique 2–bit number assigned. Any other assignment would be absurd.
Slide 4 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 5
Step 4: Derive the state transition table and the output table. There is no computed output, hence no output table. The state transition table uses the 2–bit state vectors
Present State 0 1 2 3
Slide 5 of 14 slides
00 01 10 11
Design of a Mod-4 Up Down Counter
Next State X=0 X=1 01 11 10 00 11 01 00 10
February 13, 2006
SC 5155
Chapter 7
Slide 6
Step 5:Separate the state transition table into P tables, one for each flip-flop. Flip–Flop 1 PS Y1Y0 0 0 0 1 1 0 1 1
Flip-Flop 1 Next State: Y1 X=0 X=1 0 1 1 0 1 0 0 1
PS Y1Y0 0 0 0 1 1 0 1 1
Flip-Flop 0: Y0 Next State X=0 X=1 1 1 0 0 1 1 0 0
Flip–Flop 0
Slide 6 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 7
Step 6: Decide on the types of flip-flops to use. When in doubt, use all JK’s. Here is the excitation table for a JK flip–flop
Q(T) 0 0 1 1
Slide 7 of 14 slides
Q(T+1) 0 1 0 1
Design of a Mod-4 Up Down Counter
J 0 1 d d
February 13, 2006
K d d 1 0
SC 5155
Chapter 7
Slide 8
Step 7: Derive the input table for each flip-flop Flip–Flop 1 Y1Y0 0 0 0 1 1 0 1 1
Y1 0 1 1 0
X=0 J1 0 1 d d
Y0 1 0 1 0
X=0 J0 1 d 1 d
K1 d d 0 1
Y1 1 0 0 1
X=1 J1 1 0 d d
K1 d d 1 0
Y1 1 0 1 0
X=1 J0 1 d 1 d
K0 d 1 d 1
Flip–Flop 0 Y1Y0 0 0 0 1 1 0 1 1
K0 d 1 d 1
Question: How do we produce equations for the J’s and K’s?
Slide 8 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 9
Step 8: Derive the input equations for each flip-flop The equations are based on the present state and the input. The input X produces a complication. The simplest match procedure will lead to two equations for each flip–flop input: one for X = 0 and one for X = 1. Use the “combine rule” The rule for combining expressions derived separately for X = 0 and X = 1 is X (expression for X= 0) + X(expression for X = 1). Rationale: Let F(X) = A X + BX When X = 0, F(X) = A and when X = 1, F(X) = B.
Slide 9 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 10
Input Equations for Flip–Flop 1
Y 1Y 0 0 0 0 1 1 0 1 1
Y1 0 1 1 0
X=0 J1 K1 0 d 1 d d 0 d 1 J1 = Y0 K1 = Y0
Y1 1 0 0 1
Apply the “combine rule” J1 = X’Y0 + XY0’ = X Y0 K1 = X’Y0 + XY0’ = X Y0
Slide 10 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
X=1 J1 K1 1 d 0 d d 1 d 0 J1 = Y0’ K1 = Y0 ’
SC 5155
Chapter 7
Slide 11
Input Equations for Flip–Flop 0
Y 1Y 0 0 0 0 1 1 0 1 1
Y0 1 0 1 0
X=0 J0 K0 1 d d 1 1 d d 1 J0 = 1 K0 = 1
Y1 1 0 1 0
X=1 J0 1 d 1 d J0 = 1 K0 = 1
K0 d 1 d 1
Apply the “Combine Rule” J0 = X’1 + X1 = 1 K0 = X’1 + X1 = 1 Neither J0 nor K0 depend on X. But Y0 does not depend on X.
Slide 11 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 12
Step 9: Summarize the equations by writing them in one place. Here they are. J 1 = X Y0 J0 = 1
Slide 12 of 14 slides
K1 = X Y0 K0 = 1
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 13
Step 10: Draw the Circuit As designed, it is:
Slide 13 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 14
Step 10: Draw the Circuit Implemented with T Flip–Flops, it is:
One could also use a 4– “one hot” design, with the input X used to determine the direction of the shift.
Slide 14 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
Chapter 7
Slide 1
Modulo–4 Up–Down Counter This is a counter with input. If X = 0, the device counts up: If X = 1, the device counts down:
0, 1, 2, 3, 0, 1, 2, 3, etc. 0, 3, 2, 1, 0, 3, 2, 1, etc.
Step 1a: Derive the state diagram and state table for the circuit.
Note two transitions between the state pairs: one is up and one is down.
Slide 1 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 2
Step 1b: Derive the State Table Present State 0 1 2 3
Next State X=0 X=1 1 3 2 0 3 1 0 2
This is just a restatement of the state diagram. Note: Two columns for the “Next State”.
Slide 2 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 3
Step 2: Count the States and Determine the Flip–Flop Count Count the States There are four states for any modulo–4 counter. N = 4 The states are simple: 0, 1, 2, and 3. Calculate the Number of Flip–Flops Required Let P be the number of flip–flops. Solve 2P-1 < N 2P. So 2P-1 < 4 2P and P = 2. We need two flip–flops.
