Number System Conversions, Logic Gates, Boolean Functions & Circuits

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Added on 2024/07/01

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This assignment focuses on number system conversions and logic circuit design. It includes converting hexadecimal numbers to binary, converting decimal numbers to 2's complement, converting binary numbers to hexadecimal, and converting 2's complement representations to decimal. The assignment also covers the construction of an XOR gate using AND, OR, and NOT gates, along with its truth table. Furthermore, it addresses the design of a logical circuit and Boolean function for course enrollment sessions and the development of a truth table and Boolean function for a given logic diagram. The solutions provide step-by-step explanations and diagrams for each problem.

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Question 1
A. To convert this number from hexadecimal to binary. First of all, I will convert this
number into the decimal.
AC1216 = 10*163 + 12*162 + 1*161 + 2*160;
= 40960 + 3072 + 16 + 2;
= 44050;
= 4405010
Now, I will convert this decimal number into the binary number by using proper
conversion process which is given below:
2 44050 0
2 22025 1
2 11012 0
2 5506 0
2 2753 1
2 1376 0
2 688 0
2 344 0
2 172 0
2 86 0
2 43 1
2 21 1
2 10 0
2 5 1
2 2 0
2 1 1
0
Now binary is 10101100000100102.
B. To convert −10710 to 8-bit 2’s Complement. First of all, this number will convert to the
decimal number.

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2 107 1
2 53 1
2 26 0
2 13 1
2 6 0
2 3 1
2 1 1
0
Now, the number is 11010112
Then determines the MSB of the given number (-107) whether the number is negative or
positive. Here, the given number is negative so MSB of this number will also negative.
Step 1:
For representing the negative number, 1 is used as MSB in a binary number of 107.
-10710 = 1 11010112 Here, MSB 1 represents the negative number.
Step 2:
Now, convert all zero into one and one into zero but don’t convert MSB. Then
1 0010100
Step 3:
Now, add 1 in above output
10010101
So −10710 to 8-bit 2’s Complement is 10010101.
C. To convert 11001101012 to Hexadecimal. First of all, this number will convert into
decimal no.
11001101012 = 1*29 + 1*28 + 0*27 + 0*26 + 1*25 +1*24 + 0*23 + 1*22 + 0*21 + 1*20
= 512 + 256 + 0 + 0 + 32 + 16 + 0 + 4 + 0 + 1
= 82110
Then convert this decimal number (821) into hexadecimal number.
16 821 5
16 51 3
16 3 3
0

Then 11001101012 to Hexadecimal are 33516.
D. For convert 10011110 (8-bit 2’s complement representation) to decimal. Some steps are
followed which are given below:
Step 1:
First of all, check the left most bit that it is 1 or not if it is 1 then I will negative otherwise
it will positive. Here, leftmost bit is 1 so it will negative.
Step 2:
Then subtract 1 from given number
10011110 – 1 = 10011101
Step 3:
Then replace all zeros with one and all ones with zero in 10011101.
01100010.
Step 4:
01100010 = 0*27 + 1*26 + 1*25 + 0*24 + 0*23 + 0*22 + 1*21 + 0*20
= 0 + 64 + 32 +0 + 0 + 0 + 2 + 0
= 9810
Because leftmost bit is 1 so number will negative so 10011110 (8-bit 2’s complement
representation) to decimal is (- 98)10.
E. Convert -1.5 number into IEEE
Step 1:
First of all, the number 1.5 converts into binary. For that purpose, number 1.5 splits into
two parts. 1 is the whole number and .5 is fraction number.
1(10) = 1(2)
.5(10) =?
.5*2 = 1.0
.5(10) = 1(2)
Therefore 1.5 in binary is (1.1)2
Step 2:
Now, check the sign bit of – 1.5 which is negative therefore sign bits is 1(2).
Step 3:
Then normalize the (1.1)2 binary number which is 1.1 * 20.

Then calculate mantissa M = fraction of 1.1 = (1)2. We need to add zeroes in mantissa to
make bits 23 long,
M = 10000000000000000000000
Calculate exponent (E) = 0 + 127 = 12710 = (01111111)2
Signed bit (1 bit) Exponent (8 bits Mantissa or Fraction (23 bits)
1 01111111 10000000000000000000000

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Question 2:
A. The XOR gate is constructed by using AND, OR and NOT gates. So the logical equation
of the XOR gate is C = ͞ BA + ͞ AB of the constructed XOR gate which is given below in
the diagram.
R U
Figure 1: Logical Diagram of 2.a
T V
The truth table for the above diagram:
Inputs Outputs
A B C R T U V
0 0 0 1 0 1 0
0 1 0 0 1 1 1
1 0 1 1 1 0 0
1 1 0 0 0 0 0
The truth table is prepared with the help of the inputs and outputs of the circuit diagram.
By using a truth table, the logical equation will prepare which are BA + ͞ AB.
So C = ͞ BA + ͞ AB.
This circuit is designed by using two AND, two NOT and OR gates which are shown in the
above circuit diagram.
B. Here, a logical circuit and Boolean function are prepared for the three sessions in college
and students can enroll for 3 courses in every session.

According to the given scenario, a truth table is created. Then, prepare the K-map and
construct the logical equation on the basis of K-map.
Inputs Outputs Sessions
A B C D Q Months
0 0 0 0 1 Jan
0 0 0 1 0 Feb
0 0 1 0 1 March
0 0 1 1 1 April
0 1 0 0 1 May
0 1 0 1 0 June
0 1 1 0 1 July
0 1 1 1 1 Aug
1 0 0 0 1 Sep
1 0 0 1 0 Oct
1 0 1 0 1 Nov.
1 0 1 1 1 Dec.
1 1 0 0 x X
1 1 0 1 x X
1 1 1 0 x X
1 1 1 1 x X
After, analysis the K-map value, logical equation can be constructed which is given in
below: ͞
D + C + AB
So the output Q is ͞ D + C + AB.
On the basis of the logical equation, a logical diagram is prepared which is given below.

Figure 2: Logical diagram of 2.b

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Question 3:
Prepare the Boolean function and truth table for given the logic diagram.
Figure 3: Logical diagram of 3
To create a truth table of a given logical diagram. First of all, logical inputs (0, 1) are applied at
the given inputs i.e. H, L, E. Then the output will receive at the output point P which is given in
the truth table so truth table is given below:
Inputs Output
H L E P
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1

After, construct the truth table of given design, K-map is created by using this truth table. After
that, analysis the K-map, the final logical equation is obtained for the given logical diagram
which is given below:
Logical Equation is LE + HE + HL