<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/49cd2de9-9ce3-46e1-a6f7-51bd167dd6b3/bg1.png" alt="Pré-visualização de imagem de arquivo"><div class="t m0 x1 h2 y1 ff1 fs0 fc0 sc0 lsd">DRAFT</div><div class="t m1 x2 h3 y2 ff1 fs1 fc1 sc0 ls0">E<span class="fs2 ls1 ws0">NGENHEIR<span class="blank _0"></span>O<span class="fs1 ls2">(</span><span class="ls3">A<span class="fs1 ls4">)</span><span class="lsd ws1">D E<span class="blank _1"> </span><span class="fs1 ls5">E</span></span></span><span class="ws2">QUIP<span class="blank _2"></span>AMENT<span class="blank _3"></span>OS <span class="fs1 ls6">J</span><span class="ls7">R<span class="fs1 lsd ws3">-<span class="blank _1"> </span>E </span></span><span class="ws0">LETR<span class="blank _0"></span>ÔNICA<span class="blank _4"></span></span></span></span></div><div class="t m1 x3 h3 y3 ff1 fs1 fc1 sc0 ls5">E<span class="fs2 ls1 ws0">NGENHEIR<span class="blank _0"></span>O<span class="fs1 ls8">(</span><span class="ls9">A<span class="fs1 ls4">)</span><span class="lsd ws1">D E<span class="blank _1"> </span><span class="fs1 ls0">E</span></span></span><span class="ws4">QUIP<span class="blank _2"></span>AMENT<span class="blank _3"></span>OS <span class="fs1 ls6">J</span><span class="ls7">R<span class="fs1 lsd ws5">-<span class="blank _1"> </span>E </span></span><span class="ws0">LÉTRICA<span class="blank _4"></span></span></span></span></div><div class="t m1 x4 h3 y4 ff1 fs1 fc1 sc0 ls0">E<span class="fs2 ls1 ws0">NGENHEIR<span class="blank _0"></span>O<span class="fs1 ls2">(</span><span class="ls3">A<span class="fs1 ls4">)</span><span class="lsd ws1">D E<span class="blank _1"> </span><span class="fs1 lsa">A</span>U T<span class="blank _5"> </span>O M A<span class="blank _5"> </span>Ç Ã<span class="blank _6"> </span>O<span class="blank _1"> </span><span class="fs1 ls6">J</span>R</span></span></span></div><div class="t m1 x5 h4 y5 ff1 fs3 fc1 sc0 lsd ws6">Análise de Sinais</div><div class="t m1 x6 h5 y6 ff1 fs4 fc1 sc0 lsd ws6">Questões Resolvidas</div><div class="t m1 x7 h6 y7 ff1 fs5 fc1 sc0 lsb">Q<span class="fs6 lsd ws7">U E S TÕ E S<span class="blank _7"> </span>R E T I R A DA S<span class="blank _7"> </span>D E<span class="blank _7"> </span>P RO<span class="blank _5"> </span>VA S<span class="blank _8"> </span>DA<span class="blank _7"> </span>B A N C A<span class="blank _8"> </span></span><span class="lsc ws8">CESGRANRIO<span class="blank _3"></span></span></div><div class="t m1 x8 h3 y8 ff1 fs1 fc1 sc0 lsd ws6">Eng.<span class="blank _9"> </span>Roni Gabr<span class="blank _5"> </span>iel Rigoni</div><div class="t m1 x9 h7 y9 ff2 fs7 fc1 sc0 lsd ws9">www.concursopetrobraseng.com.br</div><a class="l"><div class="d m2" style="border-width:1.000000px;border-style:solid;border-color:rgb(0,255,255);position:absolute;left:229.264000px;bottom:66.701000px;width:163.094000px;height:9.292000px;background-color:rgba(255,255,255,0.000001);"></div></a></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"><div class="t m0 x1 h2 y1 ff1 fs0 fc0 sc0 lsd">DRAFT</div><div class="t m1 x0 h8 ya ff1 fs8 fc1 sc0 lsd">Introdução</div><div class="t m1 xa h9 yb ff1 fs7 fc1 sc0 lsd wsa">P<span class="blank _3"></span>ara a utilização deste material é recomendado que o<span class="blank _a"> </span>leitor já tenha estudado todo o conteúdo</div><div class="t m1 x0 h9 yc ff1 fs7 fc1 sc0 lsd wsb">ref<span class="blank _0"></span>erente a Análise de Sinais, como consta no edital do concurso<span class="blank _0"></span>.<span class="blank _b"> </span>P<span class="blank _3"></span>or este motivo<span class="blank _3"></span>, não é explicado</div><div class="t m1 x0 h9 yd ff1 fs7 fc1 sc0 lsd wsc">detalhadamente cada método<span class="blank _3"></span>, teorema ou definição<span class="blank _a"> </span>utilizados durante as resoluções.<span class="blank _7"> </span>P<span class="blank _3"></span>orém fazemos</div><div class="t m1 x0 h9 ye ff1 fs7 fc1 sc0 lsd wsd">questão de sempre deixar e<span class="blank _0"></span>xplícito qual método/teorema/definição está sendo utilizado<span class="blank _3"></span>,<span class="blank _7"> </span>para o leitor</div><div class="t m1 x0 h9 yf ff1 fs7 fc1 sc0 lsd ws6">poder cunsultá-lo na bibliog<span class="blank _0"></span>rafia que pref<span class="blank _3"></span>er<span class="blank _5"> </span>ir<span class="blank _3"></span>.</div><div class="t m1 xa h9 y10 ff1 fs7 fc1 sc0 lsd wse">Não será dado nenhum tipo de assistência pós-venda par<span class="blank _3"></span>a compradores deste mater<span class="blank _5"> </span>ial,<span class="blank _c"> </span>ou</div><div class="t m1 x0 h9 y11 ff1 fs7 fc1 sc0 lsd wsf">seja, qualquer dúvida ref<span class="blank _0"></span>erente às resoluções de<span class="blank _0"></span>ve se sanada por iniciativ<span class="blank _3"></span>a própr<span class="blank _5"> </span>ia do comprador<span class="blank _4"></span>, seja</div><div class="t m1 x0 h9 y12 ff1 fs7 fc1 sc0 lsd ws10">consultando docentes da área ou a bibliog<span class="blank _3"></span>rafia.<span class="blank _d"> </span>Apenas serão considerados casos em que o leitor</div><div class="t m1 x0 h9 y13 ff1 fs7 fc1 sc0 lsd ws11">encontrar algum erro (conceitual ou de digitação) e desejar inf<span class="blank _3"></span>or<span class="blank _5"> </span>mar ao autor tal erro a fim de ser</div><div class="t m1 x0 h9 y14 ff1 fs7 fc1 sc0 lsd ws12">corrigido.