Digital Signal Conditioning & Digital Fundamentals

Instruments 3.1

Imron Rosyadi

Learning Objectives

By the end of this session, you should be able to:

1. Explain what digital signal conditioning is and why it is needed in modern instrumentation and control.
2. Describe how analog process variables (e.g., temperature, level) are represented using digital words.
3. Convert between binary, decimal, octal, and hexadecimal representations (including fractional numbers).
4. Interpret and construct simple Boolean algebra expressions for control logic.
5. Map Boolean expressions into digital gate circuits (AND/OR and NAND/NOR).
6. Describe the basic role of PLCs and computer interfaces (buses and tri‑state buffers) in control systems.

Emphasize that this lecture is about the “digital side” of instrumentation: once we’ve measured some physical quantity as an analog signal, we often want to process, transmit, or use it in digital logic or a computer.

We are not trying to turn them into digital design experts in one lecture; we’re giving just enough digital background so that later we can talk sensibly about A/D, D/A, and computer‑based control.

Why Digital Signal Conditioning?

• Modern systems are dominated by digital electronics and computers.
• Common ECE/control examples:
  • Automotive ECUs (engine, ABS, airbags)
  • Home appliances (washing machines, microwave ovens)
  • Building automation (HVAC, security, lighting)
  • Industrial process control (chemical plants, refineries)
• All these systems sense analog variables (temperature, pressure, motion, etc.) but internally use digital data.

Use this slide to motivate the topic. Point out that almost every “smart” device they know has a microcontroller or microprocessor inside, but the world around it is analog.

Digital signal conditioning is the bridge between analog signals coming from sensors and the digital domain where computers and digital logic operate.

Differentiate: - “Signal conditioning” in a narrow sense usually means analog conditioning (amplifiers, filters, isolation). - “Digital signal conditioning” is about representing and handling the information in a digital format after A/D conversion, and designing the digital logic structures that use that data.

What Is Digital Signal Conditioning?

• In process control, digital signal conditioning means:
  • Representing analog process information in a digital format.
  • Ensuring the digital representation is compatible with:
    • Digital logic families and voltage levels
    • Microcontrollers, PLCs, and computers
    • Communication buses and networks
• Typically involves:
  • A/D conversion (analog to digital)
  • Scaling, encoding, or formatting digital data
  • Error detection/correction, filtering in the digital domain

Note

Digital representation rarely increases raw accuracy; in fact, quantization and finite word length usually reduce it slightly. But the big win is robustness to noise, drift, and long‑distance transmission.

Emphasize: we are not magically improving measurement accuracy just by going digital.

Accuracy can even get worse because: - We quantize the signal to a finite number of bits. - The A/D converter has its own errors (offset, gain, INL, DNL, etc.).

However, once the signal is digital: - Noise added on a digital line is much easier to tolerate (as long as it doesn’t cross the logic thresholds). - Digital data can be copied, stored, transmitted, and processed with very little additional error.

Why Use Digital Techniques in Control?

Beyond noise immunity, digital control offers:

1. Multivariable control
   • One computer can coordinate many variables and loops.
2. Linearization of sensor outputs
   • Nonlinear sensor curves (e.g., thermocouples) can be corrected in software.
3. Complex control algorithms
   • PID with scheduling, model predictive control, adaptive control, etc.
4. Easy modification and upgrades
   • Change control laws or tuning by updating software rather than rewiring.
5. Networking and integration
   • Multiple controllers, HMIs, and databases share data over a plant network.

Relate to a simple example: - A chemical reactor controlled by a PLC: it might monitor temperature, pressure, and level, compute flows, implement PID loops, manage interlocks, and log data.

All of this is much easier, cheaper, and more maintainable in digital form than with analog controllers and relays.

