{"id":7741,"date":"2021-09-15T18:46:23","date_gmt":"2021-09-15T23:46:23","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=7741"},"modified":"2026-04-05T21:07:03","modified_gmt":"2026-04-06T02:07:03","slug":"s3eva2021paoi_t3-edo-respuesta-a-entrada-cero-en-un-sistema-ltic","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s3eva30\/s3eva2021paoi_t3-edo-respuesta-a-entrada-cero-en-un-sistema-ltic\/","title":{"rendered":"s3Eva2021PAOI_T3 EDO Respuesta a entrada cero en un sistema LTIC"},"content":{"rendered":"\n<p><em><strong>Ejercicio<\/strong><\/em>: <a href=\"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-3eva30\/3eva2021paoi_t3-edo-respuesta-a-entrada-cero-en-un-sistema-ltic\/\" data-type=\"post\" data-id=\"7708\">3Eva2021PAOI_T3 EDO Respuesta a entrada cero en un sistema LTIC<\/a><\/p>\n\n\n\n<p>la ecuaci\u00f3n a resolver es:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\">\\frac{\\delta^2 y(t)}{\\delta t^2}+3 \\frac{\\delta y(t)}{ \\delta t}+2 y(t) =0<\/span>\n\n\n\n<p>con valores iniciales: y<sub>0<\/sub>(t)=0 , y\u2019<sub>0<\/sub>(t) =-5<\/p>\n\n\n\n<p>se puede escribir como:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> y&quot;+3 y'+2y = 0<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> y&quot; = -3y'-2y<\/span>\n\n\n\n<p>sustituyendo las expresiones de las derivadas como las funciones f(x) por z y g(x) por z':<\/p>\n\n\n\n<p>y' = z = f(x)<\/p>\n\n\n\n<p>y'' = z'= -3z-2y = g(x)<\/p>\n\n\n\n<p>Los valores iniciales de t0=0, y0=0, z0=-5 se usan en el algoritmo.<\/p>\n\n\n\n<p>En este caso tambi\u00e9n se requiere conocer un intervalo de tiempo a observar [0,t<sub>n<\/sub>=6] y definir el tama\u00f1o de paso o resoluci\u00f3n del resultado<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> h = \\delta t = \\frac{b-a}{n} = \\frac{6-0}{60} = 0.1 <\/span>\n\n\n\n<p>t<sub>0<\/sub> = 0, y<sub>0<\/sub> = 0,&nbsp; y\u2019<sub>0<\/sub> = z<sub>0<\/sub> = -5<\/p>\n\n\n\n<p><strong>iteraci\u00f3n 1<\/strong><\/p>\n\n\n\n<p>K1y = h * f(ti,yi,zi) = 0.1 (-5) = -0.5<\/p>\n\n\n\n<p>K1z = h * g(ti,yi,zi) ) = 0.1*(-3(-5)-2(0)) = 1.5<\/p>\n\n\n\n<p>K2y = h * f(ti+h, yi + K1y, zi + K1z) = 0.1(-5+1.5) = -0.35<\/p>\n\n\n\n<p>K2z = h * g(ti+h, yi + K1y, zi + K1z) = 0.1 ( -3(-5+1.5) - 2(0-0.5)) = 1.15<\/p>\n\n\n\n<p>yi = yi + (K1y+K2y)\/2 =0+(-0.5-0.35)\/2=-0.425<\/p>\n\n\n\n<p>zi = zi + (K1z+K2z)\/2 = -5+(1.5+1.15)\/2 = -3.675<\/p>\n\n\n\n<p>ti = ti + h = 0+0.1 = 0.1<\/p>\n\n\n\n<p><strong>iteraci\u00f3n 2<br><\/strong><\/p>\n\n\n\n<p>K1y = 0.1 (-3.675) = -0.3675<\/p>\n\n\n\n<p>K1z = 0.1*(-3(-3.675)-2(-0.425)) = 1.1875<\/p>\n\n\n\n<p>K2y = 0.1(-3.675+ 1.1875) = -0.24875<\/p>\n\n\n\n<p>K2z = 0.1 ( -3(-3.675+ 1.1875) - 2(-0.425-0.3675)) = 0.90475<\/p>\n\n\n\n<p>yi = -0.425+(-0.3675-0.24875)\/2=-0.7331<\/p>\n\n\n\n<p>zi = -3.675+( 1.1875+0.90475)\/2 = -2.6288<\/p>\n\n\n\n<p>ti =0.1+0.1 = 0.2<\/p>\n\n\n\n<p><strong>iteraci\u00f3n 3<\/strong><\/p>\n\n\n\n<p>continuar como ejercicio<\/p>\n\n\n\n<p>El algoritmo permite obtener la gr\u00e1fica y la tabla de datos.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"640\" height=\"480\" src=\"http:\/\/blog.espol.edu.ec\/algoritmos101\/files\/2021\/09\/ltic_respuestaentradacero01.png\" alt=\"ltic respuesta entrada cero 01\" class=\"wp-image-18637\" \/><\/figure>\n\n\n\n<p>los valores de las iteraciones son:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>t, y, z\n&#091;&#091; 0.        0.       -5.      ]\n &#091; 0.1      -0.425    -3.675   ]\n &#091; 0.2      -0.733125 -2.628875]\n &#091; 0.3      -0.949248 -1.807592]\n &#091; 0.4      -1.093401 -1.167208]\n &#091; 0.5      -1.18168  -0.67202 ]\n &#091; 0.6      -1.226984 -0.293049]\n &#091; 0.7      -1.239624 -0.006804]\n &#091; 0.8      -1.227806  0.205735]\n &#091; 0.9      -1.19804   0.359943]\n &#091; 1.       -1.155465  0.468225]\n &#091; 1.1      -1.104111  0.540574]\n &#091; 1.2      -1.047121  0.585021]\n &#091; 1.3      -0.986923  0.608001]\n &#091; 1.4      -0.925374  0.614658]\n &#091; 1.5      -0.863874  0.609087]\n &#091; 1.6      -0.803463  0.594537]\n &#091; 1.7      -0.744893  0.573574]\n &#091; 1.8      -0.68869   0.548208]\n &#091; 1.9      -0.635205  0.520011]\n &#091; 2.       -0.584652  0.490193]\n &#091; 2.1      -0.53714   0.459683]\n &#091; 2.2      -0.492695  0.42918 ]\n &#091; 2.3      -0.451288  0.399206]\n &#091; 2.4      -0.412843  0.370135]\n &#091; 2.5      -0.377253  0.342233]\n &#091; 2.6      -0.34439   0.315674]\n &#091; 2.7      -0.314114  0.290567]\n &#091; 2.8      -0.286275  0.266966]\n &#091; 2.9      -0.26072   0.244887]\n &#091; 3.       -0.237297  0.224314]\n &#091; 3.1      -0.215858  0.205211]\n &#091; 3.2      -0.196256  0.187526]\n &#091; 3.3      -0.178354  0.171195]\n &#091; 3.4      -0.162019  0.156149]\n &#091; 3.5      -0.147126  0.142312]\n &#091; 3.6      -0.133558  0.129611]\n &#091; 3.7      -0.121206  0.117969]\n &#091; 3.8      -0.109966  0.107312]\n &#091; 3.9      -0.099745  0.097569]\n &#091; 4.       -0.090454  0.08867 ]\n &#091; 4.1      -0.082013  0.080549]\n &#091; 4.2      -0.074346  0.073146]\n &#091; 4.3      -0.067385  0.066401]\n &#091; 4.4      -0.061067  0.06026 ]\n &#091; 4.5      -0.055334  0.054673]\n &#091; 4.6      -0.050134  0.049591]\n &#091; 4.7      -0.045417  0.044972]\n &#091; 4.8      -0.04114   0.040776]\n &#091; 4.9      -0.037263  0.036964]\n &#091; 5.       -0.033748  0.033503]\n &#091; 5.1      -0.030563  0.030362]\n &#091; 5.2      -0.027677  0.027512]\n &#091; 5.3      -0.025062  0.024926]\n &#091; 5.4      -0.022692  0.022581]\n &#091; 5.5      -0.020546  0.020455]\n &#091; 5.6      -0.018602  0.018527]\n &#091; 5.7      -0.016841  0.01678 ]\n &#091; 5.8      -0.015246  0.015196]\n &#091; 5.9      -0.013802  0.013761]\n &#091; 6.       -0.012494  0.012461]]<\/code><\/pre>\n\n\n\n<p>Instrucciones en Python<\/p>\n\n\n<div class=\"wp-block-syntaxhighlighter-code \"><pre class=\"brush: python; title: ; notranslate\" title=\"\">\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,3),dtype=float)\n    # incluye el punto &#x5B;x0,y0]\n    estimado&#x5B;0] = &#x5B;x0,y0,z0]\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&#x5B;i] = &#x5B;xi,yi,zi]\n    return(estimado)\n\n# PROGRAMA\nf = lambda t,y,z: z\ng = lambda t,y,z: -3*z -2*y\n\nt0 = 0\ny0 = 0\nz0 = -5\n\nh = 0.1\ntn = 6\nmuestras = int((tn-t0)\/h)\n\ntabla = rungekutta2_fg(f,g,t0,y0,z0,h,muestras)\nti = tabla&#x5B;:,0]\nyi = tabla&#x5B;:,1]\nzi = tabla&#x5B;:,2]\n\n# SALIDA\nnp.set_printoptions(precision=6)\nprint('t, y, z')\nprint(tabla)\n\n# GRAFICA\nplt.plot(ti,yi, label='y(t)')\n\nplt.ylabel('y(t)')\nplt.xlabel('t')\nplt.title('Runge-Kutta 2do Orden d2y\/dx2 ')\nplt.legend()\nplt.grid()\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"<p>Ejercicio: 3Eva2021PAOI_T3 EDO Respuesta a entrada cero en un sistema LTIC la ecuaci\u00f3n a resolver es: con valores iniciales: y0(t)=0 , y\u20190(t) =-5 se puede escribir como: sustituyendo las expresiones de las derivadas como las funciones f(x) por z y g(x) por z': y' = z = f(x) y'' = z'= -3z-2y = g(x) Los [&hellip;]<\/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":[52],"tags":[58,54],"class_list":["post-7741","post","type-post","status-publish","format-standard","hentry","category-mn-s3eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/7741","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=7741"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/7741\/revisions"}],"predecessor-version":[{"id":23943,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/7741\/revisions\/23943"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=7741"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=7741"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=7741"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}