</div><div class="t m1 xa h9 y15 ff1 fs7 fc1 sc0 lsd ws13">O autor deste material não tem nenhum tipo de vínculo com a empresa CESGRANRIO<span class="blank _0"></span>, e as</div><div class="t m1 x0 h9 y16 ff1 fs7 fc1 sc0 lsd ws14">resoluções aqui apresentadas são de autoria exclusiv<span class="blank _3"></span>a de Roni Gabr<span class="blank _5"> </span>iel Rigoni, formado pela Univer-</div><div class="t m1 x0 h9 y17 ff1 fs7 fc1 sc0 lsd ws15">sidade F<span class="blank _0"></span>ederal de Santa Catarina e atualmente Engenheiro de A<span class="blank _0"></span>utomação da P<span class="blank _3"></span>etrobras T<span class="blank _e"></span>ranspor<span class="blank _5"> </span>tes -</div><div class="t m1 x0 h9 y18 ff1 fs7 fc1 sc0 lsd ws16">T<span class="blank _e"></span>ranspetro<span class="blank _3"></span>.</div><div class="t m1 xa h9 y19 ff1 fs7 fc1 sc0 lsd ws17">Este material é de uso exclusiv<span class="blank _3"></span>o do Comprador Cód.<span class="blank _f"> </span>T34TRJ59YNKS. Sendo v<span class="blank _3"></span>edada,<span class="blank _10"> </span>por</div><div class="t m1 x0 h9 y1a ff1 fs7 fc1 sc0 lsd ws18">quaisquer meios e a qualquer título<span class="blank _3"></span>,<span class="blank _11"> </span>a sua reprodução<span class="blank _3"></span>, cópia,<span class="blank _11"> </span>divulgação e distribuição<span class="blank _3"></span>.<span class="blank _c"> </span>Sujeitando-se</div><div class="t m1 x0 h9 y1b ff1 fs7 fc1 sc0 lsd ws6">o infrator à responsabilização civil e criminal.</div><div class="t m1 xa h9 y1c ff1 fs7 fc1 sc0 lsd ws6">F<span class="blank _3"></span>aça um bom uso do material, e que ele possa ser muito útil na conquista da sua vaga.</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 y1d w2 ha" alt src="https://files.passeidireto.com/49cd2de9-9ce3-46e1-a6f7-51bd167dd6b3/bg3.png" alt="Pré-visualização de imagem de arquivo"><div class="t m3 xb hb y1e ff1 fs9 fc2 sc0 lsd ws19">T34TRJ59YNKS T34TRJ59YNKS T34TRJ59YNKS</div><div class="t m1 x0 h8 ya ff1 fs8 fc1 sc0 lsd ws6">Análise de Sinais</div><div class="t m1 xc hc y1f ff3 fs5 fc1 sc0 lsd ws6">Questão 1<span class="blank _12"> </span><span class="ff1 fs7 v1">(Eng.<span class="blank _11"> </span>de Automação Jr - T<span class="blank _e"></span>r<span class="blank _0"></span>anspetro 2006)</span></div><div class="c xd y20 w3 hd"><div class="t m4 xe he y21 ff4 fsa fc3 sc0 lsd ws45">PROVA 35 - ENGENHEIRO(A) JÚNIOR - ÁREA AUTOMAÇÃO</div><div class="t m1 xf hf y22 ff4 fsb fc3 sc0 lsd">8</div><div class="t m1 xe h10 y23 ff4 fsc fc3 sc0 ls26">27</div><div class="t m1 xe h11 y24 ff5 fsd fc3 sc0 ls27 ws46">A figura acima mostra uma fonte de tensão contínua alimen-</div><div class="t m1 xe h11 y25 ff5 fsd fc3 sc0 ls28 ws47">tando um circuito RC. Com o capacitor descarregado, a chave</div><div class="t m1 xe h11 y26 ff5 fsd fc3 sc0 lse ws48">fecha-se no instante inicial, isto é, em t=0. <span class="blank _3"></span>A<span class="blank _3"></span> expressão</div><div class="t m1 xe h12 y27 ff5 fsd fc3 sc0 lsf ws49">matemática do tempo total (<span class="ff4 ls10">t</span><span class="ls11 ws4a">), contado a <span class="ls29 ws4b">partir do instante</span></span></div><div class="t m1 xe h11 y28 ff5 fsd fc3 sc0 ls12 ws4c">inicial até o capacitor se carregar com <span class="ls13 ws4d">1/5 da tensão da</span></div><div class="t m1 xe h11 y29 ff5 fsd fc3 sc0 ls2a ws47">fonte, é:</div><div class="t m1 xe h11 y2a ff5 fsd fc3 sc0 ls2b ws1a">(A) <span class="ff6 fse fc1 ls2c ws1b v0">RC<span class="blank _13"></span><span class="lsd ws1c">,<span class="blank _4"></span><span class="ff7 ws1d">2<span class="blank _14"></span>0</span></span></span></div><div class="t m1 xe h13 y2b ff5 fsd fc3 sc0 ls2b ws1e">(B) <span class="ff8 fsf fc1 lsd v2"></span></div><div class="t m1 x10 h14 y2c ff8 fsf fc1 sc0 lsd"></div><div class="t m1 x10 h14 y2d ff8 fsf fc1 sc0 lsd"></div><div class="t m1 x11 h14 y2e ff8 fsf fc1 sc0 lsd"></div><div class="t m1 x11 h14 y2f ff8 fsf fc1 sc0 lsd"></div><div class="t m1 x11 h14 y30 ff8 fsf fc1 sc0 lsd"></div><div class="t m1 x12 h15 y31 ff7 fsf fc1 sc0 lsd">3</div><div class="t m1 x12 h15 y32 ff7 fsf fc1 sc0 lsd">5</div><div class="t m1 x13 h16 y31 ff7 fsf fc1 sc0 ls14">2<span class="ff6 ls2d v3">ln</span></div><div class="t m1 x14 h17 y33 ff6 fsf fc1 sc0 ls2e">RC</div><div class="t m1 xe h11 y34 ff5 fsd fc3 sc0 ls2f">(C)</div><div class="t m5 x15 h18 y35 ff9 fs10 fc1 sc0 ls30">()</div><div class="t m1 x16 h15 y36 ff7 fsf fc1 sc0 lsd ws1f">2<span class="blank _14"></span>0<span class="ff6 ws20">,<span class="blank _15"></span><span class="ls2d ws21">ln<span class="blank _16"></span><span class="ls2e">RC</span></span></span></div><div class="t m1 xe h19 y37 ff5 fsd fc3 sc0 ls2f ws22">(D) <span class="ffa fs11 fc1 lsd v2"></span></div><div class="t m1 x10 h1a y38 ffa fs11 fc1 sc0 lsd"></div><div class="t m1 x10 h1a y39 ffa fs11 fc1 sc0 lsd"></div><div class="t m1 x11 h1a y3a ffa fs11 fc1 sc0 lsd"></div><div class="t m1 x11 h1a y38 ffa fs11 fc1 sc0 lsd"></div><div class="t m1 x11 h1a y39 ffa fs11 fc1 sc0 lsd"></div><div class="t m1 x12 h1b y3b ff7 fs11 fc1 sc0 lsd">5</div><div class="t m1 x12 h1b y3c ff7 fs11 fc1 sc0 lsd">3</div><div class="t m1 x13 h1c y3b ff7 fs11 fc1 sc0 ls15">2<span class="ff6 ls31 v3">ln</span></div><div class="t m1 x14 h1d y3d ff6 fs11 fc1 sc0 ls32">RC</div><div class="t m1 xe h11 y3e ff5 fsd fc3 sc0 ls2b ws23">(E) <span class="ff6 fs12 fc1 ls33 ws24 v4">RC<span class="blank _13"></span><span class="lsd ws25">,<span class="blank _e"></span><span class="ff7 ws26">6<span class="blank _14"></span>0</span></span></span></div><div class="t m1 xe h10 y3f ff4 fsc fc3 sc0 ls26">28</div><div class="t m1 xe h11 y40 ff5 fsd fc3 sc0 ls34 ws47">Considere a figura abaixo.