1) 0 01111111111 0000000000000000000000000000000000000000000000000000 sign bit is 0(+ve) exp bits are 01111111111 Converting 01111111111 to decimal 01111111111 => 0x2^10+1x2^9+1x2^8+1x2^7+1x2^6+1x2^5+1x2^4+1x2^3+1x2^2+1x2^1+1x2^0 => 0x1024+1x512+1x256+1x128+1x64+1x32+1x16+1x8+1x4+1x2+1x1 => 0+512+256+128+64+32+16+8+4+2+1 => 1023 in decimal it is 1023 so, exponent/bias is 1023-1023 = 0 frac bits are IEEE-754 Decimal value is 1.frac * 2^exponent IEEE-754 Decimal value is 1. * 2^0 1. in decimal is 1 => 1. => 1x2^0 => 1x1 => 1 => 1 so, 1 * 2^0 in decimal is 1 so, 0011111111110000000000000000000000000000000000000000000000000000 in 64-bit format is 1 Answer: 1 2) 0 01111111111 0000000000000000000000000000000000000000000000000001 sign bit is 0(+ve) exp bits are 01111111111 Converting 01111111111 to decimal 01111111111 => 0x2^10+1x2^9+1x2^8+1x2^7+1x2^6+1x2^5+1x2^4+1x2^3+1x2^2+1x2^1+1x2^0 => 0x1024+1x512+1x256+1x128+1x64+1x32+1x16+1x8+1x4+1x2+1x1 => 0+512+256+128+64+32+16+8+4+2+1 => 1023 in decimal it is 1023 so, exponent/bias is 1023-1023 = 0 frac bits are 0000000000000000000000000000000000000000000000000001 IEEE-754 Decimal value is 1.frac * 2^exponent IEEE-754 Decimal value is 1.0000000000000000000000000000000000000000000000000001 * 2^0 1.0000000000000000000000000000000000000000000000000001 in decimal is 1.0000000000000002 => 1.0000000000000000000000000000000000000000000000000001 => 1x2^0+0x2^-1+0x2^-2+0x2^-3+0x2^-4+0x2^-5+0x2^-6+0x2^-7+0x2^-8+0x2^-9+0x2^-10+0x2^-11+0x2^-12+0x2^-13+0x2^-14+0x2^-15+0x2^-16+0x2^-17+0x2^-18+0x2^-19+0x2^-20+0x2^-21+0x2^-22+0x2^-23+0x2^-24+0x2^-25+0x2^-26+0x2^-27+0x2^-28+0x2^-29+0x2^-30+0x2^-31+0x2^-32+0x2^-33+0x2^-34+0x2^-35+0x2^-36+0x2^-37+0x2^-38+0x2^-39+0x2^-40+0x2^-41+0x2^-42+0x2^-43+0x2^-44+0x2^-45+0x2^-46+0x2^-47+0x2^-48+0x2^-49+0x2^-50+0x2^-51+1x2^-52 => 1x1+0x0.5+0x0.25+0x0.125+0x0.0625+0x0.03125+0x0.015625+0x0.0078125+0x0.00390625+0x0.001953125+0x0.0009765625+0x0.00048828125+0x0.000244140625+0x0.0001220703125+0x6.103515625e-05+0x3.0517578125e-05+0x1.52587890625e-05+0x7.62939453125e-06+0x3.814697265625e-06+0x1.9073486328125e-06+0x9.5367431640625e-07+0x4.76837158203125e-07+0x2.384185791015625e-07+0x1.1920928955078125e-07+0x5.960464477539063e-08+0x2.9802322387695312e-08+0x1.4901161193847656e-08+0x7.450580596923828e-09+0x3.725290298461914e-09+0x1.862645149230957e-09+0x9.313225746154785e-10+0x4.656612873077393e-10+0x2.3283064365386963e-10+0x1.1641532182693481e-10+0x5.820766091346741e-11+0x2.9103830456733704e-11+0x1.4551915228366852e-11+0x7.275957614183426e-12+0x3.637978807091713e-12+0x1.8189894035458565e-12+0x9.094947017729282e-13+0x4.547473508864641e-13+0x2.2737367544323206e-13+0x1.1368683772161603e-13+0x5.684341886080802e-14+0x2.842170943040401e-14+0x1.4210854715202004e-14+0x7.105427357601002e-15+0x3.552713678800501e-15+0x1.7763568394002505e-15+0x8.881784197001252e-16+0x4.440892098500626e-16+1x2.220446049250313e-16 => 1+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+2.220446049250313e-16 => 1.0000000000000002 so, 1.0000000000000002 * 2^0 in decimal is 1.0000000000000002 so, 0011111111110000000000000000000000000000000000000000000000000001 in 64-bit format is 1.0000000000000002 Answer: 1.0000000000000002 3) 0 10000110011 0000000000000000000000000000000000000000000000000000 