bio impedance and bio electricity basics_solutions_manual

C:\Documents and Settings\bensonm\Desktop\Grimnes\Problems and solutions ch2.doc25 April 2008 12:28 PM

CH 2 PROBLEMS 1. What is electronegativity of an atom? 2. What is covalent bonding? 3. What is hydration of an ion? 4. What is the current density at a position in an electrolyte solution where the

electric field is 2 V/m and the conductivity is 0.3 S/m. 5. What is a faradaic current? 6. What is a double layer? Does it have capacitive properties? 7. With a Pt wire and a gold wire as electrodes in the same solution, what is the

voltage generated by the electrode system? 8. What are the four electrokinetic effects? 9. What is incremental resistance? 10. A sinusoidal current waveform generates potential differences which are also

sinusoidal if the system is linear. Is the current amplitude for onset of non-linearity dependent on the sine frequency?

CH 2 SOLUTIONS

1. The relative ability of an atom to gain electrons 2. Covalent bonds are formed when two atoms share valence electron pairs 3. Hydration of an ion is the formation of sheath of oriented water molecules around

the ion. 4. The current density J = σE = 0.3 [S/m] x 2 [V/m] = 0.6 [SV/m2] = 0.6 [A/m2] 5. Faradaic current is the current obeying Faradays law 6. The double layer is a molecular capacitor formed in the liquid at the surface of a

solid-liquid interface. 7. Potential difference according to Table 2.9 is about 0.3 volt. 8. Electrophoresis, electro-osmosis, sedimentation potential, streaming potential. 9. Incremental resistance R is the slope of a V-I curve, R=∆V/∆I. 10. Yes (Section 2.5.4).

C:\Documents and Settings\bensonm\Desktop\Grimnes\Problems ch3.doc25 April 2008 12:29 PM 1Sverre Grim

CH 3 PROBLEMS 1. At which frequency may copper be considered as a dielectric defined as

B=G? For copper use εr=3 and σ=6x106 . 2. What is the force between two opposite charges of 1 coulomb at distance 1

m? 3. How large is the polarisation P [Cm/m3] in a dielectric with relative

permittivity equal to 3? 4. What is the complex admittance Y of an ideal capacitor with dimensions

area 1 cm2 and thickness 1mm, and with a dielectric with εr=3 and σ= 2S/m at 1 MHz?

5. What is the complex impedance Z of the same capacitor? 6. An ideal capacitor with dimensions area 1 cm2 and thickness 1mm is

measured to have a frequency independent resistance 106 ohm and a capacitance of 15x10-12 farad. Find the complex permittivity, conductivity and resistivity of the dielectric at 1 MHz.

7. Sketch Fig.3.14 by replacing the σ sigma parameter with the εr'' parameter.

CH 3 SOLUTIONS

1. B=G at ωε=σ, then f = σ/2πε = 6x106/(2πx3x8.8x10-12) ≈ 30x106 GHz. 2. Using eq.3.1: F= 1x1 / (12 x 8.8x10-12) ≈ 1011 [N] ≈ 1010 [kilo]. Electrostatic

forces are surprisingly large! 3. Using eq.3.4: P=(3-1)x8.8x10-12xE ≈ 18 x 10-12 x E [Cm/m3]. 4. Using eq.3.12: Y=(10-4/10-3)x(2+j2π106x3x8.8x10-12) ≈ 0.2+j16x10-6 [S]. 5. Using eq.8.11: Z = R+jX = (G-jB)/Y2 = (0.2-j16x10-6)/(0.22+162x10-6x2) ≈

5-j0.4x10-3 [Ω]. 6. Using eq.3.21: ε'=Cd/A=15x10-12x10-3/10-4 = 15x10-11 [F/m] and εr'=ε'/ε0 ≈20.

σ'=d/AR=10-3/(10-4x106) = 10-5 [S/m]. ε''= σ'/ω = 10-5/2π106=1.5x10-12 [F/m]. To find the resistivity, σ'' must first be found with eq.3.21, and then σ' and σ'' are inserted in eq.3.18.

7. Using eq.3.16 and noticing that in order to have relative permittivity it must be put in the form εr''= σ'/ωε0 :


CH 4 PROBLEMS

1. From data found in Section 4.1, what is roughly the permittivity of water at 37oC? 2. Are the electrical properties of amino acids dependent on the water content? 3. With a membrane capacitance of 1uF/cm2 and a membrane thickness of 7 nm,

what is the permittivity of the membrane material? 4. How large is the conductivity anisotropy at low frequencies in muscle tissue

(Fig.4.9)? 5. Sketch roughly the corresponding ε'' curve on Fig.4.13. 6. Fig.4.25 indicates that the resistance of a finger is about 500 ohm. Calculate the

mean resistivity of the finger tissue modelled as a cylinder. Use the dimensions of your own longest finger. CH 4 SOLUTIONS

1. From Fig.4.1 the relative permittivity is about 73. 2. A strong dependence, cf. Fig.4.3. 3. Using eq.3.21: ε'=Cd/A= 10-2x7x10-9/1=7x10-11 [F/m]; and εr'=7.7. 4. Transversal/longitudinal conductivity ratio about 1:8. 5. The values of ε'' can be found by using eq.3.16, the values are simply the σ' values

divided by ω. As Fig.4.13 uses relative permittivity it would be logical to divide also by ε0 to obtain εr'' values.