</div><div class="t m1 xe h11 y41 ff5 fsd fc3 sc0 lsd ws4e">A chave S, no circuito, encontrava-se aberta por um longo</div><div class="t m1 xe h11 y42 ff5 fsd fc3 sc0 lsd ws4f">tempo, tendo o circuito alcançado o regime permanente.</div><div class="t m1 xe h11 y43 ff5 fsd fc3 sc0 ls35 ws47">Imediatamente após fechar a chave S, o valor da corrente I<span class="fs13 lsd v5">1</span></div><div class="t m1 x17 h11 y43 ff5 fsd fc3 sc0 lsd">,</div><div class="t m1 xe h11 y44 ff5 fsd fc3 sc0 ls36 ws47">em ampères, será:</div><div class="t m1 xe h11 y45 ff5 fsd fc3 sc0 ls37 ws47">(A) 0,75<span class="blank _17"> </span>(B) 1,00</div><div class="t m1 xe h11 y46 ff5 fsd fc3 sc0 ls38 ws47">(C) 1,25<span class="blank _17"> </span>(D) 1,50</div><div class="t m1 xe h11 y47 ff5 fsd fc3 sc0 ls37 ws47">(E) 2,00</div><div class="t m6 x18 h1e y48 ffb fs14 fc3 sc0 lsd">R</div><div class="t m6 x19 h1f y49 ffb fs14 fc3 sc0 ls16">E<span class="lsd v6">+</span></div><div class="t m6 x1a h20 y4a ffb fs14 fc3 sc0 ls17">C<span class="lsd v7">R</span></div><div class="t m7 x1b h21 y4b ffb fs15 fc3 sc0 lsd ws27">12 V</div><div class="t m7 x1c h21 y4c ffb fs15 fc3 sc0 ls18 ws28">2m<span class="blank _18"></span>H</div><div class="t m7 x1d h21 y4d ffb fs15 fc3 sc0 ls18 ws28">3m<span class="blank _18"></span>F</div><div class="t m7 x1e h21 y4e ffb fs15 fc3 sc0 lsd">S</div><div class="t m7 x1f h21 y4f ffb fs15 fc3 sc0 ls19">4<span class="ff9 lsd"></span></div><div class="t m7 x20 h21 y50 ffb fs15 fc3 sc0 ls1a">5<span class="ff9 lsd"></span></div><div class="t m7 x21 h21 y51 ffb fs15 fc3 sc0 lsd ws29">20 <span class="ff9"></span></div><div class="t m7 x22 h21 y52 ffb fs15 fc3 sc0 lsd ws2a">I<span class="fs16 v8">1</span></div><div class="t m7 x14 h21 y53 ffb fs15 fc3 sc0 lsd">+</div><div class="t m1 x23 h10 y54 ff4 fsc fc3 sc0 ls26">29</div><div class="t m1 x23 h11 y55 ff5 fsd fc3 sc0 lsd ws50">Uma planta industrial pode ser modelada através de uma</div><div class="t m1 x23 h11 y56 ff5 fsd fc3 sc0 ls39 ws47">Função de Transferência G(s) racional e contínua, de terceira</div><div class="t m1 x23 h11 y57 ff5 fsd fc3 sc0 ls3a ws47">ordem, estritamente própria e estável. Com relação a G(s),</div><div class="t m1 x23 h11 y58 ff5 fsd fc3 sc0 ls3b ws51">é correto afirmar que:</div><div class="t m1 x23 h11 y59 ff5 fsd fc3 sc0 ls3c ws52">(A) possui três pólos localizados no semiplano s da direita.</div><div class="t m1 x23 h11 y5a ff5 fsd fc3 sc0 ls3c ws53">(B) possui pelo menos um zero localizado no infinito.</div><div class="t m1 x23 h11 y5b ff5 fsd fc3 sc0 ls3d ws47">(C) o seu grau relativo é zero.</div><div class="t m1 x23 h11 y5c ff5 fsd fc3 sc0 ls3e ws51">(D) possui dois zeros localizados sobre o eixo imaginário no</div><div class="t m1 x24 h11 y5d ff5 fsd fc3 sc0 ls3f ws54">plano s.</div><div class="t m1 x23 h11 y5e ff5 fsd fc3 sc0 ls40 ws51">(E) todos os pólos estão localizados sobre o eixo real nega-</div><div class="t m1 x24 h11 y5f ff5 fsd fc3 sc0 ls41">tivo.</div><div class="t m1 x23 h10 y60 ff4 fsc fc3 sc0 ls26">30</div><div class="t m1 x23 h11 y61 ff5 fsd fc3 sc0 ls42 ws47">A figura acima mostra um sinal oriundo de uma descarga de</div><div class="t m1 x23 h11 y62 ff5 fsd fc3 sc0 lsd ws55">capacitor<span class="blank _4"></span>, cuja expressão é dada por:</div><div class="t m8 x25 h22 y63 ff9 fs17 fc1 sc0 ls43">()</div><div class="t m8 x0 h22 y64 ff9 fs17 fc1 sc0 ls43">()</div><div class="t m1 x26 h23 y65 ffc fs18 fc1 sc0 lsd"></div><div class="t m1 x26 h23 y66 ffc fs18 fc1 sc0 lsd"></div><div class="t m1 x26 h23 y67 ffc fs18 fc1 sc0 lsd"></div><div class="t m1 x27 h24 y68 ff9 fs18 fc1 sc0 lsd ws2b"><<span class="blank _19"></span>=</div><div class="t m1 x28 h25 y69 ff9 fs18 fc1 sc0 lsd ws2c">≥<span class="blank _1a"></span>= <span class="fs19 ws2d v9">α<span class="blank _1b"></span>−</span></div><div class="t m1 x29 h26 y68 ff7 fs18 fc1 sc0 lsd ws2b">0<span class="blank _1c"></span>0</div><div class="t m1 x2a h26 y69 ff7 fs18 fc1 sc0 lsd">0</div><div class="t m1 x2b h27 y68 ff6 fs18 fc1 sc0 lsd ws2e">t<span class="blank _1d"></span><span class="ls1b ws2f">para<span class="blank _1e"></span><span class="lsd ws2e">t<span class="blank _1f"></span>f</span></span></div><div class="t m1 x2c h28 y69 ff6 fs18 fc1 sc0 lsd ws2e">t<span class="blank _20"></span><span class="ls1b ws2f">para<span class="blank _1e"></span><span class="ls44 ws30">Ae<span class="blank _1d"></span><span class="lsd ws31">t<span class="blank _1f"></span>f <span class="fs19 v9">t</span></span></span></span></div><div class="t m1 x23 h11 y6a ff5 fsd fc3 sc0 ls45 ws56">onde A<span class="blank _3"></span> <span class="blank _6"> </span>e <span class="blank _6"> </span><span class="ff9 lsd ws32">α</span><span class="ls2a ws57"> são constantes positivas. <span class="blank _4"></span>A<span class="blank _0"></span> expressão da T<span class="blank _3"></span>rans-</span></div><div class="t m1 x23 h11 y6b ff5 fsd fc3 sc0 ls46 ws47">formada de Fourier deste sinal é:</div><div class="t m1 x23 h11 y6c ff5 fsd fc3 sc0 ls2b">(A)</div><div class="t m9 x2d h29 y6d ff9 fs1a fc1 sc0 ls47">()</div><div class="t m1 x2e h2a