Digital Fundamentals – Big Picture

To use digital techniques in process control we need a basic toolkit:

1. Digital information and words
   • Bits, binary representation of numbers.
2. Number systems
   • Decimal, binary, octal, hexadecimal; integer and fractional forms.
3. Boolean algebra
   • Represent logic conditions as equations.
4. Digital electronics
   • Logic gates implementing Boolean operations.
5. Programmable Logic Controllers (PLCs)
   • Industrial digital controllers for logic‑heavy tasks.
6. Computer interfaces and tri‑state buffers
   • How digital devices share a bus.

Explain that this is a “review” or “crash course” in digital for those who have already taken basic digital logic or computer engineering courses.

The goal isn’t to teach them logic design from scratch, but to refresh the concepts in the specific context of process control and instrumentation.

Digital Information: Bits and Logic Levels

• Digital signals are binary (two‑state) levels.
• Physical implementations can be:
  • Two voltages (e.g., 0 V and 5 V)
  • Two currents
  • Two frequencies
  • Two phases
• We label the states:
  • Logic high: H or 1
  • Logic low: L or 0

Connect to physical reality: a wire doesn’t “contain a 1” in an abstract sense; it has some voltage, current, or waveform that we interpret as logic 0 or 1 based on thresholds.

Point out that in industrial environments, digital levels and isolation (e.g., opto‑isolators) are chosen to survive noise and harsh conditions.

From Bits to Words

• A single binary signal (one wire) can carry only 1 bit of information.
• To represent richer information (e.g., a measured temperature), we use multiple bits together as a binary word.
• Example: a 6‑bit word
  • Pattern: 101011₂
  • That’s a 6‑digit base‑2 number, with each digit a bit.
• General binary whole number:
  • For bits \(a_n a_{n-1} \dots a_1 a_0\):
  • \[N_{10} = a_n 2^n + a_{n-1} 2^{n-1} + \dots + a_1 2^1 + a_0 2^0\]

Tip

In instrumentation, the number of bits (word length) determines the resolution of your measurement.

Tie this directly to A/D converters: an N‑bit A/D can output 2^N different digital codes, mapping some analog range (e.g., 0–10 V) into discrete steps.

They will see later why 8, 12, 16, or 24 bits are common in instrumentation.

Example 1 – Binary to Decimal (Whole Number)

Problem: Find the base‑10 equivalent of the binary whole number \(\mathbf{00100111_2}\).

Step 1 – Remove leading zeros: \[ 00100111_2 = 100111_2 \]

Step 2 – Write positional expansion: \[ N_{10} = (1)2^5 + (0)2^4 + (0)2^3 + (1)2^2 + (1)2^1 + (1)2^0 \]

Step 3 – Evaluate: \[ N_{10} = 32 + 0 + 0 + 4 + 2 + 1 = 39 \]

Answer: \[ \mathbf{00100111_2 = 39_{10}} \]

Walk students through how each bit corresponds to a power of 2, exactly as decimal digits correspond to powers of 10.

Ask a quick check question: “What is the decimal value of 00010000₂?” to make sure they see that leading zeros don’t matter.

Example 2 – Decimal to Binary (Whole Number)

Problem:

Find the binary equivalent of the base‑10 number \(47\).

We use successive division by 2, tracking the remainders.

\[ \frac{47}{2} = 23 \text{ with remainder } 1 \Rightarrow a_0 = 1 \]

\[ \frac{23}{2} = 11 \text{ with remainder } 1 \Rightarrow a_1 = 1 \]

\[ \frac{11}{2} = 5 \text{ with remainder } 1 \Rightarrow a_2 = 1 \]

\[ \frac{5}{2} = 2 \text{ with remainder } 1 \Rightarrow a_3 = 1 \]

\[ \frac{2}{2} = 1 \text{ with remainder } 0 \Rightarrow a_4 = 0 \]

\[ \frac{1}{2} = 0 \text{ with remainder } 1 \Rightarrow a_5 = 1 \]

Now read remainders from last to first: \(a_5 a_4 a_3 a_2 a_1 a_0 = 101111_2\).

