answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
padilas [110]
2 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]2 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.

You might be interested in
Water at 70 kPa and 1008C is compressed isentropically in a closed system to 4 MPa. Determine the final temperature of the water
Nikitich [7]

Answer:

The answer is "909.3928  KJ".

Explanation:

70 \ kPa  \ \ and \ \ 100^{\circ}C \\\\s_i= 7.56162\ \frac{kJ}{kgK}\\\\u_i= 2509.39 \ \frac{kJ}{kg}\\\\

The method is isentropic since the cylinders are shielded.

Calculating the work:

w= u_2-u_i \\\\

   = 3418.7728-2509.38 \\\\=909.3928 \ KJ

3 0
2 years ago
A spring-loaded toy gun is used to shoot a ball of mass m = 1.50 kg straight up in the air. The spring has spring constant k = 6
adell [148]

Answer:

1) a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

2) The muzzle velocity of the ball is approximately 5.272 meters per second.

3) The maximum height of the ball is 1.417 meters.

Explanation:

1) Which of the following statements are true?

a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

True, statement indicates that there is no air resistence and no friction between ball and the inside of the gun because the first never touches the latter one.

b) The forces of gravity and the spring have potential energies associated with them.

False, force of gravity do work on the ball and spring receives a potential energy at being deformated by the ball.

c) No conservative forces act in this problem after the ball is released from the spring gun.

False, the absence of no conservative forces is guaranteed for the entire system according to the statement of the problem.

2) According to the statement, we understand that spring is deformed and once released and just after reaching its equilibrium position, the muzzle velocity is reached. As spring deformation is too small in comparison with height, we can neglect changes in gravitational potential energy. By Principle of Energy Conservation, we describe the motion of the ball by the following expression:

U_{k, 1}+K_{1}=U_{k,2}+K_{2} (Eq. 1)

Where:

U_{k,1}, U_{k,2} - Initial and final elastic potential energies of spring, measured in joules.

K_{1}, K_{2} - Initial and final translational kinetic energies of the ball, measured in joules.

After using definitions of elastic potential and translational kinetic energies, we expand the equation above as:

\frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2}) = \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})

And the final velocity is cleared:

m\cdot (v_{2}^{2}-v_{1}^{2}) = k\cdot (x_{1}^{2}-x_{2}^{2})

v_{2}^{2}-v_{1}^{2} =\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})

v_{2}^{2} =v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})

v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) } (Eq. 2)

Where:

v_{1}, v_{2} - Initial and final velocities of the ball, measured in meters per second.

k - Spring constant, measured in newtons per meter.

m - Mass of the ball, measured in kilograms.

x_{1}, x_{2} - Initial and final position of spring, measured in meters.

If we know that v_{1} = 0\,\frac{m}{s}, k = 667\,\frac{N}{m}, m = 1.50\,kg, x_{1} = -0.25\,m and x_{2} = 0\,cm, the muzzle velocity of the ball is:

v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\left(\frac{667\,\frac{N}{m} }{1.50\,kg} \right)\cdot [(-0.25\,m)^{2}-(0\,m)^{2}]}

v_{2}\approx 5.272\,\frac{m}{s}

The muzzle velocity of the ball is approximately 5.272 meters per second.

3) After leaving the toy gun, the ball is solely decelerated by gravity. We construct this model by Principle of Energy Conservation:

U_{g,2}+K_{2} = U_{g,3}+K_{3} (Eq. 3)

Where:

U_{g,2}, U_{g,3} - Initial and gravitational potential energies of the ball, measured in joules.

K_{2}, K_{3} - Initial and final translational kinetic energies of the ball, measured in joules.

After applying definitions of gravitational potential and translational kinetic energies, we expand the equation above and solve the resulting for the final height:

m\cdot g \cdot (h_{3}-h_{2}) = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{3}^{2})

h_{3}-h_{2}=\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}

h_{3} = h_{2} +\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g} (Eq. 4)

h_{2}, h_{3} - Initial and final heights of the ball, measured in meters.

v_{2}, v_{3} - Initial and final velocities of the ball, measured in meters per second.

g - Gravitational acceleration, measured in meters per square second.

If we get that v_{2} = 5.272\,\frac{m}{s}, v_{3} = 0\,\frac{m}{s}, h_{2} = 0\,m and g = 9.807\,\frac{m}{s^{2}}, the maximum height of the ball is:

h_{3} = 0\,m+\frac{\left(5.272\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

h_{3} = 1.417\,m

The maximum height of the ball is 1.417 meters.

5 0
2 years ago
The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 3.7 × 104 148 1.0 × 105 13
prohojiy [21]

Answer:

i) S–N plot is attached

ii) fatigue strength = 100 MPa

iii) fatigue life = 5.62 x 10^(5) cycles

Explanation:

i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

Log(4 × 10^(6)) = 6.6

From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa

iii) The question says we should find the fatigue life for 120 MPa.

Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

Thus,the fatigue life will be the inverse log of 5.75.

Thus, fatigue life = 10^(5.75)

Fatigue life = 5.62 x 10^(5)

8 0
2 years ago
Tech A says that proper footwear may include both leather and steel-toed shoes. Tech B says that when working in the shop, you o
olga55 [171]

Answer:

Tech A is correct.

Explanation:

If a person is doing something dangerous in a shop, he should wear safety glasses to protect his eyes from danger and also wear leather shoes to protect himself from any electric shock. Leather boots will disconnect a person with direct earth and therefore he can save himself if he gets a electric shock.

6 0
2 years ago
Determine the angular acceleration of the uniform disk if (a) the rotational inertia of the disk is ignored and (b) the inertia
lukranit [14]

Answer:

α = 7.848 rad/s^2  ... Without disk inertia

α = 6.278 rad/s^2  .... With disk inertia

Explanation:

Given:-

- The mass of the disk, M = 5 kg

- The right hanging mass, mb = 4 kg

- The left hanging mass, ma = 6 kg

- The radius of the disk, r = 0.25 m

Find:-

Determine the angular acceleration of the uniform disk without and with considering the inertia of disk

Solution:-

- Assuming the inertia of the disk is negligible. The two masses ( A & B )  are hung over the disk in a pulley system. The disk is supported by a fixed support with hinge at the center of the disk.

- We will make a Free body diagram for each end of the rope/string ties to the masses A and B.

- The tension in the left and right string is considered to be ( T ).

- Apply newton's second law of motion for mass A and mass B.

                      ma*g - T = ma*a

                      T - mb*g = mb*a

Where,

* The tangential linear acceleration ( a ) with which the system of two masses assumed to be particles move with combined constant acceleration.

- g: The gravitational acceleration constant = 9.81 m/s^2

- Sum the two equations for both masses A and B:

                      g* ( ma - mb ) = ( ma + mb )*a

                      a =  g* ( ma - mb ) / ( ma + mb )

                      a = 9.81* ( 6 - 4 ) / ( 6 + 4 ) = 9.81 * ( 2 / 10 )

                      a = 1.962 m/s^2  

- The rope/string moves with linear acceleration of ( a ) which rotates the disk counter-clockwise in the direction of massive object A.

- The linear acceleration always acts tangent to the disk at a distance radius ( r ).

- For no slip conditions, the linear acceleration can be equated to tangential acceleration ( at ). The correlation between linear-rotational kinematics is given below :

                     a = at = 1.962 m/s^2

                     at = r*α      

Where,

           α: The angular acceleration of the object ( disk )

                    α = at / r

                    α = 1.962 / 0.25

                    α = 7.848 rad/s^2                                

- Take moments about the pivot O of the disk. Apply rotational dynamics conditions:

             

                Sum of moments ∑M = Iα

                 ( Ta - Tb )*r = Iα

- The moment about the pivots are due to masses A and B.

 

               Ta: The force in string due to mass A

               Tb: The force in string due to mass B

                I: The moment of inertia of disk = 0.5*M*r^2

                   ( ma*a - mb*a )*r = 0.5*M*r^2*α

                   α = ( ma*a - mb*a ) / ( 0.5*M*r )

                   α = ( 6*1.962 - 4*1.962 ) / ( 0.5*5*0.25 )

                   α = ( 3.924 ) / ( 0.625 )

                   α = 6.278 rad/s^2

6 0
2 years ago
Other questions:
  • A gas metal arc welder is also known as a _____ welder.
    5·1 answer
  • Water has a density of 1.94 slug/ft^3. What is the density expressed in SI units? Express the answer to three significant figure
    8·1 answer
  • Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's
    11·2 answers
  • The 2-Mg concrete pipe has a center of mass at point G. If it is suspended from cables AB and AC, determine the diameter of cabl
    12·1 answer
  • 6.8.1: Function pass by pointer: Transforming coordinates. Define a function CoordTransform() that transforms its first two inpu
    9·1 answer
  • Define initial set and final set. Briefly discuss one method used to determine them. The following laboratory tests are performe
    12·1 answer
  • Oil enters a counterflow heat exchanger at 600 K with a mass flow rate of 10 kg/s and exits at 200 K. A separate stream of liqui
    12·1 answer
  • The rigid beam is supported by a pin at C and an A992 steel guy wire AB of length 6 ft. If the wire has a diameter of 0.2 in., d
    14·1 answer
  • An ideal Stirling cycle filled with air uses a 75 F energy reservoir as a sink. The engine is designed to so that the maximum ai
    8·1 answer
  • The typical Canadian worker is able to produce 100 board feet (a unit of measure) of lumber or 1000 light bulbs per year. The wo
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!