Resolucion de un circuito rlc en matlab

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) S.G

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); « Nos dibuja la figura que deseamos » Title(‘q vs t') « Sirve para colocar los titulos a la figura » xlabel(‘t(s)') « Se colocan un menbrete en los ejescoordenados » ylabel(‘q(c)') figure(2) plot (t,i) title('i vs t') xlabel('t(s)') ylabel('i(A)')

[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].

GraficosresultanteS CARGA vs TIEMPO

Resolucion de un circuito rlc en matlab

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Resolucion de un circuito rlc en matlab