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.
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Answer:
The answer to the question is
The heat transferred in the process is -274.645 kJ
Explanation:
To solve the question, we list out the variables thus
R-134a = Tetrafluoroethane
Intitial Temperaturte t₁ = 100 °C
Initial pressure = 3.5 bar = 350 kPa
For closed system we have m₁ = m₂ = m
ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂
For constant pressure process we have
Work done = W = = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)
From the tables we have
State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg
State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg
Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)
= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ
Answer:
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Explanation:
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