Transcript Nervous System Stimulation Using Microwave

Nervous System Stimulation
Using Microwave
Class Project of ELCT 891B
Course Instructor:
Dr. Mohammod Ali
Prepared By:
Md. Anas Boksh Mazady
[email protected]
Date: 04/07/2009
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Outline

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Overview of human nervous system
How signal propagates in human nervous system
Review of 2 papers
◦ Model of the stimulation of a nerve fibre
◦ Design considerations of MC
Findings
Future scopes
Conclusion
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Human Nervous System
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Human Nervous System (Cont)
[www.spineuniverse.com]
1.
2.

Central Nervous System
i) Brain
ii) Spinal Cord
iii) Nerve roots
Peripheral Nervous System: 31 pairs of nerve
roots branch off the spinal cord to enable the
body to move and feel.
Fig: Vertebrae
i) Cervical – 8 pairs
ii) Thoracic – 12 pairs
iii) Lumbar – 5 pairs
iv) Sacral – 5 pairs
v) Coccyx – 1 pair
The spinal cord is nearly an inch in diameter at its widest point
and 18 inches long.
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Signal Propagation in Human Nervous System
[Paul Davidovits, “Physics in biology and medicine,” pp. 180-197, 2007]
When a neuron receives an
appropriate stimulus, it produces
electrical pulses that are propagated
along its cable like structure.
 The pulses are constant in magnitude
and duration. The strength of the
stimulus is conveyed by the number
of pulses produced.
 Each neuron consists of a cell body
to which are attached input ends called
dendrites and a long tail called axon
which propagates the signal away from
the cell.
 Some axons may be as long as 1 meter, such as the axons connecting the spine
with the fingers and toes. Some of the axons are covered with a segmented sheath
of fatty material called myelin.

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Signal Propagation in Human Nervous System (cont)

The segments are about 2 mm long, separated by gaps called the Nodes of Ranvier.
The largest axons in humans have a diameter of only about 20um. The myelin
sheath increases the speed of pulse propagation along the axon.

The neurons can be divided into 3 classes: sensory neurons, motor neurons, and
interneurons.

A stimulus from a muscle produces nerve impulses that travel to the spine. Here
the signal is transmitted to a motor neuron, which in turn sends impulses to
control the muscle. This is an example of the simplest reflex action.

In the aqueous environment of the body, salts are dissociated into ions. As a result,
body fluids are relatively good conductors, the resistivity is 100 million times
greater than that of copper.
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Signal Propagation in Human Nervous System (cont)

Fluids inside the axons are separated by a thin (50e-10 m to100e-10 m) axon
membrane. This is a relatively good but not perfect electrical insulator.

The electrical resistivities of the internal and external fluids are about the same,
but their chemical compositions are substantially different. The ionic solutes in the
external fluids are mostly sodium and chloride ions. Whereas the internal fluids are
mostly positive potassium ions and negatively charged large organic molecules.

In the resting condition, when the axon is not conducting an electrical pulse, the
axon membrane is highly permeable to potassium and only slightly permeable to
sodium, impermeable to the large organic ions.

As a result, a negative potential of 70 mV is produced inside the axon with respect
to the outside.
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Signal Propagation in Human Nervous System (cont)

Some sodium ions do in fact leak into the axon, but they are continuously removed
by a metabolic mechanism called “sodium pump”. This mechanism brings in an
equal number of potassium ions.

Physiologists have studied the properties of nerve impulses by inserting a probe
into the axon and measuring the changes in the axon voltage with respect to the
surrounding fluid.

A nerve impulse is produced only if the stimulus exceeds a certain threshold value,
typically +15 mV higher than the resting value. When this value is exceeded, an
impulse is generated at the point of stimulation and propagates down the axon.
Such a propagating impulse is called an “action potential”
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Signal Propagation in Human Nervous System (cont)

A sudden rise of potential is observed inside the axon for this impulse.
-70 mV  +30 mV  -90 mV  -70 mV

Fast-acting axons propagate the pulse at speeds up to 100 m/sec.

