The Karnaugh map for a Boolean function is given as CD АВ C'D' C'D CD C'D CD A'B A'B 1 АВ 1 1 АВ 1 1 1 The simplified Boolean equation for the above Karnaugh Map is
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- 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 1answere 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.Build a truth table and draw the output wave form for the following logic gates shown in Figure Q2. A o B Co Do E o D D Figure Q2 Z
- 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 FalseThe following waveforms are the input to the shift register you constructed in question #5. Based onthe waveforms for the CLK (clock) and D0 input (input on the leftmost Flipflop), generate thewaveforms for Q0, Q1, Q2, Q3, Q4. (Question #5 is "Using JK-Flipflops and Digital Logic Gates, build a 5-stage Shift Register")mybmsajmanac ERSITY Design My courses Logic Design General Qua 2 LD/DLD on Tue. 7/12/21-Dr. Zidan The correct state sequence of the cirtut with initial state Qo1, 01 and Q0 D. Q D, a. LSB MSB Clock Select one O a1, 2, 5.3, 7,6,4 O b.1,6, 5,7, 2.3,4 O C1,2.7,3, 5,6, 4 O d 1,3,4, 6, 7,3.2
- (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 1431.204 number ( O 1431.200 O No one of them 1341.200 O 1341.201 ) in octalQ4/ The decimal 793.251)= ( number O 1431.200 O No one of them 1431.204 number ( O 1341.200 O 1341.201 ) in octal
- 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)2Q1/The binary number ( 1011011011011.1110111001101)= ( ) in octal number *Design 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).