Analog Signal Conditioning

Instrumentation 2.1

Imron Rosyadi

Learning Objectives

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

• Explain the role of analog signal conditioning in a measurement / control system.
• Describe and apply basic signal-level changes: biasing, amplification, attenuation.
• Explain loading using Thévenin models and compute its effect on measurements.
• Use voltage dividers and bridge circuits to convert resistance/impedance changes into voltages.
• Analyze RC filters (low-pass, high-pass, band-pass, notch) and design simple filters for noise reduction.
• Connect these concepts to real ECE applications: sensors, ADC interfaces, and industrial 4–20 mA loops.

Emphasize that this lecture is about the “plumbing” between sensors and the rest of the system. Students often jump to digital / microcontroller code and ignore the analog front end, which is where many real‑world problems show up.

Big Picture: Where Does Signal Conditioning Live?

Typical Instrumentation / Control Chain

1. Physical quantity (temperature, pressure, light, force, …)
2. Sensor / transducer → electrical signal (V, I, R, f)
3. Analog signal conditioning
   • Level & bias changes
   • Linearization
   • Conversions (R→V, V→I, etc.)
   • Filtering & impedance matching
4. ADC / digital logic / microcontroller
5. Control algorithm / data logging
6. Actuator / display / network interface

Note

Key idea: Signal conditioning makes the raw sensor signal usable by the next block in the chain.

Make this diagram concrete with one example: - Thermocouple measuring temperature, producing a few mV. - We amplify, filter 50/60 Hz, shift level to match 0–5 V ADC, then digitize. Ask: what goes wrong if we connect the thermocouple directly to the ADC? (Poor resolution, noise, incorrect range).

1. What Is Analog Signal Conditioning?

• Operations on analog signals to convert them into a form suitable for:
  • Other analog circuits (amplifiers, filters, actuators).
  • ADC inputs in mixed‑signal systems.
• In this chapter, we focus on analog transformations (output still analog).
• Even if the final processing is digital, analog conditioning almost always happens first.

Real ECE examples:

• ECG front‑end: microvolt cardiac signals → 1–3 V, filtered, centered, then digitized.
• Strain gauge bridge on a load cell → mV → amplified / filtered → 4–20 mA transmitter.
• Light sensor (CdS cell) → resistance change → voltage via divider/bridge → linearized / scaled for an ADC.

Point out that students see this in lab whenever they need to scale a sensor to an Arduino’s 0–5 V analog input. Stress that this is not “extra stuff”—it’s what determines measurement accuracy and robustness.

2. Principles of Analog Signal Conditioning

2.1 Sensors & Transfer Functions

• A sensor / transducer converts a physical variable into an electrical quantity:
  • Voltage, current, resistance, capacitance, inductance, or frequency.
• The relationship between input (physical) and output (electrical) is its transfer function.

Example:

• Cadmium sulfide (CdS) photoresistor:
  • Resistance \(R_{\text{CdS}}\) varies nonlinearly and inversely with light intensity.
  • We don’t get to choose this curve; nature chose it.
• A signal conditioning block also has a transfer function, e.g.
  • Simple voltage amplifier: \(V_{\text{out}} = K V_{\text{in}}\) (gain \(K\)).

Important

When you design a measurement system, you are really designing the overall transfer function from physical variable → final electrical (or digital) output.

Stress the idea of transfer functions as “input–output relationships.” Students often see them only in Laplace / systems courses, but here it’s simpler: slope, linear vs. nonlinear, etc.

2.1 Signal-Level and Bias Changes

Scenario:

• Sensor output: \(0.2\ \text{V}\) to \(0.6\ \text{V}\) over measurement range.
• ADC or controller expects: \(0\) to \(5\ \text{V}\) for that same range.

Steps:

1. Bias (zero shift): shift 0.2 V → 0 V.
   • Subtract 0.2 V: \(V' = V_{\text{sensor}} - 0.2\ \text{V}\).
   • Now \(V'\) ranges 0 to 0.4 V.
2. Gain (amplification): scale 0–0.4 V → 0–5 V.
   • Gain \(G = 5 / 0.4 = 12.5\).
   • \(V_{\text{out}} = 12.5\,V'\).

If \(G > 1\) → amplification. If \(G < 1\) → attenuation (still called an amplifier circuit).

Design concerns:

• Frequency response: bandwidth must cover the signal spectrum.
• Input impedance: high enough not to load previous stage.
• Output impedance: low enough to drive next stage.

Relate to op‑amp circuits they know: summing amplifier can do both level shift and gain in one block. Mention that “offset and gain calibration” is exactly this.

2.2 Linearization

Figure 1: Linearization concept. (a) Nonlinear sensor output vs variable. (b) Block diagram of linearizer. (c) Resulting linear output vs variable.

• Many sensors are nonlinear: \(V_{\text{sensor}}(x)\) vs \(x\) is curved.
• But most ADCs / controllers assume a linear mapping between measured variable and digital value.

Two approaches:

1. Old school (pure analog):
   • Design nonlinear analog circuits (diodes, piecewise linear networks, log/antilog amplifiers) to reshape the curve.
   • Works but is hard and limited to narrow ranges.
2. Modern approach (mixed‑signal):
   • Feed the nonlinear analog signal into an ADC.
   • Use software to map it: table lookup, polynomial calibration, or interpolation.
   • Can handle virtually any nonlinearity, in (near) real time.

ECE example: thermistor temperature sensor + microcontroller performing table‑based linearization.

Explain that in older instrumentation texts you’ll see elaborate analog linearizers; today we mostly do “good enough” analog conditioning, then correct the rest digitally.

2.3 Conversions (R→V, V→I, Analog→Digital)

Common conversion tasks in instrumentation:

• Resistance change → voltage
  • Using voltage dividers or bridges (covered later).
• Voltage → current (V‑to‑I converter) for transmission.
• Current → voltage (I‑to‑V converter) at receiver.
• Analog → Digital (ADC input scaling).

4–20 mA current loop

• Industrial standard: transmit measurement as 4–20 mA current.
• Why current?
  • Immune to series resistance: voltage drops along long cables don’t change current (much).
  • 4 mA “live zero” distinguishes broken wire (0 mA).

Need:

• At transmitter: convert sensor R or V to 4–20 mA.
• At receiver: convert 4–20 mA to voltage, typically via a precision shunt resistor.

Analog to Digital Interface

• ADC input may require, e.g. 0–5 V, 0–3.3 V, ±2.5 V, etc.
• Sensor might yield 30–80 mV, or a resistance.
• Analog conditioning:
  • Gain + offset + possibly filtering.
  • Ensure sensor output fully uses ADC input range → better resolution.