Impulses produced by a given neuron are always of the same size and
propagate down the axon with very less attenuation as the axon is insulated.
Nevertheless, to ensure that the signal does not fail, nodes of Ranvier help
boost the signal.

Nerve impulses are produced at a rate proportional to the intensity of the
stimulus. The upper limit of the frequency is limited by the propagation speed
of the impulse.
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Axon as an electric cable

The axon may be approximated as an insulated electric cable submerged in a
conducting fluid. We’ll have to take into account the resistances of the fluids both
inside and outside the axon, as the membrane is a lossy dielectric it is
characterized by both capacitance and resistance.
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Axon as an electric cable

However this model fails to explain some of the most striking
characteristics of the axon. An electrical signal along such a circuit
propagates at nearly speed of light (3e8 m/sec), whereas a pulse along an
axon propagates at a speed of about 100 m/sec.

Furthermore, this circuit dissipates an electrical signal very quickly, yet we
know that action potentials propagate along the axon without any
attenuation.
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Propagation of action potential
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Propagation of action potential

When the voltage across a portion of the membrane is reduced below a
threshold, the permeability of the axon membrane to sodium ions increases
rapidly. As a result, potential inside axon rises to +30 mV. This potential
increases the permeability of the membrane to the Na immediately ahead of
it, which in turn produces a spike in that region. In this way the pulse is
propagated.

At the peak of the action potential, the membrane closes its gate to sodium
and opens them wide to potassium ions. The potassium ions now rush out
making the inside potential negative again.

We know,
ΔQ = C ΔV
ΔV = 70 mV - (-30) mV = 100 mV
C = 3e-7 F/m, for nonmyelinated axon
So, ΔV/q = 1.87e11, sodium ions enter per meter of axon length,
which is much lower than the number of ions at rest (7e15)
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Propagation of action potential (cont)

During the propagation of one pulse, the whole axon capacitance is successively
discharged and then must be recharged again. The energy required to recharge 1
m length of nonmyelinated axon is,
E = ½ (C ΔV)2 = ½ * 3e-7 * (0.1)2 = 1.5e-9 J/m

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

Since the duration of each pulse is 1/100 sec,
so required power to recharge its capacitance is
1.5e-7 W/m.
At the far end, the axon branches into nerve endings which extend to the cells
that are to be activated. The nerve endings are not in contact with the cells. This
gap is called “synapse” and it is about 1 nm.
When the impulse reaches the synapse, a chemical substance is released at the
nerve ending which quickly diffuses across the gap and stimulates the adjacent
cells. The chemical is released in bundles of discrete size.
In the muscle fiber, the duration of action potential is usually longer,
about 20 msec. In heart muscle it may last a quarter of second.
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Bradley J. Roth, Peter J. Basser, “A model of the stimulation of a nerve fibre by electromagnetic
induction,” IEEE Transactions on Biomedical Engineering, vol. 37, No. 6, June 1990.
Analysis: Without stimulation
 From Ohm’s Law:
riIi  
V
…………….(1)
x
From KCL: Membrane current,
im  
 Ii
x
………………(2)
V V
i
m  cm

Also,
t rm
……….(3)
 2V
V
Combining equations (1), (2), and (3) 

V


x 2
t
2
Where,
Length Constant,

rm
ri
and time constant, 
………(4)
 cm rm
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Model of a Nerve Fiber (cont)
V
E


i
• Axial component of electric field inside the fiber,
x

With electromagnetic induction by a time varying magnetic field
V
Ei  
  x ( x, t )
x
induced electric field parallel to the fiber
But the induced electric field equals the negative rate of change of the vector
magnetic potential 
A
 
t
V
r
iIi  
  x ( x, t )
Now,
x
So,
 2V
V
2  x


V




x 2
t
x
2
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Model of a Nerve Fiber (cont)

The reason that the derivative of the electric field, not the electric field
itself appears in the last equation is because the membrane current, not
the axial current along the fiber, depolarizes the membrane.

