A hidden passage is controlled by a redstone circuit for an adder which performs binary additiontwo bits at a time. He notes that it adds the inputs (x2i, y21) and (x2i+1, Y2i-1), the (2i)th and (2i+1)thbits of two binary numbers, and returns the sum as outputs s2i, S2+1. The addition is performedeach clock cycle, starting from i = 0 until i =п1, where n is the number of bits and2presumed to be even.For example, when adding x = 011010, and ynumber as x1:0 and y1:0 respectively. In this case, X1:0 is 10 and y1:0 = 01, and the carry bit co is 1.Our sum, s,0 is 00 and the next carry bit c, is 1. On the next clock cycle, we look at the next two0110012, we take the first two bits of eachbits, X3:2= 10 and y3:2 =10, so our sum s3-2 is 01, but we have a carry c41. Finally, our lastaddition to perform is on X5-4 = 01 and y5:4 = 01, with a carry bit c4 = 1. Our sum therefore is s54 =11 with carry C6 = 0. Our final result of the addition is 1101002.You are only responsible for the addition of each two-bit partition of the number and carry bits,and are not responsible for the final sum or selecting two-bit partitions of the numbers.a) Fill in the state table for the new adder, the first row has been provided.X2i+1:2iY2i+1:21C2iC2i+2S2i+1:2iX2i+1:2iY2i+1:21C2iC2i+2S2i+1:2i000000 b) Draw a schematic design using a D flip-flop, two Full Adders, and a minimal network of AND,OR, and NOT gates. You only need to(do not be concerned how the next set of bits would be selected or passed into this circuit.)thsclnaticthe addition ofgiven twe

A hidden passage is controlled by a redstone circuit for an adder which performs binary addition two bits at a time. He notes that it adds the inputs (X Ya) and (Xa. ya), the (2i)th and (2i+1)th bits of two binary numbers, and returns the sum as outputs s, Sa1. The addition is performed each clock cycle, starting from i = 0 until i =+ - 1, where n is the number of bits and presumed to be even. For example, when adding x = 011010, and y = 011001,, we take the first two bits of each number as x1, and ye respectively. In this case, x1g is 10 and yo = 01, and the carry bit c, is 1. Our sum, s,, is 00 and the next carry bit c, is 1. On the next clock cycle, we look at the next two bits, X32 = 10 and ys2 = 10, so our sum s32 is 01, but we have a carry c, = 1. Finally, our last addition to perform is on x = 01 and ys = 01, with a carry bit c, = 1. Our sum therefore is s. = 11 with carry c, = 0. Our final result of the addition is 110100,. You are only responsible for the addition of each two-bit partition of the number and carry bits, and are not responsible for the final sum or selecting two-bit partitions of the numbers. a) Fill in the state table for the new adder, the first row has been provided. Yana Sa X Vaa 00 00 00

A hidden passage is controlled by a redstone circuit for an adder which performs binary addition two bits at a time. He notes that it adds the inputs (X Ya) and (Xa. ya), the (2i)th and (2i+1)th bits of two binary numbers, and returns the sum as outputs s, Sa1. The addition is performed each clock cycle, starting from i = 0 until i =+ - 1, where n is the number of bits and presumed to be even. For example, when adding x = 011010, and y = 011001,, we take the first two bits of each number as x1, and ye respectively. In this case, x1g is 10 and yo = 01, and the carry bit c, is 1. Our sum, s,, is 00 and the next carry bit c, is 1. On the next clock cycle, we look at the next two bits, X32 = 10 and ys2 = 10, so our sum s32 is 01, but we have a carry c, = 1. Finally, our last addition to perform is on x = 01 and ys = 01, with a carry bit c, = 1. Our sum therefore is s. = 11 with carry c, = 0. Our final result of the addition is 110100,. You are only responsible for the addition of each two-bit partition of the number and carry bits, and are not responsible for the final sum or selecting two-bit partitions of the numbers. a) Fill in the state table for the new adder, the first row has been provided. Yana Sa X Vaa 00 00 00

