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Elis [28]
3 years ago
10

Vending machine controller (adapted from Katz, "Contemporary Logic Design") Design and implement a finite state machine that con

trols a vending machine. The vending machine sells packages of gum that cost 15 cents each. The machine accepts nickels (5 cents) and dimes (10 cents). After someone has inserted 15 cents, the FSM will send an output signal to a mechanical system that releases a package of gum. If a customer inserts 20 cents (two dimes, they will get a pack of gum but they will not receive any change back. In addition to standard reset and clock signals, your FSM will have two input signals: N and D. N is asserted when the customer has inserted a nickel and D is asserted when the customer inserts a dime. (You can assume that both can never be asserted at the same time. The system has one output Y, which indicates that the mechanical system should dispense the gum. After the system outputs Y = 1 for one clock cycle, it returns to its initial state for the next customer.
(a) Draw a Moore-style state diagram that shows the function of your system. (Hint: you can do it in only four states.)
(b) Choose a state encoding to use and indicate it clearly. Use full binary encoding, that is, make your four states have values 00, 01, 10, and 11. Here, give a table that shows which encoding value you will use for each of the states in your state machine.
(C) Draw the binary encoded state table. That is, make a table that shows how Q1, Q, and Y relate to Qı, Qo, N, and D.
(d) Write and simplify Boolean expressions for Q1, Q., and Y.. about the system's reset.

Engineering
1 answer:
babunello [35]3 years ago
6 0

Answer:

Check the explanation

Explanation:

A vending machine controller is that type of machine that comes with a single serial port on the same chip as the programmable processor. The controller comprises of a port arbitrator that selectively attaches or links one of a number of serially communicating devices to this single serial port.

Kindly check the attached image to get the step by step explanation to the above question.

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Air at atmospheric pressure and at 300K flows with a velocity of 1.5m/s over a flat plate. The transition from laminar to turbul
Savatey [412]

Answer:3.47 m

Explanation:

Given

Temperature(T)=300 K

velocity(v)=1.5 m/s

At 300 K

\mu =1.846 \times 10^{-5} Pa-s

\rho =1.77 kg/m^3

And reynold's number is given by

Re.=\frac{\rho v\time x}{\mu }

5\times 10^5=\frac{1.77\times 1.5\times x}{1.846\times 10^{-5}}

x=\frac{5\times 10^5\times 1.846\times 10^{-5}}{1.77\times 1.5}

x=3.47 m

5 0
3 years ago
The current in a 20 mH inductor is known to be: 푖푖=40푚푚푚푚푡푡≤0푖푖=푚푚1푒푒−10,000푡푡+푚푚2푒푒−40,000푡푡푚푚푡푡≥0The voltage across the induct
Anni [7]

Answer:

a) The expression for electrical current: i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

The expression for voltage: v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) For t<=0 the inductor is storing energy and for t > 0 the inductor is delivering energy.

Explanation:

The question text is corrupted. I found the complete question on the web and it goes as follow:

The current in a 20 mH inductor is known to be: i = 40 mA at t<=0 and i = A1*e^(-10,000*t) + A2*e^(-40,000*t) A at t>0. The voltage across the inductor (passive sign convention) is -68 V at t = 0.

a. Find the numerical expressions for i and v for t>0.

b. Specify the time intervals when the inductor is storing energy and is delivering energy.

A inductor stores energy in the form of a magnetic field, it behaves in a way that oposes sudden changes in the electric current that flows through it, therefore at moment just after t = 0, that for convenience we'll call t = 0+, the current should be the same as t=0, so:

i = A1*e^(-10,000*(0)) + A2*e^(-40,000*(0))

40*10^(-3) = A1*e^(-10,000*0) + A2*e^(-40,000*0)

40*10^(-3) = (A1)*1 + (A2)*1

40*10^(-3) = A1 + A2

A1 + A2 = 40*10^(-3)

Since we have two variables (A1 and A2) we need another equation to be able to solve for both. For that reason we will use the voltage expression for a inductor, that is:

