Mathematical Modeling

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    14-Nov-2014

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<p>Universit degli Studi di Catania Facolt di Ingegneria Corso di Laurea Specialistica in Ingegneria dellAutomazione e del Controllo dei sistemi complessiCorso di Fondamenti di Bioingegneria Elettronica</p> <p>MATHEMATICAL MODELING OF BIOLOGICAL SYSTEMSLEPISCOPO GAETANO MAZZARA BOLOGNA GIUSEPPE</p> <p>ANNO ACCADEMICO 2006-2007</p> <p>ELETTRICAL MODELS FOR BIOLOGICAL SYSTEMS Biological System</p> <p>Physical Laws</p> <p>Electrical Parameters</p> <p>State-Space Equations</p> <p>Transfer Function</p> <p>RESISTANCEResistance Resistive and dissipative properties of systemV I R</p> <p>Ohms Law V=RI</p> <p>V I</p> <p>Potential Current</p> <p>Ohms Law can be generalized, applying it to other systems</p> <p>Generalized Ohms Law y=Rz</p> <p>y z</p> <p>Generalized effort Generalized flow</p> <p>RESISTANCE EXAMPLESApplications of generalized Ohms law Mechanics damping law F=Rmv</p> <p>Fluidic Poiseuilles law P=RtQ1</p> <p>Q</p> <p>P</p> <p>Fouriers Thermal transfer law =RtQ</p> <p>2 Q</p> <p>1</p> <p>2 Q</p> <p>Chemical Ficks law of Diffusion =RcQ</p> <p>CAPACITANCECapacitance Capacitance Law Storage properties of systemV</p> <p>1 V = idt C</p> <p>VI C</p> <p>Potential Current</p> <p>I</p> <p>Also Capacitance Law can be generalized, applying it to other systems Generalized Capacitance Law</p> <p>1 y = zdt C</p> <p>y z</p> <p>Generalized effort Generalized flow</p> <p>CAPACITANCE EXAMPLESApplications of generalized Capacitance law Hookes Mechanics Compliance law 1 F= vdt CM Fluidic Compliance law V=CfPx F</p> <p>V P</p> <p>Thermal Heat storage law Q=Ct </p> <p>1</p> <p>2 =1-2</p> <p>INERTANCEInertance Inductance Law Inertial properties of systemV L</p> <p>dI V =L dt</p> <p>V I</p> <p>Potential Current</p> <p>I</p> <p>Also inertance Law can be generalized, applying it to other systems Generalized Inductance Law dz y=L dt</p> <p>y z</p> <p>Generalized effort Generalized flow</p> <p>INERTANCE EXAMPLESApplications of generalized inertance law Newtons second lawma m F</p> <p>dv F =m dt</p> <p>Fluidic inertance law</p> <p>dQ P = LF dt</p> <p>Q</p> <p>P</p> <p>There is no element that represents inertance in thermal and chemical systems</p> <p>Exercise 1: 5-element Windkessel Model of aortic and arterial hemodynamics</p> <p>Rao Lao Cao Rp//Cp</p> <p>Viscous resistance of aortic wall Inertance to flow through aorta Compliance of aortic wall Modeling of the rest of arterial vasculature</p> <p>Exercise 1: 5-element Windkessel Model of aortic and arterial hemodynamicsState space equations:dQ R V P = ao Q C + ao dt Lao Lao LaodVC Q Pao = dt Cao + C P Rao ( Cao + C P )Pao Q, VC input Q output VC</p> <p>State space variables</p> <p>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</p> <p>Exercise 2: Equivalent electrical circuit of Hodgkin-Huxley model of neuronal electrical activity</p> <p>C Rk,Na,C1 Ek,Na,C1</p> <p>Membrane capacitance Resistance of membrane to K,Na,C1 Nernst Potential of membrane for K,Na,C1</p> <p>Exercise 2: Equivalent electrical circuit of Hodgkin-Huxley model of neuronal electrical activityEquations are:dV 1 1 1 E E E V + K Na + Cl I =C + + + dt RK RNa RCl RK RNa RCl IK = V + EK RK</p> <p>I Na</p> <p>V E Na = RNa V + ECl RCl</p> <p>I Cl =</p> <p>Exercise 3: Analysis of the respiratory mechanics model with effect of inertance to gas flow in central airwaysRP RC, LC, CS</p> <p>Cw CL</p> <p>Rc Lc Cs</p> <p>Resistance of central airways Compliance of central airways</p> <p>Rp Cw CL</p> <p>Resistance of peripheral airways Compliance of chest-wall Compliance of lung</p> <p>Inertance through central airways</p> <p>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</p> <p> ( Q Q ) dtA</p> <p> 1 1 1 QA dt = RP QA + + C CS L CW </p> <p> ( Q Q ) dtA</p> <p>Reducing two equations to one, we obtain:d 2 Pao 1 dPao d 3Q LC d 2Q 1 R dQ 1 1 1 2 + + C + + = LC 3 + RC + 2 dt C R C dt R C C + C Q dt RPCT dt dt RP CT P T P S W L SWhere: 1 1 1 CT = + + C CW CS L 1</p> <p>Exercise 3: Analysis of respiratory mechanics model with effect of inertance to gas flow in central airwaysUsed Simulink model is: Input Pao=sin(2b*60-1*t) cm H2O where b = breaths/min Output Q and Volume</p> <p>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</p> <p>Exercise 3: Analysis of respiratory mechanics model with effect of inertance to gas flow in central airwaysSimulation for LC=0 cm H2O s2 L-1 Neglecting inertance</p> <p>Peak to peak amplitudes at 15 breaths/min: Q=0.127 L/s Volume=0.502 L</p> <p>Peak to peak amplitudes at 60 breaths/min: Q=0.504 L/s Volume=0.496 L</p> <p>Exercise 3: Analysis of respiratory mechanics model with effect of inertance to gas flow in central airwaysSimulation for LC=0.01 cm H2O s2 L-1 Taking inertance into account</p> <p>Peak to peak amplitudes at 15 breaths/min: Q=0.129 L/s Volume=0.515 L</p> <p>Peak to peak amplitudes at 60 breaths/min: Q=0.512 L/s Volume=0.509 L</p> <p>Exercise 3: Analysis of respiratory mechanics model with effect of inertance to gas flow in central airwaysSimulation 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)</p> <p>Peak to peak amplitudes at 15 breaths/min: Q=0.166 L/s Volume=0.661 L</p> <p>Peak to peak amplitudes at 60 breaths/min: Q=0.457 L/s Volume=0.496 L</p> <p>Exercise 3: Analysis of respiratory mechanics model with effect of inertance to gas flow in central airwaysSummary of all simulations</p> <p>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.</p>