Closed Loop Control System

Closed Loop Control System

Solution 

Report

Problem a:

Steady state right atrial pressure: 4.2850 mm Hg

Cardiac output: 5.4270 L/min

Problem b:

Steady state right atrial pressure: 3.4540 mm Hg

Cardiac output: 3.5442 L/min

Problem c:

Steady state right atrial pressure: 5 mm Hg

Cardiac output: 7 L/min

%%Cardiac Output Curve Data

Pra = 0:13; %%in mm Hg

Co = [0 0.3 1.0 2.5 4.8 7.0 9.0 10.5 11.0 11.3 11.5 …

11.6 11.6 11.6]; %%in L/min

%%Venous return characteristics

symsQr(P);

Qr(P)= piecewise(P <=0, 14, 0 < P < 7 , 14 – 2*P, 0);

Q = Qr(Pra);

%%plot

figure(1)

plot(Pra,Co,’o-‘,Pra,Q);

grid on;

xlabel(‘Right Atrial Pressure,Pra (mm Hg)’)

ylabel(‘Cardiac Output of Venous Return(L/min)’)

legend(‘Cardiac Output Curve’,’Venous Return Curve’)

title(‘Problem 1’)

%%From the plot we get that the intesection lies between

%% Pra=4 and Pra=5, so we now search for the exact point

%%of intersection

%%We use linear interpolation with Pra=4 and Pra=5

tol = 0.005;

x1 = Pra(5);x2 = Pra(6);y1 = Co(5);y2 = Co(6);

for Pressure=x1:0.001:x2

Co_linear = (y2-y1)/(x2-x1)*(Pressure-x1) + y1;

if abs(Qr(Pressure)- Co_linear)<tol

Co_result = Co_linear;

break;

end

end

%%Part 1 solution

steady_state_pressure_Prob1 = Pressure;

cardiac_output_Prob1 = Co_result;

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%%From eq. 3.35b we get that Qr is inversely proportional to total

%%resistance. This implies if the total resistance is doubled Qr will be

%%halved.

symsQr_doubleResistance(P);

Qr_doubleResistance(P)= piecewise(P <=0, 14/2, 0 < P < 7 , (14 – 2*P)/2, 0);

Q_doubleResistance = Qr_doubleResistance(Pra);

%%plot

figure(2)

plot(Pra,Co,’o-‘,Pra,Q_doubleResistance);

grid on;

xlabel(‘Right Atrial Pressure,Pra (mm Hg)’)

ylabel(‘Cardiac Output of Venous Return(L/min)’)

legend(‘Cardiac Output Curve’,’Venous Return Curve’)

title(‘Problem 2’)

%%From the plot we get that the intesection lies between

%% Pra=3 and Pra=4, so we now search for the exact point

%%of intersection

%%We use linear interpolation with Pra=3 and Pra=4

tol = 0.005;

x1 = Pra(4);x2 = Pra(5);y1 = Co(4);y2 = Co(5);

for Pressure=x1:0.001:x2

Co_linear = (y2-y1)/(x2-x1)*(Pressure-x1) + y1;

if abs(Qr_doubleResistance(Pressure)- Co_linear)<tol

Co_result = Co_linear;

break;

end

end

%%Part 2 solution

steady_state_pressure_Prob2 = Pressure;

cardiac_output_Prob2 = Co_linear;

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%%From eq. 3.35b we get that Qr=(Pms-Pra)/(Rv+(Ra*Ca)/(Ca+Cv))

%%If mean systemic pressure is increased by 5mm Hg Pms in the expression

%%will be replaced by Pms+5

symsQr_sysPressure(P);

Qr_sysPressure(P)= piecewise(P <=0, 7 + 5, 0 < P < 12 , (14 – 2*P)/2 + 5, 0);

Q_sysPressure = Qr_sysPressure(Pra);

%%plot

figure(3)

plot(Pra,Co,’o-‘,Pra,Q_sysPressure);

grid on;

xlabel(‘Right Atrial Pressure,Pra (mm Hg)’)

ylabel(‘Cardiac Output of Venous Return(L/min)’)

legend(‘Cardiac Output Curve’,’Venous Return Curve’)

title(‘Problem 3’)

%%From the plot we get that the intesection lies between

%% Pra=5 and Pra=6, so we now search for the exact point

%%of intersection

%%We use linear interpolation with Pra=5 and Pra=6

tol = 0.005;

x1 = Pra(6);x2 = Pra(7);y1 = Co(6);y2 = Co(7);

for Pressure=x1:0.001:x2

Co_linear = (y2-y1)/(x2-x1)*(Pressure-x1) + y1;

if abs(Qr_sysPressure(Pressure)- Co_linear)<tol

Co_result = Co_linear;

break;

end

end

%%Part 3 solution

steady_state_pressure_Prob3 = Pressure;

cardiac_output_Prob3 = Co_result;

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