Answer: \[ \mathbf{47_{10} = 101111_2} \]

Tip

Algorithm:

• Divide by 2.
• Record remainder (0 or 1).
• Use the quotient as new dividend.
• Stop when quotient is 0.
• Read bits from last remainder to first.

If time permits, ask students to convert 13₁₀ as a quick in‑class exercise. Write steps live and confirm: 13₁₀ = 1101₂.

Mention that many programming languages and calculators support binary prefix (e.g., 0b101111) to represent binary literals.

Octal and Hexadecimal – Why Bother?

• Long binary strings are hard for humans to read.
• Octal (base 8) and hexadecimal (base 16) are convenient shorthand because:
  • 1 octal digit = 3 binary bits.
  • 1 hex digit = 4 binary bits.
• Octal digits: 0–7.
• Hex digits: 0–9, A, B, C, D, E, F.

Binary → Octal

• Group bits in 3s from the right.
• Example:
  • \(101011_2 = 101\ 011_2\)
  • \(101_2 = 5_8,\ 011_2 = 3_8\)
  • So \(101011_2 = 53_8\).

Binary → Hex

• Group bits in 4s from the right.
• Example:
  • \(10110110_2 = 1011\ 0110_2\)
  • \(1011_2 = B_{16},\ 0110_2 = 6_{16}\)
  • So \(10110110_2 = \text{B6H}\).

Emphasize that hex is extremely common in microcontrollers, PLCs, and communication protocols (e.g., Modbus frames, CAN IDs).

Also remind them of notation: - Subscript 8, 10, 16 for octal, decimal, hex. - “H” appended is common for hex (e.g., B6H).

Fractional Binary Numbers

So far, we’ve handled whole numbers. What about fractions?

Decimal fractional numbers use negative powers of 10: \[ 0.473_{10} = 4 \cdot 10^{-1} + 7 \cdot 10^{-2} + 3 \cdot 10^{-3} \]

Binary fractional numbers use negative powers of 2: \[ N_{10} = b_1 2^{-1} + b_2 2^{-2} + \dots + b_m 2^{-m} \]

where:

• \(b_1 b_2 \dots b_m\) is the binary fraction (bits to the right of the “binary point”).
• \(m\) is the number of fractional bits.

Note

Conceptually, it’s exactly like decimal fractions, but every step to the right halves the weight (divide by 2) instead of dividing by 10.

Connect to fixed‑point representation: many embedded controllers use fixed‑point arithmetic where a certain number of bits is reserved for the fractional part.

This is also related to the resolution of an A/D converter: the LSB often corresponds to a fraction of the full‑scale analog range.

Example 3 – Binary Fraction to Decimal

Problem: Find the base‑10 equivalent of the binary number \(\mathbf{0.11010_2}\).

We use: \[ N_{10} = b_1 2^{-1} + b_2 2^{-2} + \dots + b_m 2^{-m} \]

Here, \(m = 5\), and the bits are: \[ b_1 b_2 b_3 b_4 b_5 = 1\ 1\ 0\ 1\ 0 \]

So: \[ N_{10} = (1)2^{-1} + (1)2^{-2} + (0)2^{-3} + (1)2^{-4} + (0)2^{-5} \]

\[ N_{10} = \frac{1}{2} + \frac{1}{4} + 0 + \frac{1}{16} + 0 \]

\[ N_{10} = 0.5 + 0.25 + 0.0625 = 0.8125_{10} \]

Answer: \[ \mathbf{0.11010_2 = 0.8125_{10}} \]

Point out how the first bit after the binary point is 1/2, the next is 1/4, then 1/8, etc.

You can check quickly by converting 0.5₁₀, 0.25₁₀, 0.75₁₀ to binary as mini‑exercises.

Example 4 – Decimal Fraction to Binary / Octal / Hex

Problem: Find the binary, octal, and hex equivalents of \(0.3125_{10}\).