y6e ff9 fsb fc1 sc0 lsd ws33">ω<span class="blank _21"></span>+<span class="blank _22"></span>α</div></div><div class="c x2f y6f w4 h2b"><div class="t m1 x30 h2a y70 ff9 fsb fc1 sc0 lsd">ω</div></div><div class="c xd y20 w3 hd"><div class="t m1 x31 h2a y71 ff9 fsb fc1 sc0 lsd ws34">=<span class="blank _23"></span>ω <span class="ff6 va">j</span></div><div class="t m1 x32 h2c y72 ff6 fsb fc1 sc0 lsd">A</div><div class="t m1 x33 h2c y73 ff6 fsb fc1 sc0 lsd">F</div><div class="t m1 x23 h11 y74 ff5 fsd fc3 sc0 ls2b">(B)</div><div class="t m9 x2d h29 y75 ff9 fs1a fc1 sc0 ls47">()</div><div class="t m1 x2e h2a y76 ff9 fsb fc1 sc0 lsd ws33">ω<span class="blank _21"></span>+<span class="blank _22"></span>α</div><div class="t m1 x31 h2a y77 ff9 fsb fc1 sc0 lsd ws34">=<span class="blank _23"></span>ω <span class="ff6 va">j</span></div><div class="t m1 x34 h2c y78 ff6 fsb fc1 sc0 lsd">A</div><div class="t m1 x33 h2c y77 ff6 fsb fc1 sc0 lsd">F</div><div class="t m1 x23 h11 y79 ff5 fsd fc3 sc0 ls2f">(C)</div><div class="t m9 x2d h29 y7a ff9 fs1a fc1 sc0 ls47">()</div><div class="t m1 x2e h2a y7b ff9 fsb fc1 sc0 lsd ws33">ω<span class="blank _21"></span>−<span class="blank _22"></span>α</div><div class="t m1 x31 h2a y7c ff9 fsb fc1 sc0 lsd ws35">=<span class="blank _23"></span>ω <span class="ff6 va">j</span></div><div class="t m1 x34 h2c y7d ff6 fsb fc1 sc0 lsd">A</div><div class="t m1 x33 h2c y7c ff6 fsb fc1 sc0 lsd">F</div><div class="t m1 x35 h11 y7e ff5 fsd fc3 sc0 ls2f">(D)</div><div class="t ma x2c h2d y7f ff9 fs1b fc1 sc0 ls48">()</div><div class="t m1 x36 h2e y80 ff7 fs1c fc1 sc0 lsd ws36">2<span class="blank _24"></span>2 <span class="ff9 fs1d ws37 vb">ω<span class="blank _25"></span>+<span class="blank _21"></span>α</span></div><div class="t m1 x37 h2f y81 ff9 fs1d fc1 sc0 lsd ws38">=<span class="blank _23"></span>ω <span class="ff6 v6">A</span></div><div class="t m1 x2b h30 y81 ff6 fs1d fc1 sc0 lsd">F</div><div class="t m1 x35 h11 y82 ff5 fsd fc3 sc0 ls2b">(E)</div><div class="t mb x2c h31 y83 ff9 fs1e fc1 sc0 ls49">()</div><div class="t m1 x38 h32 y84 ff7 fs1f fc1 sc0 lsd ws39">2<span class="blank _24"></span>2 <span class="ff9 fs20 ws3a vb">ω<span class="blank _25"></span>+<span class="blank _21"></span>α</span></div><div class="t m1 x37 h33 y85 ff9 fs20 fc1 sc0 lsd ws3b">=<span class="blank _23"></span>ω <span class="ff6 v6">A</span></div><div class="t m1 x2b h34 y85 ff6 fs20 fc1 sc0 lsd">F</div><div class="t mc x39 h35 y86 ffd fs21 fc3 sc0 lsd">f(t)</div><div class="t mc x3a h36 y87 ffd fs21 fc3 sc0 ls1c">0<span class="lsd vc">t</span></div><div class="t mc x3b h35 y88 ffd fs21 fc3 sc0 lsd">A</div><div class="t m1 x3c h37 y89 ffe fs22 fc4 sc0 ls4a">www.pciconcursos.com.br</div></div><div class="t m1 x3d hc y8a ff3 fs5 fc1 sc0 lsd">Resolução:</div><div class="t m1 x3d h6 y8b ff1 fs5 fc1 sc0 lsd ws6">A T<span class="blank _e"></span>r<span class="blank _3"></span>ansformada de Fourier de um sinal é dada pela e<span class="blank _3"></span>xpressão:</div><div class="t m1 x3e h38 y8c fff fs5 fc1 sc0 ls1d">F<span class="ff10 lsd ws3c">(</span><span class="ls1e">ω<span class="ff10 lsd ws3d">) = <span class="ff11 ls1f vd">Z</span><span class="ff12 fs23 ls20 ve">+<span class="ff13 lsd">∞</span></span></span></span></div><div class="t m1 x3f h39 y8d ff13 fs23 fc1 sc0 lsd">−∞</div><div class="t m1 x40 h3a y8c fff fs5 fc1 sc0 ls21">f<span class="ff10 ls22">(</span><span class="lsd ws3e">t<span class="ff10 ws3c">)</span><span class="ls23">e</span><span class="ff13 fs23 ws3f v9">−<span class="ff14 ws40">j ω t<span class="blank _6"> </span></span></span>dt</span></div><div class="t m1 x3d h6 y8e ff1 fs5 fc1 sc0 lsd ws41">Mas como <span class="fff ls24">f</span><span class="ff10 ws3c">(<span class="fff ws3e">t</span><span class="ls25 ws42">)=0</span></span><span class="ws43">par<span class="blank _0"></span>a <span class="fff ws44">t < <span class="ff10 ws3c">0</span></span><span class="ws41">, podemos integrar a função de z<span class="blank _0"></span>ero a infinito</span></span></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 y8f w2 h3b" alt src="https://files.passeidireto.com/49cd2de9-9ce3-46e1-a6f7-51bd167dd6b3/bg4.png" alt="Pré-visualização de imagem de arquivo"><div class="t m3 xb hb y1e ff1 fs9 fc2 sc0 lsd ws19">T34TRJ59YNKS T34TRJ59YNKS T34TRJ59YNKS</div><div class="t m1 x0 h9 y90 ff1 fs7 fc1 sc0 lsd ws6">ANÁLISE DE SINAIS<span class="blank _26"> </span><span class="ff2 ws58">www.concursopetrobraseng.com.br </span>2</div><div class="t m1 x41 h6 y91 ff1 fs5 fc1 sc0 lsd">apenas:</div><div class="t m1 x42 h38 y92 fff fs5 fc1 sc0 ls4b">F<span class="ff10 lsd ws3c">(</span><span class="ls1e">ω<span class="ff10 lsd ws3d">) = <span class="ff11 ls4c vd">Z</span><span class="ff12 fs23 ws59 ve">+<span class="ff13">∞</span></span></span></span></div><div class="t m1 x43 h3c y93 ff12 fs23 fc1 sc0 lsd">0</div><div class="t m1 x44 h3a y92 fff fs5 fc1 sc0 ls21">f<span class="ff10 ls22">(</span><span class="lsd ws3e">t<span class="ff10 ws3c">)</span><span class="ls23">e</span><span class="ff13 fs23 ws3f v9">−<span class="ff14 ws40">j ω t<span class="blank _6"> </span></span></span>dt</span></div><div class="t m1 x42 h38 y94 fff fs5 fc1 sc0 ls4b">F<span class="ff10 lsd ws3c">(</span><span class="ls1e">ω<span class="ff10 lsd ws3d">) = <span class="ff11 ls4c vd">Z</span><span class="ff12 fs23 ws59 ve">+<span class="ff13">∞</span></span></span></span></div><div class="t