6. From Section 3.3 ρ=RA/L and finger length L=9 cm, diameter 2,4 cm: ρ= =500xπx1.22x10-4/9x10-2 ≈ 210 [Ωm]


CH 5 PROBLEMS

CHAPTER 5 SOLUTIONS

1. Nernst equation 2.24 in the form of eq.5.1: V = 61 x log(ce/ci) = - 43 [mV]. The K+ concentration is higher on the outside, so the inside is negative. The Nernst equation is dealing with one ion species, and in a real case there are more than one species. Then the Goldman equation is useful, but the permeability (P) of each ion must be known.

2. The area of the cell membrane is 4πr2=4π(5x10-6)2 = π10-10 [m2] and the capacitance of the membrane 10-2 [F/m2]. The total membrane capacitance is then π [pF]. Q=CU and therefore ∆Φ=∆q/C=2x106x1.6x10-19/π10-12 = 0.1 [V].

3. Eq.5.3 (Weiss) gives I=1.1x I0. Eq.5.4 (Lapicque) gives 1.00004xI0. 4. The action potential is the potential waveform measured with fixed electrode

positions when an excitable cell or group of cells are triggered. It refers to any fixed PU electrode leads inside or outside the cell


CH 6 PROBLEMS 1. A sphere of radius 1 cm is positioned in an infinite medium of resistivity 1

ohmmeter. a. Calculate the resistance between the sphere and an infinite large and

distant counter electrode. b. Calculate the resistance with a spherical counter electrode at distance 10

cm. c. Calculate the resistance with an infinite area plate counter electrode at

distance 10 cm. 2. Consider a large plastic cylinder of diameter 80 cm filled with saline of ρ=2

Ωmeter. In the centre is placed an electrical equivalent heart dipole of moment 10 mAcm. Calculate the measured voltage in two pick-up electrodes 10 cm apart, parallel to the heart dipole and at distance 10 cm. Consider the case to be ideal according to Section 6.4.2 .

3. Make a comparison between the 2-electrode system (the resistance of two hemispheres far away from each other according to eq.6.1 is R=ρ/πa) and the 4-electrode system of eq.6.35.

4. Find the dipole current of a heart vector with the following parameters given: u=1mV, Lpu=0.3m, Lcc=1 cm, r: 5 cm, ρ=2Ωm. Consider eq.6.29 valid.

CH 6 SOLUTIONS

1. a. Eq.6.1 is valid for a hemisphere, for a total sphere the resistance R is halved: R=ρ/4πa=1/4πx10-2= 8 [Ω]. b. Eq.6.2 is valid for a hemisphere, for a complete sphere the resistance R is halved: R= (ρ/4π)x(1/a-1/r)=(1/4 π)x(1/10-

2-1/10-1)= 7.2 [Ω]. c. Using eq.6.9: G=1x4xπx10-2x(1+10-2/5x10-2+10-

4/25x10-4) = 0.156 [S] and R=6.4 [Ω]. 2. Using eq.6.15: ∆Φ=2x10x10-5x10-1/4π10-1x3 = 1.5 mV. 3. The resistance between the two spheres is ρ/πa. The transfer resistance in the

4-electrode system is according to eq.6.35 ρ/2πd. When d>>a it is clear that the 4-electrode system measures a much lower resistance. The two systems are not at all compatible: the resistance of the two spheres is dominated by the zone near the electrodes, the 4-electrode system by the volume between the electrodes.

4. Using eq.6.29: I=4πr3u / ρLpu Lcc=4πx5-2x3x10-3 / 2x0.3x10-2= 0.13 [mA]

C:\Documents and Settings\bensonm\Desktop\Grimnes\Problems ch 7.doc25 April 2008 12:30 PM 1Sverre Grim

CH 7 PROBLEMS

1. What is transfer resistance? 2. What is reciprocity? 3. Why is eq.7.24 important for understanding lock-in amplifiers and Fourier

analysis? 4. What is a polarized electrode? Give an example. 5. What is a non-polarizable electrode? Give an example. 6. What is a typical dry skin impedance modulus at 10 Hz? 7. What is a typical wet skin impedance modulus at 10 Hz? 8. What is a typical wet gel electrode polarization impedance modulus at 10 Hz? 9. What is a typical solid gel electrode polarization impedance modulus at 10 Hz?

CHAPTER 7 SOLUTIONS

1. The transfer impedance is Ztr = u/i and refers to black box two port systems where i is the input current to port 1 and u the voltage difference at port 2.

2. Reciprocity in black box two port systems is the swapping of the current injection and voltage reading ports.

3. Eq.7.24 shows that if two sine waveforms of equal frequency are multiplied with each other, the product contains a dc component (½cosφ). A lock-in amplifier therefore works as a rectifying filter, only waveforms containing the reference signal contributes to the dc output. Fourier analysis: By multiplying a known sinusoidal reference signal with an unknown waveform, the dc level of the product tells us whether a reference frequency signal is contained in the unknown waveform.