Step 1 – Decimal fraction to binary (successive multiplication by 2)

\[ 2(0.3125) = 0.6250 \Rightarrow b_1 = 0 \]

\[ 2(0.6250) = 1.250 \Rightarrow b_2 = 1 \]

\[ 2(0.250) = 0.500 \Rightarrow b_3 = 0 \]

\[ 2(0.500) = 1.000 \Rightarrow b_4 = 1 \]

The whole parts (0, 1, 0, 1) give bits: \[ 0.3125_{10} = 0.0101_2 \]

We can also write this as \(0.010100_2\) (trailing zeros do not change the value).

Step 2 – Binary fraction to octal

Group bits in 3s: \[ 0.010100_2 = 0.010\ 100_2 \]

\[ 010_2 = 2_8,\quad 100_2 = 4_8 \]

So: \[ 0.010100_2 = 0.24_8 \]

Step 3 – Binary fraction to hex

Group bits in 4s: \[ 0.010100_2 = 0.0101\ 0000_2 \]

\[ 0101_2 = 5_{16},\quad 0000_2 = 0_{16} \]

So: \[ 0.010100_2 = 0.50\mathrm{H} \]

Final Answer: \[ 0.3125_{10} = 0.0101_2 = 0.24_8 = 0.50\mathrm{H} \]

Tip

Algorithm for decimal fraction → binary: - Multiply fraction by 2. - Record whole part (0 or 1) as next bit. - Use new fractional part and repeat.

Note that not all decimal fractions have a finite binary representation (e.g., 0.1₁₀ is repeating in binary), just as 1/3 is repeating in decimal.

0.3125₁₀ was chosen because it is exactly representable with a finite number of bits.

Boolean Algebra – Logic with True/False

• In control systems, many decisions are yes/no or true/false:
  • Is the tank level high?
  • Is pressure above its limit?
  • Has the emergency stop button been pressed?
• Boolean algebra lets us:
  • Represent such conditions as variables (1 = true, 0 = false).
  • Combine them using logical operations:
    • AND (·), OR (+), NOT (overbar).

Note

Boolean algebra is the language for specifying logic‑based control (interlocks, alarms, safety systems).

Clarify the mapping of true/false to 1/0: - 1 usually means condition is true (e.g., level is high, switch is closed). - 0 means false (e.g., level is not high).

This is logical truth, not necessarily voltage 1 or 0, although they are related.

Example – Alarm Logic in a Mixing Tank

Consider a mixing tank with three monitored variables:

• \(A\): Level (1 = high, 0 = low)
• \(B\): Pressure (1 = high, 0 = low)
• \(C\): Temperature (1 = high, 0 = low)

We want an alarm output \(D\) (1 = alarm ON) with these conditions:

1. Low level with high pressure.
2. High level with high temperature.
3. High level with low temperature and high pressure.

Mixing tank with level, pressure, and temperature sensors

Make the problem concrete: - A: Level switch, threshold at “normal operating” level. - B: Pressure switch, threshold at safe maximum pressure. - C: Temperature switch, threshold at safe maximum temperature.

“Low level with high pressure” might indicate a blocked outlet or compressible gas, so it’s dangerous. “High level with high temperature” could be overflow plus overheating. “High level with low temperature and high pressure” could indicate some abnormal mixing or upstream fault.

Constructing the Boolean Expression

For each condition, write a Boolean term:

1. Low level with high pressure
   • Low level means \(A = 0 \Rightarrow \bar{A} = 1\).
   • High pressure means \(B = 1\).
   • Condition 1 term: \(\bar{A} \cdot B\).
2. High level with high temperature
   • High level: \(A = 1\).
   • High temperature: \(C = 1\).
   • Condition 2 term: \(A \cdot C\).
3. High level with low temperature and high pressure
   • High level: \(A = 1\).
   • Low temperature: \(C = 0 \Rightarrow \bar{C} = 1\).
   • High pressure: \(B = 1\).
   • Condition 3 term: \(A \cdot \bar{C} \cdot B\).