m1 x43 h3c y95 ff12 fs23 fc1 sc0 lsd">0</div><div class="t m1 x44 h3a y94 fff fs5 fc1 sc0 lsd ws3e">Ae<span class="ff13 fs23 ws3f v9">−<span class="ff14 ws5a">αt </span></span><span class="ls23">e</span><span class="ff13 fs23 ws3f v9">−<span class="ff14 ws5b">j ω t<span class="blank _6"> </span></span></span>dt</div><div class="t m1 x42 h38 y96 fff fs5 fc1 sc0 ls4b">F<span class="ff10 lsd ws3c">(</span><span class="ls1e">ω<span class="ff10 lsd ws3d">) = <span class="ff11 ls4c vd">Z</span><span class="ff12 fs23 ws59 ve">+<span class="ff13">∞</span></span></span></span></div><div class="t m1 x43 h3c y97 ff12 fs23 fc1 sc0 lsd">0</div><div class="t m1 x44 h3d y96 fff fs5 fc1 sc0 lsd ws3e">Ae<span class="ff13 fs23 ws3f v9">−<span class="ff12 ls4d">(<span class="ff14 lsd ws5c">α</span><span class="ls20">+<span class="ff14 lsd ws5d">j ω </span></span>)<span class="ff14 ls4e">t</span></span></span>dt</div><div class="t m1 x42 h3e y98 fff fs5 fc1 sc0 ls4b">F<span class="ff10 lsd ws3c">(</span><span class="ls1e">ω<span class="ff10 lsd ws5e">) =<span class="blank _27"> </span><span class="ff15 ws5f vf">−<span class="fff">A</span></span></span></span></div><div class="t m1 x45 h3f y99 fff fs5 fc1 sc0 ls4f">α<span class="ff10 ls50">+</span><span class="lsd ws60">j ω<span class="blank _8"> </span><span class="ff11 ls51 v10"></span><span class="ws3e vf">e</span><span class="ff13 fs23 ls20 ve">−<span class="ff12 lsd ws59">(<span class="ff14 ls52">α</span>+<span class="ff14 ws61">j ω </span>)<span class="ff14 ls4e">t</span></span></span><span class="ff11 ws62 v10"></span><span class="ff12 fs23 ls20 v11">+<span class="ff13 lsd">∞</span></span></span></div><div class="t m1 x46 h3c y9a ff12 fs23 fc1 sc0 lsd">0</div><div class="t m1 x42 h40 y9b fff fs5 fc1 sc0 ls4b">F<span class="ff10 ls22">(</span><span class="ls53">ω<span class="ff10 lsd ws5e">) =<span class="blank _27"> </span><span class="ff15 ws5f vf">−<span class="fff">A</span></span></span></span></div><div class="t m1 x45 h41 y9c fff fs5 fc1 sc0 ls4f">α<span class="ff10 ls50">+</span><span class="lsd ws60">j ω<span class="blank _8"> </span><span class="ff10 ws63 vf">[0 <span class="ff15 ls54">−</span>1]</span></span></div><div class="t m1 x42 h3e y9d fff fs5 fc1 sc0 ls4b">F<span class="ff10 ls22">(</span><span class="ls53">ω<span class="ff10 lsd ws5e">) =<span class="blank _28"> </span><span class="fff vf">A</span></span></span></div><div class="t m1 x45 h42 y9e fff fs5 fc1 sc0 ls4f">α<span class="ff10 ls50">+</span><span class="lsd ws60">j ω</span></div><div class="t m1 x47 h43 y9f ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x47 h43 ya0 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 y9f ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 ya0 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x49 hc ya1 ff3 fs5 fc1 sc0 lsd ws6">Alternativa (B)</div><div class="t m1 x0 h44 ya2 ff1 fs23 fc1 sc0 lsd ws6">Material de uso exclusiv<span class="blank _3"></span>o do Comprador Cód.<span class="blank _a"> </span>T34TRJ59YNKS<span class="blank _3"></span>. Sendo vedada, por quaisquer meios e a qualquer título<span class="blank _0"></span>, a sua</div><div class="t m1 x0 h44 ya3 ff1 fs23 fc1 sc0 lsd ws6">reprodução<span class="blank _3"></span>, cópia, divulgação e distr<span class="blank _5"> </span>ibuição<span class="blank _3"></span>.<span class="blank _a"> </span>Sujeitando-se o infrator à responsabilização civil e criminal.</div><a class="l"><div class="d m2" style="border-width:1.000000px;border-style:solid;border-color:rgb(0,255,255);position:absolute;left:229.264000px;bottom:779.309000px;width:163.094000px;height:12.948000px;background-color:rgba(255,255,255,0.000001);"></div></a></div><div class="pi" data-data="{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}"></div></div> <div id="pf5" class="pf w0 h0" data-page-no="5"><div class="pc pc5 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 ya4 w2 h45" alt src="https://files.passeidireto.com/49cd2de9-9ce3-46e1-a6f7-51bd167dd6b3/bg5.png" alt="Pré-visualização de imagem de arquivo"><div class="t m3 xb hb y1e ff1 fs9 fc2 sc0 lsd ws19">T34TRJ59YNKS T34TRJ59YNKS T34TRJ59YNKS</div><div class="t m1 x0 h9 y90 ff1 fs7 fc1 sc0 lsd ws6">ANÁLISE DE SINAIS<span class="blank _26"> </span><span class="ff2 ws58">www.concursopetrobraseng.com.br </span>3</div><div class="t m1 xc hc ya5 ff3 fs5 fc1 sc0 lsd ws6">Questão 2<span class="blank _12"> </span><span class="ff1 fs7 v1">(Eng.<span class="blank _11"> </span>de Equipamentos Jr Eletrônica - P<span class="blank _3"></span>etrobras 2012/1)</span></div><div class="c x4a ya6 w5 h46"><div class="t m1 x23 h47 ya7 ff17 fs24 fc3 sc0 lsd ws6">A<span class="blank _3"></span> Figura mostra o espectro de Fourier de um sinal x(t). </div><div class="t m1 x3 h48 ya8 ff18 fs25 fc3 sc0 lsd ws64">X( )<span class="blank _29"></span><span class="ff19">w</span></div><div class="t m1 x4b h49 ya9 ff19 fs25 fc3 sc0 lsd">w</div><div class="t m1 x27 h4a yaa ff18 fs25 fc3 sc0 ls62 ws65">2x<span class="blank _2a"></span>1<span class="blank _2b"></span>0<span class="blank _2c"></span><span class="ff19 ls55">p<span class="ff18 fs26 lsd v7">4</span></span></div><div class="t m1 x23 h4b yab ff17 fs24 fc3 sc0 lsd ws6">A<span class="blank _3"></span> taxa de amostragem de Nyquist do sinal x<span class="fs27 ls56 v7">3</span>(t) é </div><div class="t m1 x23 h47 yac ff17 fs24 fc3 sc0 lsd ws6">(A) <span class="blank _6"> </span> 3.33 