Hazards (static and dynamic) in combinational logic: Exercises

Before we start, four plain-word definitions we will lean on the whole page:

Definition The vocabulary of this page

Output name. Throughout we call the circuit's single output $F$ (or $Y$ when we are talking only about timing through wires, not a specific Boolean formula — they mean the same thing: "the one signal coming out of the circuit").
A prime implicant is the biggest possible rectangle of $1$ s you can legally draw on a Karnaugh map (a "K-map" — the grid of every input combination). Each rectangle becomes one AND-term.
The consensus of two products ( $X Y$ and $\overset{ˉ}{X} Z$ ) is a third AND-term, $Y Z$ : you drop the variable ( $X$ ) that appears true in one and complemented in the other, and AND the leftovers. It is automatically $1$ whenever both originals could hand off to each other. See Consensus Theorem and Redundancy.
The consensus sum factor is the exact dual for OR-terms: the consensus of two sums $(X + Y)$ and $(\overset{ˉ}{X} + Z)$ is the sum $(Y + Z)$ — again drop the split variable $X$ , but this time OR the leftovers. It is automatically $0$ whenever both originals could hand off. This is what fixes glitches in Product-of-Sums form (see Sum of Products and Product of Sums Forms).

Level 1 — Recognition

Recall Solution

Before $= 1$ , after $= 1$ , so $before = after$ → expected transitions $= 0$ (rule above). Actual $= 2$ . Holding-1-but-dipping is a static-1 hazard. Surplus $= 2$ (even ✔), so the output ends correctly at $1$ .

Recall Solution

Before $= 0$ , after $= 1$ , so $before \neq = after$ → expected transitions $= 1$ . Actual $= 3$ (odd, $\geq 3$ ). That is a dynamic hazard (matches the up-front definition). The two extra transitions form one spike (surplus $= 2$ , even), so the final value $1$ is still the correct "changed once" result.

Recall Solution

All three have $before = after = 0$ → expected $= 0$ . Only (b) has actual $= 2 > 0$ : a static-0 (held $0$ but spiked to $1$ ). (a) and (c) have actual $= 0$ — no glitch. A slow-but-clean edge is propagation delay, not a hazard — see Propagation Delay and Gate Timing.

Level 2 — Application

Recall Solution

$B = 1$ : $F = (1) (1) + (0) (1) = 1$ .
$B = 0$ : $F = (1) (0) + (1) (1) = 1$ . Same value $\Rightarrow$ static-1 candidate. In the bad window $A B$ has already fallen to $0$ but $\overset{ˉ}{B} C$ has not yet risen (the inverter that makes $\overset{ˉ}{B}$ is one gate-delay late), so both product terms are $0$ and $F$ momentarily $= 0$ . Glitch: $1 \to 0 \to 1$ .

Recall Solution

Consensus of $A B$ and $\overset{ˉ}{B} C$ (drop the variable $B$ ): $A C$ . $F = A B + \overset{ˉ}{B} C + A C .$ During L2.1, $A = C = 1$ , so $A C = (1) (1) = 1$ regardless of $B$ . The output is pinned at $1$ the whole time — no handoff, no dip.

Recall Solution

$B = 0$ : $G = (0 + 0) (1 + 0) = 0$ . $B = 1$ : $G = (0 + 1) (0 + 0) = 0$ . Holds $0$ $\Rightarrow$ static-0.
Mid-transition both factors can briefly be $1$ (the delayed $\overset{ˉ}{B}$ ), so $G$ spikes to $1$ .
Consensus sum factor (from the up-front definition): the two factors are $(A + B)$ and $(\overset{ˉ}{B} + C)$ ; the split variable is $B$ ; drop it and OR the leftovers $\Rightarrow (A + C)$ . Fixed: $G = (A + B) (\overset{ˉ}{B} + C) (A + C) .$ When $A = C = 0$ , $(A + C) = 0$ pins $G$ at $0$ for any $B$ . See Sum of Products and Product of Sums Forms.

Level 3 — Analysis

The figure below is the answer diagram: read the K-map (rows are $A$ , columns are the pair $B C$ in Gray order $00, 01, 11, 10$ ). The purple loop is the prime implicant $A B$ ; the coral loop is $\overset{ˉ}{B} C$ ; the dashed green loop is the consensus $A C$ that bridges them. The arrow marks the two adjacent $1$ -cells that live in different loops — the handoff.

Figure — Hazards (static and dynamic) in combinational logic

Figure s01 — K-map of $F = A B + \overset{ˉ}{B} C$ . Observe: the two $1$ -cells in the top row with $C = 1$ (at $B C = 11$ and $B C = 01$ ) are neighbours but are boxed by two different coloured loops; the dashed green $A C$ loop covers both, so adding $A C$ removes the handoff.

Recall Solution

The minterms of $F$ are where $F = 1$ : $A B C$ combos ${A = 1, B = 1, C = 0}, {A = 1, B = 1, C = 1}, {A = 0, B = 0, C = 1}, {A = 1, B = 0, C = 1}$ . The hazard pair is ${A = 1, B = 1, C = 1}$ (covered by $A B$ , the purple loop) and ${A = 1, B = 0, C = 1}$ (covered by $\overset{ˉ}{B} C$ , the coral loop). They are adjacent (only $B$ differs) yet sit in different prime implicants with no single term covering both — the handoff point marked by the arrow. The dashed-green consensus loop $A C$ covers both cells and removes the gap.

