a) Convert the flowing number (226)10 to binary, octal, and hexadecimal.
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- The numbers from 0-9 and a no characters is the Basic 1 digit seven segment display * .can show False True In a (CA) method of 7 segments, the anodes of all the LED segments are * "connected to the logic "O False True Some times may run out of pins on your Arduino board and need to not extend it * .with shift registers True FalseDesign a code converter that converts a decimal digit from BCD to excess-3 code, the input variables are organized as (A BC D) respectively with A is the MSB, the output variables are organized as (W XY Z) respectively with W is the MSB, put the invalid decimal numbers as don't care. X= BCD'+B'D+B'C X= BC'D'+B'D+BC X= BC'D'+B'D+B'C X= BC'D'+BD+B'CDesign a code converter that converts a decimal digit from BCD to excess-3 code, the input variables are organized as (A BC D) respectively with A is the MSB, the output variables are organized as (W X Y Z) respectively with W is the MSB, put the invalid decimal numbers as don't care. X= BCD'+B'D+B'C X= BC'D'+B'D+BC X= BC'D'+B'D+B'C X= BC'D'+BD+B'C
- (c) Figure Q3(c)(i) shows a register and Figure Q3(c)(ii) shows the input waveforms (CLOCK and Data in) to the circuit. A1 A9 A10 A2 Function generator A3 A11 A12 AS A13 A6 A14 A7 A15 Data in Bop.7) ip.r 82p.7) Logic analyser U1 U2 U3 U4 UO 6. 1. 6 1 6 INVERTER 3 CLK 3 CLK oCLK CLK 5 K K 5 K K 4027 Clock Function generator Figure Q3(c)(i) (i) Determine the type of register as shown in Figure Q3(c)(i).Q4/ The decimal 793.251)= ( number O 1431.200 O No one of them 1431.204 number ( O 1341.200 O 1341.201 ) in octalWhich function performs the following operation? Give the assembly instruction and show your work. Before A= 11011011, CF=1 After A= 10111101, CF=1
- a) Convert the decimal number - 239 into a signed binary number using (1) sign-magnitude method (2) 1's complement method (3) 2's complement method b) If the above decimal number is positive , that is + 239, then convert it to a signed binary number using 1 's complement method c) Do the following binary subtraction: 101111111.01111- 001011010.10110 d)Explain Types of RAM?USE DIGITAL LOGIC AND DESIGN Part 1: In Figure_4; we have 4-bit Comparator using 2-bit Comparators block. You have to satisfy given condition by applying all data on figure 4. At the end, given condition should produce HIGH output and other two should be LOW. A3 A2 A1 A0 = 1101 and B3 B2 B1 B0 = 1110 Figure_4 Part 2: The serial data-input waveform (Data in) and data-select inputs (S0 and S1) are shown in Figure_5. Determine the data-output waveforms from D0 through D3. Figure_5 Part 3: Decoder can be useful when we have to decode some specific numbers from their equivalent code. Figure 6 has a concept of 3 to 8 line decoder from which you have to generate output waveform from D0 to D7 with proper relationship to input. Figure_6 Part 4: The data-input and…procedure to the instructor. 6. Convert (-00101001.0011)2 to its equivalent octal and hexadecimal representation. Show all calculation procedure to the instructor.
- For a ((A+B)' + (A'B')) Boolean equation, with the input waveforms as shown in Figure 2, which output waveform is correct? INPUT A INPUT B OUTPUT a OUTPUT b OUTPUT C OUTPUT d- Figure 2 Output b Output a Output d Output c A full adder logic circuit has Three inputs and three outputs. Three inputs and two outputs. Two inputs and one output. Two inputs and two outputs.a. (697)10 into Octal and Hexa decimal b. (2D9)16 into Decimal and Octal1. An arithmetic system operates with 3-digit decimal numbers with sign presented in BCD (12 bits), negative numbers utilize the 10's complement form, one decimal digit (4 bits) is occupied by sign: 0 for "+" and 9 for "-". 1a) Show the most positive and the most negative numbers in both decimal and BCD forms. 1b) Present numbers A=+36 and B= -42 in BCD. 1c) Present the 10's complements of A and B in BCD. 1d) Perform the following operations bit-by-bit, show carries, apply BCD correction (add 6) when necessary (the sum of two decimal digits is greater than 9). 1d1) A+B in BCD. 1d2) A - B = A+(-B) in BCD by addition of A and the 10's complement of B. 1d3) B-A= B + (-A) in BCD by addition of B and the 10's complement of A.