{"id":2259,"date":"2018-01-14T08:05:42","date_gmt":"2018-01-14T13:05:42","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/telg1001\/?p=2259"},"modified":"2026-04-05T21:28:59","modified_gmt":"2026-04-06T02:28:59","slug":"s2eva2016tii_t2-lti-ct-circuito-rc-respuesta-de-frecuencia-h%cf%89-impulso-ht","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/ss-s2eva\/s2eva2016tii_t2-lti-ct-circuito-rc-respuesta-de-frecuencia-h%cf%89-impulso-ht\/","title":{"rendered":"s2Eva2016TII_T2 LTI CT Circuito RC respuesta de frecuencia H(\u03c9), impulso h(t)"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 2Eva2016TII_T2 LTI CT Circuito RC respuesta de frecuencia H(\u03c9), impulso h(t)<\/a><\/p>\n\n\n\n

\"2E2016TII_T2<\/figure>\n\n\n\n

literal a<\/h2>\n\n\n v_1 (t) = v_R (t) +v_C (t)<\/span>\n\n\n v_1 (t) = R i(t) +v_2 (t)<\/span>\n\n\n i(t) = C \\frac{\\delta v_2 (t)}{\\delta t} <\/span>\n\n\n v_2(t) = RC \\frac{\\delta v_2 (t)}{\\delta t} +v_2 (t)<\/span>\n\n\n V_1 (\\omega) = j \\omega RC V_2 (\\omega) + V_2(\\omega) = V_2(\\omega) [1+j\\omega RC]<\/span>\n\n\n H(\\omega) = \\frac{V_2 (\\omega)}{V_1(\\omega)} = \\frac{1}{1+j\\omega RC} = \\frac{1}{1+j\\frac{\\omega}{\\omega_c}}<\/span>\n\n\n \\omega _c = \\frac{1}{RC}<\/span>\n\n\n H(\\omega) =\\frac{V_2(\\omega)}{V_1(\\omega)} = \\frac{1}{1+j\\frac{\\omega}{2}}<\/span>\n\n\n \\begin{cases} |H(\\omega) = \\frac{1}{\\sqrt{1+\\big( \\frac{\\omega}{2}\\big)^2}} \\\\ \\theta_{H(\\omega)} = -tg^{-1} \\big( \\frac{\\omega}{2}\\big) \\end{cases}<\/span>\n\n\n \\omega _c = \\frac{1}{RC} <\/span>\n\n\n\n

literal c<\/h2>\n\n\n\n

La respuesta impulso del filtro LPF, se obtiene mediante:<\/p>\n\n\n h(t) = \\mathcal{F}^{-1} \\Big[ H(\\omega) \\Big] = \\mathcal{F}^{-1} \\Big[ \\frac{1}{1+j(\\omega \/2)} \\Big] <\/span>\n\n\n =2\\mathcal{F}^{-1} \\Big[ \\frac{1}{j\\omega +2} \\Big] <\/span>\n\n\n h(t)=2e^{2t}\\mu (t) <\/span>\n\n\n\n

literal d<\/h2>\n\n\n\n

M\u00e9todo 1: usando la respuesta de frecuencia<\/p>\n\n\n \\begin{cases} |H(\\omega) = \\frac{1}{\\sqrt{1+\\big( \\frac{\\omega}{2}\\big)^2}} \\\\ \\theta_{H(\\omega)} = -tg^{-1} \\big( \\frac{\\omega}{2}\\big) \\end{cases}<\/span>\n\n\n \\begin{cases} |H(50) = \\frac{1}{\\sqrt{1+\\big( \\frac{50}{2}\\big)^2}} = \\frac{1}{\\sqrt(626)} \\\\ \\theta_{H(\\omega)} = -tg^{-1} \\big( \\frac{50}{2}\\big) = -87.7\\end{cases}<\/span>\n\n\n v_2(t) = |H(50)| \\sin \\big(50t+\\theta_{H(50)} \\big)<\/span>\n\n\n v_2(t) = \\frac{1}{\\sqrt(626)} \\sin \\big(50t-87.7) \\big) <\/span>\n\n\n\n

