Output PL1, of the logic rung shown, will be true when the value of the number stored in word N7:10 is: Ladder logic program PL1 ਆਮ EQU EQUAL Source A Source 8 N7 10 O equal to 80. O equal to or greater than 80. greater than 80. O less than 80. PLI Output
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- (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).Select a suitable example for for combinational logic circuit. O a. None of the given choices O b. De-multiplexer O c. PLA O d. LatchesDetermine the truth table for the following logic circuit. Then identify the type of logic gate using Boolean algebra. O +Vcc R23 ER O O/P Q1 Q2 BO
- Question1: Read the following table, and design a logic circuit that can convert the binary code from 2421 to Excess-3. Four Different Binary Codes for the Decimal Digits Decimal ВCD 8421 Digit 2421 Excess-3 8, 4, –2, –1 0000 0000 0011 0000 1 0001 0001 0100 0111 0010 0010 0101 0110 3 0011 0011 0110 0101 4 0100 0100 0111 0100 5 0101 1011 1000 1011 0110 1100 1001 1010 7 0111 1101 1010 1001 8 1000 1110 1011 1000 1001 1111 1100 1111 1010 0101 0000 0001 Unused 1011 0110 0001 0010 bit 1100 0111 0010 0011 combi- 1101 1000 1101 1100 nations 1110 1001 1110 1101 1111 1010 1111 1110Electrical Engineering Draw 2, 1 bit ALUS to create a basic 2 bit ALU. the carry out and carry in bits must ripple across. The ALU should subtract/add, logical NOR, logical AND, and logical OR. Draw out the adding logic circuitThe 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 False
- 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).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 ZFor 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.
- Using truth table for 7-segment logic:1. Determine the minimum logic for segment b2. Determine the minimum logic for segment c3. Determine the minimum logic for segment d4. Determine the minimum logic for segment e5. Determine the minimum logic for segment f6. Determine the minimum logic for segment gmybmsajmanac 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, 7,3.216. The following serial data are applied to the flip-flop through the AND gates as indicated in Figure 7-85 Ⓒ. Determine the resulting serial data that appear on the Qoutput. There is one clock pulse for each bit time. Assume that Q is initially 0 and that and PRE are HIGH. Rightmost bits are applied first. J₁: 1010011; J₂:0111010; J: 1111000; K: 0001110; K 1101100, K: 1010101 CLK K₁ CLR Figure 7-85 C K PRE -Q CLR