Now the alarm \(D\) should be true if any of these conditions is true:

\[ D = \bar{A} \cdot B + A \cdot C + A \cdot \bar{C} \cdot B \tag{2} \]

Tip

AND (·) collects requirements that must all be true. OR (+) combines alternative ways to trigger the alarm.

Walk them through the logic: - Each term is one scenario. - OR means “if this OR this OR this happens, then alarm.”

You can ask them to propose an additional scenario (e.g., both high pressure and high temperature, regardless of level) and add a term to the expression.

Digital Electronics – Logic Gates

• Boolean operations are implemented in hardware using logic gates:

  • AND gate → multiplication (·).
  • OR gate → addition (+).
  • NOT gate (inverter) → overbar.

• Gates are built from transistors and belong to families (TTL, CMOS, etc.) with specific voltage levels.

• Complex logic functions can be composed from:

  • Basic AND, OR, NOT, or
  • Universal gates: NAND and NOR.

Mention that in integrated circuits, NAND and NOR are often preferred because they are easier to implement and pack densely in silicon.

In industrial PCBs for instrumentation, students may see both discrete logic and “system on chip” microcontrollers.

Example 5 – Implementing Equation (2) with AND/OR Gates

We want to implement: \[ D = \bar{A} \cdot B + A \cdot C + A \cdot \bar{C} \cdot B \tag{2} \]

The gate‑level realization uses:

• Inverters to generate \(\bar{A}\) and \(\bar{C}\).
• AND gates for each product term.
• An OR gate to combine the term outputs.

AND/OR gate implementation of Boolean alarm logic

Note

A direct translation from Boolean equation to gates:

• Products → AND gates.
• Sums → OR gates.
• Inversions → NOT gates (inverters).

Explain the signal flow in the diagram:

• Input A passes through an inverter for \(\bar{A}\) and also directly to AND gates that need A.
• Input C passes through an inverter for \(\bar{C}\).
• Each AND gate outputs one of the terms.
• The OR gate outputs D as the logical OR of all terms.

Mention that digital logic simplification (Boolean algebra, Karnaugh maps) can reduce gate count.

Simplifying the Logic

The original expression: \[ D = \bar{A} \cdot B + A \cdot C + A \cdot \bar{C} \cdot B \]

can be simplified (using Boolean algebra) to: \[ D = A \cdot C + B \]

Interpretation:

• If pressure is high (\(B = 1\)), the alarm is ON regardless of level or temperature.
• Otherwise (\(B = 0\)), the alarm is ON if both level and temperature are high (\(A = 1, C = 1\)).

Important

Simplifying logic reduces cost, power, and failure points in physical implementations.

You do not need to fully derive the simplification on this slide, but it’s worth sketching:

• Group terms with \(B\) and factor, then apply absorption rules.

Stress that simplification is especially valuable in relay logic and PLC ladder logic, where too many contacts or rungs can be confusing and error‑prone.

Example 6 – Implementing with NAND/NOR

We can also implement the logic using NAND and NOR gates.

One direct but inefficient way:

• Build the AND/OR implementation.
• Then replace each AND and OR with NAND/NOR gates plus inverters.

NAND/NOR gate implementation of alarm logic (direct way)

A more efficient approach:

• First simplify the expression.
• Then use Boolean theorems (e.g., DeMorgan’s) to match NAND/NOR structures.

For the simplified expression: \[ D = A \cdot C + B \]

We can rewrite using DeMorgan’s theorem: \[ D = \bar{\bar {(A \cdot C)} \cdot \bar {B}} \]

This form suggests a realization using two NAND gates and an inverter.

You may not want to walk through every inversion step in class, but emphasize the design principle:

• NAND and NOR are “universal” — you can build any logic function using just one of them plus inverters.
• Real‑world digital ICs (like 7400 series) often give you multiple NAND gates in one package.

Encourage students to verify equivalence by constructing a truth table or using a logic simulator.

