Question 1: Sequential Circuit Design
The ABC Saw Mill is the smallest Canadian producer of fire wood. They wish to
automate their log-cutting efforts and have decided to design the conveyor belt
system shown in Figure 1. You have been commissioned to develop the
synchronous sequential saw controller that can detect the length of wood logs
and cut only the logs that are long. For simplicity, you may assume that all logs
are of only two sizes, either LONG or SHORT. Using two sensors (S 1 and S2),
the SHORT logs will never pass under both sensors at the same time, however,
the LONG logs will. The sensor input will be set to logic high when a log is under
the sensor, otherwise the input it will remain at zero. After a LONG log is
detected (i.e. the end of the log passes sensor S2), the SAW output must be
asserted, which will pause the conveyor belt and lower the saw to cut the log.
You may also assume that there will always be sufficient space between logs
such that two logs will never be under the sensors at the same time.
For all input sequences that are not physically possible, ensure that your system
returns to the “initial” state. Furthermore, all unused states should return your
system to the “initial” state.
1) Draw the State Transition Diagram for the sequential saw controller.
2) Derive the State Transition Table for the sequential saw controller.
3) Draw the sequential circuit diagram for the sequential saw controller using
only positive edge-triggered JK flip-flops. Minimize any combination logic
using K-maps.
Figure 1: Circuit used in Question 1: Sequential Circuit Design
SOLUTION to Question 1
PRESENT
STATE
A
B
0
0
0
0
0
0
0
0
0
1
0
1
0
1
0
1
1
0
1
0
1
0
1
0
1
1
1
1
1
1
1
1
INPUTS
S1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
S2
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
NEXT
STATE
A+
B+
0
0
0
0
0
1
0
0
0
0
0
0
0
1
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
0
FF INPUTS
JA
0
0
0
0
0
0
0
1
X
X
X
X
X
X
X
X
KA
X
X
X
X
X
X
X
X
1
0
1
0
1
1
1
1
JB
0
0
1
0
X
X
X
X
0
0
0
0
X
X
X
X
OUTPUT
KB
X
X
X
X
1
1
0
1
X
X
X
X
1
1
1
1
SAW
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
SOLUTION to Question 1
S1S2
AB
00
00
01
11
10
1
X
X
X
X
01
11
10
X
X
01
X
X
11
1
1
10
1
KA = B + S2‘
X
X
1
X
X
1
1
11
10
X
X
1
X
X
11
10
01
X
X
X
X
JA = BS1S2
11
10
S1S2
AB
00
00
S1S2
AB
00
00
01
11
X
X
01
X
X
10
JB = A’S1S2‘
S1S2
AB
00
00
01
X
X
X
X
01
1
1
1
0
11
1
1
1
1
10
X
X
X
X
KB = A’ + B + S1 + S2‘
Question 2: Sequential Circuit Analysis
The sequential circuit in Figure 2 is a sequence detector that asserts its output Z
when a 4-bit serial input sequence is detected. The serial input is denoted X.
This circuit is synchronized with respect to the clock signal CLK. The 16x3 ROM
has the following programming table:
A3
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
INPUTS
A2 A1
0
0
0
0
0
1
0
1
1
0
1
0
1
1
1
1
0
0
0
0
0
1
0
1
1
0
1
0
1
1
1
1
A0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
OUTPUTS
Y2
Y1
Y0
0
0
0
0
1
0
1
1
0
1
0
0
0
1
0
1
1
0
0
1
0
0
1
1
0
0
0
1
0
1
1
1
0
0
1
1
1
0
0
1
1
1
1
0
0
1
0
1
a) Derive the state transition table.
b) Draw the state transition diagram.
c) What four bit sequence does this circuit detect? (Hint: State 00 is the initial
state.)
