<div id="pf1" class="pf w0 h0" data-page-no="1"><div class="pc pc1 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/4c1a8c95-92aa-4fdd-8530-2b263ed548ca/bg1.png"><div class="t m0 x1 h2 y1 ff1 fs0 fc0 sc0 lsa ws7"> </div><div class="t m0 x2 h2 y2 ff1 fs0 fc0 sc0 lsa ws7">1 </div><div class="c x1 y3 w2 h3"><div class="t m0 x0 h2 y4 ff1 fs0 fc0 sc0 lsa ws7"> </div></div><div class="t m0 x1 h4 y5 ff2 fs1 fc1 sc0 lsa ws7">Vibrações Mecânicas \u2013 <span class="_0 blank"> </span>Resumo </div><div class="t m0 x1 h4 y6 ff2 fs1 fc1 sc0 lsa ws7"> </div><div class="t m0 x1 h5 y7 ff3 fs2 fc0 sc0 lsa ws7">SISTEMAS MECÂNICOS VIBRACIONAIS </div><div class="t m0 x1 h5 y8 ff3 fs2 fc0 sc0 lsa ws7">COM MDOF \u2013 ANÁLISE MODAL </div><div class="t m0 x1 h5 y9 ff3 fs2 fc0 sc0 lsa ws7">ANALÍTICA \u2013 VIBRAÇÕES FORÇADAS<span class="fs3 fc2"> </span></div><div class="t m0 x1 h6 ya ff3 fs3 fc1 sc0 lsa ws7"> Introdução: </div><div class="t m0 x1 h7 yb ff2 fs4 fc0 sc0 lsa ws7">Um sistema mecânico vibratório <span class="_0 blank"> </span>com vibração forçada <span class="_0 blank"> </span>pode se<span class="_0 blank"> </span>r descrito pela seguinte </div><div class="t m0 x1 h7 yc ff2 fs4 fc0 sc0 lsa ws7">equação diferencial: </div><div class="t m0 x3 h8 yd ff4 fs4 fc0 sc0 lsa ws0">\ue879\ue89e</div><div class="t m0 x4 h8 ye ff4 fs4 fc0 sc0 ls0">\u0308<span class="lsa ws1 v0">+ \ue86f\ue89e</span></div><div class="t m0 x5 h8 ye ff4 fs4 fc0 sc0 ls0">\u0307<span class="lsa ws7 v0">+ \ue877\ue89e<span class="_1 blank"> </span>=<span class="_1 blank"> </span>\ue872 (\ue887)<span class="ff2"> </span></span></div><div class="t m0 x1 h9 yf ff2 fs4 fc0 sc0 lsa ws7">Sendo <span class="ff3 ws2">F</span><span class="ls1"> o </span><span class="fc1">vetor força de excitação</span><span class="ff3 ws2">.</span> </div><div class="t m0 x1 h7 y10 ff2 fs4 fc0 sc0 lsa ws7"> Temos <span class="_0 blank"> </span>vários <span class="_0 blank"> </span>mé<span class="_0 blank"> </span>todos <span class="_0 blank"> </span>para <span class="_0 blank"> </span>resolução, <span class="_0 blank"> </span>o<span class="_0 blank"> </span>s <span class="_0 blank"> </span>mais <span class="_0 blank"> </span>tradicionais, <span class="_2 blank"> </span>os <span class="_0 blank"> </span>métodos <span class="_0 blank"> </span>numéricos <span class="_2 blank"> </span>já </div><div class="t m0 x1 h7 y11 ff2 fs4 fc0 sc0 lsa ws7">mencionados <span class="_3 blank"> </span>na <span class="_3 blank"> </span>aula <span class="_3 blank"> </span>14, <span class="_3 blank"> </span>como <span class="_3 blank"> </span>o <span class="_3 blank"> </span>método <span class="_3 blank"> </span>de <span class="_3 blank"> </span>Runge-Kutta, <span class="_3 blank"> </span>o <span class="_3 blank"> </span>método <span class="_3 blank"> </span>de <span class="_3 blank"> </span>Newmark, </div><div class="t m0 x1 h7 y12 ff2 fs4 fc0 sc0 lsa ws7">entre outros. </div><div