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Resolucion de un circuito rlc en matlab

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Resolucion de una EDO de segundo grado aplicado a un circuito RLC

Resolucion de una EDO de segundo grado aplicado a un circuito RLC

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  • 1. UNIVERSIDA PÓLITECNICA SALESIANA
    Ingeniería Electrónica
    Ecuaciones Diferenciales
    Resolución de un circuito RLC
    David Basantes
    Israel Campaña
    Vinicio Masabanda
    Juan Ordoñez
  • 2. PROBLEMA
    Encuentre la carga al tiempo t=0.92s en el circuito LCR donde L=1.52H, R=3Ὠ, C=0.20f y E(t)=15sen (t) + 5eᶺ(t) V
    La carga y la corriente son nulas.
  • 3. Procedimiento
    • 1.-Buscamos la fórmula para resolver el circuito RCL.
    *L(d²q/dt²) + R(dq/dt) + 1/C(q)= E(t)
    • 2.-Reemplazamos los datos.
    1.51(d²q/dt²) + 3(dq/dt) + 1/0.20(q)= 15sen (t) + 5eᶺ(t)
    • 3.-Igualamos a cero la parte homogénea
    1.51(d²q/dt²) + 3(dq/dt) + 1/0.20(q)= 0
  • 4.
    • 4.-Reconocer el grado de la EDO y remplazar en la fórmula general.
    1.51m² + 3m + 5 = 0
    Dónde: a=1.51 ; b=3; c=5
    • 5.- Reconocemos α y β y lo aplicamos a la fórmula y reemplazamos
    α= -0.99 y β=1.52
    Y= eᶺ(αt) [A cos(βt) + B sen(βt)]
    qh= eᶺ(-0.99t) [A cos(1.52t) + B sen(1.52t)]
    • 6.- Separamos qh en q1 y q2
    q1= eᶺ (-0.99t) cos(1.52t)
    q2= eᶺ (-0.99t) sen(1.52t)
  • 5.
    • 7.- Escribimos el Wronskiano y calculamos W1 Y W2.
    q1= eᶺ (-0.99t) cos(1.52t)
    q´1= -0.99eᶺ (-0.99t)*cos(1,52) – 1.52 eᶺ (-0.99t)* sen(1.52t)
    q2= eᶺ (-0.99t)* sen(1.52t)
    q´2= -0.99eᶺ (-0.99t) *sen(1.52t) + 1.52 eᶺ (-0.99t) *cos(1.52t)
  • 6.
    • W= [ eᶺ (-0.99t) cos(1.52t)][ -0.99eᶺ (-0.99t) sen(1.52t) + 1.52 eᶺ (-0.99t) cos(1.52t)]
    –[ eᶺ (-0.99t) sen(1.52t)][ -0.99eᶺ (-0.99t)cos(1,52) – 1.52 eᶺ (-0.99t) sen(1.52t)]
    W= eᶺ (-1.98t)[-0.99 cos(1.52t)* sen(1.52t) + 1.52cos²(1.52t)] - eᶺ (-1.98t) [-0.99 sen(1.52t)* cos(1.52t) - 1.52sen²(1.52t)]
    W= eᶺ (-1.98t){-0.99 cos(1.52t)* sen(1.52t) + 1.52cos²(1.52t) – {-0.99 sen(1.52t)* cos(1.52t)- 1.52sen²(1.52t)}
    W= eᶺ (-1.98t)(-0.99 cos(1.52t)* sen(1.52t) + 1.52cos²(1.52t) + 0.99 sen(1.52t)* cos(1.52t) + 1.52sen²(1.52t)}
    W= eᶺ (-1.98t)(1.52cos²(1.52t) + 1.52sen²(1.52t))
    W= eᶺ (-1.98t)[ 1.52(cos²(1.52t) + sen²(1.52t)]
    W= eᶺ (-1.98t)*(1.52)
  • 7.
    • Donde: f(t)= 15sen (t) + 5eᶺ(t)
    W1= 0 - [15sen (t) + 5eᶺ(t)][ eᶺ (-0.99t) sen(1.52t)]
    W1= -[ 15sen (t). eᶺ (-0.99t) sen(1.52t)) + (5eᶺ(t). eᶺ (-0.99t) sen(1.52t)]
    W1= -15 eᶺ (-0.99t).sen (t). sen(1.52t) - 5 eᶺ (0.01t) sen(1.52t)
    W1= - sen(1.52t) [15 eᶺ (-0.99t) sen (t) + 5 eᶺ (0.01t)]
  • 8. W2= [ eᶺ (-0.99t) cos(1.52t)][ 15sen (t) + 5eᶺ(t)] – 0
    W2= [(eᶺ (-0.99t) cos(1.52t)( 15sen (t)) + (eᶺ (-0.99t) cos(1.52t). 5eᶺ(t)]
    W2= [15 eᶺ (-0.99t) cos(1.52t). sen (t) + 5eᶺ(0.01t) cos(1.52t)
    W2= cos(1.52t).[ 15 eᶺ (-0.99t) sen (t) + 5eᶺ(0.01t)]
  • 9.
    • 8.- Calculamos U1 Y U2
    U´1= W1/W= (- sen(1.52t) [15 eᶺ (-0.99t) sen (t) + 5 eᶺ (0.01t)])/( eᶺ (-1.98t)*(1.52)
    U´1= - sen(1.52t)[9.87 eᶺ (0.99t) + 3.29 eᶺ (1.99t)]
    U´1= - sen(1.52t). 9.87 eᶺ (0.99t) - 3.29 eᶺ (1.99t). sen(1.52t)
    U1=ʃ -9.87 eᶺ (0.99t) sen(1.52t) - 3.29 eᶺ (1.99t). sen(1.52t)
    U1= -9.87ʃ eᶺ (0.99t). sen(1.52t) - 3.29ʃ eᶺ (1.99t). sen(1.52t)
    ʃ eᶺ(a)(u) senbu.du= eᶺ(a)(u)/(a²+b²).(a senbu – b cosbu) + C
    U1= -9.87[(eᶺ (0.99t)/(0.99)²+(1.52)²)* (0.99 sen(1.52t) – 1.52 cos(1.52t))]
    -3.29 [eᶺ (1.99t)/(1.99)²+(1.52)² (1.99 sen(1.52t) – 1.52 cos(1.52t))]
    U´2= W2/W = cos(1.52t).[ 15 eᶺ (-0.99t) sen (t) + 5eᶺ(0.01t)]/ ( eᶺ (-1.98t)*(1.52)
    U´2= cos(1.52t).[9.87 eᶺ (0.99t ) + 3.29 eᶺ (1.99t)]
    U´2= 9.87 eᶺ (0.99t ) cos(1.52t) + 3.29 eᶺ (1.99t) cos(1.52t)
    U2= ʃ 9.87 eᶺ (0.99t ) cos(1.52t) + 3.29 eᶺ (1.99t) cos(1.52t)
    U2= 9.87 ʃ eᶺ (0.99t ) cos(1.52t) + 3.29 ʃ eᶺ (1.99t) cos(1.52t)
    ʃ eᶺ(a)(u) cosbu.du= eᶺ(a)(u)/(a²+b²).(b senbu + b cosbu) + C
    U2= 9.87 [(eᶺ (0.99t)/(0.99)²+(1.52)²)* (1.52 sen(1.52t) + 0.99 cos(1.52t))]
    + 3.29 [eᶺ (1.99t)/(1.99)²+(1.52)² (1.52 sen(1.52t) + 1.99 cos(1.52t))]
  • 10.
    • 9.- La solución general de la EDO es:
    Y= Yh + Yp
    Yp= U1Y1 + U2Y2
    Yh= eᶺ(-0.99t) [A cos(1.52t) + B sen(1.52t)]
     
