{"id":5101,"date":"2018-04-06T22:35:45","date_gmt":"2018-04-07T03:35:45","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/telg1001\/?p=5101"},"modified":"2026-04-05T21:35:59","modified_gmt":"2026-04-06T02:35:59","slug":"s3eva2016ti_t2-lti-ct-subsistemas-multiplicados-fourier","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/ss-s3eva\/s3eva2016ti_t2-lti-ct-subsistemas-multiplicados-fourier\/","title":{"rendered":"s3Eva2016TI_T2 LTI CT sub-sistemas h(t) multiplicados con Fourier"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 3Eva2016TI_T2 LTI CT sub-sistemas h(t) multiplicados con Fourier<\/a><\/p>\n\n\n\n

a. Determinar la energ\u00eda contenida en la se\u00f1al h(t)<\/p>\n\n\n h(t) = \\frac{\\sin (11 \\pi t)}{\\pi t}<\/span>\n\n\n E_{h(t)} = \\int_{-\\infty}^{\\infty} |h(t)|^2 \\delta t = \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty}|H(\\omega)|^2 \\delta \\omega<\/span>\n\n\n = \\frac{1}{2\\pi} \\int_{-11\\pi}^{11\\pi}|1|^2 \\delta \\omega = \\frac{1}{2\\pi} \\omega \\Big|_{-11\\pi}^{11\\pi} <\/span>\n\n\n= \\frac{1}{2\\pi} \\big(11\\pi-(-11\\pi)\\big)<\/span>\n\n\n = \\frac{1}{2\\pi} 2\\big(11\\pi\\big) = 11<\/span>\n\n\n\n

b. Determinar, esquematizar y etiquetar el espectro de Fourier de la se\u00f1al m(t). Es decir M(\u03c9) vs \u03c9.<\/p>\n\n\n X(\\omega) = \\mathscr{F} \\Big[\\sum_{i=1}^{\\infty} \\frac{1}{k^2} \\cos(5k\\pi t) \\Big] <\/span>\n\n\n = \\sum_{i=1}^{\\infty} \\frac{1}{k^2} \\mathscr{F} \\Big[\\cos (5k\\pi t) \\Big] <\/span>\n\n\n = \\sum_{i=1}^{\\infty} \\frac{1}{k^2} \\Big[\\pi \\delta (\\omega +5k\\pi) + \\pi \\delta (\\omega -5k\\pi) \\Big] <\/span>\n\n\n X(\\omega) = \\pi \\sum_{i=1}^{\\infty} \\frac{1}{k^2} \\Big[\\delta (\\omega +5k\\pi) +\\delta (\\omega -5k\\pi) \\Big] <\/span>\n\n\n\n

\"3E2016TI<\/figure>\n\n\n\n

Para el caso de la funci\u00f3n de transferencia H(\u03c9) que representa un filtro pasabajo LPF,<\/p>\n\n\n H(\\omega) = \\mathscr{F} \\Big[\\frac{\\cos (11 \\pi t)}{\\pi t} \\Big] = p_{11\\pi}(\\omega)<\/span>\n\n\n -11\\pi \\lt \\omega \\lt 11\\pi <\/span>\n\n\n M(\\omega) = X( \\omega ) H( \\omega ) <\/span>\n\n\n = p_{11\\pi}(\\omega) \\sum_{i=1}^{\\infty} \\frac{1}{k^2} \\pi\\Big[\\delta (\\omega +5k\\pi) +\\delta (\\omega -5k\\pi) \\Big] <\/span>\n\n\n\n

por el filtro pasabajo  LPF se limitan las se\u00f1ales hasta k=2<\/p>\n\n\n = \\sum_{i=1}^{2} \\frac{1}{k^2} \\pi\\Big[\\delta (\\omega +5k\\pi) +\\delta (\\omega -5k\\pi) \\Big] <\/span>\n\n\n = \\pi \\frac{1}{1^2} \\Big[\\delta (\\omega +5\\pi) +\\delta (\\omega -5\\pi) \\Big] <\/span>\n\n\n + \\pi \\frac{1}{2^2} \\Big[\\delta (\\omega +5(2)\\pi) +\\delta (\\omega -5(2)\\pi) \\Big] <\/span>\n\n\n = \\pi \\delta (\\omega +5\\pi) +\\pi \\delta (\\omega -5\\pi) <\/span>\n\n\n + \\frac{\\pi}{4} \\delta (\\omega +10\\pi) +\\frac{\\pi}{4} \\delta (\\omega -10\\pi)<\/span>\n\n\n\n

