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Exam with Model Exam with Model Answer Benha University June 2011 High Institute of Technology Electrical Dep artment Biomedical Instruments (E372 ) 3rd year (control) Dr.Wael Abdel-Rahman Mohamed Time: 3 Hrs Exam with Model Answer Answer the following questions with the aid of drawing and equations as possible . Question (1): [14 Marks] Differentiate between: a) The rest potential obtained by Nernst and Goldman. The Nernst equation has a physiological application when used to calculate the potential of an ion of charge z across a membrane. This potential is determined using the concentration of the ion both inside and outside the cell: When the membrane is in thermodynamic equilibrium (i.e., no net flux of ions), the membrane potential must be equal to the Nernst potential. However, in physiology, due to active ion pumps, the inside and outside of a cell are not in equilibrium. In this case, the resting potential can be determined from the Goldman equation: • Em = The membrane potential (in volts, equivalent to joules per coulomb) • Pion = the permeability for that ion (in meters per second) • [ion ]out = the extracellular concentration of that ion (in moles per cubic meter, to match the other SI units, though the un its strictly don't matter, as the ion concentration terms become a dimensionless ratio) • [ion ]in = the intracellular concentration of that ion (in moles per cubic meter) • R = The ideal gas constant (joules per kelvin per mole) • T = The temperature in kelvin • F = Faraday's constant (coulombs per mole) The potential across the cell membrane that exactly opposes net diffusion of a particular ion through the membrane is called the Nernst potential for that ion. As seen above, the magnitude of the Nernst potentia l is determined by the ratio of the concentrations of that specific ion on the two sides of the membrane. The greater this ratio the greater the tendency for the ion to diffuse in one direction, and therefore the greater the Nernst potential required to pr event the diffusion. In brief: Nernst solves for the potential due to one type of ions but Goldman gives the total effect of all ions and takes into consideration the ratio of permeabilities. b) Bonded and unbonded strain gages. Unbonded strain gages can be constructed so that they are linear over a wide range of applied force but are very delicate. The bonded strain gage is generally more rugged but is linear over a smaller range of forces. c) Operational and instrumentation amplifiers. The main difference is: we can use the operational amplifier in any configuration as inverting, non-inverting, comparator …etc. where the instrumentation amplifier is only a differential amplifier configuration. d) Piezoresistive and piezoelectric elements. The direct piezoelectric effect consists of an electric polarization in a fixed direction while the piezoelectric crystal is deformed. The polarization is proportional to the deformation and causes an electrical potential difference over the crystal. The inverse piezoelectric effect, on the other hand, means that an applied potential difference induces a deformation of the crystal - a useful property in actuators. The piezoresistive effect is closely related to the piezoelectric effect. In this case, the material's conductivity changes when the material is subjected to an external force. Such materials can also be used in sensor applications. e) A-mode and B-mode scanners. Pulse Fires Reflection Reflection Near Scatter Amplitude Transducer Transducer Signal Path Signal Path Target Target Tissue Tissue Boundary Boundary A-Mode B-Mode In the A-mode, the amplitude of the received signals deflects the display beam vertically. In the B-mode, the brightness of the beam is modulated by this amplitude. f) Linear and convex array of ultrasound transducers. Linear Array Curved Array Shape Image shape Rectangle Sector Imaging range Small Large Application Peripheral Vascular, Small Abdominal, Obstetrician / Parts, Musculoskeletal Nerves, Gynecologist, Vascular Pediatrics g) Specular and diffuse reflection of ultrasound beam. Solve by yourself. [Hint: see fig .17-4 in your text book] Question (2): [12 Marks] a) Draw the circuit model for two biomedical electrodes produce a differential voltage and attached to a differential amplifier. b) Draw the action potential and explain its main parts. The action potential is an electrical event occurring when a stimulus of sufficient intensity is applied to a neuron or muscle cell, allowing sodium to move into th e cell and reverse the polarity. Normally neurons (nerve cells) maintain a resting potential of - 70mV across their membrane by the active pumping of 3Na + ions out of the cell for every 2 K + ions pumped into the cell by a Na +/K + pump. When the neuron is stimulated, sodium ion channels open in the membrane and sodium ions flood in to the cell down an electrochemical gradient by diffusion, increasing the potential of the cell to +40mV. This is