class="t m0 x1 h7 y13 ff2 fs4 fc0 sc0 lsa ws7">Uma <span class="_4 blank"> </span>outra <span class="_4 blank"> </span>maneira, <span class="_4 blank"> </span>que <span class="_4 blank"> </span>nós <span class="_4 blank"> </span>abordaremos <span class="_4 blank"> </span>agora, <span class="_4 blank"> </span>é <span class="_4 blank"> </span>através <span class="_4 blank"> </span>do <span class="_4 blank"> </span>método <span class="_4 blank"> </span>das </div><div class="t m0 x1 h7 y14 ff2 fs4 fc0 sc0 lsa ws7">transformadas Laplace e/ou Fourier, que nós também já vimos, na aula 12 desse c<span class="_5 blank"></span>urso.<span class="_0 blank"> </span> </div><div class="t m0 x1 h6 y15 ff3 fs3 fc1 sc0 lsa ws7">Método das Transformadas de Laplace e Fourier<span class="_0 blank"> </span> </div><div class="t m0 x1 ha y16 ff2 fs4 fc0 sc0 lsa ws7">Assumindo condições inicias nulas: <span class="ff4 ls2">x<span class="lsa ws0 v1">(</span><span class="ls1">0<span class="lsa ws0 v1">)</span><span class="lsa"> e \ue754<span class="_6 blank"></span>\u0307<span class="_7 blank"> </span>(0)<span class="ff2">, assim: </span></span></span></span></div><div class="t m0 x6 ha y17 ff4 fs4 fc0 sc0 ls3">\ue73a<span class="lsa ws0 v1">(</span><span class="ls4">\ue74f<span class="ls5 v1">)</span><span class="lsa ws7">=<span class="_1 blank"> </span> \u2112<span class="_0 blank"></span><span class="ws0 v1">{</span><span class="ws3">\ue754 (\ue750 )<span class="ls6 v1">}</span></span><span class="ff2"> </span></span></span></div><div class="t m0 x6 ha y18 ff4 fs4 fc0 sc0 ls7">\ue728<span class="lsa ws0 v1">(</span><span class="ls4">\ue74f<span class="ls5 v1">)</span><span class="lsa ws7">=<span class="_1 blank"> </span> \u2112<span class="_0 blank"></span><span class="ws0 v1">{</span><span class="ws4">\ue728 (\ue750 )<span class="ws0 v1">}</span></span><span class="ff2"> </span></span></span></div><div class="t m0 x1 h7 y19 ff2 fs4 fc0 sc0 lsa ws7">Substituindo essas expressões em (a), fica: </div><div class="t m0 x7 h8 y1a ff4 fs4 fc0 sc0 lsa ws0">[<span class="ws5 v2">\ue879\ue74f </span><span class="fs5 ls8 v3">\ueb36</span><span class="ws1 v2">+ \ue86f\ue74f + \ue891</span>]<span class="ls6 v2">\ue884</span>(<span class="ls4 v2">\ue74f</span><span class="ls5">)</span><span class="ws6 v2">= \ue872(\ue74f<span class="_0 blank"> </span>)<span class="ff2 ws7"> </span></span></div><div class="t m0 x1 h7 y1b ff2 fs4 fc0 sc0 lsa ws7">Essa equação pode ser escrita como: </div><div class="t m0 x8 ha y1c ff4 fs4 fc0 sc0 lsa ws0">\ue886(\ue74f<span class="_0 blank"></span>)\ue884<span class="v1">(</span><span class="ls4">\ue74f<span class="ls5 v1">)</span></span><span class="ws6">= \ue872(\ue74f)<span class="ff2 ws7"> </span></span></div><div class="t m0 x1 h9 y1d ff2 fs4 fc0 sc0 lsa ws7">Onde <span class="ff3 ws2">Z</span>(s) a matriz de <span class="fc1">impedância mecânica </span>ou matriz de <span class="fc1">rigidez