Recall Solution

Path of $A B$ falling: $B \to AND$ = $1$ ns. Path of $\overset{ˉ}{B} C$ rising: $B \to inverter (2) \to AND (1) = 3$ ns. The OR sees $A B$ fall at $1$ ns but $\overset{ˉ}{B} C$ rise at $3$ ns. Glitch window $\approx 3 - 1 = 2$ ns — exactly the extra delay the inverter injects on one path.

The waveform below is the answer: time runs left to right, the purple staircase is $Y$ , and the three dashed vertical lines mark the moments each path's signal arrives and toggles the output.

Figure s02 — Dynamic hazard timing. Observe: $Y$ starts at $0$ ; at $1$ ns it flips to $1$ , at $2$ ns back to $0$ , at $3$ ns up to $1$ . Three flips = three transitions on a "change once" output.

Recall Solution

Start $Y = 0$ . Arrival at $1$ ns flips $\to 1$ ; at $2$ ns flips $\to 0$ ; at $3$ ns flips $\to 1$ . Sequence $0 \to 1 \to 0 \to 1$ : 3 transitions (odd, $\geq 3$ ) $\Rightarrow$ dynamic hazard. Expected was $1$ transition (before $= 0$ , after $= 1$ ); the two extras came as a pair, so the final value is still the correct "changed once" result.

Level 4 — Synthesis

The K-map below is the working: purple loop $\overset{ˉ}{A} C$ (minterms $1, 3$ ), coral loop $A B$ (minterms $6, 7$ ), and the dashed-green consensus $B C$ that bridges the adjacent handoff cells $3$ and $7$ .

Figure s03 — Hazard-free cover for minterms {1,3,6,7}. Observe: minterms $3$ and $7$ (both have $B = 1, C = 1$ ) are neighbours but boxed by different loops; the dashed-green $B C$ loop covers both, closing the handoff.

Recall Solution

Group on the K-map. Prime implicants: $\overset{ˉ}{A} C$ (covers minterms $1, 3$ ) and $A B$ (covers $6, 7$ ). Base cover $F = \overset{ˉ}{A} C + A B$ . Now check adjacencies between the two groups: minterm $3 (0, 1, 1)$ and minterm $7 (1, 1, 1)$ are adjacent (only $A$ differs) but in different implicants — a hazard handoff. Add their consensus: consensus of $\overset{ˉ}{A} C$ and $A B$ (drop the split variable $A$ , AND the leftovers $C$ and $B$ ) $= B C$ . $F_{safe} = \overset{ˉ}{A} C + A B + B C .$ The $B C$ loop covers both $3$ and $7$ , closing the gap. Every pair of adjacent 1s now shares a covering term.

Recall Solution

Use the up-front consensus sum factor rule on the two sums $(A + C)$ and $(\overset{ˉ}{A} + C)$ :

Find the split variable. $A$ appears true in the first factor and complemented ( $\overset{ˉ}{A}$ ) in the second → the split variable is $A$ .
Drop it, OR the leftovers. Remove $A$ and $\overset{ˉ}{A}$ ; the leftover of the first factor is $C$ , the leftover of the second is $C$ . OR them: $C + C$ .
Simplify. $C + C = C$ (idempotence). So the consensus sum factor is just $(C)$ . $G_{safe} = (A + C) (\overset{ˉ}{A} + C) (C) .$ Why it cures the glitch: whenever the two original factors could hand off (both briefly $1$ during an $A$ transition), the new factor $(C)$ is $0$ exactly there (it is $0$ whenever $C = 0$ , which is the only time both original factors evaluate the way that risks a spike), pinning $G = 0$ . This is the exact dual of the SOP fix: sums-and-zeros instead of products-and-ones.

Recall Solution

Take $F = A$ (output equals a single input). One prime implicant, one path, no reconvergence — nothing to hand off, so no consensus term is ever needed. More generally: if every pair of adjacent 1-cells lies inside a single prime implicant of your chosen cover, the two-level circuit is already hazard-free. The extra term is needed only when a transition crosses between two different terms.

Level 5 — Mastery

Recall Solution

Static hazards need one input reaching the output through a direct path and an inverted (delayed) path. Only $B$ appears both true ( $A B$ ) and complemented ( $\overset{ˉ}{B} C$ ). So only a $B$ transition (with $A = C = 1$ , holding $F = 1$ ) can cause the static-1 glitch. Changing $A$ or $C$ alone routes through a single polarity — no reconvergent disagreement, no glitch.

Recall Solution

Minimal: $F = A B + \overset{ˉ}{B} C$ → 2 AND, 1 OR, 1 inverter. Hazard-free: $F = A B + \overset{ˉ}{B} C + A C$ → 3 AND, 1 OR (now 3-input, to accept the extra $A C$ term), 1 inverter. Cost of glitch-freedom: one extra AND gate (for $A C$ ) plus widening the OR to 3 inputs. Cheap insurance against a transient failure.

Recall Solution

A flip-flop samples one instant; if you sample after the circuit settles (respecting setup/hold, see Metastability and Setup-Hold Time), the glitch is invisible — but only masked, not removed. It still causes real failures when the glitchy signal drives an asynchronous input: a clock, a latch enable, an edge-triggered reset, or a combinational feedback loop — see Synchronous vs Asynchronous Circuits. There, the spike can toggle state at the wrong time.

Recall Solution

A dynamic hazard comes from three or more reconvergent unequal-delay paths, so the output toggles multiple times as arrivals stagger. A single consensus term patches one handoff between two terms; it cannot guarantee the monotonic (single-step) switching that $\geq 3$ paths violate. Cure: flatten to a two-level network with all consensus terms (so every adjacency is covered and there are no internal multi-path reconvergences), or otherwise ensure the multi-level structure is provably hazard-free.