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Bioelectromagnetism Exercise #3 – Answers TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q1: Measurement Lead (Lead) Vector Three electrodes (a, b and c) are on the surface of a volume conductor. Inside the conductor is a dipole source. The lead vectors of this dipole defined at the three locations (a, b and c) are: The dipole is parallel to the unit vector i. What is the ratio of the voltages measured between electrodes a and b to a and c? ref A B C

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q1: Measurement Lead (Lead) Vector c ab & c ac c ab = c b – c a = 2i + 5j + k c ac = 6i + 3j + 3k V ab & V ac p = xi(parallel to the x-axis =i) V ab = 2*x [V] V ac = 6*x [V] V ab / V ac = 1/3 ref A B C

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q2: 12-lead ECG Lead Vectors Derive the lead vectors of the limb leads I, II, and III, leads VR, VF, VL, and the Goldberger leads aVR, aVL ja aVF in a spherical homogeneous volume conductor. Preconditions –dipole in a fixed location –homogeneous spherical volume conductor Equilateral triangle |C I | = |C II | = |C III | (=1) C I = j sin60°=|a| / |C II | ==> |a| = sin60°= |C II | = 3/2 C II = 1/2 j - 3/2 k C III = -1/2 j - 3/2 k x(i) y(j) z(k) I II III a 60° LA RA LL

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q2: 12-lead ECG Lead Vectors Goldberger leads aVR, aVL ja aVF |C aVR | = |C aVF | = |C aVL | = |a| = 3/2 (*|I|) C aVF = -|a| k = -3/2 k |c| = cos30° * |aVR| = 3/2 * 3/2 = 3/4 |b| = sin30° * |aVR| = 1/2 * 3/2 = 3/4 C aVL = |c|j + |b|k = 3/4 j + 3/4 k C aVR = -|c|j + |b|k = -3/4j + 3/4 k I II III a aVF 30° LA RA LL aVL aVR c b x(i) y(j) z(k)

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q2: 12-lead ECG Lead Vectors Leads VR, VF and VL |C VR | = |C VF | = |C VL | =? 1/2*|C I | = |C VR |*cos30° |C VR | = 1/2*|C I | /cos30° = 1/2 / 3/2 = 1/ 3 |C VF |/|C aVF |=1/3 / 3/2 = 2/3 C VF = C aVF * 2/3 = -3/2 k * 2/3 = -3/3 k C VL = C aVL * 2/3 = (3/4 j + 3/4 k )* 2/3 = 1/2 j + 3/6 k C VR = C aVR * 2/3 = (-3/4 j + 3/4 k )* 2/3 = -1/2 j + 3/6 k I II III VF 30° LA RA LL VL VR x(i) y(j) z(k)

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q3: 12-lead ECG Lead Vectors During a QRS-complex at some time instant t the following potentials were measured: III-lead+1,1 mV aVR-lead-3,4 mV V1 -lead-3,5 mV Approximate the potentials in the leads I, II, aVL, aVF, V6 and V4. We need c & p to solve the potentials (V=cp) Frontal plane: III & aVR identify p |C I | = 2/3*|C aVR | =>V aVR = mV |p| = ( ) = 4.1 mV tan = 1.1/3.93 => = 15.6° V I = 4.1 * cos(30+ ) = 2.9 mV V II = 4.1 * cos(30- ) = 4.0 mV V aVL = 4.1 * cos(60+ ) = 1.0 mV V aVF = 4.1 * cos(60- ) = 2.9 mV Augmented leads scaling: V aVL = V aVL * 3/2 = 0.9 mV V aVF = V aVF * 3/2 = 2.5 mV aVF IIIII aVR I aVL 1.1 mV -3.93

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q3: 12-lead ECG Lead Vectors Potentials V6 and V4? Transveral plane: V 1 & V 5 /I identify p V V1 = -3.5 mV V I = V V5 = 2.85 mV x*cos(30- ) = 3.5 x*cos(30+ ) = 2.85 = 10.1° (HOW?, next slide) |p| = 2.8/cos(30+ ) = 3.7 mV V V6 = 3.6 * cos( ) = 3.7 mV V V4 = 3.6 * cos(60+ ) = 1.3 mV V2V2 V3V3 V1V1 I V6V6 V4V4 V5V5 assumption: |c I | = |c V1…6 | -3.5 mV 2.85 mV

