Discussion
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?
(Figure: the code word has three fields, M, S-address and R-address. The S-address field feeds P, whose output goes to S; the R-address field feeds Q, whose output goes to R. The outputs of S and R are the two inputs of T, and M is connected to T. The output of T goes to X, which feeds the DDS.)
Answer: C
P is a 10:\(2^{10}\) decoder, Q is a 5:\(2^{5}\) decoder and T is a 2:1 multiplexer.
S holds 1024 = \(2^{10}\) words, so its address field has 10 bits, and P must turn those 10 bits into one select line out of 1024: that is a 10-to-\(2^{10}\) decoder. R holds 32 = \(2^{5}\) words, so Q turns 5 address bits into one of 32 select lines, a 5-to-\(2^{5}\) decoder. T has two data inputs, the word from S and the word from R, and the mode bit M chooses which one is passed to X; choosing one of two inputs under a select signal is a 2:1 multiplexer.
A 10:1 multiplexer picks one of ten inputs, which does not address 1024 words. An encoder converts one active line into a code, which is not what T does. A 1:10 de-multiplexer routes one input to ten outputs, again not an address decoder for 1024 locations.
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