dinâmica</span>. </div><div class="t m0 x9 ha y1e ff4 fs4 fc0 sc0 lsa ws0">\ue886<span class="v1">(</span><span class="ls4">\ue74f<span class="ls5 v1">)</span><span class="ls9">=</span></span><span class="v1">[</span><span class="ws5">\ue879\ue74f <span class="fs5 ls8 v4">\ueb36</span><span class="ws1">+ \ue86f\ue74f<span class="_8 blank"> </span>+ \ue891</span></span><span class="v1">]</span><span class="ff2 ws7"> </span></div><div class="t m0 x1 h9 y1f ff2 fs4 fc0 sc0 lsa ws7">A solução do sistema pode ser obtida invertendo-se a matriz de impedância <span class="ff3 ws2">Z</span>(s): </div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf2" class="pf w0 h0" data-page-no="2"><div class="pc pc2 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/4c1a8c95-92aa-4fdd-8530-2b263ed548ca/bg2.png"><div class="t m0 x1 h2 y1 ff1 fs0 fc0 sc0 lsa ws7"> </div><div class="t m0 x2 h2 y2 ff1 fs0 fc0 sc0 lsa ws7">2 </div><div class="c x1 y3 w2 h3"><div class="t m0 x0 h2 y4 ff1 fs0 fc0 sc0 lsa ws7"> </div></div><div class="t m0 x4 ha y20 ff4 fs4 fc0 sc0 ls6">\ue884<span class="lsa ws0 v1">(</span><span class="ls4">\ue74f<span class="ls5 v1">)</span><span class="lsa ws6">= \ue886<span class="ff2 ws8">(s)</span><span class="fs5 ws9 v5">\ueb3f\ueb35 </span><span class="ws0">\ue872(\ue74f)<span class="ff2 ws7"> </span></span></span></span></div><div class="t m0 x1 h7 y21 ff2 fs4 fc0 sc0 lsa ws7">A inversa <span class="_8 blank"> </span>da matriz <span class="_8 blank"> </span>de impedância <span class="_9 blank"> </span>é c<span class="_0 blank"> </span>hamada de <span class="_8 blank"> </span><span class="fc1">receptância <span class="_9 blank"> </span></span><span class="ls1">ou <span class="_9 blank"> </span></span><span class="fc1">compliância </span>do<span class="_0 blank"> </span> </div><div class="t m0 x1 h7 y22 ff2 fs4 fc0 sc0 lsa ws7">sistema: </div><div class="t m0 xa ha y23 ff4 fs4 fc0 sc0 lsa ws0">\ue6f6<span class="v1">(</span><span class="ls4">\ue74f<span class="ls5 v1">)</span></span><span class="ws6">= \ue886<span class="ff2 ws8">(s)</span><span class="fs5 wsa v5">\ueb3f\ueb35 </span><span class="ff2 ws7"> </span></span></div><div class="t m0 x1 h9 y24 ff2 fs4 fc0 sc0 lsa ws7">Obtendo <span class="_5 blank"></span><span class="ff3 ws2">X<span class="ff2 ws7">(s) aplica-se <span class="_5 blank"></span>a <span class="_5 blank"></span>transformada <span class="_5 blank"></span>inversa <span class="_5 blank"></span>de <span class="_5 blank"></span>Laplace <span class="_5 blank"></span>obtendo <span class="_5 blank"></span>assim <span class="_5 blank"></span>a resposta <span class="_5 blank"></span>do </span></span></div><div class="t m0 x1 h7 y25 ff2 fs4 fc0 sc0 lsa ws7">sistema no domínio temporal. </div><div class="t m0 x1 h6 y26 ff3 fs3 fc1 sc0 lsa ws7">Exercício </div><div class="t m0 x1 h7 y27 ff2 fs4 fc0 sc0 lsa ws7">Considere <span class="_9 blank"> </span>o <span class="_9 