Programmable Logic Controllers (PLCs)

• PLCs are industrial computers designed for:
  • Digital logic control (replacement for relay panels).
  • Harsh environments (temperature, vibration, electrical noise).
• Key characteristics:
  • Rugged I/O modules for analog and digital signals.
  • Programmable using ladder logic, function block diagrams, etc.
  • Real‑time operation with deterministic scan cycles.

Note

In practice, Boolean equations and gate diagrams for process control are often implemented as PLC ladder logic, not discrete gates.

Make the historical link: PLCs evolved from relay‑based sequence control.

Mention common vendors (Allen‑Bradley, Siemens, Schneider) without going into marketing.

Stress that understanding Boolean algebra and digital representations helps interpret and design PLC programs.

Computer Interface: The Bus

• A typical computer (or microcontroller) connects to external devices through a bus:
  • Data lines: carry data values.
  • Address lines: select specific devices/locations.
  • Control lines: indicate read/write, interrupts, etc.
• All attached devices share these lines.

Generic model of a computer bus system

Warning

Sharing the bus safely requires that only one device drive a line at a time. Otherwise, you get bus contention and possibly hardware damage.

Explain that in instrumentation, external A/D and D/A converters, digital I/O modules, and memory‑mapped registers are often connected via busses like:

• Parallel data busses in older systems.
• Serial busses (SPI, I²C, CAN) in modern embedded systems (conceptually similar).

The key idea: multiple devices share physical lines, so we must control when each device is allowed to drive a line.

Tri‑State Buffers – Sharing a Bus Line

To prevent devices from fighting on a shared line, we use tri‑state buffers.

A tri‑state buffer has:

• An input (logic 0 or 1).
• An output.
• An enable control signal.

It can put the output into three states:

1. Logic 0
2. Logic 1
3. High‑impedance (effectively disconnected)

Two signals sharing a bus via tri-state buffers

• When \(E_1\) is active, \(A\) is placed on the bus; other buffers must be disabled.
• When \(E_2\) is active, \(B\) is placed on the bus.

Tip

High‑impedance state = “not driving the line” → like opening a switch.

Use an analogy: think of the bus as a single PA system microphone line. Only one person should speak at a time; if two shout conflicting words, the result is garbled. Tri‑state buffers are like giving each speaker a push‑to‑talk button.

Also connect back to digital signal conditioning: A/D converters often present their digital outputs through tri‑state buffers so that multiple converters can share the same parallel data bus to a microcontroller.

Worked Problem 1 – Encoding a Sensor Reading

Scenario: You have a temperature sensor whose output after analog conditioning is mapped to a digital word between 0 and 63 (6 bits). At a given moment, the reading is \(101011_2\).

1. What is the decimal value of this reading?
2. If 0 corresponds to 0 °C and 63 corresponds to 63 °C, what temperature does this code represent?
3. What is the hex equivalent of this binary code?

Solution

1. Convert \(101011_2\) to decimal: \[ 101011_2 = (1)2^5 + (0)2^4 + (1)2^3 + (0)2^2 + (1)2^1 + (1)2^0 = 32 + 0 + 8 + 0 + 2 + 1 = 43_{10} \]

2. Mapping 0–63 to 0–63 °C is 1:1, so \[ T = 43\ ^\circ\mathrm{C} \]

3. Convert \(101011_2\) to hex:

   • Pad to 8 bits: \(00101011_2\).
   • Group: \(0010\ 1011_2\).
   • \(0010_2 = 2_{16}\); \(1011_2 = B_{16}\).
   • So: \(101011_2 = \text{2BH}\).

Answers: - Decimal value: \(\mathbf{43_{10}}\). - Temperature: \(\mathbf{43\,^\circ\mathrm{C}}\). - Hex value: \(\mathbf{2BH}\).

Point out how natural it becomes to move between these number systems once you practice. This is exactly what microcontrollers/PLCs are doing internally when they read an A/D converter output word and interpret it as an engineering unit like °C or psi.

