Sistemas de segundo orden

El filtro RLC consiste de un inductor, un capacitor y un resistor en serie, alimentado por un voltaje de entrada. El voltaje de salida se mide en el resistor y esto forma un filtro pasabanda de segundo orden.

La función de transferencia de este sistema es tal y como sigue:

num = [R/L 0];den = [1 R/L 1/(L*C)];

H(s) = (R/L)*s/(s^2+(R/L)*s+(1/(L*C)))

Representación del sistema LTI en MATLAB

R = 1; % ohmsL = 50e-3; % henryC = 620e-6; % farad

num = [R/L 0];den = [1 R/L 1/(L*C)];RLC_sys = tf(num,den)

Respuesta del sistema al escalón y al impulso

figureimpulse(RLC_sys)

%% System Step Responsefigurestep(RLC_sys)

Características de respuesta transitoria del sistema

%% System Transient Response Characteristics

% Resonant angular frequencywo = 1/sqrt(L*C) % Q-FactorQ = sqrt(L/C)/R

Respuesta del sistema a una entrada Senoidal

%% System Sine Input Responset = 0:0.0001:0.5;figureu = sin(0.5*wo*t);lsim(RLC_sys, u, t)title('Reponse to Excitation at 0.5x Natural Frequency') figureu = sin(wo*t);lsim(RLC_sys, u, t)title('Reponse to Excitation at Natural Frequency') figureu = sin(2*wo*t);lsim(RLC_sys, u, t)title('Reponse to Excitation at 2x Natural Frequency')

Respuesta en frecuencia del sistema

%% System Frequency Response of Varying Q-Factors % Filter Circuit with Intermediate Q-factorQ_int = 0.5;R_int = sqrt(L/C)/Q_int;RLC_sys_int = tf([R_int/L 0],[1 R_int/L 1/(L*C)]); % Filter Circuit with Low Q-factorQ_low = 0.05;R_low = sqrt(L/C)/Q_low;RLC_sys_low = tf([R_low/L 0],[1 R_low/L 1/(L*C)]); % Obtain the Bode diagram of multiple systems with different Q-factorsfigurebode(RLC_sys,RLC_sys_int,RLC_sys_low)grid onlegend('High-Q (9)','Intermediate-Q (0.5)','Low-Q (0.05)')title('Bode Diagram of an RLC Bandpass Filter with Different Q-factors')

Análisis de sistema de segundo orden sin amortiguamiento

El sistema de masa-resorte es un

ejemplo de un sistema de segundo orden sin amortiguamiento. El sistema puede modelarse con la siguiente función de transferencia:

H(s)= 1/(m*s^2+k)

Representación del sistema LTI en MATLAB

m = 1;k = 5;num = 1;den = [m 0 k];massSpring_sys = tf(num, den)

Respuesta al escalón/polos del sistema

%% Unit Step Responsefigurestep(massSpring_sys,10)[y,t] = step(massSpring_sys,10); %% System Characteristics - Polesr = roots(den)p = pole(massSpring_sys) % Poles of the systemdisp('Poles of a system are equivilant to the roots of

the denominator of the transfer function') abs(p)disp('Magnitude of the poles gives their frequency

(rad/s)')

Parámetros del sistema/respuesta al impulso

%% Imaginary Poles indicate that the system is oscillating

wn = sqrt(k/m) % Natural Frequency [rad/s]fn = wn/(2*pi) % Natural Frequency [Hz]T = 1/fn % Period %% Impulse Response% Notice that the time between consecutive peaks gives the period.

figureimpulse(massSpring_sys,10)

Respuesta del sistema a entradas senoidales

%% Sine Input Responset = 0:0.001:15;figureu = sin(wn/2*t);lsim(massSpring_sys, u, t)title('Reponse to Excitation at 0.5x Natural Frequency')ylim([-3 3]) figureu = sin(wn*t);lsim(massSpring_sys, u, t)title('Reponse to Excitation at Natural Frequency')ylim([-3 3]) figureu2 = sin(wn*2*t);lsim(massSpring_sys, u2, t)title('Reponse to Excitation at 2x Natural Frequency')

Respuesta en frecuencia del sistema

%% Frequency Response figurebode(massSpring_sys)grid on

Análisis de un sistema amortiguado de segundo irden

El sistema de masa-resorte-amortiguador es un ejemplo de un sistema amortiguado de segundo orden.

El sistema puede ser representado con esta función de transferencia.

H(s)= 1/(m*s^2+b*s+k)

Representando al sistema LTI en MATLAB

%% Representing a Linear Time Invariant System in MATLAB

% System Parametersm = 1;b = 0.6;k = 5; % Creating a Transfer Function Representationnum = 1;den = [m b k];MSD_sys = tf(num, den)

Respuesta transitoria del sistema

%% Unit Step Responsefigurestep(MSD_sys)[y,t] = step(MSD_sys); %% Transient Response System Characteristicswn = sqrt(k/m) % Natural Frequency [rad/s]fn = wn/(2*pi) % Natural Frequency [Hz]z = b/(2*sqrt(k*m)) % Damping Coefficientwd = wn * sqrt(1-z^2) % Damped Natural Frequency [rad/s] p = pole(MSD_sys)abs(p)damp(MSD_sys) ts = 4/(z*wn) % Settling TimeOS = exp(-z*pi*sqrt(1-z^2)) * 100 % Percentage Overshoot stepResponseCharacteristics = stepinfo(MSD_sys)

Comparando respuesta de sistemas críticamente amortiguados, subamortiguados y sobreamortiguados

% Underdamped SystemUnderdamped = MSD_sys; % Critically Damped Systemb_critical = 2*sqrt(k*m);Critically_Damped = tf([1], [m b_critical k]); % Overdamped Systemb_over = 10;Overdamped = tf([1], [m b_over k]); ltiview({'step', 'pzmap'}, Underdamped,

Critically_Damped, Overdamped)

Respuesta a entradas senoidales

%% Sine Input Responsefigureu1 = sin(0.5*wd*t);lsim(MSD_sys, u1, t)legend('System Response','Location','SE')title('Reponse to Excitation at 0.5x Damped Natural Frequency') t = 0:0.001:15;figureu = sin(wd*t);lsim(MSD_sys, u, t)legend('System Response','Location','SE')title('Reponse to Excitation at Damped Natural Frequency') figureu2 = sin(2*wd*t);lsim(MSD_sys, u2, t)legend('System Response','Location','SE')title('Reponse to Excitation at 2x Damped Natural Frequency')

Respuesta en frecuencia

%% Frequency Responsefigurebode(MSD_sys)grid on

Efecto de variar el amortiguamiento en la respuesta en frecuencia

%% Effect of Varying Damping on the Frequency Responsefigurehold on % Initialize fixed variablesm = 1;k = 5; % Iterate over different damping coefficientsfor b = [0.1 0.5 1 2 4] bode(tf(1,[m b k]),{0.1,10}) pause(0.5) drawnowendgrid onlegend('b = 0.1','b = 0.5','b = 1','b = 2','b = 4','Location','SW')title('Effect of Varying Damping on the Frequency Response')hold off

Efecto de variar la masa en la respuesta en frecuencia

%% Effect of Varying Mass on the Resonant Frequency and Frequency Response

figurehold on b = 0.6;k = 5; % Iterate over different massesfor m = 2:6 bode(tf(1,[m b k])) pause(0.5) drawnowendgrid onlegend('m = 2','m = 3','m = 4','m = 5','m = 5','Location','SW')title('Effect of Varying Mass on Resonant Frequency and Frequency

Response')hold off