blank"> </span>Sistema <span class="_9 blank"> </span>mecânico <span class="_8 blank"> </span>com <span class="_9 blank"> </span>dois <span class="_9 blank"> </span>graus <span class="_9 blank"> </span>de <span class="_8 blank"> </span>Liberdade <span class="_9 blank"> </span>mostrada <span class="_9 blank"> </span>na <span class="_9 blank"> </span>figura. </div><div class="t m0 x1 h7 y28 ff2 fs4 fc0 sc0 lsa ws7">Considerando <span class="_5 blank"></span>que <span class="_5 blank"></span>m<span class="fs6 lsb v6">1</span> = <span class="_6 blank"></span>1 (kg), <span class="_5 blank"></span>m<span class="fs6 lsc v6">2</span> <span class="_5 blank"></span>= <span class="_5 blank"></span>2 <span class="_5 blank"></span>(kg), <span class="_5 blank"></span>k<span class="fs6 lsb v6">1</span> = <span class="_5 blank"></span>k<span class="_5 blank"></span><span class="fs6 lsc v6">2<span class="fs4 lsa v7"> = <span class="_5 blank"></span>100 <span class="_5 blank"></span>(N/s), <span class="_5 blank"></span>c <span class="_5 blank"></span>= <span class="_5 blank"></span>2 <span class="_5 blank"></span>(N.s/m), <span class="_5 blank"></span> <span class="_5 blank"></span>as <span class="_5 blank"></span>condições </span></span></div><div class="t m0 x1 h7 y29 ff2 fs4 fc0 sc0 lsa ws7">iniciais sejam nulas e a excitação na massa 1 seja F = F<span class="fs6 lsd v6">o</span><span class="ws8">.sen(<span class="ff5 wsb">\uf077</span></span>t), com <span class="ff5 wsb">\uf077</span> = 10 rad/s. </div><div class="t m0 x1 h7 y2a ff2 fs4 fc0 sc0 lsa ws7"> <span class="_a blank"> </span> </div><div class="t m0 x1 h7 y2b ff2 fs4 fc0 sc0 lsa ws7">Figura 01: Exercício: Vibrações Forçadas </div><div class="t m0 x1 h9 y2c ff3 fs4 fc1 sc0 lsa ws7">Resolução: </div><div class="t m0 x1 h9 y2d ff2 fs4 fc0 sc0 lsa ws7">Primeiro passo: Obter as Matrizes <span class="ff3 ws2">M</span>, <span class="ff3 ls1">C<span class="ff2"> e </span><span class="lsa ws2">K</span></span>: </div><div class="t m0 x1 h7 y2e ff2 fs4 fc0 sc0 lsa ws7">A partir do método das energias e equações de Lagrange, obtemos: </div><div class="t m0 xb hb y2f ff4 fs4 fc0 sc0 lsa wsc">\ue879 = \uf242<span class="ws7 v8"> 1<span class="_b blank"> </span>0 </span></div><div class="t m0 xc hc y30 ff4 fs4 fc0 sc0 lsa wsd">0<span class="_c blank"> </span>2 <span class="lse v9">\uf243<span class="fs7 fc3 lsf ws7"> </span><span class="ls9 wse">\ue86f=\uf242<span class="_d blank"></span><span class="lsa ws7 v8"> \ueadb<span class="_b blank"> </span>\uead9 </span></span></span></div><div class="t m0 xd hd y30 ff4 fs4 fc0 sc0 lsa ws7">\uead9<span class="_e blank"> </span>\uead9 <span class="v9">\uf243<span class="_f blank"> </span> \ue877<span class="_1 blank"> </span>=<span class="_1 blank"> </span>\uf242 <span class="_10 blank"> </span></span><span class="ls1 wsf va">\ueadb\uead9\uead9 \u2212\ueada\uead9\uead9</span></div><div class="t m0 xe he y30 ff4 fs4 fc0 sc0 