electronic and digital applications
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Electrical/Electronicand Digital
applications
Digital Control 1 Jim Flounders
SOUD2235 - Design and simulate logic circuits to a given specification
Phil Illingworth
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Question 1
1. Simplify the following Boolean expressions using De Morgans theorem and / or Boolean
algebra:
a. AC + ACD+ ABC+ ABCD
We can apply DeMorgans laws to Boolean Equations to help simplify them, the two rules (AQA,
2013), shown below, allow us to build equations only involving one sort of gate.
A way of remembering the rule is that if you break the bar, you change the function
We also need to use some of the rules of Boolean algebra; these are shown in the table below (CSCI
2150, 2001)
We are asked to simplify using DeMorgans laws and Boolean algebra
First we need to factorise the equation
A C + C D + A B C + C D we can now look through the table above and find a suitable rule for
progressing through the equation. We can apply rule 11 to both sets of brackets to simplify.
C + C D = C + D C + C D = C + D we can now add this back in
A C + D + AB C + D now we can factorise this
C + D A + A B we can now apply rule 11 from the table to simplify the second set of brackets
A + A B = A + B we can now add this back in
C + D A + B Now we need to multiply out these brackets
C A + C B + D A + D B and to finish off we can alphabetise
A C + B C + A D + B D
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We can start of by applying DeMorgans law to the equation, this will separate the long bar between
the two brackets and change the function from OR to AND
. .
= A B + A B we can replace this into our equation and use DeMorgans rule again on the
second bracket
. A B + C this simplifies again
. A B + C
2. Simplify the following Boolean equation, in sum-of-products form, using a Karnaugh map:
a. f(A,B,C,D) =(4,5,6,12,14,15), with a dont care at 10.
The first step towards setting up a Karnaugh map is to create a truth table for the inputs, in this case
we have four inputs, A, B, C and D and we only need the first Ten values as we have a dont care at
ten but as we are asked to go up to 15 I will fill the table in, we build the truth table by converting
the decimal value into binary and finding the values given in the equation above
Now I have completed my truth table the information can
be added to a Karnaugh map, for simplicity I will use
Karnaugh Map Explorer 2.0 (Techie Buzz, 2013). This
confirms my truth table is correct and relays the
information into a Karnaugh map with a dont careat
10.
Dec A B C D X
0 0 0 0 0 0
1 0 0 0 1 0
2 0 0 1 0 0
3 0 0 1 1 0
4 0 1 0 0 1
5 0 1 0 1 1
6 0 1 1 0 1
7 0 1 1 1 0
8 1 0 0 0 0
9 1 0 0 1 0
10 1 0 1 0 0
11 1 0 1 1 0
12 1 1 0 0 1
13 1 1 0 1 0
14 1 1 1 0 1
15 1 1 1 1 1
8 4 2 1
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From the information in the Karnaugh map I can now simplify the original Equation
f(A,B,C,D) =(4,5,6,12,14,15)
F (A, B, C,D) = B+ B
3. Simplify the following Boolean equation, in product-of-sums form, using a Karnaugh map:
As we are using the product of sums method we use the inverse of the information put into the
Karnaugh map to find out answer, I drew out a Karnaugh map in rough using the information and
then plugged this into Karnaugh Explorer 2.0, shown below. As we are using product of sums we
group the zeros
F(ABCD) =B +++ D
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4. assuming that all numbers are 16 bits wide, complete the missing entries which are not shaded
in the following table. (Note: no marks will be awarded unless you show how the solution is
derived):
My first step is to look up some conversion tables for converting between decimal, hex, binary and
octal and find information on moving between each set of data. Octal is also known as the Base 8
number system, the values used in the base 8 system are 0,1,2,3,4,5,6 and 7. The value 8 isrepresented as one 8 and zero 1s, or 108. I am given the octal value 1336, I can convert this into
decimal like so.
Digits 1 3 3 6
Numbering 3 2 1 0
I now multiply the top value by 8 to the power of the bottom value and add each corresponding
value
1x83
+ 3x82+ 3x8
1+ 6x8
0
(512)+(192)+ (24) + (6) = 734 so octal 1336 = decimal 734 I can work back the other way using long
division
Binary is to base 2, we can make a small table to help convert from decimal into binary, we have
been told that all numbers are 16 bits wide so we take binary values up to the 16th
from right to left,
then we take our decimal value 7955 and work from left to right marking either a 0 or a 1 in each
corresponding box to make up our value and give us our complete binary number
32763 16384 8192 4096 2048 1024 512 256 128 64 32 16 8 4 2 1
0 0 0 1 1 1 1 1 0 0 0 1 0 0 1 1
Our binary value is 0001 1111 0001 0011
As a check we can add each value with a 1 together to get our original value
4096 + 2048 + 1024 + 512 +256 + 16 + 2 + 1 = 7955
Hexadecimal is to base 16 so we can use the same method as with base 8 to convert from hex to
decimal but replace the power 8 with 16.
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The table to the left (Learn44, 2011), shows the
conversions from decimal to hex to binary. We have the
Hex value 0D71
I can make up a table to illustrate the digits I u will use
and the numbering
Digits 0 D 7 1
Numbering - 2 1 0
The table shows D represents 13
13x162+ 7x16
1+ 1x16
0
(3328) + (112) + (1) = 3441
I have now shown how to work between all methods and can complete the table
Decimal Hexadecimal Binary Octal
7955 1F13 0001-1111-0001-0011 17423
734 2DE 0000-0010-1101-1110 13363441 0D71 0000-1101-0111-0001 6561
33025 8101 1000-0001-0000-0001 100401
5. An 8 x 2 ROM is to be used to implement simultaneously the following Boolean functions:
f(A,B,C) = AC + ABC g(A,B,C) = BC + ABC
a.
Complete the missing stored data values (in hexadecimal) in the table below to implement
functions f and g. Also, state how you would connect the logic variables, and functions f
and g, to the ROM
I will draw up my own table to clearly show all address lines and data lines and the hexadecimal
output
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Hex A B C F G Hex
0 0 0 0 0 0 0
1 0 0 1 1 0 2
2 0 1 0 0 1 1
3 0 1 1 0 0 0
4 1 0 0 1 1 3
5 1 0 1 0 0 0
6 1 1 0 1 1 3
7 1 1 1 0 0 0
6. Sketch the Moore state diagram for the circuit below, where A is the input variable. You may
assume that initially, Q[2:0] = 000:
I first need to draw up a truth table showing all possible outcomes, I willneed to consider the rule shown in the table to the left.
Clock A K Q0 J1 Q1 Q2
0 1 0 0 0 0 0
1 1 0 1 1 0 0
2 1 0 1 1 1 0
3 1 1 1 0 0 1
4 1 1 0 0 0 0
5 1 0 0 0 0 0
6 1 0 1 1 0 0
I can now take another table from this to clearly show my output data before putting into a Moore
state diagram
3 address lines A,B,C 2 data lines F,G
J K
0>0 0 x
0>1 1 x
1>0 x 1
1>1 x 0
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We can stop at clock 4 as we have returned back to 0 0 0, I can now
put this into a Moore state diagram
A=1
101 001
D B
011
C
Clock Q2 Q1 Q0 Location
0 0 0 0 A
1 0 0 1 B
2 0 1 1 C
3 1 0 1 D4 0 0 0 E
5 - - - -
000
A