{"id":1156,"date":"2016-12-27T08:09:10","date_gmt":"2016-12-27T13:09:10","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/estg1003\/?p=1156"},"modified":"2026-04-04T10:56:28","modified_gmt":"2026-04-04T15:56:28","slug":"autocovarianza-en-am","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/stp-u02eva\/autocovarianza-en-am\/","title":{"rendered":"Autocovarianza - AM y PM"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\">1. Autocovarianza AM<\/h2>\n\n\n\n<p><strong><em>Referencia<\/em><\/strong>: Le\u00f3n Garc\u00eda Ejemplo 9.9 p495, Gubner Ej10.35 p496<\/p>\n\n\n\n<figure class=\"wp-block-image aligncenter\"><img decoding=\"async\" src=\"http:\/\/blog.espol.edu.ec\/estg1003\/files\/2017\/07\/autocovarianza_Ej_LeonGarcia9_9.png\" alt=\"\" \/><\/figure>\n\n\n\n<p>Sea X(t) = A cos(2\u03c0t), donde A es una variable aleatoria, con un comportamiento semejante a la figura.<br><\/p>\n\n\n\n<p>Encontrar el valor esperado , la autocorrelaci\u00f3n y autocovarianza de de X(t).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<p>El valor esperado se calcula a continuaci\u00f3n, note que la media varia respecto a t y que el valor es cero para valores de t donde cos(2\u03c0t) =0.<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> E[X(t)] = E[A cos(2\\pi t)] <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> E[X(t)] = E[A] cos(2\\pi t) <\/span>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<p>La autocorrelaci\u00f3n es:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> R_X(t_1,t_2) = E[A cos(2\\pi t_1) A cos(2\\pi t_2)] <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = E[A^{2} cos(2\\pi t_1) cos(2\\pi t_2)] <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = E[A^{2}] cos(2\\pi t_1) cos(2\\pi t_2) <\/span>\n\n\n\n<p>usando:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> 2 cos(x)cos(y) = cos(x-y) + cos(x+y) <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> cos(x)cos(y) = \\frac{ cos(x-y) + cos(x + y)}{2} <\/span>\n\n\n\n<p>se reemplaza:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = E[A^{2}] \\frac{1}{2}[cos(2\\pi t_1 - 2\\pi t_2) + cos(2\\pi t_1 + 2\\pi t_2)] <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> R_X(t_1,t_2) = E[A^{2}] \\frac{[cos(2\\pi (t_1 - t_2)) + cos(2\\pi (t_1 + t_2))]}{2} <\/span>\n\n\n\n<p>se observa que el valor de autocorrelaci\u00f3n depende de las diferencias de tiempo t<sub>1<\/sub> y t<sub>2<\/sub>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<p>La autocovarianza es:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> Cov_X(t_1,t_2) = R_X(t_1,t_2) - E[X(t_1)]E[X(t_2)] <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = E[A^{2}] cos(2\\pi t_1) cos(2\\pi t_2) - E[A] cos(2\\pi t_1)E[A] cos(2\\pi t_2) <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = E[A^{2}] cos(2\\pi t_1) cos(2\\pi t_2) - E[A]^2 cos(2\\pi t_1)cos(2\\pi t_2) <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = (E[A^{2}] - E[A]^2) cos(2\\pi t_1)cos(2\\pi t_2) <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = Var[A] cos(2\\pi t_1)cos(2\\pi t_2) <\/span>\n\n\n\n<p>con el mismo procedimiento de cos(x)cos(y):<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> Cov_X(t_1,t_2) = Var[A] \\frac{[cos(2\\pi (t_1 - t_2)) + cos(2\\pi (t_1 + t_2))]}{2} <\/span>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<h2 class=\"wp-block-heading\">2. Autocovarianza PM<\/h2>\n\n\n\n<p><strong><em>Referencia<\/em><\/strong>: Le\u00f3n Garc\u00eda Ejemplo 9.10 p495, Gubner Ejemplo 10.8 p389, Gubner Ejemplo 10.17 p396<\/p>\n\n\n\n<p>Sea X(t) = cos(\u03c9 t + \u03a6), donde \u03a6 es uniforme en el intervalo (-\u03c0,\u03c0) Encontrar la autocovarianza de X(t).<br><img decoding=\"async\" src=\"http:\/\/blog.espol.edu.ec\/algoritmos101\/files\/2017\/07\/autocovarianza_Ej_LeonGarcia9_10.png\"><\/p>\n\n\n\n<p>la variable aleatoria \u03a6 tiene distribuci\u00f3n uniforme en el intervalo, por lo que la funci\u00f3n f<sub>\u03a6<\/sub>(\u03c6) es constante = 1\/[\u03c0 - (-\u03c0)] = 1\/2\u03c0.