kHz <span class="blank _2d"> </span> </div><div class="t m1 x23 h47 yad ff17 fs24 fc3 sc0 lsd ws70">(B) 10 <span class="blank _4"></span>kHz <span class="blank _2e"> </span> </div><div class="t m1 x23 h47 yae ff17 fs24 fc3 sc0 lsd ws71">(C) 20 <span class="blank _0"></span>kHz <span class="blank _2f"> </span> </div><div class="t m1 x23 h47 yaf ff17 fs24 fc3 sc0 lsd ws71">(D) 30 <span class="blank _0"></span>kHz <span class="blank _2f"> </span> </div><div class="t m1 x23 h47 yb0 ff17 fs24 fc3 sc0 lsd ws70">(E) 60 <span class="blank _4"></span>kHz</div></div><div class="t m1 x3d hc yb1 ff3 fs5 fc1 sc0 lsd">Resolução:</div><div class="t m1 x3d h6 yb2 ff1 fs5 fc1 sc0 lsd ws66">Do gráfico f<span class="blank _3"></span>or<span class="blank _5"> </span>necido podemos identificar diretamente a máxima frequência</div><div class="t m1 x41 h6 yb3 ff1 fs5 fc1 sc0 lsd ws6">do espectro<span class="blank _3"></span>, que é <span class="fff ls57">ω</span><span class="ff10 ws5e">= 2<span class="fff ls58">π</span><span class="ws3c">10<span class="ff12 fs23 ls59 v12">4</span><span class="fff ws67">r ad/s</span></span></span>.</div><div class="t m1 x3d h6 yb4 ff1 fs5 fc1 sc0 lsd ws6">Das propriedades da T<span class="blank _e"></span>ransf<span class="blank _3"></span>or<span class="blank _5"> </span>mada de F<span class="blank _3"></span>our<span class="blank _5"> </span>ier podemos inferir que:</div><div class="t m1 x4c h6 yb5 ff1 fs5 fc1 sc0 lsd ws68">Se <span class="fff ls5a">x</span><span class="ff10 ws3c">(<span class="fff ws3e">t</span><span class="ls5b">)<span class="ff15 ls5c">⇔<span class="fff ls1d">F</span></span></span>(<span class="fff ls53">ω</span><span class="ls22">)</span></span><span class="ws69">, então <span class="fff ls5a">x<span class="ff12 fs23 ls5d v9">3</span></span><span class="ff10 ws3c">(<span class="fff ws3e">t</span><span class="ls5b">)<span class="ff15 ls5c">⇔<span class="fff ls1d">F</span></span></span>(3<span class="fff ls1e">ω</span>)</span></span></div><div class="t m1 x3d h6 yb6 ff1 fs5 fc1 sc0 lsd ws6a">Ou seja, para o sinal <span class="fff ls5a">x<span class="ff12 fs23 ls5d v12">3</span></span><span class="ff10 ws3c">(<span class="fff ws3e">t</span><span class="ls5e">)</span></span>a máxima frequência do espectro<span class="blank _3"></span>, que chamare-</div><div class="t m1 x41 h6 yb7 ff1 fs5 fc1 sc0 lsd ws6b">mos de <span class="fff ws3e">ω<span class="ff12 fs23 ls5d v13">3</span></span><span class="ws6">, será:</span></div><div class="t m1 x4d h42 yb8 fff fs5 fc1 sc0 ls5f">ω<span class="ff12 fs23 ls60 v13">3</span><span class="ff10 lsd ws5e">= 3<span class="fff">ω</span></span></div><div class="t m1 x4e h3d yb9 ff10 fs5 fc1 sc0 lsd ws3c">2<span class="fff ws6c">π f<span class="ff12 fs23 ls60 v13">3</span></span><span class="ws6d">=<span class="blank _11"> </span>3 <span class="ff15 ls54">×</span></span>(2<span class="fff ls58">π</span>10<span class="ff12 fs23 ls59 v9">4</span>)</div><div class="t m1 x4f h42 yba fff fs5 fc1 sc0 ls61">f<span class="ff12 fs23 ls60 v13">3</span><span class="ff10 lsd ws5e">= 30</span><span class="lsd ws6e">k H<span class="blank _30"> </span>z</span></div><div class="t m1 x3d h6 ybb ff1 fs5 fc1 sc0 lsd ws6f">Como sabemos, a taxa de amostr<span class="blank _0"></span>agem de Nyquist de<span class="blank _3"></span>ve ter o dobro da má-</div><div class="t m1 x41 h6 ybc ff1 fs5 fc1 sc0 lsd ws6">xima frequência do espectro<span class="blank _3"></span>, então:</div><div class="t m1 x4d h42 ybd fff fs5 fc1 sc0 ls61">f<span class="ff14 fs23 lsd ws5b v13">ny q<span class="blank _7"> </span></span><span class="ff10 lsd ws6d">=<span class="blank _a"> </span>2 <span class="ff15 ls54">×</span></span>f<span class="ff12 fs23 lsd v13">3</span></div><div class="t m1 x4d h42 ybe fff fs5 fc1 sc0 ls61">f<span class="ff14 fs23 lsd ws5b v13">ny q<span class="blank _7"> </span></span><span class="ff10 lsd ws6d">=<span class="blank _a"> </span>2 <span class="ff15 ls54">×</span><span class="ws3c">30<span class="fff">k</span></span></span></div><div class="t m1 x4d h42 ybf fff fs5 fc1 sc0 ls61">f<span class="ff14 fs23 lsd ws5b v13">ny q<span class="blank _7"> </span></span><span class="ff10 lsd ws5e">= 60</span><span class="lsd ws6e">k H<span class="blank _30"> </span>z</span></div><div class="t m1 x47 h43 yc0 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x47 h43 yc1 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 yc0 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 yc1 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x49 hc yc2 ff3 fs5 fc1 sc0 lsd ws6">Alternativa (E)</div><div class="t m1 x0 h44 ya2 ff1 fs23 fc1 sc0 lsd ws6">Material de uso exclusiv<span class="blank _3"></span>o do Comprador Cód.<span class="blank _a"> </span>T34TRJ59YNKS<span class="blank _3"></span>. Sendo vedada, por quaisquer meios e a qualquer título<span class="blank _0"></span>, a sua</div><div class="t m1 x0 h44 ya3 ff1 fs23 fc1 sc0 lsd ws6">reprodução<span class="blank _3"></span>, cópia, divulgação e distr<span class="blank _5"> </span>ibuição<span class="blank _3"></span>.<span class="blank _a"> </span>Sujeitando-se o infrator à responsabilização civil e criminal.