4.3 Logic Diagrams, Boolean Algebra, and Control Logic

4.3 Logic Diagrams, Boolean Algebra, and Control Logic

The NCEES Instrumentation and Controls specification lists exactly three sub-topics: sensors, data acquisition, and logic diagrams. The first two are covered in the preceding two sections. This one is the most frequently skipped and the cheapest to learn — logic items are pure reasoning with no unit conversions and no handbook lookups, so they are among the fastest points available anywhere on the exam.

Gate Definitions and Truth Tables

A logic diagram represents a binary decision network. Each gate maps binary inputs to one binary output.

Two-input truth table for the core gates:

A network with $n$ inputs requires $2^n$ truth-table rows: 3 inputs give 8 rows, 4 inputs give 16.

Boolean Algebra Identities

De Morgan's theorems are the ones the exam tests. Read them in words: "not (both)" is the same as "either not", and "not (either)" is the same as "neither". Every NAND/NOR simplification question is a De Morgan question in disguise.

Worked Example: Simplify a Logic Expression

Simplify $Y = \overline{(\bar{A} + B)} + AB$.

Apply De Morgan to the first term: $\overline{(\bar{A} + B)} = A \cdot \bar{B}$.

$Y = A\bar{B} + AB = A(\bar{B} + B) = A \cdot 1 = \boxed{A}$

A five-gate network collapses to a wire. Verify with the truth table: when $A=0$, $\bar{A}+B$ is 1 so its complement is 0, and $AB = 0$, giving $Y=0$. When $A=1$ and $B=0$: complement of $(0+0)$ is 1, so $Y=1$. When $A=1$ and $B=1$: complement of $(0+1)$ is 0, but $AB=1$, so $Y=1$. Output tracks $A$ exactly. ✓

From Truth Table to Expression: Sum of Products

To read a Boolean expression off a truth table, take every row where the output is 1, write the AND term (minterm) that makes that row true, and OR the minterms together.

Consider a pump that should run when the tank level is low ($L$), the manual switch is on ($S$), and no fault is present ($F$):

Only one row produces a 1, so the sum-of-products expression is a single minterm:

$\text{Run} = L \cdot S \cdot \bar{F}$

That is a three-input AND with one inverted input — the canonical structure of a permissive chain.

Interlocks, Permissives, and Industrial Control Logic

Process control distinguishes two functions that map onto the two basic gates:

The safety logic asymmetry. Permissives are ANDed and trips are ORed, and both choices push the system toward the safe state. This is the same structure as the AND/OR gates in Fault Tree Analysis covered in the Safety, Health, and Environment chapter: an AND gate needs every input event to occur, so it models redundancy and yields a low combined probability; an OR gate needs only one, so it models single-point failure and yields a high combined probability.

Fail-Safe vs. Fail-Danger and Normally Closed Contacts

Safety-critical inputs are wired through normally closed contacts, so that a broken wire, a lost signal, or a power failure looks identical to a genuine trip and shuts the process down. A normally open contact would fail silently — a severed wire and a healthy safe condition are indistinguishable.

$\text{Enable} = \overline{\text{E-Stop}} \cdot \overline{\text{HighPressure}} \cdot \overline{\text{LowLube}} \cdot \text{StartCmd}$

Reading Ladder Logic

Programmable logic controllers use ladder diagrams, in which each horizontal rung is one Boolean expression:

• Contacts in series on a rung = AND
• Contacts in parallel (branches) = OR
• A slashed contact = NOT (normally closed)
• The rightmost coil = the output

So a rung with three series contacts and one parallel branch around the middle contact reads $Y = A(B + C)D$. Series is AND, parallel is OR — that single mapping answers most ladder-logic items.

Worked Example: Two-out-of-Three Voting

A reactor trips on high temperature if at least two of three independent sensors read high. Write the logic and evaluate the trip probability if each sensor has a 2% chance of a spurious high reading.

Two-out-of-three voting ORs the three pairwise ANDs:

$\text{Trip} = AB + BC + AC$

For a spurious trip, at least two must falsely read high. With $p = 0.02$ each and independence, the probability that exactly two do is $3p^2(1-p) = 3(0.0004)(0.98) = 0.001176$, and that all three do is $p^3 = 0.000008$:

$P(\text{spurious trip}) = 0.001176 + 0.000008 = 0.00118 \approx 0.12\%$

Compare to a single sensor tripping the reactor directly, at $2%$. Two-out-of-three voting cuts nuisance trips by roughly a factor of 17 while still tripping if any two real sensors agree — which is why it is the standard architecture for high-consequence shutdowns.