All Practice Exams
100+ Free HKIE Professional Assessment — Biomedical Discipline Practice Questions
HKIE Professional Assessment — Biomedical Discipline (Hong Kong Institution of Engineers) practice questions are available now; exam metadata is being verified.
✓ No registration✓ No credit card✓ No hidden fees✓ Start practicing immediately
100+ Questions
100% Free
Loading practice questions...
Same family resources
Explore More HKIE Professional Assessment
Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.
2026 Statistics
Key Facts: HKIE Professional Assessment — Biomedical Discipline Exam
Portfolio + Interview + Essay
Assessment Format
HKIE Professional Assessment Documentation
2 Hours
Written Essay Duration
HKIE PA Guidelines
HK$ 3,100
Total Examination Fee
HKIE Fee Schedule 2026
4 Areas
HKIE Competence Standard
HKIE Competence Standards
MDACS Class I-IV
HK Medical Device Control
HK Dept of Health MDD
IEC 60601-1
Electrical Safety Standard
Clinical Engineering Standards
The HKIE PA Biomedical Discipline assessment evaluates candidate competence via portfolio, interview, and essay. This 100-question practice bank provides an English-language MCQ adaptation covering technical principles, MDACS regulations, IEC 60601, biomaterials, biomechanics, clinical engineering, and HKIE ethics.
Sample HKIE Professional Assessment — Biomedical Discipline Practice Questions
Try these sample questions to test your HKIE Professional Assessment — Biomedical Discipline exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1An electrocardiogram (ECG) biopotential amplifier records a differential cardiac signal of 1.0 mV alongside a common-mode power-line interference voltage of 2.0 V. If the differential gain is 1,000 and the Common Mode Rejection Ratio (CMRR) is 80 dB, what is the amplitude of the common-mode interference appearing at the amplifier output?
A.0.20 V
B.0.02 V
C.2.00 V
D.0.002 V
Explanation: A CMRR of 80 dB corresponds to a linear common-mode rejection ratio of 10^(80/20) = 10,000. The common-mode gain Ac is calculated as Ad / CMRR = 1,000 / 10,000 = 0.10. Therefore, the common-mode output voltage is Vout,cm = Ac * Vcm = 0.10 * 2.0 V = 0.20 V.
2What primary operational objective is served by incorporating a Right Leg Drive (RLD) active feedback circuit into a clinical ECG monitoring system?
A.To active-drive common-mode displacement currents back into the patient, reducing common-mode noise and maintaining patient ground safety
B.To provide high-voltage isolation (>4 kV) protecting the patient against defibrillator discharge surges
C.To digitize analog biopotential signals directly at the skin surface before differential amplification
D.To compensate for electrode polarization potential by applying a DC offset voltage across Lead II
Explanation: The Right Leg Drive (RLD) circuit senses the common-mode voltage on the patient via summing resistors, amplifies and inverts this signal, and drives it back onto the patient through the right leg electrode. This active negative feedback reduces patient common-mode voltage by a factor of (1 + A_rld), eliminating 50/60 Hz power-line interference while limiting fault current for patient safety.
3An electroencephalogram (EEG) signal contains clinical diagnostic spectral components up to 70 Hz (gamma band). According to the Nyquist-Shannon sampling theorem, what is the absolute minimum theoretical sampling rate required, and what practical anti-aliasing cut-off strategy should be employed?
A.Minimum sampling rate 140 Hz; anti-aliasing low-pass filter set at or below 70 Hz before digitization
B.Minimum sampling rate 70 Hz; anti-aliasing high-pass filter set at 70 Hz before digitization
C.Minimum sampling rate 280 Hz; anti-aliasing bandpass filter set between 70 Hz and 140 Hz
D.Minimum sampling rate 100 Hz; anti-aliasing notch filter centered at 50 Hz power mains frequency
Explanation: The Nyquist-Shannon sampling theorem dictates that the sampling frequency f_s must be at least twice the maximum signal frequency component f_max (f_s >= 2 * 70 Hz = 140 Hz). An anti-aliasing low-pass filter with a cutoff frequency at or below 70 Hz must be placed prior to the ADC to attenuate any frequency components above f_s/2 that would otherwise alias into the EEG bandwidth.
4A semiconductor piezoresistive strain gauge with a Gauge Factor (GF) of 120 is bonded to an orthopedic bone plate. If the plate experiences an axial mechanical strain of 500 microstrain (500 x 10^-6), what is the fractional change in electrical resistance (delta_R / R)?
A.0.060 (6.0%)
B.0.006 (0.6%)
C.0.240 (24.0%)
D.0.0012 (0.12%)
Explanation: The fundamental formula relating gauge factor (GF), strain (epsilon), and resistance change is GF = (delta_R / R) / epsilon. Rearranging yields delta_R / R = GF * epsilon = 120 * (500 x 10^-6) = 0.060 (or 6.0%). Piezoresistive semiconductor gauges exhibit high gauge factors compared to metallic foil gauges (GF ~ 2).
