Question:
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?
\includegraphics[width=0.5\linewidth]{image40.png}
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?
\includegraphics[width=0.5\linewidth]{image40.png}
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Decoders are used to select specific words or registers based on the address. Multiplexers are used for selecting between two options, typically controlled by a selection bit.
Updated On: Jan 11, 2026
- P is 10:1 multiplexer; Q is 5:1 multiplexer; T is 2:1 multiplexer
- P is \( 10^{10} \) decoder; Q is \( 5 \times 25^5 \) decoder; T is 2:1 encoder
- P is \( 10^{10} \) decoder; Q is \( 5 \times 25^5 \) decoder; T is 2:1 multiplexer
- P is 1:10 de-multiplexer; Q is 1:5 de-multiplexer; T is 2:1 multiplexer
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The Correct Option is C
Solution and Explanation
From the diagram and the explanation in the problem, we have:
- P: The S-address requires a decoder to address one of the 1024 words, which can be done using a \( 10^{10} \) decoder, since \( 2^{10} = 1024 \).
- Q: The R-address is from a 32-word register file. A decoder is used to choose one of the 32 registers, so Q is a \( 5 \times 25^5 \) decoder since \( 2^5 = 32 \).
- T: The T block functions as a 2:1 multiplexer, based on the value of the mode bit \( M \), selecting either from the register or memory.
Thus, the correct answer is (C).
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