AMME1705 · Introduction to Electromechanical Systems
Signals, Sensors & Noise
This chapter of AMME1705 Introduction to Electromechanical Systems at the University of Sydney is where a circuit first meets the real world: how a physical quantity becomes an electrical signal, and how you describe and clean that signal. You will separate analog (continuous) from digital (discrete-level) signals, put a number on how fast a signal repeats with the frequency relation f = 1/T, judge signal quality with the signal-to-noise ratio (SNR), and read a real-world quantity with an analog sensor wired as a voltage divider. It is the Week 3 foundation the op-amp, filter and AC-waveform chapters all build on.
What this chapter covers
- 01Analog (continuous) vs digital (discrete-level) signals, and why systems convert between them (ADC/DAC)
- 02Frequency and period: f = 1/T in hertz, with the millisecond-to-second conversion
- 03Range vs precision, and why digital sampling trades precision for noise immunity
- 04Digital logic voltage bands: HIGH, LOW and the forbidden band on a 5 V input
- 05Noise, and the signal-to-noise ratio SNR = P_signal / P_noise
- 06Why amplifying a noisy signal does not improve its SNR
- 07Common analog sensors: LDR, thermistor, photodiode, microphone, strain gauge
- 08Passive vs active sensors, calibration and signal conditioning
- 09Reading a resistive sensor with a two-element voltage divider into a logic input or ADC
An LDR light switch: divider voltage vs the logic threshold
- +1Set up the divider. With the LDR on top and R on the bottom, Vout = Vin · R/(RLDR + R) = 5 · 10/(RLDR + 10), with resistances in kΩ (the kΩ cancel).
- +1(a) Darkness — voltage. RLDR = 90 kΩ: Vout = 5 · 10/(90 + 10) = 5 · 10/100 = 0.5 V.
- +1(a) Darkness — level. 0.5 V is below 0.8 V, so the input reads LOW (0).
- +1(b) Bright light — voltage. RLDR = 2 kΩ: Vout = 5 · 10/(2 + 10) = 50/12 = 4.17 V.
- +1(b) Bright light — level. 4.17 V is above 2.0 V, so the input reads HIGH (1).
- +1(c) Dusk — the trap. RLDR = 20 kΩ: Vout = 5 · 10/(20 + 10) = 50/30 = 1.67 V, which lies in the 0.8–2.0 V forbidden band — the input cannot reliably decide and needs conditioning (a comparator).
Key terms
- Analog signal
- A continuous voltage that may take any value within a range, so it mirrors a real-world quantity (temperature, pressure, sound, position) exactly.
- Digital (discrete) signal
- A voltage restricted to a finite set of allowed levels. The smallest case is binary — two levels, {0, 1}, i.e. LOW/HIGH or OFF/ON.
- Frequency and period
- The period T is the time for one full cycle (seconds); the frequency f = 1/T is how many cycles occur per second, measured in hertz (Hz). Convert milliseconds to seconds before taking the reciprocal.
- Signal-to-noise ratio (SNR)
- The ratio of signal power to noise power, SNR = Psignal/Pnoise (dimensionless). Because power is proportional to voltage squared, it also equals (Vsignal/Vnoise)². A larger SNR means a cleaner signal.
- Logic voltage bands
- A digital input classifies a real voltage into bands: on a typical 5 V input, below VIL (about 0.8 V) reads LOW and above VIH (about 2.0 V) reads HIGH; the gap between is a forbidden/indeterminate band. Exact levels are device-specific — confirm on the data sheet.
- Passive vs active sensor
- A passive sensor changes a property (resistance, capacitance, inductance) or itself generates a small measurable energy (LDR, thermistor, strain gauge). An active sensor needs supplied energy to produce or amplify its signal (ultrasonic transducer, phantom-powered microphone).
- Voltage divider (sensor interface)
- Two series elements across a supply whose node voltage is Vout = Vin·R/(Rsensor + R). It converts a resistive sensor's changing resistance into a readable voltage for a logic input or ADC.
Signals, Sensors & Noise FAQ
Why doesn't amplifying a weak, noisy sensor signal fix it?
An amplifier multiplies the signal and the noise by the same gain, so both powers rise by the same factor and the signal-to-noise ratio is unchanged — a louder copy of a noisy signal is still noisy. You raise SNR at the source instead: a better or better-shielded sensor, or a filter that removes noise sitting outside the signal's frequency band.
How do I turn a resistive sensor like an LDR or thermistor into something a microcontroller can read?
Put it in a two-element voltage divider with a fixed resistor across the supply. The node voltage Vout = Vin·R/(Rsensor + R) swings between 0 V and the supply as the sensor resistance changes, giving a voltage a logic input or ADC can read. Whether the sensor goes on the top or bottom of the divider sets the direction of the swing.
Can AI help me with signals, sensors and noise in AMME1705?
Yes — a tutor like Sia can explain the ideas step by step: how to get from a period in milliseconds to a frequency in hertz, why amplifying does not improve SNR, or how a voltage divider maps a sensor resistance to a node voltage and then to a HIGH/LOW logic level. Use it to understand the method and to check your own reasoning; it will not hand you exam answers or guarantee a mark, and you should always confirm formulae and device levels against your current Canvas materials and data sheets.
Studying with AI? Sia — free AI electrical engineering tutor works through AMME1705 step by step.
Exam move
Put three moves on your note sheet and drill them until they are automatic. First, f = 1/T with the units nailed down — always convert the period to seconds before taking the reciprocal, since a millisecond left un-converted makes the answer 1000 times too small. Second, SNR = Psignal/Pnoise together with the one-line reason that gain never improves it. Third, the voltage-divider node voltage Vout = Vin·R/(Rsensor + R), plus the 5 V logic bands (LOW below about 0.8 V, HIGH above about 2.0 V, forbidden in between). The final exam is a paper-based 2-hour paper (plus 10 minutes reading) in the end-of-Semester-1 exam period, around June 2027 — confirm the date on Canvas — and is worth 33%. Bring a calculator and, per the exam cover sheet, one A4 double-sided handwritten note sheet, but the assessment table has described it as closed book, so confirm the open/closed-book status on your current Canvas page. The paper is out of about 80 marks over 120 minutes, so budget roughly 1.5 minutes per mark and answer to the stated precision, always naming the logic level explicitly.