Tricky Problem: Can Unequal Path Delays Glitch the Output Even When the Steady-State Result Is Unchanged?

Distinguish a circuit's steady-state Boolean correctness from its dynamic (timing) behavior, recognizing that mismatched gate delays along logically redundant paths can cause a real, physical glitch even when the algebra says nothing changed.

Two logically equivalent paths have different propagation delays. Question: can the output briefly change even though the steady-state Boolean result is unchanged? The expected concept: yes — this is a timing/hazard issue rather than a Boolean-equation issue.

Why This Problem Is Tricky

It is natural to assume that if a Boolean equation proves an output stays at the same logical value across some input transition, the physical circuit's output signal must also stay constant throughout that transition. That assumption ignores that AND, OR, and NOT gates are real physical devices with nonzero, and generally UNEQUAL, propagation delays — the algebra describes only the circuit's value after everything has settled, not what happens during the brief window while signals are still propagating through gates at different speeds.

Step-by-Step Worked Solution

Consider a concrete case of this general scenario: F = AC + A'B, with B=C=1 held constant, and A transitioning from 1 to 0. Check the Boolean value before and after: before (A=1): F = 1x1 + 0x1 = 1 + 0 = 1. After (A=0): F = 0x1 + 1x1 = 0 + 1 = 1. The steady-state value of F is 1 both before and after the transition — algebraically, nothing should change.

Now consider the physical circuit. Before the transition, F is being held at 1 by the AC term (since A=1). After the transition, F must be held at 1 by the A'B term instead (since A=0 makes AC=0). But A' is produced by an inverter acting on A, so A' does not become 1 the instant A becomes 0 — it becomes 1 only after the inverter's propagation delay has elapsed. During the gap between 'A falls, so AC starts dropping toward 0' and 'A' rises, so A'B starts rising toward 1,' BOTH product terms can be near 0 simultaneously.

Since F = AC + A'B is produced by an OR gate combining these two terms, if both terms are momentarily near 0 at the same time, the OR gate's output can dip to 0 for a brief instant before recovering to its correct steady-state value of 1. This transient dip is called a static-1 hazard (static because the steady-state value doesn't actually change; '1' because the value it briefly glitches away FROM is 1). So yes — the physical output CAN glitch, purely because of the delay mismatch between the falling AC path and the (inverter-delayed) rising A'B path, even though the Boolean equation guarantees F=1 on both sides of the transition.

Why the Obvious Wrong Answer Is Wrong

The wrong answer is 'No, F cannot change, because the Boolean equation proves F=1 both before and after the transition.' This conflates two different things: the STEADY-STATE (fully settled, after all gate delays have resolved) value of F, which the Boolean equation correctly describes, and the TRANSIENT (mid-switching) behavior of the real circuit, which depends on gate propagation delays that the Boolean equation does not model at all. A Boolean equation is a description of a circuit's behavior only once every signal has finished propagating; it makes no claims about what happens in between.

Code Examples

The example runs both checks side by side: the Boolean steady-state check (which shows no algebraic change) and the physical timing check (which shows why a glitch is still possible), making explicit that these are two different questions with two different answers. Expected result: Static-1 hazard confirmed possible due to unequal AC/A'B path delays.

Yes -- a static-1 hazard can cause a momentary glitch despite F=1 before and after.

Steady-state check:
  Before (A=1): F = A*C + A'*B = 1*1 + 0*1 = 1 + 0 = 1
  After  (A=0): F = A*C + A'*B = 0*1 + 1*1 = 0 + 1 = 1
  F = 1 both before and after -- no net Boolean change.

Physical timing check:
  A falls immediately.       -> AC term starts dropping toward 0 right away.
  A' rises only after the inverter's propagation delay.
                              -> A'B term starts rising toward 1 LATE.
  During the gap between these two events, BOTH terms can be near 0
  at the same time -> the OR gate output (F) can glitch down to 0
  before settling back to its correct steady-state value of 1.

Conclusion: YES, a momentary glitch (static-1 hazard) can occur.

Static-1 Hazard Timing Diagram

A          ----+
               +---------------

A' (delayed by inverter)    +-----
        --------------------+

AC term    ----+
               +---------------------

A'B term                  +-----------
        ----------------- +

F = AC+A'B ----+       +----------
               +-------+
               ^ glitch: both AC and A'B briefly near 0,
                 so the OR gate output dips even though
                 the settled value of F is 1 on both sides

Why Hazards Matter in Real Circuits

Static hazards are a serious concern in asynchronous circuit design and in any situation where a combinational output feeds directly into a clock, reset, or other asynchronous control input — a momentary glitch on such a signal can be misinterpreted by downstream logic as a real, intentional transition. This is exactly why designers add a redundant 'consensus' term to a hazard-prone expression (turning F=AC+A'B into the hazard-free F=AC+A'B+BC) and why raw, unprotected combinational outputs are never used to directly clock a flip-flop in a well-designed system.

No, F cannot change, because Boolean algebra proves F=1 before and after the transition

Yes, F can briefly glitch to 0 — this is a static-1 hazard caused by the falling AC path and the inverter-delayed rising A'B path having different propagation delays, even though the final settled value of F is correctly 1

Boolean algebra only describes a circuit's fully-settled (steady-state) output; it says nothing about the transient period while signals are still propagating through gates of different speeds. A momentary glitch during that transient window is a real physical event that the algebra alone cannot rule out.

Two logically equivalent paths in a circuit have different propagation delays. Can the output briefly change even though the steady-state Boolean result is unchanged? A. No, Boolean algebra guarantees the output never changes if the steady-state value is unchanged B. Yes, this is a possible static hazard caused by mismatched gate propagation delays, distinct from the Boolean equation itself C. Only if the circuit uses sequential (clocked) elements D. No, this can only happen if the Boolean simplification was done incorrectly

B. Yes, this is a possible static hazard caused by mismatched gate propagation delays, distinct from the Boolean equation itself — Boolean algebra describes only the fully-settled output value; it does not model gate propagation delay. When two paths that together keep the steady-state output unchanged have different delays, the output can briefly glitch during the transition before settling back to its correct value — this is a static hazard, a timing phenomenon, not a Boolean-equation error.

What is the standard fix for a static-1 hazard in an SOP expression like F=AC+A'B, caused by two product terms that together cover an input transition without any overlapping term? A. Increase the circuit's clock frequency B. Add a redundant consensus product term that covers the shared boundary between the two groups (e.g., adding BC) C. Remove one of the two AND gates entirely D. Add an extra inverter to the final output

B. Add a redundant consensus product term that covers the shared boundary between the two groups (e.g., adding BC) — The standard fix is to add the algebraically redundant consensus term (here, BC) that stays asserted throughout the transition, bridging the gap between when one original term drops and the other rises. This consensus term does not change the function's steady-state truth table — it is redundant in that sense — but it removes the physical timing gap that caused the glitch.

For F = AB + A'C, does a static hazard exist when B=C=1 and A transitions? If so, what redundant term eliminates it?

Before (A=1): F=1x1+0x1=1. After (A=0): F=0x1+1x1=1. F=1 both before and after, so this is a static-1 hazard candidate on the A transition — the falling AB path and the inverter-delayed rising A'C path can momentarily both be near 0. The consensus term BC eliminates the hazard: the hazard-free form is F = AB + A'C + BC.

Check the steady-state value of F before and after the A transition, then apply the same consensus-term technique used above.