Consider a digital display system (DDS) shown in the figure that displays the…
2022
Consider a digital display system (DDS) shown in the figure that displays the contents of register X. A 16-bit code word is used to load a word in X, either from S or from R. S is a 1024-word memory segment and R is a 32-word register file. Based on the value of mode bit M, T selects an input word to load in X. P and Q interface with the corresponding bits in the code word to choose the addressed word. Which one of the following represents the functionality of P, Q, and T?
Answer: C. P is 10:210 decoder; Q is 5:25 decoder; T is 2:1 multiplexer — Key idea: addresses must be decoded to select one word from S or R, and the mode bit chooses which source drives X. P: S is a 1024-word memory so its address…
- A.
P is 10:1 multiplexer; Q is 5:1 multiplexer; T is 2:1 multiplexer
- B.
P is 10:210 decoder; Q is 5:25 decoder; T is 2:1 encoder
- C.
P is 10:210 decoder; Q is 5:25 decoder; T is 2:1 multiplexer
- D.
P is 1:10 de-multiplexer; Q is 1:5 de-multiplexer; T is 2:1 multiplexer
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Correct answer: C
Key idea: addresses must be decoded to select one word from S or R, and the mode bit chooses which source drives X.
P: S is a 1024-word memory so its address field is 10 bits. P must convert those 10 address bits into one of 1024 select lines — a 10-to-1024 decoder.
Q: R is a 32-word register file so its address field is 5 bits. Q must convert those 5 bits into one of 32 select lines — a 5-to-32 decoder.
T: The mode bit M selects whether the word loaded into X comes from S or from R. T therefore is a 2:1 multiplexer that selects the output of S or the output of R based on M.
Conclusion: P should be a 10-to-1024 decoder, Q should be a 5-to-32 decoder, and T should be a 2:1 multiplexer.
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