Make this concrete with: - 4–20 mA loop driving a 250 Ω resistor → 1–5 V input. - Show quick mental calculation: 20 mA × 250 Ω = 5 V.

2.4 Filtering and Impedance Matching

Filtering

• Real environments have noise:
  • 50/60 Hz line interference.
  • Motor start transients, switching spikes.
  • High‑frequency EMI from radios, inverters, etc.

Use filters to:

• Pass desired band of frequencies.
• Reject low / high / narrow‑band noise.

Types (passive or active):

• Low-pass: pass low f, block high f (e.g., DC sensor + 60 Hz noise).
• High-pass: pass high f, block low f (e.g., remove DC offset / drift).
• Band-pass: pass midband (e.g., audio channel).
• Band-reject / notch: reject narrow band (e.g., 60 Hz notch).

Impedance Matching

• If sensor or line impedance is not matched to the receiving circuit:
  • You get loading → amplitude errors.
  • Possible reflections (at high frequency / transmission lines).
• Use passive or active networks to:
  • Present high input impedance to the sensor.
  • Provide low output impedance to whatever you drive.

Connect this to their previous circuits course: source/load model, maximum power transfer vs. measurement accuracy. Here, we usually care about small loading, not max power.

2.5 Concept of Loading

Figure 2: Thévenin model of a sensor and loading by a finite load resistance.

• Any two‑terminal source can be modeled (Thévenin) as:
  • Ideal voltage source \(V_x\) in series with internal resistance \(R_x\).
• When open‑circuit (no load): output voltage = \(V_x\).

• When you connect a load \(R_L\):
  • Current flows, some voltage drops across \(R_x\).
  • Loaded output voltage:

\[ V_y = V_x\left(1 - \frac{R_x}{R_L + R_x}\right) \]

• If \(R_L\) is not >> \(R_x\), \(V_y\) is significantly < \(V_x\).

Goal in measurement systems: make \(R_L \gg R_x\) to minimize loading.

Warning

Loading changes the apparent transfer function of your sensor + conditioning block. If your measurement depends on amplitude, loading matters a lot.

Remind them: in frequency or digital (0/1) detection, exact amplitude may not matter, but for analog magnitude (temperature, pressure), it does.

Example 1 — Loading Between Sensor and Amplifier

Figure 3: (a) Naive model ignoring loading. (b) Correct model including sensor output resistance and amplifier input resistance.

Given:

• Amplifier gain \(= 10\), input resistance \(R_L = 10\ \text{k}\Omega\).
• Sensor transfer function: \(20\ \text{mV}/^\circ\text{C}\), output resistance \(R_x = 5.0\ \text{k}\Omega\).
• Temperature \(T = 50^\circ\text{C}\).

Naive (no loading) answer:

• Sensor open‑circuit: \(V_T = (20\ \text{mV}/^\circ\text{C})(50^\circ\text{C}) = 1.0\ \text{V}\).
• Amplifier output: \(V_{\text{out}} = 10\cdot 1.0\ \text{V} = 10\ \text{V}\).

Correct answer (include loading):

• Sensor and amplifier form a divider:

\[ V_{\text{in}} = V_T\left(1 - \frac{R_x}{R_x + R_L}\right) = 1.0\text{ V}\left(1 - \frac{5.0\text{ k}\Omega}{5.0\text{ k}\Omega + 10\text{ k}\Omega}\right) \]

• Numerically: \(V_{\text{in}} \approx 0.67\ \text{V}\).
• Amplifier output: \(V_{\text{out}} = 10 \cdot 0.67\ \text{V} = 6.7\ \text{V}\).

Error from ignoring loading: \(10\ \text{V}\) vs \(6.7\ \text{V}\) → 33% error.

Use this as a wake‑up example: 33% error is huge. Ask students to think: how could we reduce it? (Increase amplifier input resistance; use a buffer amplifier).

3. Passive Circuits in Signal Conditioning

• Two essential passive building blocks:
  • Voltage dividers.
  • Bridge circuits (Wheatstone, AC bridges).
• RC networks used for filtering.

Even though op‑amps are cheap, passive circuits still matter:

• High accuracy (e.g., precision bridges).
• Simple, low‑power, and often very low noise.

3.1 Voltage Divider Circuits

Figure 4: Simple voltage divider used to convert resistance change into voltage change.

• Two resistors, \(R_1\) and \(R_2\), in series across supply \(V_s\).
• Output is taken across \(R_2\).

Divider equation:

\[ V_D = \frac{R_2}{R_1 + R_2} V_s \tag{2} \]

Often, either \(R_1\) or \(R_2\) is the sensor whose resistance varies with the measured variable.

Issues to consider:

1. Nonlinearity:
   • Even if sensor resistance varies linearly with variable, \(V_D\) vs variable is nonlinear (rational function).
2. Output impedance:
   • \(Z_{\text{out}} = R_1 \parallel R_2\).
   • Might not be high enough → loading by next stage.
3. Power dissipation:
   • Current through divider: \(I = V_s / (R_1 + R_2)\).
   • Both resistors dissipate \(P = I^2R\); sensor must not be overheated.

Connect this to using a potentiometer as a sensor: it is exactly a divider. Mention that in low‑power sensor applications, we often increase resistances to reduce power, but must then be careful about noise and input bias currents.

Example 2 — Divider with Variable Sensor Resistance

Given (Figure 4):

• \(R_1 = 10.0\ \text{k}\Omega\), \(V_s = 5.00\ \text{V}\).
• \(R_2\) is a sensor varying from \(4.00\ \text{k}\Omega\) to \(12.0\ \text{k}\Omega\).

(a) Range of \(V_D\)

Use Equation (2).

For \(R_2 = 4\ \text{k}\Omega\):

\[ V_D = \frac{(5\ \text{V})(4\ \text{k}\Omega)}{10\ \text{k}\Omega + 4\ \text{k}\Omega} = 1.43\ \text{V} \]

For \(R_2 = 12\ \text{k}\Omega\):

\[ V_D = \frac{(5\ \text{V})(12\ \text{k}\Omega)}{10\ \text{k}\Omega + 12\ \text{k}\Omega} = 2.73\ \text{V} \]

So \(V_D\) ranges from 1.43 V to 2.73 V.

(b) Output impedance range

\(Z_{\text{out}} = R_1 \parallel R_2\).