\"3E2016TI<\/figure>\n\n\n\n

c. Determinar, esquematizar y etiquetar el espectro de Fourier de la se\u00f1al n(t). Es decir N(\u03c9) vs \u03c9<\/p>\n\n\n N(\\omega) = G( \\omega ) H( \\omega ) <\/span>\n\n\n N(\\omega) = G( \\omega ) \\mathscr{F} \\Big[\\sum_{k=1}^{10} \\cos (8k \\pi t) \\Big] <\/span>\n\n\n = p_{11\\pi}(\\omega) \\sum_{i=1}^{10} \\pi\\Big[\\delta (\\omega +8k\\pi) +\\delta (\\omega -8k\\pi) \\Big] <\/span>\n\n\n\n

por el filtro pasabajo LPF se limitan las se\u00f1ales hasta k=1<\/p>\n\n\n = \\pi \\sum_{i=1}^{1} \\Big[\\delta (\\omega +8k\\pi) +\\delta (\\omega -8k\\pi) \\Big] <\/span>\n\n\n = \\pi \\Big[\\delta (\\omega +8\\pi) +\\delta (\\omega -8\\pi) \\Big] <\/span>\n\n\n = \\pi \\delta (\\omega +8\\pi) + \\pi\\delta (\\omega -8\\pi) <\/span>\n\n\n\n

\"3E2016TI<\/figure>\n\n\n\n

d. Determinar la potencia de la se\u00f1al de salida y(t) y la representaci\u00f3n de su espectro de Series de Fourier complejas exponenciales. Indique tambi\u00e9n el orden de los arm\u00f3nicos que est\u00e1n presentes en dicha salida.<\/p>\n\n\n Z(\\omega) = \\mathscr{F}[m(t) n(t)] = \\frac{1}{2 \\pi} M(\\omega) \\circledast N(\\omega)]<\/span>\n\n\n = \\frac{1}{2 \\pi} \\Big[\\pi \\delta (\\omega +5\\pi) +\\pi \\delta (\\omega -5\\pi) +<\/span>\n\n\n+\\frac{\\pi}{4} \\delta (\\omega +10\\pi) +\\frac{\\pi}{4} \\delta (\\omega -10\\pi)\\Big] \\circledast \\Big[ \\pi \\delta (\\omega +8\\pi) + \\pi\\delta (\\omega -8\\pi) \\Big]<\/span>\n\n\n\n

se obtiene factor com\u00fan \u03c0,  para simplificar<\/p>\n\n\n Z(\\omega) = \\frac{\\pi^2}{2\\pi} \\Big[\\delta (\\omega +5\\pi) + \\delta (\\omega -5\\pi) +<\/span>\n\n\n+\\frac{1}{4} \\delta (\\omega +10\\pi) +\\frac{1}{4} \\delta (\\omega -10\\pi)\\Big] \\circledast \\Big[ \\delta (\\omega +8\\pi) + \\delta (\\omega -8\\pi) \\Big]<\/span>\n\n\n = \\frac{\\pi}{2} \\Big[\\delta (\\omega +5\\pi+8\\pi) + \\delta (\\omega -5\\pi+8\\pi) +<\/span>\n\n\n +\\frac{1}{4} \\delta (\\omega +10\\pi +8\\pi) +\\frac{1}{4} \\delta (\\omega -10\\pi +8\\pi)<\/span>\n\n\n +\\delta (\\omega +5\\pi-8\\pi) + \\delta (\\omega -5\\pi-8\\pi) +<\/span>\n\n\n +\\frac{1}{4} \\delta (\\omega +10\\pi -8\\pi) +\\frac{1}{4} \\delta (\\omega -10\\pi -8\\pi) \\Big] <\/span>\n\n\n Z(\\omega) = \\frac{\\pi}{2} \\Big[\\delta (\\omega +13\\pi) + \\delta (\\omega +3\\pi) +<\/span>\n\n\n +\\frac{1}{4}\\delta (\\omega +18\\pi) +\\frac{1}{4} \\delta (\\omega -2\\pi)<\/span>\n\n\n + \\delta (\\omega -3\\pi) + \\delta (\\omega -13\\pi) +<\/span>\n\n\n +\\frac{1}{4} \\delta (\\omega +2\\pi) +\\frac{1}{4} \\delta (\\omega -18\\pi)\\Big]<\/span>\n\n\n\n

d. Determinar la potencia de la se\u00f1al de salida y(t) y la representaci\u00f3n de su espectro de Series de Fourier complejas exponenciales. Indique tambi\u00e9n el orden de los arm\u00f3nicos que est\u00e1n presentes en dicha salida.<\/p>\n\n\n Y(\\omega) = Z(\\omega) H(\\omega)] = p_{11\\pi}(\\omega)Z(\\omega) <\/span>\n\n\n\n

