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

UNIVERSIDA PÓLITECNICA SALESIANA Ingeniería Electrónica Ecuaciones Diferenciales Resolución de un circuito RLC David Basantes Israel Campaña Vinicio Masabanda Juan Ordoñez

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.

Procedimiento ,[object Object], *L(d²q/dt²) + R(dq/dt) + 1/C(q)= E(t) ,[object Object], 1.51(d²q/dt²) + 3(dq/dt) + 1/0.20(q)= 15sen (t) + 5eᶺ(t) ,[object Object], 1.51(d²q/dt²) + 3(dq/dt) + 1/0.20(q)= 0

[object Object], 1.51m² + 3m + 5 = 0 Dónde: a=1.51 ; b=3; c=5 ,[object Object], 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)] ,[object Object], q1= eᶺ (-0.99t) cos(1.52t) q2= eᶺ (-0.99t) sen(1.52t)

[object Object], 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)

[object Object], –[ 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)

[object Object], 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)]

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)]

[object Object], 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))]

[object Object], 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)

Solucion del ejerecicio por medio de MATLAB ,[object Object],((15*sin(t))+(5*(e^(t)))-(A(1)/0.20)-(3*B(1)))/1.51

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;

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)')

[object Object], 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].

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

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