Combinational Logic Circuit #2 Logic gates are used to make decisions. More complex decisions can be made by using more than 1 logic gate. This type of circuit is known as a combinational logic circuit. They have no memory as the output is completely dependent on the input. Using the 74LS00 NAND gate chip, construct the circuit in the schematic and complete the accompanying truth table. B A B 0 3010 0 1 1 1 out Based on the truth table generated, what logic function is performed by this circuit? Explain.
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1) Create both circuits on tinkercad
2) Make a truth ta
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- Build a circuit that takes four bits as input: W, X, Y, Z. Treat WX as a 2-bit unsigned binary number, and treat YZ as a second 2-bit unsigned binary number. Your circuit should generate the output corresponding to the product of WX and YZ. You will need 4 bits of output for this problem.For example, if your input was 1011, your inputs correspond to 2 and 3. That product is 6, so your output will be 0110.Create a truth table for this problem, show all k-maps and minimizations, and build the corresponding (minimized) circuit. Use XOR, XNOR, NAND, and NOR as appropriate if it reduces the number of gates used.Task 1: 1-bit Full Adder For this task, you will be building a 1-bit Full Adder circuit. To begin, consider how you do binary addition on paper, but only look at what happens in a single bit. Ultimately, for each bit, you have 3 inputs, and 2 outputs. The inputs are A and B, the two bits you are adding, along with a potential Carry In (Cin, some sort of carry from the previous bit). The outputs are the sum and a Carry Out (Cout) that goes onto the next bit. Consider all of the possible permutations of A, B, and Cin being 0 and 1, and fill out what Sum and Cout would be in the table below A B Cin Sum Cout 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 Now that you have a table for the 1-bit Full Adder, you should be able to directly come up with the sum of product (SOP) equation for Sum and Cout (separately, in terms of A, B, and Cin). Use a K-map to minimize Cout…Task 1: 1-bit Full Adder For this task, you will be building a 1-bit Full Adder circuit. To begin, consider how you do binary addition on paper, but only look at what happens in a single bit. Ultimately, for each bit, you have 3 inputs, and 2 outputs. The inputs are A and B, the two bits you are adding, along with a potential Carry In (Cin, some sort of carry from the previous bit). The outputs are the sum and a Carry Out (Cout) that goes onto the next bit. Consider all of the possible permutations of A, B, and Cin being 0 and 1,in the table below A B Cin Sum Cout 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Now that you have a table for the 1-bit Full Adder, you should be able to directly come up with the sum of product (SOP) equation for Sum and Cout (separately, in terms of A, B, and Cin). Use a K-map to minimize Cout (Sum would be requested too, but it doesn’t…
- We want to design a circuit that counts the number of 1s present on 3 binary inputs a, b, c and outputs that number in binary using 2 outputs y and z. For example an input of a=1, b=1, c=0 has two ones, so the output would be 210 = 102 i.e., y=1, z=0. a) Express y and z as sum of min‐terms of variables a, b and cb) Simplify the expressions for y and z using K maps.c) Draw the logical circuits for y and z using gates.In this problem, you should design a two-bit comparator. This circuit should have three outputs named l, g, and eq. The circuit should get two digits binary numbers (00, 01, 10, 11), and the output should change based on these rules:• If first number > second number then g = 1, l = 0, and eq = 0• If first number < second number then g = 0, l = 1, and eq = 0• If first number = second number then g = 0, l = 0, and eq = 1Your circuit will have 4 input (2 bit for the first number, and 2 bits for the second number)a. Draw the truth table for the comparator for unsigned numbers b. Show the circuit.We will construct a circuit that multiplies a double-digit binary number by three, using onlyhalf-adders, which were described in class. Please use a dotted line for the result digit, and asolid line for the carry digit. (See attached image) (a) Write out algebraicly what the calculation above represents.(b) Construct the circuit using half-adders only.
- For same input of n=7, it gives the output like a pyramid. Write the code. 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5 1 2 3 4 5 6 1 2 3 4 5 6 7A Gray code is a sequence of binary numbers with the property that no more than 1 bit changes in going from one element of the sequence to another. For example, here is a 3-bit binary Gray code: 000, 001, 011, 010, 110, 111, 101, and 100. Using three D flip-flops and a PLA, construct a 3-bit Gray code counter that has two inputs: reset, which sets the counter to 000, and inc, which makes the counter go to the next value in the sequence. Note that the code is cyclic, so that the value after 100 in the sequence is 000.In 16-bit binary code create a program for LC-3 that prints out the letters: ZYX..DCBAABCD....XYZ. There should be a total of 52 letters. Use one or two loops. (STRICTLY ONLY USE 1'S AND 0'S)
- In MIPS assembly, implement integer division using rounding (rather than truncation). This is accomplished by taking the remainder of the division and dividing it by the original divisor. If the new quotient is higher than or equal to one, multiply it by one.Otherwise, leave the original quotient alone.In MIPS assembly, implement integer division using rounding (rather than truncation). This is accomplished by taking the remainder of the division and dividing it by the original divisor. If the new quotient is higher than or equal to one, multiply it by one.Otherwise, leave the original quotient alone.In MIPS assembly, implement integer division using rounding (rather than truncation). This is accomplished by taking the remainder of the division and dividing it by the original divisor. If the new quotient is higher than or equal to one, multiply it by one.Otherwise, leave the original quotient alone.In MIPS assembly, implement integer division using rounding (rather than truncation).…. Implement a circuit for the following problem using Logisim. The input to the circuit are 3 4-bit numbers, A,B,C. The sum of A and B is subtracted from C. The difference (result of subtraction) is compared with A. The circuit has 3 outputs lines depending on comparison. Use appropriate chips in Logisim for the operations mentioned Answer step by stepDesign a circuit that takes three bits, X2, X1, X0 as input and produces one output, F. F is 1 if and only if 2<=X<=5 when X = (X2, X1, X0) is read as an unsigned integer. For example, if X2=1, X1=0, and X0=0, then the unsigned binary value is 100, which is 4, so the output would be 1. Your Assignment For This Problem Includes the Following Design the necessary circuit using Logisim to implement the situation described above. Use Kmaps for simplification. Be VERY careful to get the correct functions for your output before simplifying and designing the circuit with Logisim. You should minimize the circuit. Your circuit should have three inputs and one LED output. All inputs (X2, X1, X0) and output (F) should be labeled (in Logisim, not by hand). Please use these names to indicate the inputs and output so all projects are consistent. You should also include your name as a label on the circuit. Test your circuit to be sure it is working correctly.