Identify the pictured device and outcome – given a typical digital routing symbol with select lines and labeled outputs D0…D7, determine whether the device is a demultiplexer and, for the shown timing of selects and data, which output (e.g., D3) is driven HIGH at node X while all others remain LOW.

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:

  • The symbol shows one data input, multiple labeled outputs (D0…DN−1), and select inputs S that choose which output is active.
  • At time X, the select code corresponds to line D3.
  • Data input is asserted (logic HIGH) at the same instant.
  • Nonselected outputs remain inactive (logic LOW for active-high style).


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:

Recognize topology: 1 input → N outputs plus select signals → DEMUX.Read select code at time X: S selects line index 3 → target output D3.Propagate data: with Din = HIGH, the selected output mirrors Din.Conclude: D3 = HIGH; all other outputs = LOW at X.


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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