• For \(R_2 = 4\ \text{k}\Omega\): \(Z_{\text{out}} = 2.86\ \text{k}\Omega\).
• For \(R_2 = 12\ \text{k}\Omega\): \(Z_{\text{out}} = 5.45\ \text{k}\Omega\).

(c) Power dissipated in \(R_2\)

Use \(P = V_D^2 / R_2\):

• At 4 kΩ: \(P \approx 0.51\ \text{mW}\).
• At 12 kΩ: \(P \approx 0.62\ \text{mW}\).

Point out: divider output is only ~1.4–2.7 V from 5 V supply → not using full dynamic range. Later you can mention: we may want to add gain to map this to 0–5 V for ADC usage.

3.2 Wheatstone Bridge Circuits

Figure 5: Basic DC Wheatstone bridge.

• Purpose: convert small resistance changes into measurable voltage changes.
• Very commonly used with:
  • Strain gauges, RTDs, pressure sensors, etc.
• Arrangement: four resistors in a diamond, excited by a voltage \(V\).
• Detector \(D\) senses voltage difference between points \(a\) and \(b\).

Define:

• \(V_a =\) potential at node \(a\) w.r.t. node \(c\).
• \(V_b =\) potential at node \(b\) w.r.t. node \(c\).
• Offset: \(\Delta V = V_a - V_b\).

Assuming infinite detector impedance (no current through D):

\[ V_a = \frac{V R_3}{R_1 + R_3} \tag{4} \]

\[ V_b = \frac{V R_4}{R_2 + R_4} \tag{5} \]

So

\[ \Delta V = V_a - V_b = V\frac{R_3R_2 - R_1R_4}{(R_1 + R_3)(R_2 + R_4)} \tag{7} \]

Null (balanced bridge):

• When \(\Delta V = 0\), condition is:

\[ R_3 R_2 = R_1 R_4 \tag{8} \]

Key advantage: null condition is independent of supply voltage \(V\).

Show that in many sensors, one or more arms are active (change with variable); others are fixed or used for temperature compensation.

Example 3 — Finding an Unknown Resistance from Bridge Null

Given:

• Bridge nulls (ΔV = 0) with \(R_1 = 1000\ \Omega\), \(R_2 = 842\ \Omega\), \(R_3 = 500\ \Omega\).
• Find \(R_4\).

Use Equation (8):

\[ R_1 R_4 = R_3 R_2 \Rightarrow R_4 = \frac{R_3R_2}{R_1} = \frac{(500\ \Omega)(842\ \Omega)}{1000\ \Omega} = 421\ \Omega \]

So the unknown resistance is 421 Ω.

This is essentially how precise ohmmeters using bridges worked historically. Now replaced by more integrated circuitry but the principle is the same.

Example 4 — Small Offset from Nearly Balanced Bridge

Given:

• \(R_1 = R_2 = R_3 = 120\ \Omega\), \(R_4 = 121\ \Omega\), \(V = 10.0\ \text{V}\).
• Find voltage offset ΔV.

Using Equation (7):

\[ \Delta V = 10\ \text{V}\cdot \frac{(120\ \Omega)(120\ \Omega) - (120\ \Omega)(121\ \Omega)} {(120\ \Omega + 120\ \Omega)(120\ \Omega + 121\ \Omega)} \]

Result:

\[ \Delta V = -20.7\ \text{mV} \]

Negative sign → \(V_b > V_a\).

Highlight sensitivity: 1 Ω change (out of 120 Ω) in one resistor yields ~20 mV offset. With an amplifier, we can detect very small resistance changes.

Bridges with Galvanometer Detectors (Finite Detector Resistance)

Figure 6: Thévenin equivalent of a bridge seen by a galvanometer detector.

If the detector is a galvanometer or any finite resistance \(R_G\):

• We replace bridge (between nodes \(a\) and \(b\)) by its Thévenin equivalent:
  • Thévenin voltage \(V_{\text{Th}}\) (open‑circuit ΔV).
  • Thévenin resistance \(R_{\text{Th}}\).

From earlier:

\[ V_{\text{Th}} = V\frac{R_3R_2 - R_1R_4}{(R_1 + R_3)(R_2 + R_4)} \tag{9} \]

Thévenin resistance between \(a\) and \(b\):

\[ R_{\text{Th}} = \frac{R_1R_3}{R_1 + R_3} + \frac{R_2R_4}{R_2 + R_4} \tag{10} \]

Detector current:

\[ I_G = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_G} \tag{11} \]

Bridge null condition is still Equation (8).

This is more historical but useful because the same math applies if D is, say, a low‑input‑impedance amplifier; you must then consider loading of the bridge arms.

Example 5 — Offset Current with Galvanometer

Given:

• Bridge: \(R_1 = R_2 = R_3 = 2.00\ \text{k}\Omega\), \(R_4 = 2.05\ \text{k}\Omega\), \(V = 5.00\ \text{V}\).
• Galvanometer internal resistance \(R_G = 50.0\ \Omega\).
• Find offset current \(I_G\).

1. Find \(V_{\text{Th}}\) from Equation (9):

   \(V_{\text{Th}} = -30.9\ \text{mV}\).

2. Find \(R_{\text{Th}}\) from Equation (10):

   \(R_{\text{Th}} \approx 2.01\ \text{k}\Omega\).

3. Detector current (Equation 11):

\[ I_G = \frac{-30.9\ \text{mV}}{2.01\ \text{k}\Omega + 50\ \Omega} \approx -15.0\ \mu\text{A} \]

Negative → current flows from \(b\) to \(a\).

Bridge Resolution

• Detector resolution (minimum measurable ΔV or ΔI) translates to minimum detectable resistance change in the bridge.
• For high‑input‑impedance detector (e.g., DVM): use Equation (7) or (6) and solve for \(\Delta R\) such that \(\Delta V\) equals detector resolution.

Example 6 — Smallest Measurable Resistance Change

Given:

• Bridge: \(R_1 = R_2 = R_3 = R_4 = 120.0\ \Omega\), \(V = 10.0\ \text{V}\).
• Detector: 3½‑digit DVM on 200 mV range → resolution = \(0.1\ \text{mV} = 100\ \mu\text{V}\).
• Find resolution (smallest \(\Delta R_4\)) for measurement of \(R_4\).

Approach:

• Bridge is initially nulled (ΔV = 0) when all resistors are 120 Ω.
• Now assume \(R_4\) changes to some value giving ΔV = 100 μV.
• Use Equation (6) with unknown \(R_4\) and solve.

Result:

• \(R_4 = 119.9952\ \Omega\).
• \(\Delta R_4 = 120.0000 - 119.9952 = 0.0048\ \Omega\).