V = L*di/dt

We have the voltage drop across the inductor at t=0 and we know that the current at t=0 and the following moments after that should be equal, so we can use the current equation for t > 0 to find the derivative on that point, so:

di/dt = d(A1*e^(-10,000*t) + A2*e^(-40,000*t))/dt

di/dt = [d(-10,000*t)/dt]*A1*e^(-10,000*t) + [d(-40,000*t)/dt]*A2*e^(-40,000*t)

di/dt = -10,000*A1*e^(-10,000*t) -40,000*A2*e^(-40,000*t)

By applying t = 0 to this expression we have:

di/dt (at t = 0) = -10,000*A1*e^(-10,000*0) - 40,000*A2*e^(-40,000*0)

di/dt (at t = 0) = -10,000*A1*e^0 - 40,000*A2*e^0

di/dt (at t = 0) = -10,000*A1- 40,000*A2

We can now use the voltage equation for the inductor at t=0, that is:

v = L di/dt (at t=0)

68 = [20*10^(-3)]*(-10,000*A1 - 40,000*A2)

68 = -400*A1 -800*A2

-400*A1 - 800*A2 = 68

We now have a system with two equations and two variable, therefore we can solve it for both:

A1 + A2 = 40*10^(-3)

-400*A1 - 800*A2 = 68

Using the first equation we have:

A1 = 40*10^(-3) - A2

We can apply this to the second equation to solve for A2:

-400*[40*10^(-3) - A2] - 800*A2 = 68

-1.6 + 400*A2 - 800*A2 = 68

-1.6 -400*A2 = 68

-400*A2 = 68 + 1.6

A2 = 69.6/400 = 0.174

We use this value of A2 to calculate A1:

A1 = 40*10^(-3) - 0.174 = -0.134

Applying these values on the expression we have the equations for both the current and tension on the inductor:

i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

v = [20*10^(-3)]*[-10,000*(-0.134)*e^(-10,000*t) -40,000*(0.174)*e^(-40,000*t)]

v = [20*10^(-3)]*[1340*e^(-10,000*t) - 6960*e^(-40,000*t)]

v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) The question states that the current for the inductor at t > 0 is a exponential powered by negative numbers it is expected that its current will reach 0 at t = infinity. So, from t =0 to t = infinity the inductor is delivering energy. Since at time t = 0 the inductor already has a current flow of 40 mA and a voltage, we can assume it already had energy stored, therefore for t<0 it is storing energy.

8 0
3 years ago
What are the two main what are the two main concerns in the research of fluid power efficiency?
Galina-37 [17]

Answer:

The correct option is;

Materials and Components

Explanation:

The efficiency of fluid power is influenced by the components and the materials used to deliver the power of the fluid as such fluid power control are focused on

1) Advances in fluid power

2) Making use of the advantages

3) Making use of the other externally available technological advantages

4) Giving allowance for disadvantages

Areas of interest in advances in fluid power are;

a. Computer optimized flow

b. The use of new and improved materials/coatings

c. The use of components that save energy, such as intelligent supply pressure adapting systems

3 0
2 years ago
Read 2 more answers
Relay contacts that are defined as being normally open (n.o.) have contacts that are:_____.
Reptile [31]

Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if  the relay coil is known to have de-energized.

<h3>What is meant by normally open contacts?</h3>

Normally open (NO) are  known to be open if there is no measure of current that is flowing through a given coil but it often close as soon as the coil is said to be energized.

Note that  Normally closed (NO) contacts are said to be closed only if the coil is said to be de-energized and open only if the coil is said to carry current or is known to have energized.

The role of relay contact is wide. The Relays are tools that are often used in the work of  switching of control circuits and it is one that  a person cannot used for power switching that has relatively bigger ampacity.

Therefore, Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if  the relay coil is known to have de-energized.

Learn more about Relay contacts from

brainly.com/question/15334861
#SPJ1

8 0
1 year ago
Carbon nanotubes can be a considered toxic agent due to: (a)- Its ability to produce toxic gases (b)Its ability to penetrate ski
SpyIntel [72]

Answer: b)Its ability to penetrate skin quickly due to its very small diameter

Explanation: Carbon nano tubes(CNT) are the material widely used in the medical field due to the atomic structure of it ans also have small size. Toxicity in the carbon nano tubes is because their small sized atomic particles which can enter the skin by penetration or inhalation. But are still preferred in the medicine because having unique properties like mechanical property, chemical property,surface property etc.

4 0
3 years ago
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