The membrane current is a maximum where the spatial gradients of the
axial currents and the induced electric field are the largest.

Interestingly, at the location along the fiber where the electric field is
maximum the axial derivative of the electric field must necessarily zero.
So, we expect only a little or no stimulation to occur where the electric
field is the largest.
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Hodgkin-Huxley Model

The three conductances represent the sodium, potassium, and leakage
channels.

With this modifications the cable equation becomes
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Hodgkin-Huxley Model (cont)

In these equations it has been assumed that
the resting potential is -65 mV.
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Hodgkin-Huxley Model (cont)

The induced electric field can be calculated from the coil current and its
geometry by
dI (t ) 0 N
 (r , t ) 
(
dt
4

dl 
)
r  r
Where, I (t )  coil current
r  position where the electric field is calculated
r  position of the differential element of the coil dl
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Electric field and its gradient

The coil is at a distance of 1 cm from the nerve

The coil is assumed to be 64-sided polygon

Coil radius rc = 2.5 cm

No. of turns, N = 30

The nerve is parallel to x axis and is tangent to the coil at y = rc
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Electric field and its gradient (cont)
Electric field (dI/dt = 1 A/us)

Electric field gradient
The maximum of the derivative is not below the center of the coil nor at
its edge.
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Electric field and its gradient (cont)

This is the contour plot of the
derivative of the axial component
of electric field

The bold circle represents the
position of the coil, with the arrow
pointing to the direction of the rate
of change of current.

The dotted line indicates the location
of the nerve fiber.

The minus signs indicate the regions
where we expect a nerve that is
parallel to the x-axis to be maximally
depolarized, i.e. the location of
stimulation.

The plus signs indicate the regions of hyperpolarization.
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The current in the coil

The current pulse is generated when a capacitor C, initially charged to a
voltage V0, is discharged through a coil whose inductance is L and
resistance is R

The inductance L of a circular coil of radius rc wound with N turns of wire
having radius rw is
L   0 rc N 2 (ln(
8rc
)  1.75)
rw
A coil of radius 2.5 cm with 30 turns wound from 1.0 mm radius wire has
an inductance of 0.165 mH.
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The current in the coil

With R = 3 Ω, L=0.165 mH, C = 200 µF
R2 / 4L2 – 1/LC = 5.2342e+007 >> 1
So, the coil current I(t) would have an overdamped response.
Where,
ω1 = R/2L = 9.07 ms-1
= 7.21 ms-1
V0 = 200 V

So the resulting I(t) would be 
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Electric Field Gradient
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Findings

The nerve fiber is stimulated by the gradient of the component of the
electric field that is parallel to the fiber.

This electric field component hyperpolarizes or depolarizes the membrane
and may stimulate an action potential.

A non linear Hodgkin-Huxley cable model describes the response of the
nerve fiber to induced electric field.

Maximum stimulus is neither below the center of the coil nor below it’s
edges.
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MC design considerations
[Vernon Weh-Hau Lin, Ian N. Hsiao, and Vijay Dhaka, “Magnetic coil design considerations for
functional magnetic stimulation,” IEEE Trans. on Biomed. Eng., Vol. 47, No. 5, May 2000]
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Theoretical Computations
E generated by a MC can be calculated by
A
E  V 
t
I
dl 
 V 
(  0 / 4 ) 
t
r
 A typical current waveform through the discharging of a capacitor with a V0
to the coil (LRC circuit)

I  (V 0 / L) sin( t ) exp( at )
I / t  (V 0 / L)[cos(t )  (a /  ) sin( t )] exp( at )
E  V  {(V 0 / L)[cos(t )  (a /  ) sin( t )]  exp( at )}(  0 / 4 )  dl  / r
Where,
and
a  R /( 2 L)
  (1 / LC )  a 2
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Theoretical Computations (cont)

By assuming a uniform current distribution along the coils V becomes
insignificant compared to the other factors. So,
dl 
E 
r
1
E
L
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MC design considerations (cont)
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MC design considerations (cont)
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Experimental E Measurement
Dimension of the plastic container = 30 * 30 * 25 cm3
 The plastic container was filled with a saline solution of conductivity 0.002
S/cm (0.8% NaCl).
 The electrodes of the electric field probe were 0.2-0.5 cm apart.
 Frequency = 20 Hz.

