{"id":2646,"date":"2017-09-21T07:20:40","date_gmt":"2017-09-21T12:20:40","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/telg1001\/?p=2646"},"modified":"2026-04-05T21:20:09","modified_gmt":"2026-04-06T02:20:09","slug":"s1eva2009tii_t5-lti-dt-bloques-hz-en-serie","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/ss-s1eva\/s1eva2009tii_t5-lti-dt-bloques-hz-en-serie\/","title":{"rendered":"s1Eva2009TII_T5 LTI DT bloques H[z] en serie"},"content":{"rendered":"\n

Referencia<\/strong><\/em>: 1Eva2009TII_T5 LTI DT bloques H[z] en serie<\/a><\/p>\n\n\n\n

1. Las respuestas impulso de cada subsistema<\/p>\n\n\n\n

\"1E2009TII<\/figure>\n\n\n H_1[z] = \\frac{z}{z-(0.7)} = \\frac{z}{z-0.7}<\/span>\n\n\n\n

usando la tabla de transformadas z<\/a><\/p>\n\n\n h_1[n] = (0.7)^n \\mu[n] <\/span>\n\n\n\n

Continuando con el subsistema de la derecha<\/p>\n\n\n H_2[z] = \\frac{z}{z-(-0.5)} = \\frac{z}{z+0.5}<\/span>\n\n\n h_2[n] = (-0.5)^n \\mu[n] <\/span>\n\n\n\n

El sistema total:<\/p>\n\n\n H[z] = H_1[z] H_2[z] = \\frac{z}{(z-0.7)} \\frac{z}{(z+0.5)} = \\frac{z^2}{(z-0.7)(z+0.5)}<\/span>\n\n\n\n

fracciones parciales modificadas, multiplica ambos lados por 1\/z<\/p>\n\n\n \\frac{H[z]}{z} = \\Big( \\frac{1}{z} \\Big) \\frac{z^2}{(z-0.7)(z+0.5)}= \\frac{z}{(z-0.7)(z+0.5)}<\/span>\n\n\n \\frac{H[z]}{z} = \\frac{z}{(z-0.7)(z+0.5)} = \\frac{k_1}{z-0.7} +\\frac{k_2}{z+0.5}<\/span>\n\n\n k_1 = \\frac{z}{\\cancel{(z-0.7)}(z+0.5)} \\Big|_{z=0.7} = \\frac{0.7}{(0.7+0.5)} = 0.5833 <\/span>\n\n\n k_2 = \\frac{z}{(z-0.7)\\cancel{(z+0.5)}} \\Big|_{z=-0.5} = \\frac{-0.5}{(-0.5-0.7)} = 0.4166 <\/span>\n\n\n \\frac{H[z]}{z} = \\frac{0.5833}{z-0.7} +\\frac{0.4166}{z+0.5}<\/span>\n\n\n\n

Restaura fracciones parciales, multiplica ambos lados por z<\/p>\n\n\n H[z] = \\frac{0.5833 z}{z-0.7} +\\frac{0.4166z}{z+0.5}<\/span>\n\n\n\n

usando la tabla de transformadas z<\/a><\/p>\n\n\n h[n] = 0.5833 \\Big(0.7 \\Big)^n \\mu[n] +0.4166 \\Big(-0.5\\Big)^n \\mu[n]<\/span>\n\n\n\n

factor com\u00fan \u03bc[n]<\/p>\n\n\n h[n] = \\Bigg( 0.5833 \\Big(0.7 \\Big)^n +0.4166 \\Big(-0.5\\Big)^n \\Bigg) \\mu[n]<\/span>\n\n\n\n

