|     |        Assignment No. 01 CS302: Digital Logic Design  |        Total   Marks: 10 Due   Date: 18/04/2011  |   |||||||||||||||||
|     Instructions: Please read the   following instructions carefully before submitting assignment: It   should be clear that your assignment will not get any credit if: §    The assignment is   submitted after due date. §    The assignment is   submitted via email. §    The submitted assignment   does not open or file is corrupt. §    All types of plagiarism   are strictly prohibited.  Objectives: ·          Understanding the number system. ·          Understanding the conversion of   numbers into various number systems. Guidelines: ·           Perform all steps while converting   the numbers.  |   |||||||||||||||||||
|     Assignment     |        |   ||||||||||||||||||
|     Problem Statement: 1) Convert 10101100 binary numbers to its decimal equivalent. (2 Marks) Solution: 10101100   = 1x27 + 0x26 + 1x25   + 0x24 + 1x23+ 1x22 + 0x21 + 0x20     = 1x128 + 0x64   + 1x32+ 0x16 + 1x8 + 1x4 + 0x2 + 0x1     =   128+0+32+0+8+4+0+0     = 172 2) Convert decimal number 374 to octal number. (2 Marks) Solution: We   will use repeated division method. 
 Now writing the reminders in the reverse order 566.  3) Convert the following 0110100000111001 (BCD) in to decimal number. (2 Marks) Solution: We   first of all with will write 0110100000111001 in the groups of four bits, and   then will write down the decimal equivalent of each group. = 0110 1000 0011 1001       6        8       3       9            Our required decimal number is 6839 4) Convert B2A16 to Octal (2 Marks) Solution: First   we will convert B2A to Binary  1011   0010 1010 Now   grouping into 3 bit  101   100 101 010  5        4     5    2 We   will get  5452  5) Convert decimal number 3085 to hex Solution: 
 12   0     13 C 0 D We   will get C0D16  |   |||||||||||||||||||
|     Deadline: Your assignment must be   uploaded/submitted on or before 18th April, 2011  |   |||||||||||||||||||
Sunday, April 24, 2011
CS302 Assignment No. 01 Solution
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