write a V HDL code for the following Boolean expression using function Statements fi(A,B.C) = AB + Ac A Fz (A, B, c) = (AC) +Bc Y= f, C %3D
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- 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…IH.W: Draw a logic eircuit of the following Boolean expression before and after simplification using karnough map and Boolean expression. Y-AB+ AB A B Y 1 1Write a VHDL code for the following simple logic circuit. D- X1 X2 f X3
- 5. Assume d3d₂d₁do is a BCD number and derive Boolean expression for segments b,c,e,f,g in a seven segment display in terms of d3d₂d₁do g 4 d Based on your boolean expressions derive the values of b,c,e,f,g segments when i. d3d₂d₁do= (0100) 2 ii. d3d₂d₁do = (0111)₂ iii. d3d2d₁do (1000)2(m. Simplify the given expression using Boolean algebra and implement the simphifie expression using basic logic gates. Y = A.B.C+ A.B.C + A.B.C + A.B.CA three input logic functipn will provide a logic high output only when twp (and two only) of the inputs are logic highs. For all other input possibilities, a logic zero is provided on the output. What is the logic expression? A. Y=A'B'C+A'BC'+AB'C' B. Y=AB'C+A'BC+ABC' C. Y=AB'C+ABC+A'B'C' D. Y=ABC'+AB'C+A'B'C'
- Design 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'CWith explainationDesign 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
- Draw the logic circuit of Boolean expressionConsider two 8-bit inputs, A = $52 and B = $C3 to the arithmetic and logic unit (ALU). Compute R =A + B. Express R in the hexadecimal form $-- : -61 Express N-Z-V-C bits in the form ----:Simplify the following Boolean expressions using Karnaugh Map and draw the logic circuits. f = wxyz + wxyz + wxyż + wxỹz + wxyz + wxyz + wxỹz + wãyz