So resolution: about \(4.8\ \text{m}\Omega\).

Tip

Resolution here also tells you the measurement uncertainty (best‑case) using this bridge and detector.

Lead Compensation in Bridges

Figure 7: Lead compensation in a remote bridge sensor.

When sensor \(R_4\) is remotely located, long wires add resistance and can vary with:

• Temperature.
• Mechanical stress.
• Corrosion / chemical vapors.

Lead compensation idea:

• Route the wiring so that lead resistance changes are mirrored in two arms (\(R_3\) and \(R_4\)) of the bridge.
• If both arms change by same amount, condition \(R_3R_2 = R_1R_4\) still holds → no change in ΔV.

ECE analogy: 3‑wire RTD configuration widely used in industrial temperature measurement.

Clarify that wire (3) in the figure is just a supply lead, so changes in its resistance don’t affect the bridge balance.

Current-Balance Bridge

Figure 8: Current balance bridge.

Motivation:

• Mechanical feedback (motors adjusting resistors) is slow and noisy.
• Better to electronically null the bridge using a current source.

Construction:

• Split one arm into \(R_4\) and \(R_5\).
• Inject current \(I\) at their junction (into \(R_5\) mostly).

Under conditions \(R_4 \gg R_5\) or \((R_2 + R_4) \gg R_5\):

• Voltage at \(b\):

\[ V_b = \frac{V(R_4 + R_5)}{R_2 + R_4 + R_5} + I R_5 \tag{14} \]

• Voltage at \(a\): still Equation (4).
• Bridge offset:

\[ \Delta V = V_a - V_b = \frac{V R_3}{R_1 + R_3} - \frac{V(R_4 + R_5)}{R_2 + R_4 + R_5} - I R_5 \tag{15} \]

To null the bridge: adjust \(I\) so that \(\Delta V = 0\).

This is conceptually similar to DAC or feedback current that “balances” an analog bridge — useful in self‑balancing instrumentation.

Example 7 — Current Required to Renull a Current-Balance Bridge

Given (Figure 8):

• \(R_1 = R_2 = 10\ \text{k}\Omega\), \(R_3 = 1\ \text{k}\Omega\), \(R_4 = 950\ \Omega\), \(R_5 = 50\ \Omega\).
• High‑impedance detector, supply \(V = 10\ \text{V}\).
• Initially, with \(R_3 = 1000\ \Omega\), bridge is nulled at \(I = 0\).
• Now \(R_3\) increases by 1 Ω to 1001 Ω. Find \(I\) required to renull.

Steps:

1. Compute \(V_a\) with new \(R_3\):

   \(V_a = \dfrac{10\ \text{V}\cdot 1001}{10000 + 1001} = 0.9099\ \text{V}\).

2. Originally (with \(I = 0\)), \(V_a = V_b = 0.9091\ \text{V}\).

   • So now \(V_b\) must also be 0.9099 V to null.
   • Required increase: \(0.8\ \text{mV}\).

3. Using Equation (15) with \(\Delta V = 0\) and only \(IR_5\) as adjustable term:

   \(IR_5 = 0.8\ \text{mV} \Rightarrow I = 0.8\ \text{mV} / 50\ \Omega = 16\ \mu\text{A}\).

Bridges for Potential Measurements

Figure 9: Using a bridge to measure an unknown potential without drawing current from it.

• Place unknown voltage \(V_x\) in series with the detector between \(c\) and \(b\).
• Detector measures \(\Delta V = V_c - V_b\).
• At null (\(\Delta V = 0\)): no current flows through \(V_x\).

For conventional bridge:

• \(V_c = V_x + V_a\), and \(V_b\) from Equation (5).
• Null condition:

\[ V_x + \frac{R_3 V}{R_1 + R_3} - \frac{V R_4}{R_2 + R_4} = 0 \tag{17} \]

So \(V_x\) can be found by choosing resistors to null the detector.

Example 8 — Measuring an Unknown Potential with a Bridge

Given:

• \(R_1 = R_2 = 1\ \text{k}\Omega\), \(R_3 = 605\ \Omega\), \(R_4 = 500\ \Omega\), \(V = 10\ \text{V}\).
• Bridge nulls when unknown potential \(V_x\) is inserted as in Figure 9.
• Find \(V_x\).

Using Equation (17):

\[ V_x + \frac{605\cdot 10}{605 + 1000} - \frac{500\cdot 10}{1000 + 500} = 0 \]

Compute:

• \(\frac{605\cdot 10}{1605} \approx 3.769\ \text{V}\).
• \(\frac{500\cdot 10}{1500} = 3.333\ \text{V}\).

So

\[ V_x + 3.769 - 3.333 = 0 \Rightarrow V_x = -0.436\ \text{V} \]

Negative sign → polarity opposite to assumed orientation.

Example 9 — Potential Measurement with Current Balance Bridge

For current‑balance configuration (Figure 8) with \(V_x\) in series with detector:

• Null condition (Equation 18):

\[ V_x + \frac{R_3 V}{R_1 + R_3} - \frac{V(R_4 + R_5)}{R_2 + R_4 + R_5} - I R_5 = 0 \tag{18} \]

If fixed resistors are chosen so that bridge nulls with \(I = 0\), \(V_x = 0\), middle terms cancel:

\[ V_x - I R_5 = 0 \tag{19} \]

Given:

• \(R_1 = R_2 = 5\ \text{k}\Omega\), \(R_3 = 1\ \text{k}\Omega\), \(R_4 = 990\ \Omega\), \(R_5 = 10\ \Omega\), \(V = 10\ \text{V}\).
• For \(V_x = 12\ \text{mV}\), find \(I\) to null.

From Equation (19):

\[ 12\ \text{mV} - 10 I = 0 \Rightarrow I = 1.2\ \text{mA} \]

AC Bridges

Figure 10: General AC bridge.

• Same idea as DC bridge, but impedances \(Z_1, Z_2, Z_3, Z_4\) and AC excitation \(E\).
• Offset (phasor) voltage:

\[ \Delta E = E\frac{Z_3Z_2 - Z_1Z_4}{(Z_1 + Z_3)(Z_2 + Z_4)} \tag{20} \]

• Null condition:

\[ Z_3 Z_2 = Z_1 Z_4 \tag{21} \]

Special note: if the detector is phase sensitive, both in‑phase and quadrature components must be nulled.

Example 10 — Finding Unknown R and C from an AC Bridge

Figure 11: AC bridge for Example 10.

