A notch filter is employed to suppress the hum noise generated by the power supply in the ECG circuit.

Transcription

1 1. What is the frequency range of ECG signal? a Hz 150 Hz b. 500 Hz 1200 Hz c. 5 khz 10 khz d. 0.5 Hz 1 MHz Answer: a) The diagnostically useful frequency range is usually accepted as 0.05 to 150 Hz. Although the electric field generated by the heart can be best characterized by vector quantities, it is generally convenient to directly measure only scalar quantities, i.e. a voltage difference of mv order between the given points of the body. 2. How do you reduce the hum noise generated by the power supply in the ECG circuit? a. By implementing band pass filters b. By implementing high pass filters c. By implementing notch filters d. By implementing low pass filters Answer: c) A notch filter is employed to suppress the hum noise generated by the power supply in the ECG circuit. 3. Which of the following amplifier circulatory is employed to reduce the base line wandering? a. Low Pass filter b. High pass filter c. Notch filter d. Band pass filter Answer: b) Base line wander is a low-frequency component in the ECG signal arises from variety of noise sources. Its spectral content is usually confined to frequencies below 1 Hz. Hence a high pass filter cab be implemented to remove this noise. 4. Consider the circuit shown in the figure below. Let the input Vi connected is a sinusoidal input of 2 Vpp and the opamp1 is powered with ± 15 V. What is the operation of the circuit a. Half wave rectifies the input Vi b. Integrates the input voltage Vi c. Differentiates the input voltage Vi d. Produces single pulses at the zero-crossing point in every cycle Answer: d) Produces single pulses at the zero-crossing point in every cycle (Refer Q6 solution) 5. For the circuit shown in question 4, what is the configuration of Opamp1

2 a. Inverting configuration b. Schmitt trigger c. Zero crossing detector d. Differentiator Answer: c) Zero crossing detector (refer Q6 solutiom) 6. For the circuit shown in figure 4, select the V output signal a. b. c. d. Answer: d)

3 7. Consider a pair of thermocouples used for measuring the reference temperature and the temperature of the specimen. It has been noticed that the reference temperature is at 2 o C. If the thermocouple output employed for measuring the specimen temperature shows as 48 µv, compute the actual temperature of the specimen. Note: Table given below represents the measurements taken from the thermocouple employed for measuring the specimen temperature when the reference temperature is at 0 o C Temperature ( o C) Output (µv)

4 a. 13 o C b. 46 o C c. 48 o C d. 50 o C Solution: 15 o C (This question will be not be considered for evaluation) The sensitivity of the thermocouple can be find using the table, Sensitivity = (change in output)/ (change in input) Let s take any two readings to find the sensitivity S = (55-45)/ (20 10) = 1 µv/ o C Given, when the reference temperature is at 2 o C, the output from the thermocouple shows 48 µv. Thus, On substitution, S * (T specimen T ref) +35 µv = 48 µv T specimen = 13 o C + 2 o C = 15 o C 8. Consider the circuit shown below. If the capacitor C is initially uncharged and at t = 0 the switch is closed, compute the voltage across the capacitor at t = 1 ms Note: Consider the op-amp is supplied with ± 15 V VA a. 15 V b. 10 V c. 6.3 V d. -15 V Ans: b) Solution: Voltage at point A, VA = 10 V (Virtual short concept) Apply KCL rule at inverting terminal, k = C dvc dt dvc 6 = 1 10 dt

5 Vc = m dt 0 = [1 m 0] = 10 V 9. A voltage generated using an AD590 (temperature) sensor is used to compensate for ambient temperature error for the iron-constantan thermocouple in the circuit shown in the figure such that calculate the output voltage Vo if the values of RL = 5.3 kω, and E = V. Consider the output from the thermocouple is mv when Ta = 30 o C and Tm = 100 o C a V b V c V d V Ans: a) Solution: Given Vy = mv at Ta = 30 o C and Tm = 100 o C Therefore, Vx = (1 + (99k/1k)) * Vy = 100 * m = V The voltage drop across RL is written as V RL = I (273+Ta) µ * RL = ( ) µ * 5.3 k = V and V RL = E + VM = VM VM = mv Op-amp C is a voltage follower. Hence Vz = VM = mv Therefore, Vo = -(100k/1k) * (Vx + Vz) = - 10 * ( m) = - 10 * = V 10. Consider a copper constantan thermocouple which is used to measure an unknown temperature. If the thermocouple with unknown temperature as 20 o C and reference temperature as 0 o C shows an output voltage of mv, then compute the unknown temperature if the output voltage is mv when reference temperature is at 20 o C. Note: The sensitivity of temperature sensor is 0.04 mv/ o C a. 46 o C b o C

6 c o C d o C Answer: d) 91.8 o C Given E 20 E 0 = mv and Sensitivity S = 0.04 mv/ o C Also given, E unknown E 20 = mv (E unknown E 0) = (E unknown E 20) + (E 20 E 0) = = mv 1 o C > 0.04 mv? < mv T = / (0.04) = o C = 91.8 o C