sign bit is 0(+ve) exp bits are 10000110011 Converting 10000110011 to decimal 10000110011 => 1x2^10+0x2^9+0x2^8+0x2^7+0x2^6+1x2^5+1x2^4+0x2^3+0x2^2+1x2^1+1x2^0 => 1x1024+0x512+0x256+0x128+0x64+1x32+1x16+0x8+0x4+1x2+1x1 => 1024+0+0+0+0+32+16+0+0+2+1 => 1075 in decimal it is 1075 so, exponent/bias is 1075-1023 = 52 frac bits are IEEE-754 Decimal value is 1.frac * 2^exponent IEEE-754 Decimal value is 1. * 2^52 1. in decimal is 1 => 1. => 1x2^0 => 1x1 => 1 => 1 so, 1 * 2^52 in decimal is 4503599627370496 so, 0100001100110000000000000000000000000000000000000000000000000000 in 64-bit format is 4503599627370496 Answer: 4503599627370496 4) 0 10000110011 0000000000000000000000000000000000000000000000000001 sign bit is 0(+ve) exp bits are 10000110011 Converting 10000110011 to decimal 10000110011 => 1x2^10+0x2^9+0x2^8+0x2^7+0x2^6+1x2^5+1x2^4+0x2^3+0x2^2+1x2^1+1x2^0 => 1x1024+0x512+0x256+0x128+0x64+1x32+1x16+0x8+0x4+1x2+1x1 => 1024+0+0+0+0+32+16+0+0+2+1 => 1075 in decimal it is 1075 so, exponent/bias is 1075-1023 = 52 frac bits are 0000000000000000000000000000000000000000000000000001 IEEE-754 Decimal value is 1.frac * 2^exponent IEEE-754 Decimal value is 1.0000000000000000000000000000000000000000000000000001 * 2^52 1.0000000000000000000000000000000000000000000000000001 in decimal is 1.0000000000000002 => 1.0000000000000000000000000000000000000000000000000001 => 1x2^0+0x2^-1+0x2^-2+0x2^-3+0x2^-4+0x2^-5+0x2^-6+0x2^-7+0x2^-8+0x2^-9+0x2^-10+0x2^-11+0x2^-12+0x2^-13+0x2^-14+0x2^-15+0x2^-16+0x2^-17+0x2^-18+0x2^-19+0x2^-20+0x2^-21+0x2^-22+0x2^-23+0x2^-24+0x2^-25+0x2^-26+0x2^-27+0x2^-28+0x2^-29+0x2^-30+0x2^-31+0x2^-32+0x2^-33+0x2^-34+0x2^-35+0x2^-36+0x2^-37+0x2^-38+0x2^-39+0x2^-40+0x2^-41+0x2^-42+0x2^-43+0x2^-44+0x2^-45+0x2^-46+0x2^-47+0x2^-48+0x2^-49+0x2^-50+0x2^-51+1x2^-52 => 1x1+0x0.5+0x0.25+0x0.125+0x0.0625+0x0.03125+0x0.015625+0x0.0078125+0x0.00390625+0x0.001953125+0x0.0009765625+0x0.00048828125+0x0.000244140625+0x0.0001220703125+0x6.103515625e-05+0x3.0517578125e-05+0x1.52587890625e-05+0x7.62939453125e-06+0x3.814697265625e-06+0x1.9073486328125e-06+0x9.5367431640625e-07+0x4.76837158203125e-07+0x2.384185791015625e-07+0x1.1920928955078125e-07+0x5.960464477539063e-08+0x2.9802322387695312e-08+0x1.4901161193847656e-08+0x7.450580596923828e-09+0x3.725290298461914e-09+0x1.862645149230957e-09+0x9.313225746154785e-10+0x4.656612873077393e-10+0x2.3283064365386963e-10+0x1.1641532182693481e-10+0x5.820766091346741e-11+0x2.9103830456733704e-11+0x1.4551915228366852e-11+0x7.275957614183426e-12+0x3.637978807091713e-12+0x1.8189894035458565e-12+0x9.094947017729282e-13+0x4.547473508864641e-13+0x2.2737367544323206e-13+0x1.1368683772161603e-13+0x5.684341886080802e-14+0x2.842170943040401e-14+0x1.4210854715202004e-14+0x7.105427357601002e-15+0x3.552713678800501e-15+0x1.7763568394002505e-15+0x8.881784197001252e-16+0x4.440892098500626e-16+1x2.220446049250313e-16 => 1+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+2.220446049250313e-16 => 1.0000000000000002 so, 1.0000000000000002 * 2^52 in decimal is 4503599627370497.0 so, 0100001100110000000000000000000000000000000000000000000000000001 in 64-bit format is 4503599627370497.0 Answer: 4503599627370497.0
Exercise 4.4 Find the decimal value of the DP number for each of the bit patterns...
