Exercise DET-04.04 – Hazards

Prof. Dr. Thorsten Jungmann (last updated 2026-10-25)

Relates to DET-04 Algorithmic minimization of functions with many inputs and hazard analysis

Task

Different paths through a circuit have different delays. While the inputs change, the output of a circuit whose truth table is correct can therefore show short, unwanted pulses. In each function, the first variable has the highest weight.

a) Describe the situation in which a static 1 hazard can occur, and the situation in which a static 0 hazard can occur.

b) The function $F_1(A,B,C)=\sum m(1,3,6,7)$ is built as a two-level circuit from its minimal disjunctive normal form (DNF), $F_1=\overline{A}\cdot C+A\cdot B$, with an inverter for $A$. Find the change of a single input that can cause a static 1 hazard. Which values do the other inputs have?

c) Each gate, the inverter included, has a propagation delay of 10 ns. At $t=0$, $A$ falls from 1 to 0, while the other inputs keep the values found in b). Draw the timing diagram of $A$, $\overline{A}$, $\overline{A}\cdot C$, $A\cdot B$ and $F_1$. How long does the dip last, and after what time has the output reached its final value for good?

d) Make the circuit of b) free of this hazard with an additional product term. Why is the circuit then no longer minimal?

e) The minimal DNF of $F_2(A,B,C,D)=\sum m(0,1,2,3,7,8,12,15)$ is $F_2=\overline{A}\cdot \overline{B}+B\cdot C\cdot D+A\cdot \overline{C}\cdot \overline{D}$. Find all changes of a single input that can cause a static 1 hazard in its two-level circuit, and make the circuit free of them.

f) In the circuit $F_3=A+B$, both inputs change at the same moment: $A$ from 1 to 0 and $B$ from 0 to 1. What can the output do? Can an additional term prevent it?

Solution

a) static 1 hazard: the inputs change between two combinations for which the output is 1, and the output drops to 0 for a moment; static 0 hazard: they change between two combinations for which the output is 0, and the output rises to 1 for a moment
b) a change of $A$ while $B=1$ and $C=1$, between minterm 3 and minterm 7
c) $F_1$ is 0 from 20 ns to 30 ns, so the dip lasts 10 ns; the final value holds from 30 ns on, the delay of the longest path
d) $F_1=\overline{A}\cdot C+A\cdot B+B\cdot C$; the redundant prime implicant $B\cdot C$ costs an additional AND gate
e) a change of $A$ while $B$, $C$ and $D$ are 0 (minterms 0 and 8) and a change of $B$ while $A$ is 0 and $C$ and $D$ are 1 (minterms 3 and 7); $F_2=\overline{A}\cdot \overline{B}+B\cdot C\cdot D+A\cdot \overline{C}\cdot \overline{D}+\overline{B}\cdot \overline{C}\cdot \overline{D}+\overline{A}\cdot C\cdot D$
f) if $A$ falls before $B$ rises, $F_3$ drops to 0 for a moment, a function hazard; no term can prevent it, because the function itself demands $F_3=0$ for $A=0$ and $B=0$

▶️ Full worked solution