Assume:

• \(Z_1 = R_1\), \(Z_2 = R_2\), \(Z_3 = R_3 - j/\omega C\), \(Z_x = R_x - j/(\omega C_x)\).

At null: \(Z_2 Z_3 = Z_1 Z_x\).

Real and imaginary parts must match:

• From real part:

\[ R_x = \frac{R_2 R_3}{R_1} \]

• From imaginary part:

\[ C_x = C\frac{R_1}{R_2} \]

With numbers given in the original text:

• \(R_x = 2\ \text{k}\Omega\).
• \(C_x = 0.5\ \mu\text{F}\).

Nonlinearity of Bridge Response

Figure 12a: ΔV vs R4 for a wide range — clearly nonlinear.

Figure 12b: ΔV vs R4 near the null — approximately linear.

• Equation (7) is nonlinear in any resistor.
• If sensor resistance varies over a wide range, bridge voltage vs resistance is significantly curved (Figure 12a).
• If resistance variation is small and centered on the null (e.g., 90–110 Ω around 100 Ω):
  • The curve is almost linear over that small range (Figure 12b).
  • We can amplify the small off‑null voltage to expand the scale.

Design tip: operate bridges close to null and limit sensor range to maintain approximate linearity, then do software calibration if needed.

3.3 RC Filters — Basics

Figure 13: Low-pass RC filter.

Simple first‑order filters built from one R and one C:

• Low-pass (Figure 13): passes DC/low frequencies, attenuates high frequencies.
• High-pass (Figure 15): passes high frequencies, attenuates low frequencies.

Critical / cutoff frequency:

\[ f_c = \frac{1}{2\pi RC} \tag{22} \]

At \(f = f_c\):

• \(|V_{\text{out}}/V_{\text{in}}| = 1/\sqrt{2} \approx 0.707\) → −3 dB point.

Low-Pass RC Filter

Figure 14: Magnitude response of low-pass RC filter.

Transfer magnitude:

\[ \left| \frac{V_{\text{out}}}{V_{\text{in}}} \right| = \frac{1}{\sqrt{1 + (f/f_c)^2}} \tag{23} \]

Applications:

• Measure DC or slowly varying signals while rejecting high‑frequency noise.
• Anti‑aliasing before an ADC (though often higher‑order filters are used).

Design Guidelines

1. Choose a capacitor in the μF–pF range.
2. Compute \(R\) from \(R = 1/(2\pi f_c C)\).
3. Aim for \(1\ \text{k}\Omega \le R \le 1\ \text{M}\Omega\) to avoid excessive current or noise.
4. Component tolerances (e.g., ±20% for capacitors) mean \(f_c\) is approximate.
5. If precise \(f_c\) is needed, measure \(C\) and use a trim pot for \(R\).

Example 11 — Low-Pass Filter for 1 MHz Noise

Goal:

• Measurement signal frequencies < 1 kHz.
• Unwanted noise around 1 MHz.
• Want noise at 1 MHz attenuated to 1% (\(|V_{\text{out}}/V_{\text{in}}| = 0.01\)).
• What is effect on desired 1 kHz signal?

Use Equation (23):

Set \(f = 1\ \text{MHz}\), \(|V_{\text{out}}/V_{\text{in}}| = 0.01\):

\[ 0.01 = \frac{1}{\sqrt{1 + (10^6/f_c)^2}} \Rightarrow (10^6/f_c)^2 \approx 9999 \]

So \(f_c \approx 10\ \text{kHz}\).

Pick \(C = 0.01\ \mu\text{F}\) ⇒

\[ R = \frac{1}{2\pi f_c C} \approx 1591\ \Omega \]

Use standard \(R = 1.5\ \text{k}\Omega\) ⇒ actual \(f_c \approx 10.6\ \text{kHz}\).

Noise attenuation at 1 MHz: very close to 0.01 (slightly better).

Effect on 1 kHz signal:

\[ \left| \frac{V_{\text{out}}}{V_{\text{in}}} \right| = \frac{1}{\sqrt{1 + (1000/10610)^2}} \approx 0.996 \]

Only about 0.4% loss in the desired band.

Highlight the design trade‑off: place \(f_c\) comfortably above signal band yet far below noise frequency.

High-Pass RC Filter

Figure 15: High-pass RC filter.

Figure 16: Magnitude response of high-pass RC filter.

Transfer magnitude:

\[ \left| \frac{V_{\text{out}}}{V_{\text{in}}} \right| = \frac{f/f_c}{\sqrt{1 + (f/f_c)^2}} \tag{24} \]

At \(f = f_c\) → \(|V_{\text{out}}/V_{\text{in}}| = 0.707\) (−3 dB).

Applications:

• Remove DC offsets or low‑frequency drift from signals.
• AC‑coupled amplifiers; pulse transmission.

Example 12 — High-Pass Filter for 60 Hz Noise

Goal:

• Pulses for a stepper motor at 2000 Hz.
• 60 Hz noise present.
• Want to reduce 60 Hz noise strongly, but allow pulse amplitude to be reduced by no more than 3 dB (≈0.707).

Since 3 dB corresponds to \(|V_{\text{out}}/V_{\text{in}}| = 0.707\), we choose:

• \(f_c = 2000\ \text{Hz}\).

Now effect on 60 Hz:

\[ \left| \frac{V_{\text{out}}}{V_{\text{in}}} \right| = \frac{60/2000}{\sqrt{1 + (60/2000)^2}} \approx 0.03 \]

So 60 Hz noise is reduced to about 3% (97% attenuation).

Design: pick \(C = 0.01\ \mu\text{F}\) →

\[ R = \frac{1}{2\pi f_c C} \approx 7.96\ \text{k}\Omega \]

Choose standard 7.5 kΩ or 8.2 kΩ as close approximations.

Practical Considerations for RC Filters

1. Component range:
   • Avoid too small \(R\) (large currents, loading).
   • Avoid very large \(C\) (size, leakage).
2. Loading:
   • Finite source impedance + filter input impedance.
   • Filter output impedance + load impedance.
   • May need buffer stages (op‑amp followers) to avoid loading.
3. Cascading filters:
   • Magnitudes multiply.
   • Overall transfer = (first‑order response)^N for N cascaded sections (if no loading).
   • Be careful: second stage can load first stage → more attenuation than expected.

Example 13 — 1 vs 2 High-Pass Stages

Goal:

• 2 kHz data signal contaminated by 60 Hz noise.
• Want 60 dB attenuation of 60 Hz (\(|V_{\text{out}}/V_{\text{in}}| = 10^{-3}\)).