Revisando el resultado con el algoritmo en Python<\/p>\n\n\n\n

\"1E2009TII_T5<\/figure>\n\n\n\n

<\/p>\n\n\n\n

\"1E2009TII<\/figure>\n\n\n\n
 Hz:\n          2        \n         z         \n\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n(z - 0.7)\u22c5(z + 0.5)\n\nfracciones parciales z:\n0.416666666666667\u22c5z   0.583333333333333\u22c5z\n\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n      z + 0.5               z - 0.7      \n\npolos: {0.700000000000000: 1, -0.500000000000000: 1}\nceros: {0: 2}\nh[n]:\n\u239b                      n                        n\u239e     \n\u239d0.416666666666667\u22c5-0.5  + 0.583333333333333\u22c50.7 \u23a0\u22c5\u03b8(n)\n<\/code><\/pre>\n\n\n\n
Tarea<\/strong><\/em>: 2. Su respuesta y[n]=s]n], expresada a la m\u00ednima expresi\u00f3n frente a la siguiente excitaci\u00f3n x[n]=\u03bc[n], esquematizar.<\/p>\n\n\n\n
Algoritmo en Python<\/h3>\n\n\n\n
Algoritmos desarrollados en H[z] Fracciones parciales modificadas<\/a> con Python<\/p>\n\n\n\n
y la parte gr\u00e1fica de Transformada z con Sympy-Python<\/p>\n\n\n
\n# Transformada z- Fracciones parciales\n# Polos \u00fanicos, repetidos y complejos\n# Lathi Ejercicio 5.3a pdf495\n# https:\/\/blog.espol.edu.ec\/algoritmos101\/senales\/ss-unidades\/ss-unidad-7\/\nimport numpy as np\nimport sympy as sym\nimport telg1001 as fcnm\n \n# INGRESO\nz = sym.Symbol('z')\n\nPz = z**2\nQz = (z-0.7)*(z+0.5)\n\n# Pz = z**2\n# Qz = (z-0.7)*(z+0.5)\n\n\n#Pz = 8*z-19\n#Qz = (z-2)*(z-3)\n\n#Pz = z*(2*z**2-11*z+12)\n#Qz = (z-1)*(z-2)**3\n\n#Pz = 2*z*(3*z+17)\n#Qz = (z-1)*(z**2-6*z+25)\n\nHz = Pz\/Qz\n\nmuestras = 10 #para la gr\u00e1fica\n \n# PROCEDIMIENTO \nFz  = fcnm.apart_z(Hz)\nQs2 = fcnm.Q_cuad_z_parametros(Fz)\n\n[Q_polos,P_ceros] = fcnm.busca_polosceros_z(Fz)\n\nn = sym.Symbol('n')\nhn = fcnm.inverse_z_transform(Fz,z,n)\n\n\n# SALIDA\nprint('\\n Hz:')\nsym.pprint(Hz)\nprint('\\nfracciones parciales z:')\nsym.pprint(Fz)\nprint('\\npolos:',Q_polos)\nprint('ceros:',P_ceros)\nif len(Qs2)>0:\n    print('\\nparametros cuadraticos: ')\n    for unterm in Qs2:\n        print(unterm,':')\n        for unparam in Qs2[unterm]:\n            print(unparam,':',Qs2[unterm][unparam])\nprint('h[n]:')\nsym.pprint(hn)\n\n# GRAFICAR polos y ceros en z\nimport matplotlib.pyplot as plt\n# para graficar polos y ceros\nf_nombre = 'H'    # nombre de funci\u00f3n[z]: H,X,Y, etc\n# grafica de polos y zeros\nfig_ROC = fcnm.graficar_Fz_polos(Hz,Q_polos,P_ceros,\n                      muestras=101,f_nombre=f_nombre)\n#plt.show()\n\n# h[n] usando tabla de transformadas\nfn = sym.lambdify(n,hn)\nki = np.arange(0,muestras,1)\nfi = fn(ki)\n \n# grafica h[n]\nfig_fn, graf_fn = plt.subplots()\nplt.stem(ki,fi)\nplt.xlabel('ki')\nplt.ylabel(f_nombre.lower()+'[n]')\nuntitulo = r''+f_nombre.lower()+'[n]=$'\nuntitulo = untitulo+str(sym.latex(hn))+'$'\nplt.title(untitulo)\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Referencia: 1Eva2009TII_T5 LTI DT bloques H[z] en serie 1. Las respuestas impulso de cada subsistema usando la tabla de transformadas z Continuando con el subsistema de la derecha El sistema total: fracciones parciales modificadas, multiplica ambos lados por 1\/z Restaura fracciones parciales, multiplica ambos lados por z usando la tabla de transformadas z factor com\u00fan […]<\/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":[184],"tags":[199],"class_list":["post-2646","post","type-post","status-publish","format-standard","hentry","category-ss-s1eva","tag-senalessistemas"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2646","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=2646"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2646\/revisions"}],"predecessor-version":[{"id":23951,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2646\/revisions\/23951"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=2646"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=2646"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=2646"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}