Worked Problem 2 – Designing a Simple Interlock

Problem: A motor should run only if:

• The start button \(S\) is pressed (1 = pressed).
• The over‑temperature switch \(T\) is NOT tripped (1 = tripped).
• The emergency stop \(E\) is NOT engaged (1 = pressed).

Define a Boolean expression for the motor run signal \(M\), and show a simple gate‑level implementation.

Solution

1. Define each condition logically:

   • Start pressed → \(S = 1\).
   • Over‑temp not tripped → \(\bar{T} = 1\).
   • E‑stop not pressed → \(\bar{E} = 1\).

2. Motor should run only when all are satisfied → AND: \[ M = S \cdot \bar{T} \cdot \bar{E} \]

3. Gate implementation:

   • Use inverters on \(T\) and \(E\) to produce \(\bar{T}\) and \(\bar{E}\).
   • Feed \(S\), \(\bar{T}\), and \(\bar{E}\) into a 3‑input AND gate.

Use this to connect: this is exactly the kind of logic that would be encoded in a PLC ladder diagram using contacts and coils.

You can ask students what happens to M if E becomes 1 (E‑stop pressed) regardless of S and T; they should quickly see that M goes to 0.

Summary / Key Points

• Digital signal conditioning: representing analog process variables in digital form suitable for digital logic, PLCs, and computers.
• Binary words encode measured values; word length (number of bits) impacts resolution.
• Number systems (binary, octal, hex) are crucial for interpreting and manipulating digital data.
  • Conversions between bases can be done systematically (division for integers, multiplication for fractions).
• Boolean algebra provides a compact way to express logical control conditions.
• Logic gates (AND, OR, NOT, NAND, NOR) are physical realizations of Boolean operations.
• PLCs implement digital logic for industrial control, often replacing relay panels.
• Computer interfaces use shared buses; tri‑state buffers ensure only one device drives a line at a time.

Reinforce that everything in modern digital control systems, from the simplest interlock to a complex multivariable controller, is built on these foundations: digital representation, Boolean logic, and structured interfaces.

Encourage students to practice basic conversions and Boolean expressions so that later lectures on A/D, D/A, and digital filtering are easier to follow.

Formulas Summary

1. Binary whole number (to decimal) \[ N_{10} = a_n 2^n + a_{n-1} 2^{n-1} + \dots + a_1 2^1 + a_0 2^0 \]

2. Binary fraction (to decimal) \[ N_{10} = b_1 2^{-1} + b_2 2^{-2} + \dots + b_m 2^{-m} \]

3. Successive division by 2 (decimal → binary integer)

   • Repeatedly divide the decimal number by 2.
   • Record remainders; read them in reverse order as bits.

4. Successive multiplication by 2 (decimal fraction → binary)

   • Repeatedly multiply the fractional part by 2.
   • The integer parts of each result form the bits in order.

5. Boolean expression for mixing‑tank alarm \[ D = \bar{A} \cdot B + A \cdot C + A \cdot \bar{C} \cdot B \] Simplified: \[ D = A \cdot C + B \]

6. Example interlock logic \[ M = S \cdot \bar{T} \cdot \bar{E} \]

Encourage students to keep these formulas handy. They will come up again in later topics: A/D converter specs, digital filtering, PLC logic programming, and interfacing microcontrollers to sensors and actuators.

Interactive Deck

Warmup: Binary to Decimal

Use this code cell to convert a binary string to decimal. Change the binary_str variable and click Run Code.

Ask students to try binary strings from the text:

• 00100111 → they should see 39.
• 101011 → they should see 43.

Encourage them to observe the effect of adding/removing leading zeros.

Decimal to Binary

Now convert a non‑negative integer to binary using the successive division by 2 method. Modify n and re‑run.

Relate this directly to the worked example where 47₁₀ → 101111₂.

Have students test several values and check their answers against a calculator or mental math.

Explore Binary Fractions – Manual Conversion

Use the binary fraction formula from the main deck: \[ N_{10} = b_1 2^{-1} + b_2 2^{-2} + \dots + b_m 2^{-m} \]

Edit binary_fraction (just bits after the point) and run the cell.