Single-stage high-pass

Use Equation (24):

\[ 0.001 = \frac{60/f_c}{\sqrt{1 + (60/f_c)^2}} \]

Solve → \(f_c \approx 60\ \text{kHz}\).

Effect on 2 kHz data:

\[ \left|\frac{V_{\text{out}}}{V_{\text{in}}}\right| = \frac{2000/60000}{\sqrt{1 + (2000/60000)^2}} \approx 0.033 \]

Only 3.3% of data amplitude remains → too much attenuation.

Two cascaded stages (same \(f_c\))

• Overall response ≈ (Equation 24)\(^2\).
• For same 60 dB attenuation of 60 Hz noise, solve:

\[ 0.001 = \frac{(60/f_c)^2}{1 + (60/f_c)^2} \]

→ \(f_c \approx 1896\ \text{Hz}\).

Now effect on 2 kHz signal (using squared response):

\[ \left|\frac{V_{\text{out}}}{V_{\text{in}}}\right| = \frac{(2000/1896)^2}{1 + (2000/1896)^2} \approx 0.53 \]

So about 53% of signal remains, much better than 3.3%.

Moral: multiple gentle filter stages often better than one very aggressive stage.

Example 14 — Loading Between Cascaded RC Stages

Figure 17: Cascaded high-pass RC filter.

We now consider loading for Example 13 with specific component choices.

Given:

• First stage: choose \(C = 0.001\ \mu\text{F}\); need \(f_c = 1896\ \text{Hz}\).
• From \(f_c = 1/(2\pi RC)\):

\[ R = \frac{1}{2\pi f_c C} \approx 83.9\ \text{k}\Omega \]

• Second stage uses same \(R\) and \(C\).
• Evaluate loading effect at \(f = 2\ \text{kHz}\).

Key steps:

1. Compute capacitor reactance at 2 kHz: \(X_C \approx 79.6\ \text{k}\Omega\).
2. Find equivalent impedances and Thévenin equivalent of first stage.
3. Use AC divider to find effective input to second stage.

Result:

• Without loading, two‑stage response at 2 kHz was \(\approx 0.53\).
• With loading, actual overall response ≈ 0.36.

So loading reduces signal to 36% instead of 53%.

Still better than 3.3% from single stage, but emphasizes: - Need for buffers between stages when precision matters.

Band-Pass RC Filters

Figure 19: Generic band-pass filter response.

Figure 20: RC band-pass filter = high-pass + low-pass in cascade.

• Constructed by cascading a high-pass and a low-pass filter.
• Lower critical frequency \(f_L\) from high-pass; upper critical frequency \(f_H\) from low-pass.
• Passband is \(f_L < f < f_H\).

For the RC band‑pass in Figure 20 (including loading via ratio \(r = R_H / R_L\)):

\[ \left|\frac{V_{\text{out}}}{V_{\text{in}}}\right| = \frac{f_H f} {\sqrt{(f^2 - f_H f_L)^2 + [f_L + (1 + r)f_H]^2 f^2}} \tag{25} \]

with

\[ f_H = \frac{1}{2\pi R_L C_L}, \quad f_L = \frac{1}{2\pi R_H C_H} \tag{26} \]

Design tips:

• Want wide separation between \(f_L\) and \(f_H\).
• Keep \(r = R_H / R_L \le 0.01\) to minimize loading effects.

Example 15 — Band-Pass for 6–60 kHz Data

Goal:

• Signal frequencies: 6 kHz to 60 kHz.
• Noise at 120 Hz and 1 MHz.
• Design RC band‑pass to reduce noise by 90% (|Vout/Vin| = 0.1).

Lower cutoff (high-pass part)

At 120 Hz, want |Vout/Vin| ≈ 0.1:

Use Equation (24):

\[ 0.1 = \frac{120/f_L}{\sqrt{1 + (120/f_L)^2}} \Rightarrow f_L \approx 1200\ \text{Hz} \]

Upper cutoff (low-pass part)

At 1 MHz, want |Vout/Vin| ≈ 0.1:

Use Equation (23):

\[ 0.1 = \frac{1}{\sqrt{1 + (10^6/f_H)^2}} \Rightarrow f_H \approx 100\ \text{kHz} \]

Plot of Equation (25) (Figure 21 in text) shows:

• For \(r = 0.1\) vs \(r = 0.01\), loaded passband is lower in amplitude for larger \(r\).

Design with \(r = 0.01\):

• Choose \(R_L = 100\ \text{k}\Omega\), then \(R_H = rR_L = 1\ \text{k}\Omega\).
• From Equation (26):
  • \(C_H = 0.133\ \mu\text{F}\).
  • \(C_L = 15.9\ \text{pF}\).

Effect on data:

• At 6 kHz: |Vout/Vin| ≈ 0.969 (3% reduction).
• At 60 kHz: |Vout/Vin| ≈ 0.851 (15% reduction).

Band-Reject (Notch) Filters

Figure 22: Generic band-reject (notch) filter response.

• Reject a range of frequencies (especially a narrow band).
• Useful when one particular frequency interferes, e.g., 50/60 Hz hum in audio or ECG.

Passive RC implementation with good performance is hard; inductive or active filters usually better.

Twin-T Notch Filter

Figure 23: Twin-T notch RC filter.

Figure 24: Twin-T notch response.

Configuration:

• Top T: series–parallel of R and C.
• Bottom T: R and C interchanged.
• Grounding components \(R_1\) and \(C_1\) tune the notch.

For \(R_1 = \frac{\pi R}{10}\) and \(C_1 = \frac{10C}{\pi}\):

• Critical frequency:

\[ f_n = 0.785 f_c, \quad \text{where } f_c = \frac{1}{2\pi RC} \tag{27} \]

• Frequencies for −3 dB (0.707):

\[ f_L = 0.187 f_c, \quad f_H = 4.57 f_c \tag{28} \]

At \(f_n\), |Vout/Vin| ≈ 0.002 ⇒ good but not perfect rejection.

Example 16 — 400 Hz Notch Filter (Twin-T)

Goal:

• On an aircraft, 400 Hz is the power frequency.
• Want to reject 400 Hz.
• Evaluate effect on voice band (10–300 Hz) and find upper −3 dB frequency.

