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padilas [110]
3 years ago
13

This assignment covers the sequential circuit component: Register and ALU. In this assignment you are supposed to create your ow

n storage component for two numbers using registers. Those two numbers are then passed into a custom ALU that calculates the result of one of four possible operations. Key aspect of this assignment is to understand how to control registers, how to route signals and how to design a custom ALU.
Engineering
1 answer:
Luba_88 [7]3 years ago
4 0

Answer:

The part I called command in the first diagram has been renamed to opcode, or operation code. This is a set of bits (a number) that will tell the ALU which action to perform. I can get the LC-3 opcodes for ADD and NOT and ADD from the book, so I'm not too worried.

Note the #? comment by the switch above opcode. This means I'm not sure how many switches I will need. How many bits do I need to perform all the operations I want? The textbook will tell me.

Materials

Now I make a list of all the materials you have accumulated so far. This list is just an example; yours may be different.

Two 4-bit inputs

One 4-bit output

Two keypads for 4-bit input

Three 7-segment displays (2 for input, 1 for output)

A bunch of switches for opcode (could use a keypad, I guess, but switches are so much more geeky)

A bunch of lights too

The "is zero" LED

One button for clock

One button for reset

One switch for carry-in

Include logic to perform a SUB instruction. That is, subtract the second operand from the first (out = in1 - in2). All three values -- both inputs and the output -- must be two's complement numbers (negative numbers must be represented). Your design may work in one (8 points) or two (4 points) clock cycles.

Explanation:

You are allowed to use the Logisim built-in registers.

The clear input of the register should not be used (do not connect anything to

them).

Custom ALU

Use the provided subcircuit in the template to implement your ALU. You do not have to create additional subcircuits to do this. The ALU has a total of three inputs: First number, second number and select operation input. And one output: Result. The first and second number are used as input for the operations the ALU performs. The select input decides which operation result will be on the single output of the ALU. The ALU is supposed to calculate: NumberA OPERATION NumberB. Register 1 of the storage contains NumberA and Register 2 contains NumberB. The ALU must be able to compute signals with a 4-bit width. Make sure to add labels to all inputs and outputs.

The following operations should be performed for each select input combination (s1s0): • 00: Logic Bitwise XOR

• 01: Multiplication

• 10: Division

• 11: Addition Notes:

You can change the inputs bit width / data bits of any gate to more that 1-bit.

The Logic Bitwise XOR operation can be done with a single XOR gate.

You are also allowed to use the built-in arithmetic logic components and multi- plexer provided by Logisim.

If the result is larger than 4 bits, it will be truncated (only 4 LSB will be shown). This behavior is intended for this assignment. Also, negative results do not have to be considered.

Once you have implemented the ALU circuit, connect the wires in the main circuit properly and test all four operations of your ALU in combination with the storage component.

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Two flat plates, separated by a space of 4 mm, are moving relative to each other at a velocity of 5 m/sec. The space between the
xenn [34]

Answer:

0.008

Explanation:

From the question, the parameters given are:

Velocity V = 5 m/s

Pressure = 10 pa

But pressure = F/A

10 = F/A

F = 10A

Substitute all the parameters into the formula below

Coefficient of viscosity (η) = F × r /[AV]

Where

F = tangential force,

r = distance between layers,

A = Area, and

V = velocity

(η) = 10A × 0.004 /[A × 5]

The A will cancel out

(η) = 10 × 0.004 /[5]

(η) = 0.04 /5

(η) = 0.008

Therefore, the coefficient of viscosity of the fluid is 0.008

5 0
3 years ago
The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re
Anni [7]

Answer:

a) W_cycle = 200 KW , n_th = 33.33 %  , Irreversible

b) W_cycle = 600 KW , n_th = 100 %     , Impossible

c) W_cycle = 400 KW , n_th = 66.67 %  , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

  TL = 400 K

  TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

                                 W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

                                 n_th < n_max     ...... irreversible

                                 n_th = n_max     ...... reversible

                                 n_th > n_max     ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

                                n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And,                         n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 400 = 200 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 200 / 600 = 33.33 %

   - The type of process according to second Law of thermodynamics:

               n_th = 33.333 %                n_max = 66.67 %

                                       n_th < n_max  

      Hence,                Irreversible Process  

b) QH = 600 kW, QC = 0 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 0 = 600 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 600 / 600 = 100 %

   - The type of process according to second Law of thermodynamics:

                 n_th = 100 %                 n_max = 66.67 %

                                     n_th > n_max  

      Hence,               Impossible Process              

c) QH = 600 kW, QC = 200 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 200 = 400 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 400 / 600 = 66.67 %

   - The type of process according to second Law of thermodynamics:

               n_th = 66.67 %                 n_max = 66.67 %

                                     n_th = n_max  

      Hence,                Reversible Process

7 0
3 years ago
What are the general rules for press fit allowances
Keith_Richards [23]

Explanation:

As a general rule of thumb, the large the diameter of a bearing, bushing or pin, the larger the tolerance range,” Brieschke points out. “The inverse is true for smaller-diameter pieces.”

Mike Brieschke, vice president of sales at Aries Engineering, says a 0.25-inch-diameter metal dowel that is press-fit into a mild steel hole usually has an interference of ±0.0015 inch. Parts in noncritical assemblies tend to have looser tolerances

please rate brainliest if helps and follow

4 0
1 year ago
Wattage is:
Ksju [112]

Answer:

c.Both A and B.

Explanation:

the wattage is c and d

7 0
3 years ago
What is the rated power output in ( kw) of a 8 pole motor designed to an IEC 180L motor frame ?<br>​
ra1l [238]

Answer:

P=11 kW

Explanation:

Given that

Number of poles= 8

I.E.C. 180L motor frame

From data book , for 8 poles motor at 50 Hz

Speed = 730 rpm

Power factor = 0.75

Efficiency at 100 % load= 89.3 %

Efficiency at 50 % load= 89.1 %

Output power = 11 kW

Therefore the rated output power of 8 poles motor will be 11 kW. Thus the answer will be 11 kW.

P=11 kW

4 0
3 years ago
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