{"id":8404,"date":"2022-08-31T18:00:30","date_gmt":"2022-08-31T23:00:30","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=8404"},"modified":"2026-04-05T20:46:49","modified_gmt":"2026-04-06T01:46:49","slug":"s2eva2022paoi_t2-edo-de-circuito-rlc-con-interruptor-intermedio","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2022paoi_t2-edo-de-circuito-rlc-con-interruptor-intermedio\/","title":{"rendered":"s2Eva2022PAOI_T2 EDO de circuito RLC con interruptor intermedio"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2022PAOI_T2 EDO de circuito RLC con interruptor intermedio<\/a><\/p>\n\n\n\n

La corriente del inductor y(t) para t\u22650 se deriva para tener la expresi\u00f3n solo derivadas:<\/p>\n\n\n \\frac{\\delta}{\\delta t}y(t) + 2 y(t) + 5 \\int_{-\\infty}^t y(\\tau) \\delta \\tau = 10 \\mu(t) <\/span>\n\n\n\n

Para t>0 que es donde transcurre el experimento, el escal\u00f3n es una constante, se tiene que:<\/p>\n\n\n \\frac{\\delta ^2}{\\delta t^2}y(t) + 2 \\frac{\\delta}{\\delta t}y(t) + 5 y(t) = 0 <\/span>\n\n\n\n

tomando las condiciones iniciales dadas para t=0, y(0)=2, y'(0)=-4<\/p>\n\n\n\n

literal a<\/h2>\n\n\n\n

EL resultado esperado es el planteamiento del problema. Se reescribe la ecuaci\u00f3n con la nomenclatura simplificada y se reordena seg\u00fan el modelo del m\u00e9todo:<\/p>\n\n\n y'' = - 2y' - 5 y <\/span>\n\n\n\n

luego se sustituye la variable y se convierte a las ecuaciones:<\/p>\n\n\n z =y' = f_x(t,y,z) <\/span>\n\n\n z' = - 2z - 5 y = g_z(t,y,z) <\/span>\n\n\n\n