We want \(f_n = 400\ \text{Hz}\). From Equation (27):

\[ 400 = 0.785 f_c \Rightarrow f_c \approx 510\ \text{Hz} \]

Pick \(C = 0.01\ \mu\text{F}\) →

\[ R = \frac{1}{2\pi f_c C} \approx 31.2\ \text{k}\Omega \]

Grounding components:

\[ R_1 = \frac{\pi R}{10} \approx 9800\ \Omega \]

\[ C_1 = \frac{10C}{\pi} \approx 0.0318\ \mu\text{F} \]

Effect on voice band:

• At 10 Hz: \(10/400 = 0.025\) → from Figure 24, |Vout/Vin| ≈ 0.99 (little attenuation).
• At 300 Hz: \(300/400 = 0.75\) → |Vout/Vin| ≈ 0.03 (significant attenuation).

Upper −3 dB frequency:

• From Equation (28): \(f_H = 4.57 f_c ≈ 2300\ \text{Hz}\).

So: - Very good rejection at 400 Hz. - Low voice frequencies largely unaffected, higher low frequencies more attenuated.

Mention that in practice, active notch filters using op‑amps provide deeper notches and more controllable Q factor. Twin‑T is still useful conceptually and in simple analog designs.

Summary / Key Points

• Signal conditioning bridges the gap between raw sensor outputs and the requirements of downstream electronics (amplifiers, ADCs, transmitters).
• Key tasks:
  • Signal-level & bias changes: match sensor outputs to required voltage ranges.
  • Linearization: historically analog, now often done in software.
  • Conversions: R→V, V↔︎I, analog→digital interface.
  • Filtering: low-pass, high-pass, band-pass, band-reject/notch.
  • Impedance matching & loading mitigation: maintain accuracy.
• Loading:
  • Real sources have internal resistance.
  • When load is not ≫ source resistance, measured voltage drops (Thévenin model).

• Passive circuits:
  • Voltage dividers: simple R→V conversion but nonlinear, have finite output impedance, dissipate power.
  • Wheatstone bridges: highly sensitive resistance measurement and R→V conversion; null condition independent of supply.
  • AC bridges: extend bridges to impedances (R, C, L), account for magnitude and phase.
• RC filters:
  • First‑order low-pass and high-pass with cutoff \(f_c = 1/(2\pi RC)\).
  • Easy to design but limited slope (−20 dB/decade).
  • Cascading increases slope but requires attention to loading.
  • Band-pass = high-pass + low-pass; band-reject (notch) can be implemented with twin‑T for narrow band rejection.
• In modern systems, analog conditioning and digital processing complement each other:
  • Analog front‑end: scaling, basic filtering, protecting ADC.
  • Digital side: precise calibration, linearization, advanced filtering.

Formulas Summary

Transfer function basics

• Simple voltage gain: \(V_{\text{out}} = K V_{\text{in}}\).

Loading (Thévenin model)

• Loaded output:

\[ V_y = V_x\left(1 - \frac{R_x}{R_L + R_x}\right) \]

Divider circuit

• Output voltage:

\[ V_D = \frac{R_2}{R_1 + R_2} V_s \tag{2} \]

Wheatstone bridge (DC)

• Node voltages:

\[ V_a = \frac{V R_3}{R_1 + R_3}, \quad V_b = \frac{V R_4}{R_2 + R_4} \]

• Offset voltage:

\[ \Delta V = V_a - V_b = V\frac{R_3R_2 - R_1R_4}{(R_1 + R_3)(R_2 + R_4)} \tag{7} \]

• Null condition:

\[ R_3 R_2 = R_1 R_4 \tag{8} \]

Bridge Thévenin equivalent

• Thévenin voltage:

\[ V_{\text{Th}} = \Delta V \tag{9} \]

• Thévenin resistance:

\[ R_{\text{Th}} = \frac{R_1R_3}{R_1 + R_3} + \frac{R_2R_4}{R_2 + R_4} \tag{10} \]

• Detector current:

\[ I_G = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_G} \tag{11} \]

Current-balance bridge

• High‑impedance detector:

\[ V_b = \frac{V(R_4 + R_5)}{R_2 + R_4 + R_5} + I R_5 \tag{14} \]

\[ \Delta V = V_a - V_b = \frac{V R_3}{R_1 + R_3} - \frac{V(R_4 + R_5)}{R_2 + R_4 + R_5} - I R_5 \tag{15} \]

• For potential measurement with properly chosen resistors:

\[ V_x - I R_5 = 0 \tag{19} \]

AC bridge

• Offset (phasor):

\[ \Delta E = E\frac{Z_3Z_2 - Z_1Z_4}{(Z_1 + Z_3)(Z_2 + Z_4)} \tag{20} \]

• Null condition:

\[ Z_3 Z_2 = Z_1 Z_4 \tag{21} \]

RC filters

• Cutoff frequency (low-pass & high-pass):

\[ f_c = \frac{1}{2\pi RC} \tag{22} \]

• Low-pass magnitude:

\[ \left| \frac{V_{\text{out}}}{V_{\text{in}}} \right| = \frac{1}{\sqrt{1 + (f/f_c)^2}} \tag{23} \]

• High-pass magnitude:

\[ \left| \frac{V_{\text{out}}}{V_{\text{in}}} \right| = \frac{f/f_c}{\sqrt{1 + (f/f_c)^2}} \tag{24} \]

• Band-pass RC (including loading, \(r = R_H/R_L\)):

\[ \left| \frac{V_{\text{out}}}{V_{\text{in}}} \right| = \frac{f_H f} {\sqrt{(f^2 - f_H f_L)^2 + [f_L + (1 + r)f_H]^2 f^2}} \tag{25} \]

with

\[ f_H = \frac{1}{2\pi R_L C_L}, \quad f_L = \frac{1}{2\pi R_H C_H} \tag{26} \]

Twin-T notch filter

• Core RC frequency:

\[ f_c = \frac{1}{2\pi RC} \]

• Notch frequency:

\[ f_n = 0.785 f_c \tag{27} \]

• −3 dB frequencies:

\[ f_L = 0.187 f_c, \quad f_H = 4.57 f_c \tag{28} \]

Analog Signal Conditioning – Interactive Deck

1. Loading: Sensor + Amplifier as a Divider (Pyodide)

Explore Example 1 numerically. The sensor has internal resistance \(R_x\) and the amplifier has input resistance \(R_L\). The loaded voltage is

\[ V_{\text{in}} = V_T \left(1 - \frac{R_x}{R_x + R_L}\right) \]

Try changing \(R_L\) and \(R_x\) and see how the amplifier output changes.

Ask students: - Set R_L = 10_000, then to 100_000, then 1_000. - Which case is “good instrumentation practice,” and why?