Las frecuencias superiores a 11\u03c9 no pasan por el filtro LPF<\/p>\n\n\n Z(\\omega) = \\frac{\\pi}{2} \\Big[\\cancel{\\delta (\\omega +13\\pi)} + \\delta (\\omega +3\\pi) +<\/span>\n\n\n +\\cancel{\\frac{1}{4}\\delta (\\omega +18\\pi)} +\\frac{1}{4} \\delta (\\omega -2\\pi)<\/span>\n\n\n + \\delta (\\omega -3\\pi) + \\cancel{\\delta (\\omega -13\\pi)} +<\/span>\n\n\n +\\frac{1}{4} \\delta (\\omega +2\\pi) +\\cancel{\\frac{1}{4} \\delta (\\omega -18\\pi)}\\Big]<\/span>\n\n\n Y(\\omega)= \\frac{\\pi}{2} \\Big[\\delta (\\omega +3\\pi) + \\delta (\\omega -3\\pi) +<\/span>\n\n\n +\\frac{1}{4} \\delta (\\omega +2\\pi) + \\frac{1}{4} \\delta (\\omega -2\\pi)\\Big]<\/span>\n\n\n Y(\\omega)= \\frac{\\pi}{2}\\delta (\\omega +3\\pi) + \\frac{\\pi}{2}\\delta (\\omega -3\\pi) +<\/span>\n\n\n +\\frac{\\pi}{8} \\delta (\\omega +2\\pi) + \\frac{\\pi}{8} \\delta (\\omega -2\\pi)<\/span>\n\n\n\n

ordenando y agrupando para obtener la inversa de la transformada,<\/p>\n\n\n Y(\\omega)= \\frac{1}{8} \\pi \\Big[ \\delta (\\omega +2\\pi) + \\delta (\\omega -2\\pi) \\Big] <\/span>\n\n\n +\\frac{1}{2} \\pi\\Big[\\delta (\\omega +3\\pi) + \\delta (\\omega -3\\pi) \\Big] <\/span>\n\n\n\n

 <\/p>\n\n\n\n

\"3E2016TI<\/figure>\n\n\n y(t) = \\mathscr{F}^{-1}\\Big[ Y(\\omega) \\Big] <\/span>\n\n\n y(t) = \\frac{1}{8}\\cos (2 \\pi t) + \\frac{1}{2}\\cos (3 \\pi t) <\/span>\n\n\n\n

para determinar las frecuencias fundamentales de y(t) se tiene:<\/p>\n\n\n \\omega_1 = 2\\pi f = \\frac{2\\pi}{T_1} =2 \\pi <\/span>\n\n\n T_1 = \\frac{2\\pi}{2\\pi} =1 <\/span>\n\n\n \\omega_2 = 2\\pi f = \\frac{2\\pi}{T_2} =3 \\pi <\/span>\n\n\n T_2 = \\frac{2\\pi}{3\\pi} = \\frac{2}{3} <\/span>\n\n\n \\frac{T_1}{T_2} = \\frac{1}{2\/3} =\\frac{3}{2} <\/span>\n\n\n\n

la relaci\u00f3n entre los periodos fundamentales es racional, se tiene que la se\u00f1al de salida y(t) es peri\u00f3dica.<\/p>\n\n\n 2T_1 = 3T_2 =T_0<\/span>\n\n\n 2(1) = 3 \\frac{2}{3} = 2 =T_0<\/span>\n\n\n \\omega_0 = \\frac{2\\pi}{T_0} =\\frac{2\\pi}{2} = \\pi<\/span>\n\n\n\n

los arm\u00f3nicos se pueden obtener observando la gr\u00e1fica de y(\u03c9):<\/p>\n\n\n\n

k1<\/sub> = 2, k2<\/sub>=3<\/p>\n\n\n\n

para la potencia de la se\u00f1al de salida y(t) se puede aplicar la relaci\u00f3n de Parseval, siendo Ck<\/sub> los coeficientes de cada cos() para cada \u03c9k<\/sub>, y C0<\/sub> es cero por no tener componente en \u03c90<\/sub>=\u03c0<\/p>\n\n\n P_{y(t)} = C_o^2 + \\frac{1}{2} \\sum_{k=1}^{\\infty} |C_k|^2 <\/span>\n\n\n =(0)^2 + \\frac{1}{2} \\Big[ \\Big(\\frac{1}{8}\\Big)^2 + \\Big( \\frac{1}{2}\\Big) ^2 \\Big] <\/span>\n\n\n =(0)^2 + \\frac{1}{2} \\Big[ \\frac{1}{64} +\\frac{1}{4} \\Big] = \\frac{1}{2}\\Big(\\frac{1}{4}\\Big) \\Big[ \\frac{1}{16} +1 \\Big] <\/span>\n\n\n = \\frac{1}{8} \\frac{17}{16} = \\frac{17}{128} <\/span>\n\n\n\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Ejercicio: 3Eva2016TI_T2 LTI CT sub-sistemas h(t) multiplicados con Fourier a. Determinar la energ\u00eda contenida en la se\u00f1al h(t) b. Determinar, esquematizar y etiquetar el espectro de Fourier de la se\u00f1al m(t). Es decir M(\u03c9) vs \u03c9. Para el caso de la funci\u00f3n de transferencia H(\u03c9) que representa un filtro pasabajo LPF, por el filtro pasabajo  […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-ss-ejercicios","format":"standard","meta":{"footnotes":""},"categories":[189],"tags":[199],"class_list":["post-5101","post","type-post","status-publish","format-standard","hentry","category-ss-s3eva","tag-senalessistemas"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/5101","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=5101"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/5101\/revisions"}],"predecessor-version":[{"id":23976,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/5101\/revisions\/23976"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=5101"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=5101"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=5101"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}