{"id":1223,"date":"2016-12-07T08:48:55","date_gmt":"2016-12-07T13:48:55","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/estg1003\/?p=1223"},"modified":"2026-04-04T10:52:41","modified_gmt":"2026-04-04T15:52:41","slug":"psd-densidad-espectral-potencia-pm","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/stp-u02eva\/psd-densidad-espectral-potencia-pm\/","title":{"rendered":"PSD Densidad espectral de Potencia de PM"},"content":{"rendered":"

Referencia<\/strong><\/em>: Le\u00f3n-Garc\u00eda Ejemplo 10.2 p581, Gubner Ejemplo 10.21 p399<\/p>\n

Ejercicio<\/h2>\n

Sea X(t) = a cos(\u03c9 t + \u0398), donde \u0398 es uniforme en el intervalo (0,2\u03c0) Encontrar la autocovarianza de X(t).
\nEncuentre la densidad espectral de potencia de SX<\/sub>(f):
\n
\nSoluci\u00f3n<\/em><\/strong><\/p>\n S_X(f) = Fourier\\{ R_X (\\tau) \\} <\/span>\n

\n

Autocorrelaci\u00f3n<\/p>\n R_X (t_1,t_2) = E[(a \\cos(\\omega t_1 + \\Theta)) (a \\cos(\\omega t_2 +\\Theta))] <\/span>\n

Recordando que:<\/p>\n E[g(x)] = \\int_{-\\infty}^{\\infty} g(x) f(x) dx <\/span>\n cos(x) cos(y) = \\frac{cos(x-y) + cos(x+y) }{2} <\/span>\n

se tiene que:<\/p>\n = \\int_{-\\pi}^{\\pi} [a\\cos(\\omega t_1 + \\Theta) a\\ cos(\\omega t_2 +\\Theta)] \\frac{1}{2\\pi} d\\Theta<\/span>\n = \\int_{-\\pi}^{\\pi} a^2 \\frac{\\cos(\\omega (t_1 - t_2))+\\cos(\\omega (t_1 + t_2)+ 2\\Theta)}{2} \\frac{1}{2\\pi} d\\Theta<\/span>\n = a^2 \\int_{-\\pi}^{\\pi} \\frac{cos(\\omega (t_1 - t_2))}{2} \\frac{1}{2\\pi} d\\Theta + a^2\\int_{-\\pi}^{\\pi} \\frac{cos(\\omega (t_1 + t_2 )+ 2\\Theta)}{2} \\frac{1}{2\\pi} d\\Theta<\/span>\n

El primer integral, el coseno no depende de \u0398, mientras que el segundo integral es semejante al intergral de la media y cuyo resultado es cero.<\/p>\n = a^2 \\left. \\frac{cos(\\omega (t_1 - t_2))}{2} \\frac{\\Theta}{2\\pi} \\right|_{-\\pi}^{\\pi} + 0 <\/span>\n R_X (t_1,t_2) = \\frac{a^2}{2} cos(\\omega (t_1 - t_2)) <\/span>\n R_X (t_1,t_2) = \\frac{a^2}{2} cos(\\omega \\tau) <\/span>\n

\n

La densidad espectral de potencia entonces es:<\/p>\n S_X(f) = Fourier\\{ R_X (\\tau) \\} <\/span>\n = Fourier\\{ \\frac{a^2}{2} cos(\\omega \\tau) \\} <\/span>\n = \\frac{a^2}{2} Fourier\\{ cos(\\omega \\tau) \\} <\/span>\n = \\frac{a^2}{2} Fourier\\{ cos(2\\pi f_0 \\tau) \\} <\/span>\n

usando las tablas de transformadas de Fourier:<\/p>\n = \\frac{a^2}{2} (\\frac{1}{2}\\delta (f-f_0) + \\frac{1}{2} \\delta (f+f_0)) <\/span>\n = \\frac{a^2}{4} \\delta (f-f_0) + \\frac{a^2}{4} \\delta (f+f_0)) <\/span>\n

El promedio de potencia de la se\u00f1al es RX<\/sub>(0) = a2<\/sup>\/2.
\nDe toda esta potencia, se concentra en las frecuencias en f0<\/sub> positiva y negativa, por lo que la densidad espectral de potencia en esta frecuencias es infinita.<\/p>\n","protected":false},"excerpt":{"rendered":"

Referencia: Le\u00f3n-Garc\u00eda Ejemplo 10.2 p581, Gubner Ejemplo 10.21 p399 Ejercicio Sea X(t) = a cos(\u03c9 t + \u0398), donde \u0398 es uniforme en el intervalo (0,2\u03c0) Encontrar la autocovarianza de X(t). Encuentre la densidad espectral de potencia de SX(f): Soluci\u00f3n Autocorrelaci\u00f3n Recordando que: se tiene que: El primer integral, el coseno no depende de \u0398, […]<\/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-1223","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\/1223","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=1223"}],"version-history":[{"count":1,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1223\/revisions"}],"predecessor-version":[{"id":22143,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1223\/revisions\/22143"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=1223"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=1223"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=1223"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}