Consider a processor that contains eight 16-bit registers. Design a complete ALU with the following functionalities for this processor.
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Consider a processor that contains eight 16-bit registers.
Design a complete ALU with the following functionalities for this processor.
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- Simplify the function F represented by the following Kmap into 4 terms 9 literals SOP expression: AB 00 01 11 10 CD 00 01 1 1 11 1 10 1 1. 1. 1. 1. 1. 8.1. P VI V2 V3 V4 V5 V6 V7 V8 V9 Using the matrix P from the Floyd's II algorithm V2 9 VI 0 0 9 0 09 0: 9 8 8 8 6 0 0 8 8 8 0 9 0 V3 5 5 0 9 5 0 0 9 V4 0 0 5 0 0 2 7 V6 9 0 9 0 0 9 1 0 8 8 V5 0 1 9 0 6 8 0 0 V7 9 9 9 V8 9 9 9 9 9 7 9 0 9 0 0 9 0 6 7 V9 5308OOS 3 0 Find the path from v4 to v that has the minimum cost:expression for 2sin(n)cos(n) - sin(2π) O O O O sqrt(sqrt(2*M_PI)) O 3*tan(0.2) 2*sin(M_PI)*cos(M_PI) - sin(2*M_PI) O 0.8/0.9
- X= (ACB)((A+C)+(BC)) Lalo dolo! 1 walhe Jlsl Laa X = AB · (A + C) + ĀB · A + B + C alo aolo! 1- 2/2* الرمز الرمادي (10110111101) = () ' بالرقم العشري 931 1753 867 1877 لا احد منهم 1749 1750 869 22 (1100111) (11011001 ---RGB is a color model that defines color by the combination of Red, Green, and Blue. An RGB tuple is composed of three numbers that specify the intensity of each color. Because it uses 8 bits (0s or 1s) for each color, each intensity can be transformed into a hexadecimal number with two digits at most. There are 256 possible shades for each color, since "11111111" (or hex "ff") corresponds to decimal 255, plus the number 000. The combination of all 256 possible shades for the three primary colors gives 256 cubed, or over 16 million possible colors. Write a function that takes a color in RGB or hex and returns the opposite. If it takes in the three integers in an RGB, it should return a string with the equivalent hexadecimal notation, plus a hash sign (#) at the front. If it takes in a string in hex, it should return an array with the three integers corresponding to RGB. Examples rgbHex (150, 50, 76) "#96324c" // 150 is hex 96, 50 is hex 32 and 76 is hex 4c. - rgbHex("#303749") → (48,…
- Q6/Full the following blanks (1101 1010)BCD(5211) = ( )BCD(3321)=( )EX-3=( )16 (1101 1010)BCD(5211) = (1110 0111)BCD(3321)=(1011001)EX-3= (56)16 O None of them (1101 1010)BCD(5211) = (1101 O 0111)BCD(3321)=(1011001)EX-3= (2B)16 (1101 1010)BCD(5211) = (1011 O 1110) BCD (3321)=(1000111)EX-3= (44)16 (1101 1010)BCD(5211) = (1110 O 1100) BCD (3321)=(1010110)EX-3= (A6)16A Sequence Detector is a circuit which given an input string of bits generates an output 1 whenever the target sequence (pattern) has been detected. There are two types of sequence detectors: Overlapping: The detector allows overlapping, hence the final bits of one sequence can be the start of another sequence. Non-overlapping: The detector does not allow overlapping, hence a new sequence starts only after the previous sequence is completed. Ex: Detect pattern 101 For non-overlapping case: Input :0110101011001 Output:0000100010000 For overlapping case: Input :0110101011001 Output:0000101010000 Design a sequential circuit (overlapping) with an input ‘x’. The pattern to be detected is the binary number that is equal to 1001100 For example,the circuit detects the sequence 00010 . When the sequence is detected, the detector outputs z=‘1’ For this circuit: Draw the ASM diagram. Construct the state table. Draw the logic circuit for the datapath and control in logisim.A Sequence Detector is a circuit which given an input string of bits generates an output 1 whenever the target sequence (pattern) has been detected. There are two types of sequence detectors: Overlapping: The detector allows overlapping, hence the final bits of one sequence can be the start of another sequence. Non-overlapping: The detector does not allow overlapping, hence a new sequence starts only after the previous sequence is completed. Ex: Detect pattern 101 For non-overlapping case: Input :0110101011001 Output:0000100010000 For overlapping case: Input :0110101011001 Output:0000101010000 Design a sequential circuit (overlapping) with an input 'x' the pattern to be detected is 01010 For this circuit: 1. Draw the ASM diagram. 2. Construct the state table. 3. Draw the logic circuit for the datapath and control in logisim.
- Assignment for Computer Architecture The assignment is to create a MIPS program that demonstrates that the associative law fails in addition for floating point numbers (single or double precision). You only need to demonstrate it for single precision. Remember the associative law is a + (b + c) = (a + b) + c. The program’s output should look something like the following where the xxx’s are the numbers you chose. The resulting numbers may be different than mine depending on your choice of a, b, and c. Using a = xxx, b = xxx, and c = xxx a + (b + c) = 0 (a + b) + c = 1 The key is to have two of the number large (one positively and one negatively but equal in magnitude) floating point numbers and the third floating point number very small in comparison. As a side note, the associative law will also fail for multiplication for floating point numbers, but you do not need to demonstrate that. Please don't forget to include comments in this code6-In the representation Complement to 1, we the number when Code the absolute value thus is negative: A = True B- False 7- The result in base 10' A - 44.25 B-42015 C- 51. 13 45. 75 8- The result AT 40. B + 61 q of the transformation (101001.10) 2 usi o the transformation (52) a in bose 10 is: C² 42 D-41 WWW 30 OFind the representation x of the following signed numbers listed below in the base specified on the right. a.- (01100110)2=( x )10 b.- (10110111)2=( x )10 C.- (207)10=( x )2 d.- (-135)10=(× )2 e.- (2D7)16=( x )2 f.- (10110011)2=(x )16 g.- (-5.5)10=( x )2 e.- (11000001010000000000000000001010)2=(x )10 Note: In part e the binary representation on the left corresponds to a real number representation.