5During telemetry monitoring of a pulse oximetry plethysmogram, the root-mean-square (RMS) peak signal voltage is measured at 2.5 V, while the accompanying background RMS noise voltage is measured at 25 mV. What is the Signal-to-Noise Ratio (SNR) expressed in decibels (dB)?
A.40 dB
B.20 dB
C.100 dB
D.50 dB
Explanation: Signal-to-Noise Ratio for voltage amplitudes is calculated using SNR_dB = 20 * log10(V_signal / V_noise). Here, V_signal / V_noise = 2.5 V / 0.025 V = 100. Calculating the log gives 20 * log10(100) = 20 * 2 = 40 dB.
6A clinical engineer designs an active Sallen-Key second-order low-pass Butterworth filter to eliminate high-frequency muscle artifact (EMG) from a surface ECG signal. If both timing resistors have R = 159 kOhm and both capacitors have C = 10 nF, what is the -3 dB cutoff frequency f_c of the filter?
A.100 Hz
B.50 Hz
C.159 Hz
D.500 Hz
Explanation: The cutoff frequency for an equal-component Sallen-Key second-order low-pass filter is f_c = 1 / (2 * pi * R * C). Substituting the values: f_c = 1 / (2 * pi * 159,000 * 10 x 10^-9) = 1 / (2 * pi * 1.59 x 10^-3) = 1 / (0.00999) approx 100 Hz.
7In dual-wavelength pulse oximetry, arterial oxygen saturation (SpO2) is determined by measuring the ratio of ratios (R). What optical principle explains why red light (660 nm) and near-infrared light (940 nm) are selected for this measurement?
A.Deoxyhemoglobin (Hb) has significantly higher molar extinction at 660 nm than oxyhemoglobin (HbO2), while HbO2 has higher extinction at 940 nm
B.Oxyhemoglobin (HbO2) has higher molar extinction at both 660 nm and 940 nm compared to carboxyhemoglobin
C.Light at 660 nm and 940 nm completely avoids skin melanin absorption and bone scattering
D.At 660 nm and 940 nm, blood density changes linearly with venous pulsation
Explanation: Pulse oximetry relies on the distinct absorption spectra of Hb and HbO2. At 660 nm (red), deoxygenated hemoglobin absorbs significantly more light than oxygenated hemoglobin. At 940 nm (infrared), oxygenated hemoglobin absorbs more light than deoxygenated hemoglobin. The ratio R = (AC/DC)_660 / (AC/DC)_940 correlates empirically with SpO2.
8What standard passband frequency range is recommended by ISEK (International Society of Electrophysiology and Kinesiology) for recording surface electromyography (sEMG) signals to capture muscle activity while excluding motion artifacts and high-frequency noise?
A.10 Hz to 500 Hz
B.0.05 Hz to 150 Hz
C.0.5 Hz to 70 Hz
D.100 Hz to 2000 Hz
Explanation: Surface electromyography (sEMG) energy is predominantly concentrated between 10 Hz and 500 Hz. High-pass filtering at 10 to 20 Hz eliminates low-frequency motion artifacts (movement of cables and skin-electrode interface), while low-pass filtering at 500 Hz removes high-frequency thermal noise without corrupting motor unit action potential (MUAP) waveforms.
9A classic three-op-amp instrumentation amplifier front-end utilizes input stage buffer resistors R1 = 49.4 kOhm and an external gain-setting resistor RG = 2.0 kOhm. If the second-stage difference amplifier has a unity voltage gain (R4/R3 = 1), what is the overall differential voltage gain Av of the instrumentation amplifier?
A.50.4
B.25.7
C.100.0
D.49.4
Explanation: The overall differential gain of a standard three-op-amp instrumentation amplifier is given by Av = (1 + 2*R1 / RG) * (R4 / R3). Substituting the given values: Av = (1 + 2 * 49.4 kOhm / 2.0 kOhm) * 1 = 1 + (98.8 / 2.0) = 1 + 49.4 = 50.4.
10A photoplethysmogram (PPG) optical sensor measures a pulsatile AC light intensity amplitude of 15 mV riding on a steady DC baseline intensity voltage of 1.50 V. What is the Perfusion Index (PI) percentage of this vascular bed?
A.1.0%
B.10.0%
C.0.1%
D.15.0%
Explanation: Perfusion Index (PI) is defined as the ratio of the pulsatile AC signal amplitude to the non-pulsatile static DC baseline signal, expressed as a percentage: PI = (V_AC / V_DC) * 100%. Substituting the values: PI = (0.015 V / 1.50 V) * 100% = 0.010 * 100% = 1.0%.
About the HKIE Professional Assessment — Biomedical Discipline Practice Questions
Verified exam format metadata for HKIE Professional Assessment — Biomedical Discipline (Hong Kong Institution of Engineers) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.