</div><a class="l"><div class="d m2" style="border-width:1.000000px;border-style:solid;border-color:rgb(0,255,255);position:absolute;left:229.264000px;bottom:779.309000px;width:163.094000px;height:12.948000px;background-color:rgba(255,255,255,0.000001);"></div></a></div><div class="pi" data-data="{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}"></div></div> <div id="pf6" class="pf w0 h0" data-page-no="6"><div class="pc pc6 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 yc3 w2 h4c" alt src="https://files.passeidireto.com/49cd2de9-9ce3-46e1-a6f7-51bd167dd6b3/bg6.png" alt="Pré-visualização de imagem de arquivo"><div class="t m3 xb hb y1e ff1 fs9 fc2 sc0 lsd ws19">T34TRJ59YNKS T34TRJ59YNKS T34TRJ59YNKS</div><div class="t m1 x0 h9 y90 ff1 fs7 fc1 sc0 lsd ws6">ANÁLISE DE SINAIS<span class="blank _26"> </span><span class="ff2 ws58">www.concursopetrobraseng.com.br </span>4</div><div class="t m1 xc hc ya5 ff3 fs5 fc1 sc0 lsd ws6">Questão 3<span class="blank _12"> </span><span class="ff1 fs7 v1">(Eng.<span class="blank _11"> </span>de Equipamentos Jr Eletrônica - P<span class="blank _3"></span>etrobras 2012/1)</span></div><div class="c x50 yc4 w6 h4d"><div class="t m1 x51 h4e yc5 ff1a fs24 fc3 sc0 lsd ws7a">A<span class="blank _3"></span> energia E<span class="fs27 ls63 v5">x</span> do sinal x(t) <span class="ff1b ws72">=</span> 4[u(t<span class="ff1b ws72">+</span><span class="ws6">1) <span class="ff1b ws72">−</span></span> u(t)] <span class="ff1b ws72">+</span> 4e<span class="ff1b fs27 ws73 v7">−<span class="ff1a ws6">t </span></span></div><div class="t m1 x50 h47 yc5 ff1a fs24 fc3 sc0 lsd ws7b">u(t), onde </div><div class="t m1 x51 h47 yc6 ff1a fs24 fc3 sc0 lsd ws6">u(t) é degrau unitário, é</div><div class="t m1 x51 h4f yc7 ff1a fs24 fc3 sc0 lsd ws70">(A) E<span class="fs27 ls64 v5">x</span><span class="ff1b ws6 v0">= 4<span class="ff1a ls77"> </span></span></div><div class="t m1 x51 h47 yc8 ff1a fs24 fc3 sc0 lsd ws70">(B) E<span class="fs27 ls64 v5">x</span><span class="ff1b ws6">= 6<span class="ff1a ls77"> </span></span></div><div class="t m1 x51 h47 yc9 ff1a fs24 fc3 sc0 lsd ws71">(C) E<span class="fs27 ls64 v5">x</span><span class="ff1b ws6">= 10<span class="ff1a ls78"> </span></span></div><div class="t m1 x51 h47 yca ff1a fs24 fc3 sc0 lsd ws71">(D) E<span class="fs27 ls64 v5">x</span><span class="ff1b ws6">= 12<span class="ff1a ls78"> </span></span></div><div class="t m1 x51 h47 ycb ff1a fs24 fc3 sc0 lsd ws70">(E) E<span class="fs27 ls64 v5">x</span><span class="ff1b ws6">= 24</span></div></div><div class="t m1 x3d hc ycc ff3 fs5 fc1 sc0 lsd">Resolução:</div><div class="t m1 x3d h6 ycd ff1 fs5 fc1 sc0 lsd ws74">Primeiramente gostaríamos de salientar que a questão or<span class="blank _5"> </span>iginal apresentav<span class="blank _3"></span>a</div><div class="t m1 x41 h6 yce ff1 fs5 fc1 sc0 lsd ws75">um traço abaixo do “<span class="ff15 ls65">−</span><span class="fff ws3e">t</span>” da e<span class="blank _3"></span>xponencial, que tomamos a liberdade de retirá-lo pois</div><div class="t m1 x41 h6 ycf ff1 fs5 fc1 sc0 lsd ws6">trata-se claramente de um prob<span class="blank _3"></span>lema de digitação<span class="blank _3"></span>.</div><div class="t m1 x3d h6 yd0 ff1 fs5 fc1 sc0 lsd ws6">Sabemos que a energia de um sinal é dada por<span class="blank _5"> </span>:</div><div class="t m1 x2f h38 yd1 fff fs5 fc1 sc0 ls66">E<span class="ff14 fs23 ls67 v13">x</span><span class="ff10 ls68">=<span class="ff11 ls1f vd">Z</span><span class="ff12 fs23 ls20 ve">+<span class="ff13 lsd">∞</span></span></span></div><div class="t m1 x3f h50 yd2 ff13 fs23 fc1 sc0 lsd ws76">−∞ <span class="ff15 fs5 ws5f v14">|<span class="fff ls5a">x<span class="ff10 lsd ws3c">(<span class="fff ws3e">t</span><span class="ls22">)</span></span></span>|</span><span class="ff12 ls5d v11">2</span><span class="fff fs5 v14">dt</span></div><div class="t m1 x3d h6 yd3 ff1 fs5 fc1 sc0 lsd ws6">P<span class="blank _4"></span>or<span class="blank _6"> </span>tanto a energia do sinal dado será:</div><div class="t m1 x52 h51 yd4 fff fs5 fc1 sc0 lsd ws3e">E<span class="ff14 fs23 ls67 v13">x</span><span class="ff10 ls68">=<span class="ff11 ls4c vd">Z</span></span><span class="ff12 fs23 ws59 ve">+<span class="ff13">∞</span></span></div><div class="t m1 x38 h50 yd5 ff13 fs23 fc1 sc0 lsd ws76">−∞ <span class="ff15 fs5 ws5f v14">|<span class="ff10 ws3c">4[<span class="fff ws3e">u</span>(<span class="fff ls69">t</span><span class="ws77">+ 1) <span class="ff15 ls6a">−</span><span class="fff ws3e">u</span></span>(<span class="fff ls6b">t</span><span class="ws77">)] + 4<span class="fff ls23">e</span></span></span></span><span class="ws3f v11">−<span class="ff14 ls6c">t</span></span><span class="fff fs5 ws3e v14">u<span class="ff10 ws3c">(</span>t<span class="ff10 ls22">)</span><span class="ff15 ws5f">|</span></span><span class="ff12 ls59 v11">2</span><span class="fff fs5 v14">dt</span></div><div class="t m1 x52 h38 yd6 fff fs5 fc1 sc0 lsd ws3e">E<span class="ff14 fs23 ls67 v13">x</span><span class="ff10 ls68">=<span class="ff11 ls4c vd">Z</span></span><span class="ff12 fs23 ve">0</span></div><div class="t m1 x38 h52 yd7 ff13 fs23 fc1 sc0 ls20">−<span class="ff12 ls6d">1<span class="ff15 fs5 lsd ws5f v14">|<span class="ff10 ws3c">4[<span class="fff ws3e">u</span><span class="ls22">(<span class="fff ls69">t</span></span><span class="ws77">+ 1) <span class="ff15 ls54">−</span><span class="fff ws3e">u</span><span class="ls22">(</span><span class="fff ws3e">t</span></span>)]</span>|</span><span class="ls6e v11">2</span><span class="ff10 fs5 ls6f v14">+<span class="ff11 ls1f vd">Z</span></span></span><span class="lsd v15">∞</span></div><div class="t