---
title: "Three transducers to control room: routing and S/H"
subject: "Instrumentation II"
chapter: 4
key: "semester_5/instrumentation_ii:stepper-transducers-sh"
source: https://www.bctengineer.com/q/semester_5/instrumentation_ii/stepper-transducers-sh
site: BCT Engineering
---

# Three transducers to control room: routing and S/H

Instrumentation II · Chapter 4: Interfacing A/D And D/A Converters

- Asked in: 2068 Chaitra, 2+6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/instrumentation_ii/stepper-transducers-sh
- Whole chapter: https://www.bctengineer.com/md/semester_5/instrumentation_ii/chapter-4

## Question

Signals from three different transducers (A, B and C) located 100 meters away from a control room in a factory are very important to control stepper motors to give final tune to the products. The strength of these signals ranges from 10mV to 20mV and separated at 4KHz. Transducer A, B and C generate 6KHz, 10KHz and 14KHz signals. (a) How do you want to route these signals to the control room? (b) If your A/D converter do not have S/H hold circuits, what specification of S/H chip you select for your design? Discuss also the errors associate with the converter.

## Short answer

Route signals via FDM: preamplify, band-pass (6,10,14 kHz), sum, drive balanced line over shielded pair, then separate and MUX/S/H/ADC. For 14 kHz 8-bit ADC without S/H, use LF398 with aperture <44 ns, acquisition ~2–4 µs, droop <98 V/s. Errors: aperture jitter, droop, quantization, aliasing, offset, INL/DNL.

## Answer

#### (a) Routing the signals to the control room (100 m)
The signals are very small (10-20 mV), so noise on a 100 m cable easily affects them. The three transducers use different frequencies (6, 10 and 14 kHz, separated by 4 kHz), so **frequency division multiplexing (FDM)** can be used.
1. Place a low-noise instrumentation preamplifier at each transducer (gain about 100-250) to raise the signal to a few volts. Add a band-pass filter (centres 6, 10 and 14 kHz, bandwidth below 4 kHz) to remove noise and out-of-band interference.
2. Add the three amplified signals with a summing amplifier. Send the composite signal on **one shielded twisted pair** using a balanced line driver (for example a differential driver). Ground the cable shield at one end only.
3. In the control room, use a differential receiver, then three band-pass filters (6, 10 and 14 kHz) to separate the signals.
4. An analog multiplexer and an ADC (with S/H) convert them for the processor that drives the stepper motors.

```text
 A -> [Amp][BPF 6k] --+
 B -> [Amp][BPF10k] --+--> [Summer][Line driver] ==100 m shielded pair==>
 C -> [Amp][BPF14k] --+
 ==> [Diff. receiver] -> [BPF 6k / 10k / 14k] -> [MUX] -> [S/H] -> [ADC] -> uP
```
An alternative is to digitise each signal at the transducer and send the data serially (RS-485), which gives the best noise immunity.

#### (b) S/H specification (ADC without S/H)
A signal that changes during conversion causes an error. For a sine wave of peak A and frequency f, the maximum slope is 2 pi f A.
For A = 20 mV at f = 14 kHz: slope = 2 pi (14000)(0.02) = 1759 V/s = 1.76 mV/us (before amplification).
After the amplifier (say a gain of 125, so 20 mV becomes 2.5 V peak) and a 5 V full-scale ADC:
```text
 8-bit : LSB = 5/256  = 19.5 mV  ; aperture time <= (LSB/2)/(2 pi f A)
                       = 9.77 mV / (2 pi x 14e3 x 2.5) = 44 ns
 10-bit: LSB = 5/1024 = 4.88 mV  ; aperture time <= 11 ns
```
The selected S/H (LF398 / HA5320 class) should have:
- **Aperture time** (and jitter) below the value above (about 10-40 ns).
- **Acquisition time** shorter than the time slot of each channel. With sampling above 2 x 14 kHz = 28 kHz per channel and three channels, the slot is below 12 us, so an acquisition time of 2-4 us is suitable (hold capacitor of about 1 nF).
- **Droop rate** small: during a 100 us conversion the held voltage must change by less than 1/2 LSB, i.e. 9.77 mV / 100 us = 98 V/s. Use a low-leakage polystyrene or Teflon hold capacitor.
- Low pedestal error (charge injection), low feedthrough, high input impedance and good linearity.

#### Errors associated with the converter
1. **Quantisation error** (+/-1/2 LSB).
2. **Aliasing** if the sampling rate is below 28 kHz. Use an anti-alias low-pass filter.
3. **Aperture error / jitter** and conversion-time error (the signal moves during conversion).
4. **Offset, gain error, INL, DNL** (missing codes) and temperature drift.
5. **S/H errors:** droop, pedestal, feedthrough and acquisition-time error.
6. **Noise** (thermal, power supply, crosstalk), which is serious for a 10-20 mV signal.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which method is used to send the three signals over one cable to the control room?
   A. Time division with no filters
   B. Frequency division multiplexing with preamps, band-pass filters and a shielded twisted pair
   C. Sending each as unamplified DC
   D. Using three separate unshielded wires
   Answer: B. Different frequencies (6, 10, 14 kHz) allow FDM; amplification, filtering and shielding fight noise on 100 m. (Key point: Routing method)

2. How is the maximum aperture time of the S/H chosen?
   A. Aperture time <= (LSB/2) / (2 pi f A)
   B. Equal to the conversion time
   C. Greater than the acquisition time
   D. Equal to 1/f
   Answer: A. The signal must change by less than half an LSB during the aperture; the maximum slope is 2 pi f A. (Key point: Aperture time)

3. What limit applies to droop during a 100 us conversion?
   A. It must be zero volts
   B. It must be greater than 1 V
   C. The held voltage must change less than 1/2 LSB
   D. It must equal the LSB
   Answer: C. Droop of less than half an LSB (9.77 mV in 100 us, about 98 V/s) keeps the held value accurate. (Key point: Droop rate)

4. Why must the sampling rate per channel exceed 28 kHz?
   A. Because the cable is 100 m
   B. Because the ADC has 8 bits
   C. Because the signals are 10-20 mV
   D. Nyquist: the highest signal is 14 kHz, so sampling must exceed 2 x 14 kHz to avoid aliasing
   Answer: D. Below twice the highest frequency, aliasing occurs; an anti-alias filter is also used. (Key point: Sampling rate)

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