2. Loading Error vs. Load Resistance (Reactive Plot)

Use a slider to change \(R_L\) and see how the percentage error caused by loading changes.

viewof RL_k = Inputs.range(
  [1, 200], 
  {step: 1, value: 10, label: "Amplifier input resistance RL [kΩ]"}
)

viewof Rx_k = Inputs.range(
  [0.1, 20],
  {step: 0.1, value: 5, label: "Sensor resistance Rx [kΩ]"}
)

viewof VT = Inputs.range(
  [0.1, 5],
  {step: 0.1, value: 1.0, label: "Sensor open-circuit VT [V]"}
)

viewof gain = Inputs.range(
  [1, 20],
  {step: 1, value: 10, label: "Amplifier gain"}
)

Guide students: “Move the RL slider to match the curve to your chosen Rx and VT; notice that when RL ≫ Rx, error is small.” Also highlight the log scale on RL.

3. Voltage Divider: Nonlinear Sensor-to-Voltage Mapping (Pyodide)

Revisit the divider from Example 2:

\[ V_D = \frac{R_2}{R_1 + R_2} V_s \]

Here, \(R_2\) is a sensor that changes with the physical variable.

Ask: Is V_D vs R_2 linear in R_2? If the physical variable is proportional to R_2, what does this do to linearity of measurement vs actual variable?

4. Voltage Divider Transfer Curve (Reactive Plot)

Visualize the full curve \(V_D(R_2)\) and relate it to nonlinearity.

viewof R1_k_div = Inputs.range(
  [1, 50],
  {step: 1, value: 10, label: "R1 [kΩ]"}
)

viewof Vs_div = Inputs.range(
  [1, 10],
  {step: 0.5, value: 5.0, label: "Vs [V]"}
)

viewof R2_min_k = Inputs.range(
  [1, 20],
  {step: 1, value: 4, label: "Sensor R2_min [kΩ]"}
)

viewof R2_max_k = Inputs.range(
  [2, 50],
  {step: 1, value: 12, label: "Sensor R2_max [kΩ]"}
)

Encourage students to adjust R1 and see how the slope and curvature change. Point out that any straight line they “fit” will only be accurate over a small R2 range.

5. Wheatstone Bridge Offset Voltage (Pyodide)

Let’s compute the DC Wheatstone bridge offset:

\[ \Delta V = V\frac{R_3R_2 - R_1R_4}{(R_1 + R_3)(R_2 + R_4)} \]

Link to Example 4: their calculated −20.7 mV should match what students obtain here.

6. Wheatstone Bridge ΔV vs Sensor Resistance (Reactive Plot)

Visualize bridge nonlinearity: how ΔV changes with sensor resistance \(R_4\).

viewof R1_b = Inputs.range(
  [50, 500],
  {step: 10, value: 100, label: "R1 [Ω]"}
)

viewof R2_b = Inputs.range(
  [50, 500],
  {step: 10, value: 100, label: "R2 [Ω]"}
)

viewof R3_b = Inputs.range(
  [50, 500],
  {step: 10, value: 100, label: "R3 [Ω]"}
)

viewof Vb = Inputs.range(
  [1, 20],
  {step: 1, value: 10, label: "Bridge supply V [V]"}
)

viewof R4_min = Inputs.range(
  [10, 500],
  {step: 5, value: 0, label: "R4_min [Ω] (sensor)"}
)

viewof R4_max = Inputs.range(
  [50, 1000],
  {step: 10, value: 500, label: "R4_max [Ω] (sensor)"}
)

Ask students to zoom into a small region around the null (where ΔV ≈ 0) and see that the curve looks almost linear there. This reinforces the “small‑signal linear” region concept.

7. Low-Pass RC Filter Magnitude Response (Pyodide)

Recall:

• Cutoff frequency: \(f_c = \dfrac{1}{2\pi RC}\).
• Magnitude response:

\[ \left|\frac{V_{\text{out}}}{V_{\text{in}}}\right| = \frac{1}{\sqrt{1 + (f/f_c)^2}} \]

Explore numerically.

Encourage students to modify R and C so that fc moves below or above the test frequencies; observe how |Vout/Vin| changes.

8. Low-Pass RC Bode Magnitude Plot (Reactive Plot)

Interactive plot of \(|V_{\text{out}}/V_{\text{in}}|\) vs \(f\) (log scale), with adjustable R and C.

viewof R_lp = Inputs.range(
  [100, 100_000],
  {step: 100, value: 1_500, label: "R [Ω] (low-pass)"}
)

viewof C_lp_nF = Inputs.range(
  [0.1, 1000],
  {step: 0.1, value: 10.0, label: "C [nF] (low-pass)"}
)

Highlight: - The −3 dB line is where f ≈ fc. - Slope approaches −20 dB/decade at high frequency.

9. High-Pass RC Magnitude Response (Reactive Plot)

For high-pass:

\[ \left|\frac{V_{\text{out}}}{V_{\text{in}}}\right| = \frac{f/f_c}{\sqrt{1 + (f/f_c)^2}} \]

viewof R_hp = Inputs.range(
  [100, 100_000],
  {step: 100, value: 8_000, label: "R [Ω] (high-pass)"}
)

viewof C_hp_nF = Inputs.range(
  [0.1, 1000],
  {step: 0.1, value: 10.0, label: "C [nF] (high-pass)"}
)

Relate this to Example 12 and 13: e.g., set fc ≈ 2 kHz and observe attenuation at 60 Hz vs 2 kHz.

10. Band-Pass from Cascaded High-Pass and Low-Pass (Reactive)

Model a simple band-pass as:

\[ H_{\text{BP}}(f) = H_{\text{HP}}(f) \cdot H_{\text{LP}}(f) \]

with the first-order formulas for each.

viewof fc_low = Inputs.range(
  [10, 100_000],
  {step: 10, value: 1_200, label: "High-pass cutoff fL [Hz]"}
)

viewof fc_high = Inputs.range(
  [1_000, 500_000],
  {step: 1_000, value: 100_000, label: "Low-pass cutoff fH [Hz]"}
)

Connect to Example 15: - Try fL ≈ 1.2 kHz and fH ≈ 100 kHz. - Check attenuation at 120 Hz and 1 MHz; compare to ~0.1 (−20 dB).

11. Twin-T Notch: Frequency Mapping (Pyodide)

For a Twin‑T notch:

• \(f_c = \dfrac{1}{2\pi RC}\).
• Notch frequency: \(f_n = 0.785 f_c\).
• −3 dB frequencies: \(f_L = 0.187 f_c\), \(f_H = 4.57 f_c\).

Compute these for given R and C.

Let students set R and C so that fn ≈ 400 Hz as in Example 16, and check fL and fH.