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q3: 12-lead ECG Lead Vectors Summation equation: cos(x y) = cosxcosy sinxsiny x*cos(30- ) = 3.5 => cos30cos + sin30sin = 3.5/X x*cos(30+ ) = 2.85 => x = 2.85/(cos30cos -sin30sin ) cos30cos + sin30sin = 3.5/ 2.85 * (cos30cos -sin30sin ) (a=3.5/2.85) cos30cos -1.25*cos30cos = -a*sin30sin - sin30sin (cos30-a*cos30)*cos = -(a*sin30+sin30)*sin sin /cos = -(cos30-a*cos30)/(a*sin30+sin30) tan = => =10.1°

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q4: Lead vectors… On the outer rim of a two dimensional volume conductor at points P i, i = 1,..., 6 the potentials generated by a unit dipole oriented parallel to X or Y axes at point P o are as follows: Using these measurements is it possible to a) Derive the image surface of this source dipole location? b) Calculate the lead vector of a lead between the points 3 and 6? c) Derive the potential at the point 4 generated by a unit dipole at a point Pz? d) Derive the lead field of the volume conductor? e) Construct a VECG lead system (X and Y-leads) that would measure the X and Y components of the dipole at the point Po with similar sensitivity? ElectrodePotentials generated by X and Y dipoles P i V X V Y

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q4: Lead vectors… Is it possible to… a) Derive the image surface of this source dipole location? yes ElectrodePotentials generated by X and Y dipoles P i V X V Y P1P1 P2P2 P3P3 P5P5 P4P4 POPO P6P6

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q4: Lead vectors… Is it possible to… b) Calculate the lead vector of a lead between the points 3 and 6? yes ElectrodePotentials generated by X and Y dipoles P i V X V Y P1P1 P2P2 P3P3 P6P6 P5P5 P4P4 POPO

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q4: Lead vectors… Is it possible to… c) Derive the potential at the point 4 generated by a unit dipole at a point Pz? no ElectrodePotentials generated by X and Y dipoles P i V X V Y P1P1 P2P2 P3P3 P6P6 P5P5 P4P4 POPO PZPZ

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q4: Lead vectors… Is it possible to… d) Derive the lead field of the volume conductor? no ElectrodePotentials generated by X and Y dipoles P i V X V Y P1P1 P2P2 P3P3 P6P6 P5P5 P4P4 POPO

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q4: Lead vectors… Is it possible to… e) Construct a VECG lead system (X and Y-leads) that would measure the X and Y components of the dipole at the point Po with similar sensitivity? yes ElectrodePotentials generated by X and Y dipoles P i V X V Y P1P1 P2P2 P3P3 P6P6 P5P5 P4P4 POPO

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q5: EEG Sensitivity Figure 1 represents a potential data distribution at a model of a (half) head generated by a reciprocal current of 1.0 µA applied to electrodes C and D. Calculate the potential between electrodes C and D generated by current dipoles A and B (|P| = 4 µAcm). Fig 1. Potential data at surface of a electrolytic tank (midsagittal plane). Lines between 0.1 mV curves are spaced at equal potential differences (0.02 mV) apart. I=1.0 µA in the half-head model; fluid resistivity is 2220 Ωcm.

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q5: EEG Sensitivity Reciprocal current I CD = 1.0 µA applied to electrodes C and D forms the lead field. –now shown as potential field –shows the sensitivity of lead CD to dipolar sources –distances of the lines = lead sensitivity Golden equation –convert to E-field form reciprocal lead field source Fig 1. Potential data at surface of a electrolytic tank (midsagittal plane). Lines between 0.1 mV curves are spaced at equal potential differences (0.02 mV) apart. I=1.0 µA in the half-head model; fluid resistivity is 2220 Ωcm.

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TAMPERE UNIVERSITY OF TECHNOLOGY Ragnar Granit Institute Bioelectromagnetism Exercise 3 Q5: EEG Sensitivity One dipole source –|c A | = 43° => V CD = 25 µV –|c B | = 17° => V CD = 31 µV Fig 1. Potential data at surface of a electrolytic tank (midsagittal plane). Lines between 0.1 mV curves are spaced at equal potential differences (0.02 mV) apart. I=1.0 µA in the half-head model; fluid resistivity is 2220 Ωcm.

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