{"id":1286,"date":"2017-04-23T12:00:23","date_gmt":"2017-04-23T17:00:23","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/telg1001\/?p=1286"},"modified":"2026-04-05T22:58:18","modified_gmt":"2026-04-06T03:58:18","slug":"convolucion-integrales-tabla","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/ss-u03\/convolucion-integrales-tabla\/","title":{"rendered":"3.9.2 Convoluci\u00f3n Integrales - Tabla"},"content":{"rendered":"\n

Referencia<\/strong><\/em>: Lathi Tabla 2.1 p176<\/p>\n\n\n\n

Respuesta a estado cero y convoluci\u00f3n con Sympy-Python<\/a><\/p>\n\n\n\n

No<\/td>x1<\/sub>(t)<\/th>x2<\/sub>(t)<\/th>x1<\/sub>(t)\u2297x2<\/sub>(t) = x2<\/sub>(t)\u2297x1<\/sub>(t)<\/th><\/tr>
1<\/td>x(t)<\/td>\u03b4(t-T)<\/td>x(t-T)<\/td><\/tr>
2<\/td>e\u03bbt<\/sup> \u03bc(t)<\/td>\u03bc(t)<\/td> \\frac{1-e^{\\lambda t}}{-\\lambda} \\mu (t) <\/span><\/td><\/tr>
3<\/td>\u03bc(t)<\/td>\u03bc(t)<\/td>t \u03bc(t)<\/td><\/tr>
4<\/td>e\u03bb1<\/sub>t<\/sup> \u03bc(t)<\/td>e\u03bb2<\/sub>t<\/sup> \u03bc(t)<\/td> \\frac{e^{\\lambda _1 t}-e^{\\lambda _2 t}}{\\lambda _1 - \\lambda _2} \\mu(t)<\/span><\/td><\/tr>
 <\/td> <\/td> <\/td> \\lambda _1 \\neq \\lambda _2<\/span><\/td><\/tr>
5<\/td>e\u03bbt<\/sup> \u03bc(t)<\/td>e\u03bbt<\/sup> \u03bc(t)<\/td>t e\u03bbt<\/sup> \u03bc(t)<\/td><\/tr>
6<\/td>t e\u03bbt<\/sup> \u03bc(t)<\/td>e\u03bbt<\/sup> \u03bc(t)<\/td> \\frac{1}{2} t^2 e^{\\lambda t} \\mu(t)<\/span><\/td><\/tr>
7<\/td>t N<\/sup> \u03bc(t)<\/td>e\u03bbt<\/sup> \u03bc(t)<\/td> \\frac{N! e^{\\lambda t}}{\\lambda^{N+1}} \\mu(t) - \\sum_{k=0}^{N}\\frac{N! t^{N-k}}{\\lambda^{N+1} (N-k)!} \\mu(t)<\/span><\/td><\/tr>
8<\/td>t M<\/sup> \u03bc(t)<\/td>t N<\/sup> \u03bc(t)<\/td> \\frac{M! N!}{(M+N+1)!} t^{M+N+1} \\mu (t)<\/span><\/td><\/tr>
9<\/td>t e\u03bb1<\/sub>t<\/sup> \u03bc(t)<\/td>e\u03bb2<\/sub>t<\/sup> \u03bc(t)<\/td> \\frac{e^{\\lambda_2 t}- e^{\\lambda_1 t} + (\\lambda_1-\\lambda_2)te^{\\lambda_1 t}}{(\\lambda_1-\\lambda_2)^2} \\mu (t) <\/span><\/td><\/tr>
10<\/td>t M<\/sup> e\u03bbt<\/sup> \u03bc(t)<\/td>t N<\/sup> e\u03bbt<\/sup> \u03bc(t)<\/td> \\frac{M! N!}{(M+N+1)!} t^{M+N+1} e^{\\lambda t}\\mu (t)<\/span><\/td><\/tr>
11<\/td>t M<\/sup> e\u03bb1<\/sub>t<\/sup> \u03bc(t)<\/td>t^N e^{\\lambda_2t} \\mu (t)<\/span><\/td> \\sum_{k=0}^{M}\\frac{(-1)^k M! (N+k)! t^{M-k} e^{\\lambda_1t}}{k!(M-k)!(\\lambda_1 - \\lambda_2)^{N+k+1}} \\mu (t) <\/span><\/td><\/tr>
 <\/td>\u03bb1<\/sub> \u2260\u03bb2<\/sub><\/td> <\/td> + \\sum_{k=0}^{N}\\frac{(-1)^k N! (M+k)! t^{N-k} e^{\\lambda_2t}}{k!(N-k)!(\\lambda_2 - \\lambda_1)^{M+k+1}} \\mu (t) <\/span><\/td><\/tr>
12<\/td> e^{\\alpha t} \\cos (\\beta t + \\theta) \\mu (t) <\/span><\/td>e\u03bbt<\/sup> \u03bc(t)<\/td> \\frac{\\cos(\\theta - \\phi)e^{\\lambda t}-e^{-\\alpha t} \\cos(\\beta t + \\theta - \\phi)}{\\sqrt{(\\alpha + \\lambda)^2 + \\beta^2}} \\mu (t) <\/span><\/td><\/tr>
 <\/td> <\/td> <\/td> \\phi =tan^{-1} \\Big[\\frac{-\\beta}{(\\alpha + \\lambda})\\Big] <\/span><\/td><\/tr>
13<\/td>e\u03bb1<\/sub>t<\/sup> \u03bc(t)<\/td>e\u03bb2<\/sub>t<\/sup> \u03bc(-t)<\/td> \\frac{e^{\\lambda_1 t} \\mu (t) + e^{\\lambda_2 t} \\mu (-t)}{\\lambda_2 -\\lambda_1} <\/span><\/td><\/tr>
 <\/td> <\/td> <\/td> \\text{Re} \\lambda_2 > \\text{Re} \\lambda_1 <\/span><\/td><\/tr>
14<\/td>e\u03bb1<\/sub>t<\/sup> \u03bc(-t)<\/td>e\u03bb2<\/sub>t<\/sup> \u03bc(-t)<\/td> \\frac{e^{\\lambda_1 t} -e^{\\lambda_2 t}}{\\lambda_2 -\\lambda_1} \\mu (-t) <\/span><\/td><\/tr><\/tbody><\/table><\/figure>\n","protected":false},"excerpt":{"rendered":"

Referencia: Lathi Tabla 2.1 p176 Respuesta a estado cero y convoluci\u00f3n con Sympy-Python No x1(t) x2(t) x1(t)\u2297x2(t) = x2(t)\u2297x1(t) 1 x(t) \u03b4(t-T) x(t-T) 2 e\u03bbt \u03bc(t) \u03bc(t) 3 \u03bc(t) \u03bc(t) t \u03bc(t) 4 e\u03bb1t \u03bc(t) e\u03bb2t \u03bc(t)       5 e\u03bbt \u03bc(t) e\u03bbt \u03bc(t) t e\u03bbt \u03bc(t) 6 t e\u03bbt \u03bc(t) e\u03bbt \u03bc(t) 7 […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-ss-unidades","format":"standard","meta":{"footnotes":""},"categories":[171],"tags":[],"class_list":["post-1286","post","type-post","status-publish","format-standard","hentry","category-ss-u03"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1286","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=1286"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1286\/revisions"}],"predecessor-version":[{"id":24002,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1286\/revisions\/24002"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=1286"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=1286"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=1286"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}