EXERCISE 3 Binary-coded decimal (BCD) is a binary encoding of decimal numbers where each decimal digit is represented by its corresponding four-bit binary value, as shown in the following table Table 1 BDC Encoding 0 0 0 0 If one needs to sum two BDC digits, two 4-bit binary adders can be combined, as shown in Figure 1 4-bit adder 0 out 4-bit adder Figure 1 BDC Digit Adder Table 1 outlines the relationship between KounZs-o and CouS3-a Note that...
Convert each of the following 32 IEEE 754 single precision bit patterns to its corresponding decimal value (the bits are separated into groups of 4 to make interpretation easier). Show all of your work and include a few comments as to what you are doing at each step. 1100 0100 1011 1010 0100 1000 0000 0000 a. b. 0100 0101 1110 0010 0110 1101 0000 0000 Convert each of the following 32 IEEE 754 single precision bit patterns to its...
4.4 Find the range of values that can be represented using a 10-bit binary number. Also, how many different values can you represent using a 10-bit binary number?
Programming Exercise 4.6 Use the strategy of the decimal to binary conversion and the bit shift left operation defined in Project 5 bits = input("Enter bit string: ") bits = bits[1:] + bits[0] print ("Result:", bits) to code a new encryption algorithm. The algorithm should Add 1 to each character’s numeric ASCII value. Convert it to a bit string. Shift the bits of this string one place to the left. A single-space character in the encrypted string separates the resulting...
1. 2. Find the decimal value of the following 8-bit numbers for (i) un-signed and (ii) signed number. (a) 11010110, (b) 01011101 Express the following decimal numbers in 6-bit 2's complement representation: (a) -27, (b) 6, (c)-13, (d) -47 - 4. Convert decimal numbers 83 and 101 to 8-bit unsigned binary number. Find the sum and difference (with addition approach) of these two numbers.
. 2.1 a. Find the 16-bit 2’s complementary binary representation for the decimal number 1987. b. Find the 16-bit 2’s complementary binary representation for the decimal number −1987. What are the 16-bit 1’s and 2’s complements of the following binary numbers? a. 10000 b. 100111100001001 Convert the decimal number 19557 to floating point. Use the format SEEMMMM. All digits are decimal. The exponent is stored excess-40 (not excess-50). The implied decimal point is at the beginning of the mantissa. The sign...
2.1 a. Find the 16-bit 2’s complementary binary representation for the decimal number 1987. b. Find the 16-bit 2’s complementary binary representation for the decimal number −1987. What are the 16-bit 1’s and 2’s complements of the following binary numbers? c. 10000 d. 100111100001001 Convert the decimal number 19557 to floating point. Use the format SEEMMMM. All digits are decimal. The exponent is stored excess-40 (not excess-50). The implied decimal point is at the beginning of the mantissa. The sign...
Objective: Design a 3-bit counters based on random number patterns using D flip- flop and other gates. The pattern of number is: 011 1. List state table. (3 points - writing for the report) 2. Using K-map, find relation among current state, input, and output. (3 points - writing for the report) 3. Draw circuits. (3 points - writing for the report)
Finite state machine (FSM) counter design: Gray codes have a useful property in that consecutive numbers differ in only a single bit position. Table 1 lists a 3-bit modulo 8 Gray code representing the numbers 0 to 7. Design a 3-bit modulo 8 Gray code counter FSM. a) First design and sketch a 3-bit modulo 8 Gray code counter FSM with no inputs and three outputs, the 3-bit signal Q2:0. (A modulo N counter counts from 0 to N −...
C++ Convert a Number from Binary to Decimal using a stack: The language of a computer is a sequence of Os and 1s. The numbering system we use is called the decimal system, or base 10 system. The numbering system that the computer uses is called the binary system, or base 2 system. The purpose of this exercise is to write a function to convert a string representing a binary number from base 2 to base 10. To convert a...