{"id":1588,"date":"2016-12-22T09:28:24","date_gmt":"2016-12-22T14:28:24","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/estg1003\/?p=1588"},"modified":"2026-04-04T10:55:55","modified_gmt":"2026-04-04T15:55:55","slug":"autocorrelacion-cuadrado-triangulo","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/stp-unidades\/stp-u02eva\/autocorrelacion-cuadrado-triangulo\/","title":{"rendered":"autocorrelaci\u00f3n cuadrado y triangulo"},"content":{"rendered":"\n

Referencia<\/strong><\/em>: Problema Le\u00f3n Garc\u00eda 10.4 p.635<\/p>\n\n\n\n

\"\"<\/a><\/figure>\n\n\n\n

a) Encuentre la funci\u00f3n de autocorrelaci\u00f3n correspondiente a la densidad espectral de potencia de la figura. <\/p>\n\n\n\n

b) Encuentre el total de la potencia promedio<\/p>\n\n\n\n

c) Grafique la potencia en el rango de |f| > f0<\/sub> como funci\u00f3n de f0<\/sub>>0.<\/p>\n\n\n\n

Soluci\u00f3n propuesta<\/em><\/strong>:<\/p>\n\n\n\n

a)
S_X(f) = A \\prod \\Big(\\frac{f}{2f_2} \\Big) + (B-A) \\bigwedge \\Big( \\frac{f}{f_1} \\Big) <\/span><\/p>\n\n\n R_X(\\tau) = F^{-1}\\Big[ A \\prod \\Big(\\frac{f}{2f_2} \\Big) \\Big] + F^{-1} \\Big[ (B-A) \\bigwedge \\Big( \\frac{f}{f_1} \\Big) \\Big] <\/span>\n\n\n = 2A f_2 [Sa (2\\pi f_2 \\tau)] + (B-A) f_1 [Sa (\\pi f_1 \\tau) ]^2 <\/span>\n\n\n\n

b)
P = \\int_{-\\infty}^{\\infty} S_X(f) \\delta f <\/span><\/p>\n\n\n = A(2f_2) + (2f_1)\\frac{(B-A)}{2} <\/span>\n\n\n = 2A f_2 + (B-A)f_1 <\/span>\n\n\n\n

c)<\/p>\n\n\n\n

la potencia en funci\u00f3n de la frecuencia es par, por lo que se integra entre 0 y f0<\/sub> y se duplica para el rango entre [-f0<\/sub>, f0<\/sub>]
2\\int_{0}^{f_0} S_X(f) \\delta f = <\/span><\/p>\n\n\n\n

primera secci\u00f3n: 0 < f0<\/sub> < f1<\/sub><\/p>\n\n\n 2\\int_{0}^{f_0} \\Big[ \\Big( -\\frac{B-A}{f_1} \\Big)f +B\\Big] \\delta f <\/span>\n\n\n = 2 \\Big[ \\Big(-\\frac{B-A}{f_1}\\Big) \\frac{ f^2}{2} +Bf \\Big] \\Big|_{0}^{f_0} <\/span>\n\n\n = \\Big(-\\frac{B-A}{f_1}\\Big) f_0^2 +2Bf_0 <\/span>\n\n\n = 2Bf_0 -\\frac{B-A}{f_1} f_0^2 <\/span>\n\n\n\n

segunda secci\u00f3n: f1<\/sub> < f0<\/sub> < f2<\/sub>
= 2[ \\frac{B-A}{2}f_1 + A(f_0 - f_1) \\Big] <\/span><\/p>\n\n\n\n

\"\"<\/a><\/figure>\n\n\n\n

<\/p>\n\n\n

\n# leon- garcia 10.4\n# literal c\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# INGRESO\nA = 1\nB = 2\nf1 = 1\nf2 = 2\n\nn = 50\nfinal = 4\n# PROCEDIMIENTO\nf = np.linspace(0,final,n)\nP = np.zeros(n,dtype=float)\nfor i in range(0,n,1):\n    if f[i]= f1 and f[i]f2:\n        P[i] = 2*(((B+A)\/2)*f1 + A*(f2-f1))\n\n# SALIDA Grafica\nplt.plot(f,P)\nplt.vlines(f1,0,2.5*B, color='m', linestyle='dashed')\nplt.vlines(f2,0,2.5*B, color='m', linestyle='dashed')\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Referencia: Problema Le\u00f3n Garc\u00eda 10.4 p.635 a) Encuentre la funci\u00f3n de autocorrelaci\u00f3n correspondiente a la densidad espectral de potencia de la figura. b) Encuentre el total de la potencia promedio c) Grafique la potencia en el rango de |f| > f0 como funci\u00f3n de f0>0. Soluci\u00f3n propuesta: a) b) c) la potencia en funci\u00f3n de […]<\/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-1588","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\/1588","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=1588"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1588\/revisions"}],"predecessor-version":[{"id":22375,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1588\/revisions\/22375"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=1588"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=1588"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=1588"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}