Consider a data type called nibble its length is 4 bits. The min and max number of signed nibble is Select one: a. -8, +7 b. 0, 15 c. 0, 16 d. -8, +8
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- A binary code is a string of 0's and 1's. In particular, an n-bit binary code is a string of "n" 0's and 1's. For example, the code "01001" is a 5-bit binary code. How many 5-bit binary codes can be formed?Given are the following codewords for an error control coding scheme data 000 011 100 codeword 00011000 00000111 10101000 . a.What is the minimum Hamming distance? b. What is the coding rate for this scheme? c. what raw data rate would be required if we need a throughput of 6 Mpbs? d. in a different situation, the number of data bits to transfer is 5, and the minimum hamming distance is dmin=5. find the maximum coding rate for this situationConstruct a PDA which accepts the set of bit strings with 2/3 as many 0 bits as 1 bits. (For example, 10110 should be accepted, since there are three 1s ans two 0s, and two is 2/3 of three.)
- In the Hamming coding system (7.3), its representation is expressed as 3 control (test) bits and 7 total bits (data+control bits).In the literature, control bits are represented as (cl, c2, c3, or (pl, p2, p3, .) The bit string sent by encoding with Hamming coding (15,4) is on the receiving side, It is retrieved as 001100101100 (pl.p2.dll). According to this bit sequence;Which option is given according to the control bits (p1.p2.p3.p4) sequence of the received bit sequence?A 0110B 1100C 1001D 0011E 0010A string is called a palindrome if it reads the same when reversed. That is, the string x1 x2 ... xn is a palindrome if x1 x2 ... xn = xn ... x2 x1. The bit strings 11011 and 01011010 are palindromes.The bit strings 1011 and 01111 are not. How many bit-strings of length 7 are palindromes?Answer asap !!!! Can not use any control constructs such as if, do, while, for, switch, etc. /* * bitCount - returns count of number of 1's in the bit pattern for x * Examples: bitCount(5) = 2, bitCount(7) = 3 * Legal ops: ! ~ & ^ | + << >> * Max ops: 40 * Rating: 4 */ int bitCount(int x) { return 2; } /* * conditional - same as x ? y : z * Example: conditional(2,4,5) = 4 * Legal ops: ! ~ & ^ | + << >> * Max ops: 16 * Rating: 2 */ int conditional(int x, int y, int z) { return 2; } /* * minusTwo - return a value of -2 * Legal ops: ! ~ & ^ | + << >> * Max ops: 2 * Rating: 10 */ int minusTwo(void) { return 2; } * TMax - return maximum two's complement integer * Legal ops: ! ~ & ^ | + << >> * Max ops: 6 * Rating: 10 */ int tmax(void) { return 2; }
- 1. Compact discs record two channels (left and right) of music at a sampling frequency of ??=44.1kHz for each channel. If each sample is encoded with 16 bits, and one byte is 8 bits, how many bytes are required to store 48.1 seconds of music? 2. Consider a system that uses 8-bit ASCII codes to encode letters. How long (in microseconds) will it take to transmit the bit sequence encoding Hello-World! if we use a bit time of 4 samples per bit, and transmit samples at a rate of 1MHz? 3. The ASCII table below gives the ASCII codes for common alphanumeric characters and symbols listed from MSB to LSB. What is the bit sequence encoding the message 113? Assume that we transmit the codes of each character in sequence with the LSB first.Consider the divisor 10011, and suppose that the data-word has the value 1010100000. What is the value of CRC? What will be the sent code-word?write a sender and receiver side bit stuffing program in which a frame is to be simulated with header , payload and trailer . in that demonstrate bit stuffing concept. c/c++/java can be used for processing a given file/string of binary bits. [starting and ending flags/indicator are 01111110].
- IEEE 754-2008 contains a half precision that is only 16 bits wide. The leftmost bit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and the fraction is 10 bits long. A hidden 1 is assumed as part of the significand just like the IEEE-754 format we learnt in the class. Write down the bit pattern to represent −0.15625 × 100 in IEEE 754-2008 half-precision format. You need to find out the bits for the sign, the exponent, and the fraction. To answer this question, you need to convert the decimal fraction to its binary representation. Check out the link given in Q1. You have to show all steps including the fraction conversion (from decimal to binary).Consider a hypothetical 8 bit floating point machine representation with a sign bit, a 3 bit exponent, and a 4 bit mantissa (se1e2e3b1b2b3b4), where the exponent bias is 3 (add 3 to exponent of number to form machine representation). Recall that actual mantissa has 5 bits, since the leading 1 is not stored on the machine. (a) What is the number ?≈2.718 in this 8-bit format? (b) What is the number that (10100111)2 represents in this 8-bit format? (c) What is the upper bound of the relative error when representing a real number in this 8-bit format?Write a program that asks the user to enter and integer number and read it. Then ask him to enter a bit position (between 0 and 31) and display the value of that bit.(MIPS) srlv $s1,$s2,$s3 s2 = 3210 = 0000 0000 0000 1000 0000 0000 010 00002 s3 = 18 Bit = 10 1 = 00000000000000000000000000010000 2 = 00000000000000000000000000001000 3 = 00000000000000000000000000000100 4 = 00000000000000000000000000000010 5 = 00000000000000000000000000000001 S1 = 00000000000000000000000000000001 1 = 00000000000000000000000000000001 $t0 = 00000000000000000000000000000001 = x.1 = x AND $t0,$s1,1