called depolarization. At this point the sodium channels close and potassium ion channels open. Potassium ions flood out of the cell down their electrochemical gradient, decreasing the cell's membrane potential. This is called repolarization. There is a slight overshoot where too many potassium ions diffuse out of the cell, and there is hyper-polarization where the cell's membrane potential falls below its normal -70mV, but this is corrected and the resting potential is once again restored. This is the sequence of events that makes up a single action potential. Action potentials are transmitted by salutatory conduction in the neuron, and impulses jump from node to node along the axon of the neuron. c) Draw the block diagram of the basic design of an isolation amp. And explain the function of each block. • Input and output amplifiers to get the signal and amplify it. • The modulator to convert the electrical signal to other type that can be transferred over the isolation barrier. • The demodulator returns back the signal to its original form (electrical signal). • The isolation barrier is the region where the input and output are totally isolated. • The isolation and coupling may be optically, thermally or magnetically. d) Draw three configurations for the mechanical scanners and three configurations for the electrical scanners. Solve by yourself. [Hint: see figs. 17-16 & 17-17 & 17-18] Question (3): [12 Marks] a) List three techniques that might be used to overcome the effects of electrode offset potentials • Differential DC amplifier to acquire signal thus the DC component will cancel out. • Counter Offset-Voltage to cancel half-cell potential. • AC couple input of amplifier (DC will not pass through). b) Why the microelectrodes are used only for low frequency measurements. Explain how to null the microelectrode capacitance. Solve by yourself. [Hint: see fig.6-23 in your text book] c) Why ultrasound couldn't be used to image lungs and skeletal system? The percentage of energy that is reflected (coefficient of reflection) Г in the boundary between air and tissue or between bone and tissue is very large (approximately 99.9%) where the difference in acoustic impedance between them is very large; ͦ Γ ͔1 Ǝ ͔2 Ɣ ƴ Ƹ ͔1 ƍ ͔2 So the whole ultrasound beam is approximately reflected and no image is presented. d) Explain how the phased array is used to image a sector using the beamforming technique. Solved exactly in the last lecture. Question (4): [10 Marks] What is the relation between: • Ultrasound beam frequency and the image resolution. Directly proportional. As frequency increase; the wavelength decrease and the resolution increase. c = λ f The ultrasound wavelength in blood (for example) is 0.3mm for a 5-MHz probe and 0.6mm for a 2.5-MHz probe. The two-point discrimination (spacial resolution) of ultrasound is dependent upon wavelength as the minimum distance that can be resolved between two objects is one wavelength. A 1-MHz probe will be unable to distinguish between two structures less than 1.5mm apart, whereas a 5-MHz probe can resolve structures 0.3mm apart. Thus, as ultrasound wavelength shortens, image detail (resolution) increases. • Ultrasound beam frequency and the depth of the organ to be imaged. Inversely proportional. As the frequency increase; the attenuation increase (see fig 17-12 in your text book) hence the penetration decrease. Higher frequencies are used for superficial imaging. • Medium density and the refracted signal angle. Inversely proportional. Waves tend to have higher velocity in denser material then θR is less than θi when incident waves cross into denser material and vice versa. (sin θi / sin θR) = (V1 / V 2) • Ultrasound signal speed and the tissue density. Directly proportional. As stated before; Waves tend to have higher velocity in denser material. (see table 17-1 in your text book) • Gage factor and sensitivity. Directly proportional; GF gives relative sensitivity of a strain gauge where the greater the change in resistance per unit length the greater the sensitivity of element and the greater the gauge factor. ∆R GF = R ∆L L Question (5): [12 Marks] a) For the circuit shown, draw the output waveform if V ref = -2V and V x = 0.25V. Draw it by yourself taking into consideration that it is a comparator with hysteresis and the output will be V cc or –Vcc in a band with width equal 4Vx oriented around –Vref . b) A pneumotachograph is an instrument to measure the respiration of a patient and shown in the diagram shown. Pressure sensors P1 and P2 have again of 1, i.e. V=P (voltage output is equal to pressure input). These two transducers are placed on either side of a fine mesh with resistance to flow R. Determine the expression for Vout(t). Solved exactly in sheets. c) Consider the ECG amplifier shown. Calculate the output voltage Vo. [Note: the input voltages are 1.2mV and 3 mV] This is a differential amplifier with all resistors equal; so the gain will be unity.
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