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hw_07_sol - Vo=-K(1 K*x,1 plot(t,Vo xlabel'time ylabel'Vo...

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Homework 7-3. Integrating circuit. % Setup for homework 7-3: Setup73.m % global K K= 10000; % Case b uses K= 100 ts= [ 0.0:0.01:1.0 ]; x0= [ 0 ]; [t,x]= ode23('StEqs73',ts,x0); Vo= ( -K/(1+K))*x(:,1); plot(t,Vo); xlabel('time'); ylabel('Vo'); title('Response for const inp[ut voltage 10 v'); % File StEqs73: state equations for hw 7-3. % function xdot= StEqs73(t,x) % Parameters global K Ri= 1000; Cf= 0.0001; % Input Vi= 10.0; % Equations A= [ -1/(Ri*Cf*(1+K)) ]; B= [ 1/(Ri*Cf) ]; u= [ Vi ]; xdot= A*x +B*u; % xdot= xdot' not necessary since dim x=1
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% Setup for homework 7-3: Setup73b.m % global K K= 100; ts= [ 0.0:0.01:1.0 ]; x0= [ 0 ]; [t,x]= ode23('StEqs73',ts,x0);
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Unformatted text preview: Vo= ( -K/(1+K))*x(:,1); plot(t,Vo); xlabel('time'); ylabel('Vo'); title('Response for const inp[ut voltage 10 v'); Discussion of the results. The ideal integrator (for K equal to infinity) would have the response Vo = -(1/Ri*Cf)*Integral t (Vi). For Ri= 1000 and Cf= 0.0001, then Vo = - 10.0*Integral t (Vi). For Vi= 10.0 v constant, then Vo= -10.0*10.0*t = -100.0*t. Case a. (figure 1) At t= 1.0, the apparent value is -100.0, which is the ideal result. Case b. (figure 2) At t= 1.0, the value is approximately -94, which is drifting from ideal. Conclusion: You get what you pay for in this world....
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