m\u00e9todo 2:<\/p>\n\n\n V_2(\\omega ) = V_1(\\omega) H(\\omega) <\/span>\n\n\n V_1 (\\omega) = \\mathcal{F}[v_1(t) ] = \\mathcal{F} [ \\sin (50t) ] <\/span>\n\n\n = j \\pi \\delta (\\omega +50) - j \\pi \\delta (\\omega-50) <\/span>\n\n\n V_2 (\\omega) = V_1(\\omega) H(\\omega) <\/span>\n\n\n V_2 (\\omega) = \\Big[ j \\pi \\delta (\\omega +50) - j \\pi \\delta (\\omega-50) \\Big] \\Big[ \\frac{1}{1+j(\\omega\/2)} \\Big] <\/span>\n\n\n = j \\pi \\Big[ \\delta (\\omega + 50)\\frac{1}{1+j(\\omega \/2)} - \\delta (\\omega - 50)\\frac{1}{1+j(\\omega \/2)}\\Big] <\/span>\n\n\n = j \\pi \\Big[ \\delta (\\omega + 50)\\frac{1}{1+j(50 \/2)} - \\delta (\\omega - 50)\\frac{1}{1+j(50 \/2)}\\Big] <\/span>\n\n\n = j \\pi \\Big[ \\delta (\\omega + 50)\\frac{1+j25}{626} - \\delta (\\omega - 50)\\frac{1-j25}{626}\\Big] <\/span>\n\n\n = j \\frac{\\pi}{626} \\Big[ \\delta (\\omega + 50) - \\delta (\\omega - 50) + j25 \\delta (\\omega + 50) + j25 \\delta (\\omega - 50)\\Big] <\/span>\n\n\n = \\frac{1}{626} \\Big[ j\\pi \\delta (\\omega + 50) - j \\pi \\delta(\\omega - 50)\\Big] - \\frac{25}{626} \\Big[ \\pi \\delta (\\omega + 50) +\\pi \\delta(\\omega - 50)\\Big] <\/span>\n\n\n v_2(t) = \\mathcal{F}^{-1} [V_2 (\\omega)] <\/span>\n\n\n = \\frac{1}{626} \\mathcal{F}^{-1}\\Bigg[ \\Big[ j\\pi \\delta (\\omega + 50) - j \\pi \\delta(\\omega - 50)\\Big] - 25 \\Big[ \\pi \\delta (\\omega + 50) +\\pi \\delta(\\omega - 50)\\Big] \\Bigg] <\/span>\n\n\n v_2(t) = \\frac{1}{626} \\Big[ \\sin (50t) - 25 \\cos (50t) \\Big] <\/span>\n\n\n\n

usando fasores:<\/p>\n\n\n v_2(t) = \\frac{1}{626} \\cos (50t-177.709) = \\frac{1}{626} \\sin (50t-87.70) <\/span>\n\n\n\n

En la salida, existe un factor de atenuaci\u00f3n de 0.04 y un retardo de 87.70\u00b0.
Como la se\u00f1al de entrada se reproduce de manera exacta en su salida a pesar tener amplitud diferente y un retardo en el tiempo.<\/p>\n","protected":false},"excerpt":{"rendered":"

Ejercicio: 2Eva2016TII_T2 LTI CT Circuito RC respuesta de frecuencia H(\u03c9), impulso h(t) literal a literal c La respuesta impulso del filtro LPF, se obtiene mediante: literal d M\u00e9todo 1: usando la respuesta de frecuencia m\u00e9todo 2: usando fasores: En la salida, existe un factor de atenuaci\u00f3n de 0.04 y un retardo de 87.70\u00b0.Como la se\u00f1al […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn","format":"standard","meta":{"footnotes":""},"categories":[187],"tags":[199],"class_list":["post-2259","post","type-post","status-publish","format-standard","hentry","category-ss-s2eva","tag-senalessistemas"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2259","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=2259"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2259\/revisions"}],"predecessor-version":[{"id":23966,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2259\/revisions\/23966"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=2259"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=2259"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=2259"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}