    Yp=-9.87[(eᶺ (0.99t)/(0.99)²+(1.52)²)* (0.99 sen(1.52t) – 1.52 cos(1.52t))]
    -3.29 [eᶺ (1.99t)/(1.99)²+(1.52)² (1.99 sen(1.52t) – 1.52 cos(1.52t))] * eᶺ (-0.99t) cos(1.52t)
    + 9.87 [(eᶺ (0.99t)/(0.99)²+(1.52)²)* (1.52 sen(1.52t) + 0.99 cos(1.52t))]
    + 3.29 [eᶺ (1.99t)/(1.99)²+(1.52)² (1.52 sen(1.52t) + 1.99 cos(1.52t))] * eᶺ (-0.99t) sen(1.52t)
     
    Y= eᶺ(-0.99t) [A cos(1.52t) + B sen(1.52t)]+ =-9.87[(eᶺ (0.99t)/(0.99)²+(1.52)²)* (0.99 sen(1.52t) – 1.52 cos(1.52t))]
    -3.29 [eᶺ (1.99t)/(1.99)²+(1.52)² (1.99 sen(1.52t) – 1.52 cos(1.52t))] * eᶺ (-0.99t) cos(1.52t)
    + 9.87 [(eᶺ (0.99t)/(0.99)²+(1.52)²)* (1.52 sen(1.52t) + 0.99 cos(1.52t))]
    + 3.29 [eᶺ (1.99t)/(1.99)²+(1.52)² (1.52 sen(1.52t) + 1.99 cos(1.52t))] * eᶺ (-0.99t) sen(1.52t)
  • 11. Solucion del ejerecicio por medio de MATLAB
    • L a ecuacion que vamos a ingresar en Matlab es la siguiente:
    ((15*sin(t))+(5*(e^(t)))-(A(1)/0.20)-(3*B(1)))/1.51
  • 12. 1. En matlab creamos un nuevo documento M-file en donde ingresamos lo siguiente:
    function B=rlc(t,A)
    B=zeros(2,1);
    B(1)=A(2);
    B(2)= ((15*sin(t))+(5*(exp(t)))-(A(1)/0.20)-(3*B(1)))/1.51;
  • 13. 2. Ahora vamos a realizar las ordenes que necesitamos para obtener el dibujo del problema
    [t,A]=ode45('rlc', [-4 10], [-3 15]);
    q=A(:,1);
    i=A(:,2);
    plot(t,q);
    Title(‘q vs t')
    xlabel(‘t(s)')
    ylabel(‘q(c)')
  • 14.
    • [x,y] = ode45('función',a,b,inicial) Esta instrucción regresa un conjunto de coordenadas "x" y "y" que representan a la función y=f(x), los valores se calculan a través de métodos Runge-Kuta de cuarto y quinto orden.
    El nombre "función", define una función que representa a una ecuación diferencial ordinaria, ODE45 proporciona los valores de la ecuación diferencial y'=g(x,y).
    Los valores "a" y "b" especifican los extremos del intervalo en el cual se desea evaluar a la función y=f(x).
    El valor inicial y = f(a) especifica el valor de la función en el extremo izquierdo del intervalo [a,b].
  • 15. Grafico resultante