<\/p>\n\n\n\n<p>Recordamos que:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> E[g(x)] = \\int_{-\\infty}^{\\infty} g(x) f(x) dx <\/span>\n\n\n\n<p><strong><em>Media <\/em><\/strong>(Le\u00f3n-Garc\u00eda 4.15 p158):<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> m_X(t) = E[cos(\\omega t + \\Phi)] = <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = \\int_{-\\pi}^{\\pi} cos(\\omega t + \\Phi) \\frac{1}{2\\pi} d\\Phi <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = \\left. \\frac{-1}{2\\pi} (sin (\\omega t + \\Phi)) \\right|_{-\\pi}^{\\pi} <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = \\frac{-1}{2\\pi} [sin (\\omega t + (-\\pi)) - sin (\\omega t + \\pi)] = 0 <\/span>\n\n\n\n<p><strong><em>Autocovarianza<\/em><\/strong><\/p>\n\n\n\n<p>dado que el valor esperado es cero, la autocovarianza es igual a la autocorrelaci\u00f3n<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> C_{X} (t_1,t_2) = R_X (t_1,t_2) - E[X(t_1,t_2)] <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = R_X (t_1,t_2) = E[cos(\\omega t_1 + \\Phi) cos(\\omega t_2 +\\Phi)] <\/span>\n\n\n\n<p>Recordando que:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> E[g(x)] = \\int_{-\\infty}^{\\infty} g(x) f(x) dx <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> cos(x) cos(y) = \\frac{cos(x-y) + cos(x+y) }{2} <\/span>\n\n\n\n<p>se tiene que:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = \\int_{-\\pi}^{\\pi} [cos(\\omega t_1 + \\Phi) cos(\\omega t_2 +\\Phi)] \\frac{1}{2\\pi} d\\Phi <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = \\int_{-\\pi}^{\\pi} \\frac{cos(\\omega (t_1 - t_2))+cos(\\omega (t_1 + t_2)+ 2\\Phi)}{2} \\frac{1}{2\\pi} d\\Phi <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = \\int_{-\\pi}^{\\pi} \\frac{cos(\\omega (t_1 - t_2))}{2} \\frac{1}{2\\pi} d\\Phi + \\int_{-\\pi}^{\\pi} \\frac{cos(\\omega (t_1 + t_2 )+ 2\\Phi)}{2} \\frac{1}{2\\pi} d\\Phi <\/span>\n\n\n\n<p>El primer integral, el coseno no depende de \u03a6, mientras que el segundo integral es semejante al intergral de la media y cuyo resultado es cero.<\/p>\n\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = \\left. \\frac{cos(\\omega (t_1 - t_2))}{2} \\frac{\\Phi}{2\\pi} \\right|_{-\\pi}^{\\pi} + 0 <\/span>\n\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> C_{X} (t_1,t_2) = R_X (t_1,t_2) = <\/span>\n\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = \\frac{1}{2} cos(\\omega (t_1 - t_2))<\/span>\n","protected":false},"excerpt":{"rendered":"<p>1. Autocovarianza AM Referencia: Le\u00f3n Garc\u00eda Ejemplo 9.9 p495, Gubner Ej10.35 p496 Sea X(t) = A cos(2\u03c0t), donde A es una variable aleatoria, con un comportamiento semejante a la figura. Encontrar el valor esperado , la autocorrelaci\u00f3n y autocovarianza de de X(t). El valor esperado se calcula a continuaci\u00f3n, note que la media varia respecto [&hellip;]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-stp-unidades","format":"standard","meta":{"footnotes":""},"categories":[214],"tags":[],"class_list":["post-1156","post","type-post","status-publish","format-standard","hentry","category-stp-u02eva"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1156","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=1156"}],"version-history":[{"count":1,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1156\/revisions"}],"predecessor-version":[{"id":22147,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1156\/revisions\/22147"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=1156"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=1156"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=1156"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}