Difficulty: Medium
Correct Answer: demultiplexer, D3, HIGH, LOW
Explanation:
Introduction / Context:
Without the original figure, we apply the Recovery-First Policy and infer a standard scenario: a single data input is routed to one of many outputs via select lines, and the waveform timing indicates that output line D3 is selected at the marked instant X. Your task is to identify the device type and predict the logic level at the selected output and at the nonselected outputs.
Given Data / Assumptions:
Concept / Approach:
A demultiplexer (DEMUX) performs controlled distribution: it takes a single input and forwards it to one chosen output. If the input is HIGH and D3 is selected, the D3 output becomes HIGH, while all other outputs stay LOW. A decoder, by contrast, asserts one of many outputs based purely on the numeric value of the binary input code rather than routing a separate data input, and a multiplexer does the inverse (many inputs to one output).
Step-by-Step Solution:
Verification / Alternative check:
Cross-check with a truth table of a 1-to-8 DEMUX: when S = 011 (binary 3), Y3 follows Din and Yk≠3 are forced LOW (assuming active-high noninverted outputs and enable asserted).
Why Other Options Are Wrong:
Multiplexer choice reverses data flow (many-to-one). “LOW, HIGH” polarity contradicts selected-output mirroring. “Decoder, D12” is inconsistent with the labeling and routing role. “Encoder” converts one-of-N to binary, not relevant here.
Common Pitfalls:
Forgetting output polarity (some families use active-low outputs); neglecting the enable pin, which would force all outputs inactive if deasserted.
Final Answer:
demultiplexer, D3, HIGH, LOW
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