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plot the output c(t) using mathlab and show setting time on you graphf
A=1 B=-1 inverse laplance of 1/s=1 inverse laplance of 1/(s+5)=e^-5t c(t)= 1+e^-5t setting time is 0.7832

etwa 2 Monate vor | 0

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plot the output c(t) using mathlab and show setting time on you graphf
syms t tau; A = [0 2; -2 -5]; B = [0; 1]; C = [2 1]; x0 = [1; 2]; Phi_t = expm(A*t); x_h = Phi_t * x0; u_tau = 1; %...

etwa 2 Monate vor | 2 Antworten | 0

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