Mathematical Modeling

Università degli Studi di Catania Facoltà di Ingegneria Corso di Laurea Specialistica in Ingegneria dell’Automazione e del Controllo dei sistemi complessi Corso di Fondamenti di Bioingegneria Elettronica

MATHEMATICAL MODELING OF BIOLOGICAL SYSTEMS L’EPISCOPO GAETANO MAZZARA BOLOGNA GIUSEPPE

ANNO ACCADEMICO 2006-2007

ELETTRICAL MODELS FOR BIOLOGICAL SYSTEMS Biological System

Physical Laws

Electrical Parameters

State-Space Equations

Transfer Function

RESISTANCE Resistive and dissipative properties of system

Resistance

Ohm’s Law V=RI

V I

R

V

Potential

I

Current

Ohm’s Law can be generalized, applying it to other systems

Generalized Ohm’s Law y=Rz

y

“Generalized effort”

z

“Generalized flow”

RESISTANCE EXAMPLES Applications of generalized Ohm’s law Mechanics damping law F=Rmv

Fluidic Poiseuille’s law ΔP=RtQ

Fourier’s Thermal transfer law Δθ=RtQ

ΔP

Q

θ1

θ2 Q

φ1

Chemical Fick’s law of Diffusion ΔΦ=RcQ

φ2 Q

CAPACITANCE Capacitance

Storage properties of system V

Capacitance Law

1 V = ∫ idt C

I

C

V

Potential

I

Current

Also Capacitance Law can be generalized, applying it to other systems Generalized Capacitance Law

1 y = ∫ zdt C

y

“Generalized effort”

z

“Generalized flow”

CAPACITANCE EXAMPLES Applications of generalized Capacitance law Hooke’s Mechanics Compliance law 1 F= vdt ∫ CM

x F

ΔV

Fluidic Compliance law ΔV=CfΔP

Thermal Heat storage law Q=Ct Δθ

ΔP

θ1

θ2 Δ θ=θ1-θ2

INERTANCE Inertance

Inertial properties of system

Inductance Law

dI V =L dt

V

I

L

V

Potential

I

Current

Also inertance Law can be generalized, applying it to other systems Generalized Inductance Law dz y=L dt

y

“Generalized effort”

z

“Generalized flow”

INERTANCE EXAMPLES Applications of generalized inertance law Newton’s second law

ma

dv F =m dt

m

F

Fluidic inertance law

dQ ∆P = LF dt

Q

ΔP

There is no element that represents inertance in thermal and chemical systems

Exercise 1: 5-element Windkessel Model of aortic and arterial hemodynamics

Rao

Viscous resistance of aortic wall

Lao

Inertance to flow through aorta

Cao

Compliance of aortic wall

Rp//Cp

Modeling of the rest of arterial vasculature

Exercise 1: 5-element Windkessel Model of aortic and arterial hemodynamics VC

State space equations: dQ R V P = − ao Q − C + ao dt Lao Lao Lao dVC Q Pao = − dt Cao + C P Rao ( Cao + C P )

Pao Q, VC

input

Q

output

State space variables

Transfer function: Q S ( RP Cao + RPC P ) + 1 = 2 Pao S ( RPCao Lao + RPC P Lao ) + S ( Lao + Rao R p Cao + Rao RPC P ) + Rao + R P

Exercise 2: Equivalent electrical circuit of Hodgkin-Huxley model of neuronal electrical activity

C Rk,Na,C1 Ek,Na,C1

Membrane capacitance Resistance of membrane to K,Na,C1 Nernst Potential of membrane for K,Na,C1

Exercise 2: Equivalent electrical circuit of Hodgkin-Huxley model of neuronal electrical activity Equations are: dV  1 1 1  E K E Na ECl   I =C + + + V+ − +  dt  RK RNa RCl  RK RNa RCl IK =

I Na

V + EK RK

V − E Na = RNa

I Cl =

V + ECl RCl

Exercise 3: Analysis of the respiratory mechanics model with effect of inertance to gas flow in central airways RP

RC, LC, CS

Cw CL

Rc Lc

Resistance of central airways

Inertance through central airways

Cw

Cs

Compliance of central airways

CL

Rp

Resistance of peripheral airways Compliance of chest-wall Compliance of lung

Exercise 3: Analysis of respiratory mechanics model with effect of inertance to gas flow in central airways Equations are: Pao = RC Q + LC

dQ 1 + dt C S

∫ ( Q − Q ) dt A

 1 1  1  ∫ QA dt = RP QA +  + CS  C L CW 

∫ ( Q − Q ) dt A

Reducing two equations to one, we obtain: d 2 Pao 1 dPao d 3Q  LC  d 2Q  1 RC  dQ 1  1 1    +  Q   + = L + R + + + + C C 2 3 2     dt RPCT dt dt  RP CT  dt  CS RPCT  dt R P CS  CW CL  Where:

 1 1 1   CT =  + + C C C W S   L

−1

Exercise 3: Analysis of respiratory mechanics model with effect of inertance to gas flow in central airways Used Simulink model is: Input

Pao=sin(2πb*60-1*t) cm H2O where b = breaths/min

Output

Q and Volume

Fixed values for system parameters are: RC= 1 cm H2O L-1 RP= 0.5 cm H2O L-1 CL= 0.2 L cm H2O CW= 0.2 L cm H2O CS= 0.005 L cm H2O

Exercise 3: Analysis of respiratory mechanics model with effect of inertance to gas flow in central airways Simulation for LC=0 cm H2O s2 L-1

Neglecting inertance

Peak to peak amplitudes at 15 breaths/min: Q=0.127 L/s Volume=0.502 L

Peak to peak amplitudes at 60 breaths/min: Q=0.504 L/s Volume=0.496 L

Exercise 3: Analysis of respiratory mechanics model with effect of inertance to gas flow in central airways Simulation for LC=0.01 cm H2O s2 L-1

Taking inertance into account

Peak to peak amplitudes at 15 breaths/min: Q=0.129 L/s Volume=0.515 L

Peak to peak amplitudes at 60 breaths/min: Q=0.512 L/s Volume=0.509 L

Exercise 3: Analysis of respiratory mechanics model with effect of inertance to gas flow in central airways Simulation for LC=0.01 cm H2O s2 L-1, CL=0.4 L cm H2O-1, RP=7.5 cm H2O s L-1 Subject with emphysema (higher lung compliance and higher peripheral airway resistance)

Peak to peak amplitudes at 15 breaths/min: Q=0.166 L/s Volume=0.661 L

Peak to peak amplitudes at 60 breaths/min: Q=0.457 L/s Volume=0.496 L

Exercise 3: Analysis of respiratory mechanics model with effect of inertance to gas flow in central airways Summary of all simulations

Conclusions: Both inertance-complete model and neglecting-inertance one have quite similar trends at all input frequencies. Emphysema model, instead, has a very different trend, particularly at high frequencies; infact its peak-to-peak amplitude both for air flow and for air volume is smaller as emphysema features outline.