b. 1:11 b.0; 10 a, 1:01 a, 0:00 b. z; lz a, z Oz b. 1 A a, 0 A b, z; A Z Z Z b.0: A c. 1 1 a, z A C. 0:0 a. 1: A 90 ql q2 What is the next configuration after (q0, abca, OZ) A. q2, A, ez B. q1,bca, 00Z c. q1, bca, 0z D. q2,Z,1Z E. q2,bca, 00 OA OD OE C OB
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- b, 1:11 b, 0:10 a, 1:01 a, 0:00 b, z; lz a, z; Oz b, 1; λ a, 0; z, z; z b, z; A b0;λ c, 1; 1 a, z; A c, 0:0 a, 1;A 90 q1 92 What is the next configuration after (q0, abca, OZ) A. 92,λ, ez B. q1,bca, 00Z c. q1,bca,oz D. q2,Z,1Z E. q2,bca,00Course Code: CCCN 212 Course Name: Digital Logic Design Assignment # 2 Targeted Course Learning Outcomes: CLO 2.1 Question 1: Create a circuit that compares two 2-bit numbers A and B. It should produce a 1 if A is smaller tha B, and 0 otherwise. To accomplish this task, you can create a 1-bit circuit, then use it twice to make it a 2-bit circuit. Question 2: Given the following 5-bit signed numbers, calculate A - B. Binary Numbers Subtraction Equivalent decimal Subtraction A = 10100 B = 001006-(x +y) (x + z) =? 7- (11111)ary =(?)2 8- (A.B) + (A.C) =? 9- In BCD (0100) + (0110) =? 10- (A.B.C) + D=?
- The problem Julia code to generate the data: xdata - -2,-1.64,-1.33,-0.7,0,0.45,1.2,1.64,2.32,2.9 ydata - 0.699369,0.700462,0.695354,1.03905,1.97389,2.41143,1.91091,0.919576,-0.730975,-1.42001 and you'd like to fit the function Fip1.p2.x) p1'cos(p2"x)+p2sin(p1"x) using nonlinear least squares.This code segment read f, k. Using a define sub procedure (SU) to determine the value of S of the below * .formula S= (2=11) f=Val(Text1.Text) k = Val(Text2.Text) 1-.... 2-.... 3-... Private Sub SU(m, S) Dim I As Single S = 0 4-.... S =S+I Next End SubA=[-1 -100;1 0];B=[1;0];C=[0 100];D=[0];[num,den]=ss2tf(A,B,C,D);sys=tf(num,den);step(sys); find the system response to step input 1- Delay time (td) 2- Rise time (tr) 3- Peak time (tp) 4- Maximum overshoot (Mp%)
- write pseudo code for pso.Select the logic to implement a loop using the LOOPNZ instruction. ECX = ECX – 1 if ECX > 0 and ZF=0 , jump to destination O ECX = ECX + 1 if ECX > 0 and ZF=1, jump to destination EDX = EDX - 1 if EDX < 0 or ZF=0 , jump to destinationAbstract: the main purpose of this experiment is build real time system using PPI 8255 to control devices connected to. Problem description: assume that there are two devices are connected to port A and two sensors are connected to port B of PPI 8255. They work according to the following table Sensors (S1S2) Devices (DID2) 00 01 01 10 10 11 11 00 Write a program to control these two devices according to the values of sensors.
- 4. Reduce the following state table. NS/Z X=0 PS X=1 A. C/o D/0 H/0 B/0 E/0 F/0 c/o F/0 티/0 G/0 G/O B/0 A/1 G/1 F/O B E H. E/OI0: ADD R4 = R1 + RO; Il: SUB R9 = R3 - R4; 12: ADD R4 R5 + R6; %3D 13: LDW R2 MEM [R3 + 100]; 14: LDW R2 = MEM [R2 + 0]; 15: STW MEM [R4 + 100] 16: AND R2 = R2 & R1; R1, Target; 18: AND R9 = R9 & R1; R2; 17: BEQ R9 == Consider a pipeline with forwarding, hazard detection, and 1 delay slot for branches. The pipeline is the typical 5-stage IF, ID, EX, MEM, WB MIPS design. For the above code, complete the pipeline diagram below (instructions on the left, cycles on top) for the code. Insert the characters IF, ID, EX, MEM, WB for each instruction in the boxes. Assume that there two levels of bypassing, that the second half of the decode stage performs a read of source registers, and that the first half of the write-back stage writes to the register file. Label all data stalls (Draw an X in the,box). Label all data forwards that the forwarding unit detects (arrow between the stages handing off the data and the stages receiving the data). What is the final execution time of the code? Cycles…func: sq: ADD X20, XZR, #0x1 ADD X0, X20, #0x0 BL sq ADD X21, X0, #0x0 ADD X20, X20, #0x1 ADD X0, X20, #0x0 BL sq ADD X21, X21, X0 ADD X20, X20, #0x1 ADD X0, X20, #0x0 BL sq ADD X21, X21, X0 ADD X20, X20, #0x1 ADD X0, X20, #0x0 BL sq ADD X21, X21, X0 ADD X20, X20, #0x1 ADD X0, X20, #0x0 BL sq ADD X21, X21, X0 MUL x0, x0, x0 BR LR Assume the .text segment begins at address Ox00000000, and that the linker placed "sq" immediately after func in memory. Complete the following table (remove all leading Os and report your answer in hex): Absolute address of the function sq: Ox Byte offset of the first BL call: 0x 30 Byte offset of the second BL call: 0x 60 Byte offset of the third BL call: Ox ⁹⁰