ls1 ws10">\u2212\ueada\uead9\uead9<span class="_11 blank"> </span>\ueada\uead9\uead9 <span class="lsa ws7 v9"> \uf243<span class="ff2"> </span></span></div><div class="t m0 x1 h9 y31 ff2 fs4 fc0 sc0 lsa ws7">Montando a matriz impedância mecânica <span class="ff3 ws2">Z</span>(s): </div><div class="t m0 xf hf y32 ff4 fs4 fc0 sc0 lsa ws7">\ue886(\ue74f<span class="_0 blank"></span>)<span class="_1 blank"> </span>=<span class="_1 blank"> </span>\ued64 <span class="ls10 vb">\ue74f</span><span class="fs5 ls11 vc">\ueb36</span><span class="ws1 vb">+ 2\ue74f + 200<span class="_12 blank"> </span>\u2212100</span></div><div class="t m0 x10 h10 y33 ff4 fs4 fc0 sc0 lsa ws5">\u2212100<span class="_13 blank"> </span>2\ue74f <span class="fs5 ls11 v4">\ueb36</span><span class="ws1">+ 100<span class="ws7 vd"> \ued68<span class="ff2"> </span></span></span></div><div class="t m0 x1 ha y34 ff2 fs4 fc0 sc0 lsa ws7">Temos que <span class="ff3 ws2">Z</span><span class="ws8">(s).<span class="ff4 ls6">\ue884<span class="lsa ws0 v1">(</span><span class="ls4">\ue74f<span class="ls5 v1">)</span><span class="lsa ws6">= \ue872(\ue74f<span class="_0 blank"></span>)</span></span></span></span>, então: </div><div class="t m0 x11 hf y35 ff4 fs4 fc0 sc0 lsa ws7">\ue73c<span class="_0 blank"></span>(\ue74f<span class="_0 blank"></span>)<span class="_1 blank"> </span>=<span class="_1 blank"> </span>\ued64 <span class="ls10 vb">\ue74f</span><span class="fs5 ls11 vc">\ueb36</span><span class="ws11 vb">+ 2\ue74f + 200<span class="_12 blank"> </span>\u221210<span class="_0 blank"></span>0</span></div><div class="t m0 x12 h11 y36 ff4 fs4 fc0 sc0 lsa ws5">\u2212100<span class="_13 blank"> </span>2\ue74f <span class="fs5 ls11 v4">\ueb36</span><span class="ws11">+ 100<span class="ws7 vd"> \ued68<span class="_f blank"> </span>\ued5c</span><span class="ws0 ve">\ue884</span><span class="fs5 ls12 va">\ueada</span><span class="ws0 ve">(\ue899)</span></span></div><div class="t m0 x13 h12 y37 ff4 fs4 fc0 sc0 lsa ws0">\ue884<span class="fs5 ls12 vf">\ueadb</span>(\ue899)<span class="ls9 wse v10">\ued60=\uf244<span class="_d blank"></span><span class="ls1 v8">\ue872<span class="fs5 ls13 vf">\ueada</span><span class="lsa ws0">(\ue899)</span></span></span></div><div class="t m0 x14 h13 y38 ff4 fs4 fc0 sc0 ls14">\uead9<span class="ls1 v11">\uf245<span class="ff2 lsa ws7"> </span></span></div><div class="t m0 x1 h9 y39 ff2 fs4 fc0 sc0 lsa ws7">Obtendo <span class="ff3 ls2">H</span>(s) invertendo a <span class="ff3 ws2">Z</span>(s): </div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf3" class="pf w0 h0" data-page-no="3"><div class="pc pc3 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/4c1a8c95-92aa-4fdd-8530-2b263ed548ca/bg3.png"><div class="t m0 x1 h2 y1 ff1 fs0 fc0 sc0 lsa ws7"> </div><div class="t m0 x2 h2 y2 ff1 fs0 fc0 sc0 lsa ws7">3 </div><div class="c x1 y3 w2 h3"><div class="t m0 x0 h2 y4 ff1 fs0 fc0 sc0 lsa ws7"> </div></div><div