Figure 2: Circuit used in Question 2: Sequential Circuit Analysis
SOLUTION to Question 2
(a)
Present States
QA
QB
0
0
0
0
0
1
0
1
1
0
1
0
1
1
1
1
Input
X
0
1
0
1
0
1
0
1
FF Inputs
DA
TB
0
0
0
1
0
1
1
1
1
1
1
0
0
1
0
1
Next State
QA(t+1) QB(t+1)
0
0
0
1
0
0
1
0
1
1
1
0
0
0
0
0
Output
Z
0
0
0
0
0
0
0
1
(b)
(c) 1101
Question 1 (Arithmetic)
Let ADD4 be a 4-bit adder. We will use this adder as a building block to design other
arithmetic circuits.
1- Consider MUL3 circuit shown in Figure 1(a). This circuit has a 4-bit input A and a 6bit output P such that P = 3*A. Design this circuit using a single ADD4 component and
HA/FA cells.
2- Consider the 4-bit adder shown in Figure 1(b). We call this adder ADD4_MOD. The
behavior of the circuit depends on the values of inputs M1 and M0 as shown in Table 1 .
For instance, the circuit performs the addition of the two 4-bit numbers A and B modulo
8 when M1M0 = 01. The result of this addition appears at output S. Output Cout is the
carry-out of adding A and B when M1M0 = 00, in all other modes Cout = 0. Design
ADD4_MOD using the following components : ADD4, multiplexers and logic gates.
M1
M0
Operation
0
0 A+B
0
1 (A + B) mod 8
1
0 (A + B) mod 4
Table 1
3- Carry-select adders (CSA) are used to cope with the carry propagation delays
encountered in ripple-carry adders. A typical CSA is made of smaller adder -blocks.
Some blocks are replicated to consider the cases when the carry-in is 0 and when the
carry-in is 1. An incomplete 12-bit CSA is shown in Figure 2. Complete this figure by
adding the missing components and connections.
Solution
1-
2-
3- Straightforward.
Question 2 (PLDs)
The truth table of a binary to excess-3 encoder is shown in Table 2. We like to implement
this circuit using a PLA.
1- Derive, using K-maps, the simplified expressions for each output and its complement.
2- Deduce a programming table for the PLA. You should minimize the number of
product terms.
ABCD
WXYZ
“0000”
“0011”
“0001”
“0100”
“0010”
“0101”
“0011”
“0110”
“0100”
“0111”
“0101”
“1000”
“0110”
“1001”
“0111”
“1010”
“1000”
“1011”
“1001”
“1100”
“1010”
“0000”
“1011”
“0001”
“1100”
“0010”
“1101”
“1101”
“1110”
“1110”
“1111”
“1111”
Table 2.
Solution
1- W = BD + BC + AB'C'
W' = B'C + BC'D' + A'B'C'
X = A'BC'D' + A'B'D + A'B'C + AC'D + ABC
X' = B'C'D' + AC'D' + A'BD + A'BC + AB'C
Y = C'D' + A'CD + ABC
Y'= C'D + A'CD' + AB'C
Z = A'D' + B'C'D' + ABD + ACD
Z' = A'D + B'C'D + ABD' + ACD'
2- Functions : W, X', Y' and Z will be implemented. Total number of product terms 13.
Outputs F1,F2,F3 and F4 produce W,X,Y and Z, respectively.
Product term
Inputs
Outputs
ABCD
(T) (C) (C) (T)
F1 F2 F3 F4
BC
1 -11-
1---
BD
2 -11-
1---
AB'C'
3 100-
1---
B'C'D'
4 -000
-1-1
AC'D'
5 1–00
-1--
A'BD
6 01-0
-1--
A'BC
7 011-
-1--
AB'C
8 101-
-11-
C'D
9 --01
--1-
A'CD'
10 0 – 1 0
--1-
A'D'
11 0 - - 0
---1
ABD
12 "1 1 - 1"
---1
ACD
13 1 – 1 1
---1
Q1:
(A) Simplify the following expressions using Boolean Algebra;



V = X’ Y’ Z + X’ Y Z + X Y’ Z’ + X Y’ Z + ( X’ + Y’ + Z’ )’
P = A B C + A’ B C +A B’ C + ( A + B )’ C
R=((M(N+K)’)(M+(NK)’)’)’
(B) In the circuit shown,
12345-
Obtain the logic expression for the output (not simplified).