4. A polarized electrode has current flow through its metal surface. Example: A CC electrode.

5. A non-polarizable electrode has only small half-cell voltage dependence on current flow. Example: Silver-silver chloride electrode in a NaCl solution.

6. From Fig 7.36 and eq.3.23b: 200x3 = 600 kΩcm2. 7. From Fig 7.36 and eq.3.23b: 12x3 = 36 kΩcm2. 8. From Fig 7.39 and eq.3.23b: 113x3 = 339 Ωcm2 (X and R are found to be 80Ω

for one electrode. The modulus is therefore (802+802)0.5 = 113 Ω for 3 cm2). 9. From Fig 7.41 and eq.3.23b: 2022x5 ≈ 10 kΩcm2. (X is found to be 300Ω and R

to be 2000 Ω for one electrode. The modulus is therefore (3002+20002)0.5 = 2022 Ω for 5 cm2). This electrode can therefore not be used for skin impedance measurements at 10 Hz.


CHAPTER 8 SOLUTIONS

1. Admittance Y=G+jB=1/R+jωC=1/103+j2πx314x2.2x10-6 = 1 + j4.3 [mS]. Then |Y|2=10-6+18.5x10-6=19.5x10-6. According to eq.8.11: Impedance Z= 10-3/ 19.5x10-6-j4.3x10-3/19.5x10-6=51-j221 [Ω].

2. a), b) Complex admittance plots

B

G

0,1

1

10

100

0,01 0,1 1 10 100 1000 10000 100000Hz

B (

S)

10

100

G (

S)

m=0.7

B

G

0

5

10

15

20

25

30

35

0,01 0,1 1 10 100 1000 10000 100000Hz

B (

S)

10

20

30

40

50

60

70

80

90

100

G (

S)

Figure 1. Complex admittance. Top: Logarithmic admittance scales (2.dispersion visible).

Bottom: Linear admittance scales (2.dispersion not visible!). If B is according to eq.11.15, the frequency exponent m at low frequencies can be found from the diagram to be 0.7. If the system

were Fricke compatible, this would have corresponded to a ϕcpe=63o for the main dispersion.


c) Admittance and impedance Wessel plots.

50Hz

20

10

3

100200

5001000

0

20

40

0 20 40 60 80 100G (µ S)

jB (

S

ϕcp e

0,010,1

0,3

1Hz

0,0

0,5

1,0

1,5

2,0

2,5

3,0

3,5

4,0

22,5 23,0 23,5 24,0G (µ S)

jB (

S)

1

3

200100

5020

10Hz

0,3

-15

-10

-5

0

0 10 20 30 40 50R (k ohm )

jX (

ko

hm

)

ϕcp e

Figure 2. Wessel plots. Top: admittance. Low frequency details at the right hand side. ϕcpe is estimated to be around 80o. Comparing this result with that of Fig.1, the system under investigation is not Fricke compatible

for the main dispersion. The characteristic frequency is estimated to be fc=75Hz. Bottom: impedance. ϕcpe is estimated to be around 80o, as for admittance. However, the

characteristic frequency is estimated to be fc = 9Hz, quite different from the admittance case. d) Bode-plot, admittance.

Y

ϕ

10

100

0,01 0,1 1 10 100 1000 10000 100000Hz

degr

ees

)

0

5

10

15

20

25

30

35

Y (

S)

Figure 3. Admittance shown in a Bode plot with magnitude (log scale) and phase angle (linear scale). Note that the characteristic frequency fc is a little difficult to determine directly from the Y

curve because of the curveform change caused by the logarithmic admittance scale.


e) Capacitance and permittivity plots

0,0001

0,001

0,01

0,1

1

10

100

1000

10000

0,01 0,1 1 10 100 1000 10000 100000Hz

Cp

(nF)

10,0

100,0

G (

S)

G

Cp

Figure 4. Parallel capacitance and conductance plotted with log-scales, Table data. The LF 2.dispersion not visible in the G plot.

ε r'

ε r''

1,E-01

1,E+01

1,E+03

1,E+05

1,E+07

1,E+09

0,01 0,1 1 10 100 1000 10000 100000

Hz

ε r

Figure 5. Complex relative permittivity εr shown with log-scales. Calculated variables from the Table data, the equations in Section 3.3 and with A=1cm2 and d=1mm. The εr' curve follows the capacitance curve in Fig.4. The εr'' curve has two parallel line segments with a transition zone

around 50 Hz.


500

200

100

50Hz

20

0,0E+00

1,0E+05

2,0E+05

3,0E+05

4,0E+05

0,0E+00 1,0E+05 2,0E+05 3,0E+05

ε r'

-jε r''

0,30,1

0,03Hz

0,01

0,0E+00

1,0E+08

2,0E+08

3,0E+08

4,0E+08

5,0E+08

0,0E+00 5,0E+06 1,0E+07

ε r'

-jε r''

Figure 6. Relative permittivity Wessel plots. Data as for Fig.5. No clear arc is seen. Note different εr´ and εr´´ scales on the low frequency right diagram.