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Radda [10]
4 years ago
6

A hypothetical accumulator processor uses one of the following 16-bit instruction formats, depending on the instruction.(a) (10

points.) Draw a block diagram of this CPU, showing the sequencer, ALU, memory, and control and data registers, and use arrows to indicate how the various parts interact with each other. Assume that instructions and data are accessed using an MAR and MDR register. (This will be the Von Neumann architecture diagram of this processor, much like the one we created in class.)(b) (20 points.) Give a control unit design for this processor using D-flip-flops similar to the one we designed in the lectures for Vesp 1.0. The design should include fetch, decode and execute steps in as much detail as possible.(c) (20 points) Give a datapath design, also using D-flip-flops for the MDR register of this processor, clearly describing all the transfers into the MDR with the help of a typical flip-flop as we did in class by making sure that you completely specify both the data and clock inputs for that flip-flop.

Engineering
1 answer:
Jlenok [28]4 years ago
7 0

Answer:

Every part has been answered with complete detail. See the pictures.

Explanation:

See the pictures for detailed answer.

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Sphinxa [80]

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4 years ago
A copper block receives heat from two different sources: 5 kW from a source at 1500 K and 3 kW from a source at 1000 K. It loses
LUCKY_DIMON [66]

Answer:

a) Zero

b) the rate of entropy generation in the system's universe = ds/dt = 0.2603 KW/K

Explanation:

a) In steady state  

Net rate of Heat transfer = net rate of heat gain -  net rate of heat lost  

Hence, the rate of heat transfer = 0

b) In steady state, entropy generated  

ds/dt = - [ Qgain/Th1 + Qgain/Th2 - Qlost/300 K]

Substituting the given values, we get –  

ds/dt = -[5/1500 + 3/1000 – (5+3)/300]

ds/dt = - [0.0033 + 0.003 -0.2666]

ds/dt = 0.2603 KW/K

 

6 0
4 years ago
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Rasek [7]

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Explanation:

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7 0
3 years ago
A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed i
kogti [31]

Answer

given,

Speed of vehicle = 65 mi/hr

                            = 65 x 1.4667 = 95.33 ft/s

e = 0.07 ft/ft

f is the lateral friction, f = 0.11

central angle,Δ = 38°

The PI station is

PI = 250 + 50

   = 25050 ft

using super elevation formula

e + f = \dfrac{v^2}{rg}

0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}

r = \dfrac{95.33^2}{32.2\times 0.18}

  r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

L = \dfrac{\piR\Delta}{180}

L = \dfrac{\pi\times 1574\times 38}{180}

L = 1044 ft

Tangent of the curve calculation

  T = R tan(\dfrac{\Delta}{2})

  T = 1574 tan(\dfrac{38}{2})

      T = 542 ft

The station PC and PT are

 PC = PI - T

 PC = 25050 - 542

       = 24508 ft

       = 245 + 8 ft

PT = PC + L

     = 24508 + 1044

     =25552

     = 255 + 52 ft

the middle ordinate calculation

MO = R(1-cos\dfrac{\Delta}{2})

MO = 1574\times (1-cos\dfrac{38}{2})

     MO = 85.75 ft

degree of the curvature

D = \dfrac{5729.578}{R}

D = \dfrac{5729.578}{1574}

D = 3.64°

8 0
4 years ago
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