Obtain the truth table.
Convert the expression to all NAND form.
Implement the all NAND circuit.
Simplify the original output expression (as in 1).
A
1
2
1
3
2
1
C
3
2
1
B
3
1
2
2
1
3
2
1
2
13
A
C
1
12
2
Q2: (A) Show:
1- how two inputs NOR gate can be constructed from 2 inputs NAND
gates.
2- how you construct Y= ABCD using only 2 inputs NAND gates.
3- how 7486 XOR chip (contains 4 XOR) can be used to make an
XNOR gate using only 7486 chip.
(B) Design a logic circuit that will detect various inputs conditions for 4 inputs
(DCBA). The circuit will generate the following outputs:
- Output GT will be HIGH if the input value is greater than 9.
- Output LT will be LOW if the input value is less than 4.
- Output RNG will be HIGH if the input is greater than 6 and less than 11.
- Output signal TEN will be LOW if the input is equal to 10.
The design should include; Truth Table, K-Maps, and Logic equations.
Solutions
Q1 (A)
V = X’ Y’ Z + X’ Y Z + X Y’ Z’ + X Y’ Z + ( X’ + Y’ + Z’ )’
= X’ Z (Y +Y’ ) + X Y’ (Z+Z’) + X Y Z
= X’ Z + X(Y’ + Y Z )
Z
= X’ Z + X(Y’ + Z)
= Z + X Y’
P = A B C + A’ B C +A B’ C + ( A + B )’ C
= (A+A’ ) BC +A B’ C + A’ B’ C
= B C + C B’ (A + A’ )
= B C + C B’
=C
R=((M(N+K)’)(M+(NK)’)’)’
= (M (N + K)’ (M + (N k)’ )’ )’
= ((M N’ K’ ) ( M’ N K ))’
=1
Q1 (B)
(1) Z = A’ C + C B’ + (A B C’)’
(2)
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
Z
1
1
1
1
1
1
0
1
(3) Z = (( A’ C + C B’ + (A B C’)’)’ )’
= ( (A’ C)’ . (C B’ )’ . (A B C’)’ )’
(4)
A
B
C
U4B
U3B
1
2
4
5
4
5
U1A
74HC00N
74HC00N
3
6
6
U2B
3
4
6
5
74LS12N
U2A
U1D
1
12
2
11
13
13
74HC00N
74LS12N
U1C
9
8
10
74HC00N
(5) Z= A’ C +B’ C + A’ + B’ + C
= A’ ( C + 1) + B’ (C + 1) + C
= A’ + B’ + C
Q2 (A)
(1)
U5D
12
A
11
13
U7D
74HC00N
12
11
13
U6D
74HC00N
12
B
U8D
12
11
13
74HC00N
11
13
74HC00N
(2)
U13D
A
12
B
13
U9D
11
12
D
11
13
74HC00N
11
74HC00N
12
11
13
74HC00N
U12D
12
13
U10D
12
U11D
12
74HC00N
U14D
C
11
13
74HC00N
11
13
74HC00N
Y
12
Z
(3)
U15A
A
B
U15B
1
2
4
3
6
5
7486N
7486N
1(High)
Q2 (B) Truth Table:
Dec
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
D
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
C
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
A
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
GT
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
LT
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
RNG
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
TEN
1
1
1
1
1
1
1
1
1
1
0
1
1
1
1
1
K-Maps
BA
00 01
DC 00
01
11
10
11 10
0
1
1
1
1
DC 00
01
11
10
01
0
0
1
1
GT = DC +DB
BA
00
0
(LT)’ = D’ C’
11 10
1
1
1
1
RNG = B’ D C’ + A’ D C’ + D’ C B A’
0
(TEN)’ = D C’ B A’