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter4: Processor Technology And Architecture

Section: Chapter Questions

Problem 9VE

See similar textbooks

Similar questions

Consider a special purpose decoder that is called BCD to seven-segment decoder. The decoder takes a 4-bit Binary Coded Decimal (BCD) number X3 X2 X4 X0 (0-9) and produces 7 outputs a, b, c, d, e, f, and g The specification of this circuit is as follows: ⚫ The 7 outputs of the decoder are connected to LED segments as shown in the figure below. The LEDs are active low, ie, the LED emits light when it is connected to logic 0. • The decoder output is determined in a way so that the LED segments display the decimal representation of the input. For example, if the input X3 X2 X1 X0 is 0001 respectively (which is equivalent to decimal 1), the outputs a, b, c, d, e, f, and gare 1001111 showing the letter 1 on the display (remember the LEDs are active low). Decoder W- 4x7 e U Answer the following questions: a. Fill in the truth table of that decoder. Choose a good value for the output when the input is greater than decimal 9. b. Derive the simplified Boolean expressions for the outputs a, b, c,…
Design a circuit using a Multiplexer, (((USING A MULTIPLEXER NOT A DECODER))) that has two inputs X, and S, where X represents an 8-bit BCD number,S is a sign bit. The circuit has one output Y, which is the Binary representation of thesigned-magnitude BCD number. A negative output is represented in the Binary 2’scomplement form.
Design a sequence detector of the pattern 0100 where the circuit accepts a serial bit stream *X" as input and produces a serial bit stream *2° as outputWhenever the bit pattern *0100* appears in the input stream, it outputs 2 = 1: at all other times, 2 = Q Overlapping occurrences of the pattern are also detected,
A 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.
Design a Moore FSM for a Sequence Derector that detects five consecutive bits of "zero" in the input stream of bits, labeled x. The output z equals 1 if, during five consecutive clock cycles, the input x was equal to "zero". Once the sequence is detected, the FSM returns to the initial state S0. Add an asynchronous reset, active low, to the FSM.
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.
Design a state machine to output the binary values (include leading zeros e.g. 0101} for the following sequence each time a increase count input line changes from 0 to 1. When it reaches the end it starts over at beginning {1, 3, 9, 5, 7, 3}. It has a second input line forward/reverse which will change its direction to go in reverse direction if this line is high..
Design a sequential circuit that outputs 2’s complement of a bit sequence. The bit sequence will be given to the circuit in reverse order. For instance, if the given bit sequence is 0010101000, the circuit should output 1101011000.
Design a sequential circuit that outputs 2’s complement of a bit sequence.The bit sequence will be given to the circuit in reverse order. For instance, if the given bit sequence is 0010101000, the circuit should output 1101011000.
An 8 x 1 multiplexer has inputs A , B , and C connected to the selection inputs S2 , S1 , andS0 , respectively. The data inputs I0 through I7 are as follows:(a) I1 I2 I7 0; I3 I5 1; I0 I4 D ; and I6 D ’.(b) I1 I2 0; I3 I7 1; I4 I5 D ; and I0 I6 D ’.Determine the Boolean function that the multiplexer implements.
Design a circuit that has two inputs X, and S, where X represents an 8-bit BCD number, S is a sign bit. The circuit has one output Y, which is the Binary representation of the signed-magnitude BCD number. A negative output is represented in the Binary 2’s- complement form. You need to think of two design alternatives. Submission guidelines: 1. You should write a report that at least contains the following sections: 1. Problem definition. 2. Design alternatives : 2.1. Alternative 1 block diagram 2.2. Alternative 2 block diagram 3. Design selection criteria 4. Detailed circuit design of the selected alternative. 5. Verilog modules, and simulation results for all modules, and for the whole circuit of the selected alternative .
Design a combinational circuit that uses 4-bit adders and 2-to-1 MUXes. Given two 8-input signed 2s-complement numbers A and B and a binary input signal M, your circuit should produce an 8-bit signed 2s-complement result R, as follows: if (M == 0) R = A + 7; else R = B + 15;