Resolucion de una EDO de segundo grado aplicado a un circuito RLC
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UNIVERSIDA PÓLITECNICA SALESIANA Ingeniería Electrónica Ecuaciones Diferenciales Resolución de un circuito RLC David Basantes Israel Campaña Vinicio Masabanda Juan Ordoñez
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Transcript of Resolucion de una EDO de segundo grado aplicado a un circuito RLC
UNIVERSIDA PÓLITECNICA SALESIANA
Ingeniería Electrónica
Ecuaciones DiferencialesResolución de un circuito RLC
David BasantesIsrael CampañaVinicio MasabandaJuan Ordoñez
PROBLEMAEncuentre 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) VLa carga y la corriente son nulas.
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 homogénea1.51(d²q/dt²) + 3(dq/dt) + 1/0.20(q)= 0
4.-Reconocer el grado de la EDO y remplazar en la fórmula general.1.51m² + 3m + 5 = 0Dónde: a=1.51 ; b=3; c=5
5.- Reconocemos α y β y lo aplicamos a la fórmula y reemplazamosα= -0.99 y β=1.52Y= 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)
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)
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)
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)]
W2= [ eᶺ (-0.99t) cos(1.52t)][ 15sen (t) + 5eᶺ(t)] – 0W2= [(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)]
8.- Calculamos U1 Y U2U´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) sen bu.du= eᶺ(a)(u)/(a²+b²).(a sen bu – b cos bu) + CU1= -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) cos bu.du= eᶺ(a)(u)/(a²+b²).(b sen bu + b cos bu) + CU2= 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))]
9.- La solución general de la EDO es:
Y= Yh + YpYp= U1Y1 + U2Y2Yh= 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
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
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)')
[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].
GRAFICO RESULTANTE
