SEVEN SEGMENT HEX DECODER: 0 to F Boolean expressions: ન જ છે – 5. 6. 7. a =A+C+BD+BD B+CD+CD b = C +B+D A+Bc+BD + CD + BCD с = d = e = f = g BD +CD A+ BC +BD + CD A + BC + CD + Bc. Simplify and Give the AND logic circuit diagram of these expressions.
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- Write a VHDL code for the following simple logic circuit. D- X1 X2 f X3Design the following combinational logic circuit with a four-bit input and a three-bit output. The input represents two unsigned 2-bit numbers: A1 A0 and B1 B0. The output C2 C1.C0 is the result of the integer binary division A1 A0/B1 B0 rounded down to three bits. The 3-bit output has a 2-bit unsigned whole part C2 C1 and a fraction part CO. The weight of the fraction bit CO is 21. Note the quotient should be rounded down, i.e. the division 01/11 should give the outputs 00.0 (1/3 rounded down to 0) not 00.1 (1/3 rounded up to 0.5). A result of infinity should be represented as 11.1. A minimal logic implementation is not required. (Hint: start by producing a truth table of your design).Simplify the following Boolean expressions using Karnaugh Map and draw the logic circuits. f = wxyz + wxyz + wxyż + wxỹz + wxyz + wxyz + wxỹz + wãyz
- Write a VHDL program to implement the combinational logic circuit in the following figure. HA A>B A>E B The circuit contains three input signals r, q and s, and one output sigal z. All signals are 1 bit only. In the circuit, there are two 2-to-1 multiplexers and two 2-input comparators. The two comparators should be implemented using component.Digital logic design Solve it with drawing and simulation lab I need them both to have the full solution. And thanks Design counter that counts from 00 to 59, using the IC 74LS90 ripple counter and use two 7 segment display to display the result count. You can also use 7447 binary to 7-segment Display Decoder.Consider the multiplexer based logic circuit shown in the figure MUX MUX 1 Select one: a. W S1' S2' O b. W + S2 + S1 c. WS1 + WS2 + S1 S2 O d. WeS1es2
- Q2/A. Design a logic circuit that perform the following Boolean function, use 4-to-1 multiplexer and other logic gates you may need in the design. Use Shannor expansion 21 18 31 f(A, B, C, D, E) = ABCDE + ABCDE + ABCDE + ABCDE %3D l68421 B. Design 8-bit Adder/Subtracter using 4-bit Adder and 2-to-1 multiplexers other logic gates you may need in the design.Consider 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 ----:answere fast please question from DIGITAL LOGIC DESIGN TOPIC : Designing Combinational Logic You are designing a water level circuit using 74ALS151 (8 to 1 Multiplexer IC)* When input is 0000 that means tank is empty.* When input is 1111 that means tank is full.* When input is below 5, that means water level is low.* So, make a circuit using 74ALS151 Multiplexer IC that shows a "low water" indicator light(by setting an output L to 1) when the water level drops below level 5.
- Simplify the following expression using Karnaugh map and implement. Draw simplified logic diagram as well. Implement on Multisim software. (a) Y=A.B.C'.D+A.B'.C'.D+A'.B'.C'.D+A'.B.C'.D+A'.B'.C'.D'+A'.B.C'.D'+A'.B.C.D'+A'.B.C.D+A'.B'.C.DUSE 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…A 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'