Itc 544 Computer Organisation And Assessment Answers
- a) Determine the value of base xif (211)x= (6A)16
- b) Convert the followings:
- i) 0xBAD into a decimal number
- ii) 58810into a 3-base number
- c) Given a (very) tiny computer that has a word size of 6 bits, what are the smallest negative numbers and the largest positive numbers that this computer can represent in each of the following representations.
i) One's complement - ii) Two's complement
- a) Consider the following logic diagram of a combinational circuit where A, B, and C are inputs and Q is the output. Three 2-input AND gates and two 2-input OR gates are used in the circuit. It is possible to reduce some of the logic gates without changing the functionality of the circuit. Such component reduction results in higher operating speed (less delay time from input signal transition to output signal transition), less power consumption, less cost, and greater reliability. Construct a logic diagram of a circuit which does have the same function output with only two logic gates (instead of five).
Answer:
- Finding the radix of a number
(211) x =(6A)16
The hexadecimal notation on the right is equated to a number with an unknown radix.
Converting the hexadecimal to decimal number system,
Where A in hexadecimal is 10
Converting the unknown radix number to decimal
Using the quadratic equation to obtain the roots where a=2, b=1, and c=-105
The value of x is either x=7 or x=7.5
The final value for the unknown is x=7
- number conversion
- OxBAD
This is a C++, CPP, or C value for 2989 in decimal
The value is 298910
- 58810into a 3-base number
Dividing consistently the decimal number by 3
|
Decimal value |
Remainder |
|
588 |
0 |
|
196 |
1 |
|
65 |
2 |
|
21 |
0 |
|
7 |
1 |
|
2 |
2 |
Reading the remainder values from bottom-up,
2102103
- A word size of 6 bits.
- Ones complement
Largest positive (3110): 0111112
Smallest negative (-3110): 1000002
- Twos complement
Largest positive (3110): 0111112
Smallest negative (-3210): 1000002
- Logic diagram of a circuit which does have the same function output with only two logic gates instead of five.
The expression formed at the output Q from the inputs A, B, C is The above expression is simplified to as shown below,Therefore, the resulting simplified equation is given as,The simplified equation requires, three inputs A, B, C and two 2-input AND gates.
- Boolean algebra simplification Following the Boolean algebra identities The result of the first two terms simplification is AB.The other two terms were The final expression is, therefore.
According to the laws of Boolean Algebra, as long as there is a high, the output is a high. When both inputs are low, the output is low. This is the complement rule of the OR operator.
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