Students should confirm:

• binary_fraction = "1" → 0.5
• "01" → 0.25
• "11010" → 0.8125 (from the text).

Point out how adding more bits refines the value in smaller steps of 1/2, 1/4, 1/8, etc.

Decimal Fraction to Binary – Successive Multiplication

This implements the “repeated multiply by 2” algorithm from the text. Change x and max_bits to see more/less precision.

Have students set x = 0.3125 and max_bits = 4 to match the example 0.0101₂.

Ask them to try x = 0.1 and observe that the bits keep changing (no finite representation), similar to 1/3 in decimal.

Converting Between Binary, Octal, and Hex

This code groups bits in 3s (octal) and 4s (hex) to convert a binary string.

Check against the examples:

• binary_str = "101011" → expect 53₈.
• binary_str = "10110110" → expect B6H.

Encourage students to mentally group bits to see the octal/hex digits.

Reactive Demo: Word Length and Resolution (Sliders + Plotly)

Use the sliders to change the number of bits and the full‑scale range of an A/D converter. We’ll compute and plot the quantization staircase.

viewof n_bits = Inputs.range([2, 12], {step: 1, value: 6, label: "Number of bits (N)"})
viewof full_scale = Inputs.range([1, 20], {step: 1, value: 10, label: "Full-scale range (V)"})

Encourage students to:

• Increase n_bits and observe the staircase becoming smoother (higher resolution).
• Change full_scale and notice how the voltage per step (LSB) changes.

Connect this to the idea that more bits in a digital word allow finer representation of an analog variable.

Reactive Demo: Binary Fraction Explorer

Use the slider to select how many fractional bits to keep. We’ll display the approximate value of a fixed target number (e.g., 0.7) as a binary fraction.

viewof target = Inputs.range([0, 1], {step: 0.05, value: 0.7, label: "Target decimal value"})
viewof bits_frac = Inputs.range([1, 12], {step: 1, value: 4, label: "Number of fractional bits (m)"})

This shows how fixed‑point representation introduces quantization error for fractional values.

Ask students to:

• Fix target at 0.7 and increase bits_frac.
• Observe how the error magnitude decreases as more bits are used.

Reactive Boolean Logic: Truth Table Generator

Control three Boolean inputs with buttons, and let Python compute the alarm condition from the mixing tank example: \[ D = \bar{A} \cdot B + A \cdot C + A \cdot \bar{C} \cdot B \]

viewof A = Inputs.toggle({label: "A (Level high?)", value: false})
viewof B = Inputs.toggle({label: "B (Pressure high?)", value: false})
viewof C = Inputs.toggle({label: "C (Temperature high?)", value: false})

Show students that toggling A, B, C visually matches the conditions in the problem statement:

• Low level ((A = 0)), high pressure ((B = 1)) → D = 1.
• High level ((A = 1)), high temp ((C = 1)) → D = 1.
• High level, low temp, high pressure → (A=1, C=0, B=1) → D = 1.

The full truth table helps them verify the logic.

Interactive Exercise: Design Your Own Logic

Now define your own Boolean logic for a motor enable signal (M).

Conditions (suggestion):

• (S): Start command (1 = requested).
• (P): Pressure OK (1 = within safe range).
• (T): Temperature OK (1 = within safe range).

You can propose your own rule, e.g.: \[ M = S \cdot P \cdot T \]

Use the toggles and modify the expression inside the code.

viewof S = Inputs.toggle({label: "S (Start command)", value: false})
viewof P = Inputs.toggle({label: "P (Pressure OK)", value: true})
viewof T = Inputs.toggle({label: "T (Temperature OK)", value: true})

Invite students to change the logic expression, e.g.:

• Require either pressure OR temperature OK: M_int = S_int * (P_int or T_int) (converted appropriately to integers).
• Add a “maintenance override” variable if they wish (extend the code).

This lets them explore Boolean algebra in a concrete context.