se usa una tabla para llevar el registro de operaciones:<\/p>\n\n\n\n

Se plantea las operaciones:<\/p>\n\n\n\n

K1y = h * f(ti,yi,zi)\nK1z = h * g(ti,yi,zi)\n\nK2y = h * f(ti+h, yi + K1y, zi + K1z)\nK2z = h * g(ti+h, yi + K1y, zi + K1z)\n\nyi = yi + (K1y+K2y)\/2\nzi = zi + (K1z+K2z)\/2\nti = ti + h<\/code><\/pre>\n\n\n\n
\n\n\n\n
literal b<\/h2>\n\n\n\n
El resultado esperado es la aplicaci\u00f3n correcta de los valores en las expresiones para al menos tres iteraciones usando h=0.01<\/p>\n\n\n\n
itera = 0<\/h3>\n\n\n K1y = 0.01 y'(0) = 0.01(-4) = -0.04<\/span>\n\n\n K1z = 0.01 (- 2z(0) - 5 y(0)) = 0.01(- 2(-4) - 5 (2)) = -0.02 <\/span>\n\n\n K2y = 0.01 (-4-0.02) = -0.0402<\/span>\n\n\n K2z = 0.01 (-2(-4-0.02)-5(2-0.04)) = -0.0176<\/span>\n\n\n yi = yi + \\frac{K1y+K2y}{2} = 2+\\frac{-0.04-0.0402} {2} = 1.9599<\/span>\n\n\n zi = zi + \\frac{K1z+K2z}{2} = -4 +\\frac{-0.02-0.0176}{2} = -4.0188<\/span>\n\n\n ti = ti + h = 0+0.01 = 0.01<\/span>\n\n\n\n
itera = 1<\/h3>\n\n\n K1y = 0.01(-4.0188) = -0.040188<\/span>\n\n\n K1z = 0.01(- 2(-4.0188) - 5 (1.9599)) = -0.0176 <\/span>\n\n\n K2y = 0.01 (-4.0188-0.0176) = -0.0403<\/span>\n\n\n K2z = 0.01 (-2(-4.0188-0.0176)-5(1.9599-0.040188)) = -0.0152<\/span>\n\n\n yi = 1.9599 +\\frac{-0.040188-0.0403} {2} = 1.9196<\/span>\n\n\n zi = -4.0188 +\\frac{-0.0176-0.0152}{2} = -4.0352<\/span>\n\n\n ti = ti + h = 0.01+0.01 = 0.02<\/span>\n\n\n\n
itera = 2<\/h3>\n\n\n K1y = 0.01(-4.0352) = -0.040352<\/span>\n\n\n K1z = 0.01(- 2(-4.0352) - 5 (1.9196)) = -0.0152 <\/span>\n\n\n K2y = 0.01 (-4.0352-0.0152) = -0.0405<\/span>\n\n\n K2z = 0.01 (-2(-4.0352-0.0152)-5(1.9196-0.040352)) = -0.0129<\/span>\n\n\n yi = 1.9196 +\\frac{-0.040352-0.0405} {2} =1.8792<\/span>\n\n\n zi = -4.0352 +\\frac{-0.0152-0.0129}{2} = -4.0494<\/span>\n\n\n ti = ti + h = 0.02+0.01 = 0.03<\/span>\n\n\n\n
Resultados con el algoritmo en Python<\/h2>\n\n\n\n
   ti,   yi,    zi,      K1y,    K1z,    K2y,     K2z\n[[ 0.00  2.0000 -4.0000  0.0000  0.0000  0.0000   0.0000]\n [ 0.01  1.9599 -4.0188 -0.0400 -0.0200 -0.0402  -0.0176]\n [ 0.02  1.9196 -4.0352 -0.0401 -0.0176 -0.0403  -0.0152]\n [ 0.03  1.8792 -4.0494 -0.0403 -0.0152 -0.0405  -0.0129]\n...<\/code><\/pre>\n\n\n\n
\n\n\n\n
Literal c<\/h2>\n\n\n\n
Runge-Kutta 2do Orden tiene error de truncamiento O(h3<\/sup>)<\/p>\n\n\n\n
por lo que el error est\u00e1 en el orden de (0.01)3<\/sup> = 0.000001<\/p>\n\n\n\n
\n\n\n\n
Literal d<\/h2>\n\n\n\n
Se requiere presentar el resultado para el intervalo t entre [0,5]. Siendo el tama\u00f1o de paso h=0.01 que es peque\u00f1o, se requieren realizar (5-0)\/0.01=500 iteraciones, que es m\u00e1s pr\u00e1ctico realizarlas usando el algoritmo.<\/p>\n\n\n\n
\"Circuito<\/figure>\n\n\n\n
Instrucciones en Python<\/h2>\n\n\n
\n# Respuesta a entrada cero\n# solucion para (D^2+ D + 1)y = 0\nimport numpy as np\nimport matplotlib.pyplot as plt\n\ndef rungekutta2_fg(f,g,x0,y0,z0,h,muestras):\n    tamano = muestras + 1\n    estimado = np.zeros(shape=(tamano,7),dtype=float)\n    # incluye el punto [x0,y0]\n    estimado[0] = [x0,y0,z0,0,0,0,0]\n    xi = x0\n    yi = y0\n    zi = z0\n    for i in range(1,tamano,1):\n        K1y = h * f(xi,yi,zi)\n        K1z = h * g(xi,yi,zi)\n        \n        K2y = h * f(xi+h, yi + K1y, zi + K1z)\n        K2z = h * g(xi+h, yi + K1y, zi + K1z)\n\n        yi = yi + (K1y+K2y)\/2\n        zi = zi + (K1z+K2z)\/2\n        xi = xi + h\n        \n        estimado[i] = [xi,yi,zi,K1y,K1z,K2y,K2z]\n    return(estimado)\n\n# PROGRAMA\nf = lambda t,y,z: z\ng = lambda t,y,z: -2*z -5*y + 0\n\nt0 = 0\ny0 = 2\nz0 = -4\n\nh = 0.01\ntn = 5\nmuestras = int((tn-t0)\/h)\n\ntabla = rungekutta2_fg(f,g,t0,y0,z0,h,muestras)\nti = tabla[:,0]\nyi = tabla[:,1]\nzi = tabla[:,2]\n\n# SALIDA\nnp.set_printoptions(precision=4)\nprint('ti, yi, zi, K1y, K1z, K2y, K2z')\nprint(tabla)\n\n# GRAFICA\nplt.plot(ti,yi, color = 'orange', label='y_RK(t)')\nplt.ylabel('y(t)')\nplt.xlabel('t')\nplt.title('y(t) con Runge-Kutta 2do Orden d2y\/dx2 ')\nplt.legend()\nplt.grid()\nplt.show()\n<\/pre><\/div>\n\n\n
Nota<\/strong><\/em>: En el curso TELG1001 Se\u00f1ales y Sistemas<\/a>, la soluci\u00f3n se realiza con Transformadas de Laplace<\/p>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2022PAOI_T2 EDO de circuito RLC con interruptor intermedio La corriente del inductor y(t) para t\u22650 se deriva para tener la expresi\u00f3n solo derivadas: Para t>0 que es donde transcurre el experimento, el escal\u00f3n es una constante, se tiene que: tomando las condiciones iniciales dadas para t=0, y(0)=2, y'(0)=-4 literal a EL resultado esperado es […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[49],"tags":[58,54],"class_list":["post-8404","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8404","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=8404"}],"version-history":[{"count":5,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8404\/revisions"}],"predecessor-version":[{"id":23898,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8404\/revisions\/23898"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=8404"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=8404"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=8404"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}