Slide 3 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 4
Step 3: Assign a unique P-bit binary number (state vector) to each state. Here P = 2, so we are asg two–bit binary numbers. Vector is denoted by the binary number Y1Y0.
State 0 1 2 3
2-bit Vector Y1 Y0 0 0 0 1 1 0 1 1
Each state has a unique 2–bit number assigned. Any other assignment would be absurd.
Slide 4 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 5
Step 4: Derive the state transition table and the output table. There is no computed output, hence no output table. The state transition table uses the 2–bit state vectors
Present State 0 1 2 3
Slide 5 of 14 slides
00 01 10 11
Design of a Mod-4 Up Down Counter
Next State X=0 X=1 01 11 10 00 11 01 00 10
February 13, 2006
SC 5155
Chapter 7
Slide 6
Step 5:Separate the state transition table into P tables, one for each flip-flop. Flip–Flop 1 PS Y1Y0 0 0 0 1 1 0 1 1
Flip-Flop 1 Next State: Y1 X=0 X=1 0 1 1 0 1 0 0 1
PS Y1Y0 0 0 0 1 1 0 1 1
Flip-Flop 0: Y0 Next State X=0 X=1 1 1 0 0 1 1 0 0
Flip–Flop 0
Slide 6 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 7
Step 6: Decide on the types of flip-flops to use. When in doubt, use all JK’s. Here is the excitation table for a JK flip–flop
Q(T) 0 0 1 1
Slide 7 of 14 slides
Q(T+1) 0 1 0 1
Design of a Mod-4 Up Down Counter
J 0 1 d d
February 13, 2006
K d d 1 0
SC 5155
Chapter 7
Slide 8
Step 7: Derive the input table for each flip-flop Flip–Flop 1 Y1Y0 0 0 0 1 1 0 1 1
Y1 0 1 1 0
X=0 J1 0 1 d d
Y0 1 0 1 0
X=0 J0 1 d 1 d
K1 d d 0 1
Y1 1 0 0 1
X=1 J1 1 0 d d
K1 d d 1 0
Y1 1 0 1 0
X=1 J0 1 d 1 d
K0 d 1 d 1
Flip–Flop 0 Y1Y0 0 0 0 1 1 0 1 1
K0 d 1 d 1
Question: How do we produce equations for the J’s and K’s?
Slide 8 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 9
Step 8: Derive the input equations for each flip-flop The equations are based on the present state and the input. The input X produces a complication. The simplest match procedure will lead to two equations for each flip–flop input: one for X = 0 and one for X = 1. Use the “combine rule” The rule for combining expressions derived separately for X = 0 and X = 1 is X (expression for X= 0) + X(expression for X = 1). Rationale: Let F(X) = A X + BX When X = 0, F(X) = A and when X = 1, F(X) = B.
Slide 9 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 10
Input Equations for Flip–Flop 1
Y 1Y 0 0 0 0 1 1 0 1 1
Y1 0 1 1 0
X=0 J1 K1 0 d 1 d d 0 d 1 J1 = Y0 K1 = Y0
Y1 1 0 0 1
Apply the “combine rule” J1 = X’Y0 + XY0’ = X Y0 K1 = X’Y0 + XY0’ = X Y0
Slide 10 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
X=1 J1 K1 1 d 0 d d 1 d 0 J1 = Y0’ K1 = Y0 ’
SC 5155
Chapter 7
Slide 11
Input Equations for Flip–Flop 0
Y 1Y 0 0 0 0 1 1 0 1 1
Y0 1 0 1 0
X=0 J0 K0 1 d d 1 1 d d 1 J0 = 1 K0 = 1
Y1 1 0 1 0
X=1 J0 1 d 1 d J0 = 1 K0 = 1
K0 d 1 d 1
Apply the “Combine Rule” J0 = X’1 + X1 = 1 K0 = X’1 + X1 = 1 Neither J0 nor K0 depend on X. But Y0 does not depend on X.
Slide 11 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 12
Step 9: Summarize the equations by writing them in one place. Here they are. J 1 = X Y0 J0 = 1
Slide 12 of 14 slides
K1 = X Y0 K0 = 1
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 13
Step 10: Draw the Circuit As designed, it is:
Slide 13 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
SC 5155
Chapter 7
Slide 14
Step 10: Draw the Circuit Implemented with T Flip–Flops, it is:
One could also use a 4– “one hot” design, with the input X used to determine the direction of the shift.
Slide 14 of 14 slides
Design of a Mod-4 Up Down Counter
February 13, 2006
Related Documents 3h463d
Mod 4 Up Down Counter g2z2r
December 2019 27
Mod 5 Counter 55y4c
October 2022 0
Low Cost Design Of Mod-8 Synchronous Up/down Counter Using Reversible Logic Gate 4r6ir
December 2019 32
December 2019 56
Up-down 74191 ud5s
December 2019 55
November 2019 41
More Documents from "Giwan To Milanello" j717
Mod 4 Up Down Counter g2z2r
December 2019 27
Kutub 4 Banget.pdf 1a185r
December 2020 0
4p345f
November 2019 77
[heinrich Rickert] Zwei Wege Der Erkenntnistheorie(bookfi.org) 5gb5g
March 2023 0
Malaikat Baik 103kc
November 2019 57