class="t m0 x15 h14 y3a ff4 fs4 fc0 sc0 lsa ws12">\ue874(\ue74f<span class="_0 blank"></span>) = \ue886(\ue74f<span class="_0 blank"></span>)<span class="fs5 ws13 v4">\ueb3f\ueada </span><span class="fs7 fc3 ls15 ws7"> </span><span class="ls16">=</span><span class="v12">\ueada</span></div><div class="t m0 x8 h15 y3b ff4 fs4 fc0 sc0 lsa ws14">\ue870(\ue899) <span class="ws7 v11">\ued64 </span><span class="ws0 v13">\ueadb\ue899</span><span class="fs5 ls17 v14">\ueadb</span><span class="ls1 ws15 v13">+ \ueada\uead9\uead9<span class="_14 blank"> </span>\ueada\uead9\uead9</span></div><div class="t m0 x16 h16 y3c ff4 fs4 fc0 sc0 lsa ws16">\ueada\uead9\uead9 \ue899<span class="fs5 ls17 v4">\ueadb</span><span class="ws1">+ \ueadb\ue899 + \ueadb\uead9\uead9<span class="ws7 v15"> \ued68<span class="ff2"> </span></span></span></div><div class="t m0 x1 h7 y3d ff2 fs4 fc0 sc0 lsa ws7">D(s) <span class="_5 blank"></span>é <span class="_5 blank"></span>a <span class="_5 blank"></span><span class="fc1">equação <span class="_5 blank"></span>característica <span class="_5 blank"></span><span class="fc0">do <span class="_5 blank"></span>sistema <span class="_5 blank"></span>e <span class="_5 blank"></span>fornece <span class="_5 blank"></span>as <span class="_5 blank"></span>frequências <span class="_6 blank"></span>naturais e <span class="_5 blank"></span>os <span class="_5 blank"></span>fatores </span></span></div><div class="t m0 x1 h7 y3e ff2 fs4 fc0 sc0 lsa ws7">de amortecimento do sistema. </div><div class="t m0 x1 h7 y3f ff2 fs4 fc0 sc0 lsa ws7">Calculando fica: </div><div class="t m0 x17 ha y40 ff4 fs4 fc0 sc0 ls18">\ue726<span class="lsa ws0 v1">(</span><span class="ls4">\ue74f<span class="ls5 v1">)</span><span class="lsa ws5">=<span class="_1 blank"> </span>2\ue74f <span class="fs5 ls8 v4">\ueb38</span>+<span class="_9 blank"> </span>4\ue74f <span class="fs5 ls11 v4">\ueb37</span>+<span class="_9 blank"> </span>500\ue74f <span class="fs5 ls8 v4">\ueb36</span><span class="ws1">+ 200\ue74f + 10000<span class="ff2 ws7"> </span></span></span></span></div><div class="t m0 x1 h7 y41 ff2 fs4 fc0 sc0 lsa ws7">A transformada de Laplace do sinal de força aplicada é dada <span class="_5 blank"></span>por: </div><div class="t m0 x4 hd y42 ff4 fs4 fc0 sc0 lsa ws6">\ue872(\ue74f)<span class="_1 blank"> </span>= <span class="ls19 v0">\ued65</span><span class="ws0 va">10\ue728</span></div><div class="t m0 x18 h17 y43 ff4 fs5 fc0 sc0 lsa">\uebe2</div><div class="t m0 x19 h8 y44 ff4 fs4 fc0 sc0 ls10">\ue74f<span class="fs5 ls11 v16">\ueb36</span><span class="lsa ws1">+ 100</span></div><div class="t m0 x1a h18 y45 ff4 fs4 fc0 sc0 ls1a">0<span class="lsa ws0 v17">\ued69</span><span class="ff2 lsa ws7 v17"> </span></div><div class="t m0 x1 h7 yc ff2 fs4 fc0 sc0 lsa ws7">O <span class="_7 blank"> </span>último <span class="_7 blank"> </span>passo <span class="_7 blank"> </span>é <span class="_15 blank"> </span>aplicar <span class="_15 blank"> </span>a <span class="_7 blank"> </span>transformada <span class="_7 blank"> </span>de <span class="_15 blank"> </span>Laplace <span class="_7 blank"> </span>inversa <span class="_7 blank"> </span>a <span class="_15 blank"> </span>partir <span class="_7 blank"> </span>da <span class="_7 blank"> </span>expansão <span class="_7 blank"> </span>em </div><div class="t m0 x1 h9 y46 ff2 fs4 fc0 sc0 lsa ws7">frações parciais de <span class="ff3 ws2">X</span>(s), obtendo assim x(t): </div><div class="t m0 x1 h7 y47 ff2 fs4 fc0 sc0 lsa ws7">Graficamente, fica: </div><div class="t m0 x1 h7 y48 ff2 fs4 fc1 sc0 lsa ws7">Deslocamentos: </div><div class="t m0 x1b h7 y49 ff2 fs4 fc0 sc0 lsa ws7"> </div><div class="t m0 x1 h7 y4a ff2 fs4 fc1 sc0 lsa ws8">Velocidades<span class="fc2 ws7">: </span></div><div class="t m0 x1 h7 y4b ff2 fs4 fc0 sc0 lsa ws7"> </div><div class="t m0 x1 h7 y4c ff2 fs4 fc0 sc0 lsa ws7"> </div><div class="t m0 x1 h7 y4d ff2 fs4 fc0 sc0 lsa ws7"> </div><div class="t m0 x1 h7 y4e ff2 fs4 fc0 sc0 lsa ws7"> </div><div class="t m0 x1 h7 y4f ff2 fs4 fc0 sc0 lsa ws7"> </div><div class="t m0 x1 h7 y50 ff2 fs4 fc0 sc0 lsa ws7"> </div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf4" class="pf w0 h0" data-page-no="4"><div class="pc pc4 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/4c1a8c95-92aa-4fdd-8530-2b263ed548ca/bg4.png"><div class="t m0 x1 h2 y1 ff1 fs0 fc0 sc0 lsa ws7"> </div><div class="t m0 x2 h2 y2 ff1 fs0 fc0 sc0 lsa ws7">4 </div><div class="c x1 y3 w2 h3"><div class="t m0 x0 h2 y4 ff1 fs0 fc0 sc0 lsa ws7"> </div></div><div class="t m0 x1 h7 y51 ff2 fs4 fc0 sc0 lsa ws7">Aplicando a <span class="_0 blank"> </span>transformada de <span class="_0 blank"> </span>Fourier <span class="_0 blank"> </span>no domínio <span class="_0 blank"> </span>da frequência <span class="_0 blank"> </span>s = <span class="_0 blank"> </span>j.<span class="_2 blank"> </span><span class="ff5 wsb">\uf077</span>, poderíamos <span class="_0 blank"> </span>ter </div><div class="t m0 x1 h7 y52 ff2 fs4 fc0 sc0 lsa ws7">feito <span class="_7 blank"> </span>uma <span class="_3 blank"> </span>outra <span class="_7 blank"> </span>abordagem <span class="_7 blank"> </span>por <span class="_3 blank"> </span>meio <span class="_7 blank"> </span>da <span class="_7 blank"> </span>matr<span class="_0 blank"> </span>iz <span class="_3 blank"> </span><span class="fc1">Função <span class="_7 blank"> </span>de <span class="_3 blank"> </span>Resposta <span class="_7 blank"> </span>em <span class="_7 blank"> </span>Frequência<span class="_0 blank"> </span> </span></div><div class="t m0 x1 h7 y53 ff2 fs4 fc1 sc0 lsa ws8">(FRF)<span class="fc0 ws7">, no entanto, não iremos avançar nessa análise nesse momento. </span></div><div class="t m0 x1 h2 y54 ff1 fs0 fc0 sc0 lsa ws7"> </div><div class="t m0 x1 h2 y55 ff1 fs0 fc0 sc0 lsa ws7"> </div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div>
