EE 2000 Logic Circuit Design - Semester A, 2007/08 - Tutorial 1

EE 2000 Logic Circuit Design,
Semester A, 2007/08
Tutorial 1
Week 2 (12th September, 2007)
The questions are divided into two levels. Level 1 is the basic level question. You should
be able to answer them after completing the lecture. Level 2 is the advanced level
question. You may have to take a research before answering them. Please be prepared
before the tutorial.
Level 1
Question 1: Complete the following table by converting the below numbers to decimal,
binary, octal and hexadecimal number systems.

Decimal	Binary	Octal	Hexadecimal
27	11011	33	1B
237.25	11101101.01	355.2	ED.4
250.8	1111 1010.1100 1100…	372.6314 6314…	FA.CC…

Question 2:
(a) What is the largest binary number that can be obtained with 16 bits?
(1111 1111 1111 1111)2

(b) What is its decimal equivalent?

 2¹⁵ + 2¹⁴ + … + 2¹ + 2⁰ = 2¹⁶ – 1 = 65535

Question 3: Find the 1's and 2's complements of the following 8-digit binary numbers:

(a) 0010 1010
1101 0101 (1’s complement)
1101 0110 (2’s complement)

(b) 1111 1111
0000 0000 (1’s complement)
0000 0001 (2’s complement)


Question 4: Perform the arithmetic operations (-128) + (-1) in binary using

(a) 8-bit signed 2's complement representation for negative numbers

-128 = 1000 0000;
-1 = 1111 1111;
(-128) + (-1)
= 1000 0000 + 1111 1111
= (1) 0111 1111
Discard carry, we get 0111 1111
(Result incorrect. Overflow because –ve + -ve becomes +ve)

(b) 9-bit signed 2’s complement representation for negative numbers
Does overflow happen? Why?

-128 = 1 1000 0000;
-1 = 1 1111 1111;
(-128) + (-1)
= 1 1000 0000 + 1 1111 1111
= (1) 1 0111 1111
Discard carry, we get 1 0111 1111 (= -129)
(No overflow as –ve + -ve gives a –ve number)


Level 2
Question 5: Please calculate a + b, a - b, a · b, and a / b for the pair of binary numbers a
= 10101 and b = 1011 without converting to decimal (Show your work and the carries)

A = 10101, B = 1011

  11110

   10101

+)  1011

 100000 a + b = 100000

01010

 10101

-) 1011

   1010 a - b = 1010 (You can get the same result if using 2’s complement)

      10101

x)     1011

10101000

              0

    101010

      10101

11100111 a · b = 11100111

                 01

1011 ) 10101

             1011

               1010 a / b: quotient = 1, remainder = 1010

Question 6: Determine the value of x if (211)x = (152)8.


(211)x = (152)8
2 · x2 + 1 · x + 1 = 1 · 82 + 5 · 8 + 2
2x2 + x – 105 = 0
(2x + 15)(x – 7) = 0
∴ x = 7 (x must be a positive integer so cannot be -7.5)


Question 7: Formulate a simple procedure for converting base-4 numbers directly to
hexadecimal number. Use the procedure to convert (12321)4 to base 16. (Hints: 42 = 16)

Partition the digits into groups of 2 each (one hex digit = 2 base-4 digits)

Add necessary 0s to the left and right

Replace each group of base-4 digits by the hex equivalent

So, (12321)₄

= (01 23 21)₄ (Note: 23₄ = 2·4¹+ 3·4⁰ = 11 = B₁₆)

= (1 B 9)₁₆

Question 8: In the lecture, we only discuss the r’s-complement and (r-1)’s-complement
of an integer number. Find the 1’s complement and 2’s complement of the following
fractional binary numbers.
(a) 010.11

1’s complement: 2³ - 010.11 - 2^-2 = 101.00

                                2’s complement: 2³ - 010.11 = 101.01

(b) 11011.100

1’s complement: 2⁵ - 11011.100 - 2^-3 = 00100.011

                                 2’s complement: 2⁵ - 11011.100 = 00100.100