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Experimental E Measurement : No of Slinky

The E field distribution for coils IA-IE at 3 mm above each coil is as below

The ratio Eprimary /Esecondary increases as no. of slinky increases. Because
effective r increases, so Esecondary decreases.
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Experimental E Measurement : Circular or Rectangular

Circular (IB )VS rectangular (IF) coil E profile is shown below for same R.

The circular coil has higher primary E field but less focalization as
compared to the rectangular coil (2.6 vs 2.2).
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Experimental E Measurement : Coil Diameter

Coils IIB, IB, and IID with diameters of 11.5, 7.62, and 5.08 cm.

The ratio Eprimary /Esecondary decreases as the coil diameter increases. Eprimary
decreases because L ∞ D and E ∞ 1/L. Esecondary is higher for large dia
because of less interference from the side limb of the other coil.
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Field penetration by coils of different diameters

Coils IB (7.62 cm), IIB (11.5 cm), and IID (5.08 cm) were investigated.

Lower diameter coil has higher initial field strength, but higher diameter
coil penetrates deeper.
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Analytic computation

A computer program, FARADAY, was used to simulate the electric field.
This software uses boundary integral equation to solve for the fields.
I = 30,000 A at 4 KHz was used.
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Analytic Computations : no. of slinky

The E field distribution for coils IA-IE is as below

The ratio Eprimary /Esecondary increases as no. of slinky increases. Because
effective r increases, so Esecondary decreases.
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Analytic Computation : Coil Diameter
Varying Coil Diameter:

Coils IIB, IB, and IID with diameters of 11.5, 7.62, and 5.08 cm were
studied.

The initial field strengths are 454, 517, and 789 V/m respectively. So the
smallest coil has the largest initial field strength. Eprimary increases because L
∞ D and E ∞ 1/L. Esecondary is higher for large dia because of less
interference from the side limb of the other coil.

The ratio Eprimary /Esecondary was found to be 1.5, 2.0, and 2.3 respectively. So
the ratio decreases as the coil diameter increases agreeing to the
experimental results.

But the larger coils maintain a higher field strength after 30 mm distance.
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MC design considerations (cont)
Findings

Slinky coil design produces more focalized stimulation when compared to
the planar round coils. The primary to secondary peak ratios of induced E
from slinky 1 to 5 are 1, 2.2, 2.85, 2.62, and 3.54.

Coils with larger diameters have better penetration than those with
smaller diameters.

Coils with less number of turns have higher initial field strengths.
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Simulations
Phantom
 Homogeneous rectangular phantom
 Size: 225 mm * 150 mm * 150 mm
 Shell of 2 mm at the interface
 Tissue property:
 Conductivity = 0.97 (S/m)
 Relative permittivity = 41.5
 Density = 1000 Kg/m3
 Shell property
 Conductivity = 0 (S/m)
 Relative permittivity = 3.7
 Density = 0 Kg/m3
Coil
 Each coil had 4 turns,
 Wire length was 80 mm,
 Coil radius was 38.1 mm,
 Wire radius was 0.25 mm.
 PEC was used for both coil and wire.
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Simulations (cont)
Run parameters

The nearest coil element was 3 mm apart from the tissue.

Base cell size was 1 mm in all directions.

Sinusoidal current excitation of 30 kA at 915 MHz was used.

Liao absorbing boundary was used.
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Simulations (cont)
Results

Eprimary= 2.2e5 V/m, Esecondary = 1.1e5 V/m, ratio = 2
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Question and Answer
47
Thank you
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