m1 x53 h53 yd7 ff12 fs23 fc1 sc0 ls70">0<span class="ff15 fs5 lsd ws5f v14">|<span class="ff10 ws3c">4<span class="fff ls23">e</span></span></span><span class="ff13 lsd ws3f v11">−<span class="ff14 ls4e">t<span class="fff fs5 ls71 vb">u<span class="ff10 lsd ws3c">(<span class="fff ws3e">t</span><span class="ls22">)</span><span class="ff15 ws5f">|</span></span></span><span class="ff12 lsd">2</span></span></span></div><div class="t m1 x52 h38 yd8 fff fs5 fc1 sc0 lsd ws3e">E<span class="ff14 fs23 ls67 v13">x</span><span class="ff10 ws5e">= 4<span class="ff12 fs23 ls72 v9">2</span><span class="ff11 ls1f vd">Z</span><span class="ff12 fs23 ve">0</span></span></div><div class="t m1 x54 h52 yd9 ff13 fs23 fc1 sc0 lsd ws3f">−<span class="ff12 ls72">1</span><span class="ff15 fs5 ws5f v14">|<span class="ff10 ls73">[<span class="fff lsd ws3e">u</span><span class="ls22">(<span class="fff ls74">t</span><span class="lsd ws77">+ 1) <span class="ff15 ls54">−</span><span class="fff ws3e">u</span></span>(<span class="fff lsd ws3e">t<span class="ff10 ws3c">)]</span></span></span></span>|</span><span class="ff12 ls6e v11">2</span><span class="ff10 fs5 ws77 v14">+ 4</span><span class="ff12 ls72 v11">2</span><span class="ff11 fs5 ls1f v16">Z</span><span class="v15">∞</span></div><div class="t m1 x55 h50 yd9 ff12 fs23 fc1 sc0 ls70">0<span class="ff15 fs5 lsd ws5f v14">|<span class="fff ws3e">e</span></span><span class="ff13 ls20 v11">−<span class="ff14 ls4e">t<span class="fff fs5 lsd ws3e vb">u<span class="ff10 ls22">(</span>t<span class="ff10 ws3c">)<span class="ff15 ws5f">|</span></span></span><span class="ff12 lsd">2</span></span></span></div><div class="t m1 x52 h54 yda fff fs5 fc1 sc0 lsd ws3e">E<span class="ff14 fs23 ls67 v13">x</span><span class="ff10 ws78">=<span class="blank _11"> </span>16 <span class="ff15 ls54">×</span><span class="ws3c">1<span class="ff12 fs23 ls6e v9">2</span><span class="ls6f">+<span class="ff11 ls75 v17"></span></span><span class="ws79">16 </span></span></span><span class="vf">e</span><span class="ff13 fs23 ls20 v18">−<span class="ff12 lsd ws59">2<span class="ff14">t</span></span></span></div><div class="t m1 x56 h55 ydb ff15 fs5 fc1 sc0 ls65">−<span class="ff10 ls76">2<span class="ff11 ls75 v19"></span><span class="ff13 fs23 lsd v1a">∞</span></span></div><div class="t m1 x57 h3c ydc ff12 fs23 fc1 sc0 lsd">0</div><div class="t m1 x52 h42 ydd fff fs5 fc1 sc0 lsd ws3e">E<span class="ff14 fs23 ls67 v13">x</span><span class="ff10 ws77">=<span class="blank _11"> </span>16 + [0 <span class="ff15 ls54">−</span><span class="ws3c">(<span class="ff15 ls65">−</span>8)]</span></span></div><div class="t m1 x52 h42 yde fff fs5 fc1 sc0 lsd ws3e">E<span class="ff14 fs23 ls67 v13">x</span><span class="ff10 ws77">=<span class="blank _11"> </span>16 + 8</span></div><div class="t m1 x52 h42 ydf fff fs5 fc1 sc0 lsd ws3e">E<span class="ff14 fs23 ls67 v13">x</span><span class="ff10 ws5e">= 24</span></div><div class="t m1 x47 h43 ye0 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x47 h43 ye1 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 ye0 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 ye1 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x49 hc ye2 ff3 fs5 fc1 sc0 lsd ws6">Alternativa (E)</div><div class="t m1 x0 h44 ya2 ff1 fs23 fc1 sc0 lsd ws6">Material de uso exclusiv<span class="blank _3"></span>o do Comprador Cód.<span class="blank _a"> </span>T34TRJ59YNKS<span class="blank _3"></span>. Sendo vedada, por quaisquer meios e a qualquer título<span class="blank _0"></span>, a sua</div><div class="t m1 x0 h44 ya3 ff1 fs23 fc1 sc0 lsd ws6">reprodução<span class="blank _3"></span>, cópia, divulgação e distr<span class="blank _5"> </span>ibuição<span class="blank _3"></span>.<span class="blank _a"> </span>Sujeitando-se o infrator à responsabilização civil e criminal.</div><a class="l"><div class="d m2" style="border-width:1.000000px;border-style:solid;border-color:rgb(0,255,255);position:absolute;left:229.264000px;bottom:779.309000px;width:163.094000px;height:12.948000px;background-color:rgba(255,255,255,0.000001);"></div></a></div><div class="pi" data-data="{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}"></div></div> <div id="pf7" class="pf w0 h0" data-page-no="7"><div class="pc pc7 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 y1d w2 h56" alt src="https://files.passeidireto.com/49cd2de9-9ce3-46e1-a6f7-51bd167dd6b3/bg7.png" alt="Pré-visualização de imagem de arquivo"><div class="t m3 xb hb y1e ff1 fs9 fc2 sc0 lsd ws19">T34TRJ59YNKS T34TRJ59YNKS T34TRJ59YNKS</div><div class="t m1 x0 h9 y90 ff1 fs7 fc1 sc0 lsd ws6">ANÁLISE DE SINAIS<span class="blank _26"> </span><span class="ff2 ws58">www.concursopetrobraseng.com.br </span>5</div><div class="t m1 xc hc ya5 ff3 fs5 fc1 sc0 lsd ws6">Questão 4<span class="blank _12"> </span><span class="ff1 fs7 v1">(Eng.<span class="blank _11"> </span>de Equipamentos Jr Elétrica - P<span class="blank _0"></span>etrobras 2010/2)</span></div><div class="c x58 ye3 w7 h57"><div class="t m1 x59 h58 ye4 ff1c fs28 fc3 sc0 lsd ws6">ENGENHEIRO(A) DE EQUIP<span class="blank _3"></span>AMENTOS JÚNIOR</div><div class="t m1 x5a h58 ye5 ff1c fs28 fc3 sc0 lsd">ELÉTRICA</div><div class="t m1 x5b h59 ye6 ff1d fs29 fc3 sc0 lsd">15</div><div class="t m1 x5c h5a ye7 ff1d fs2a fc3 sc0 lsd">59</div><div class="t m1 x5c h5b ye8 ff1e fs2b fc3 sc0 lsd ws88">A<span class="blank _3"></span> figura acima mostra um diagrama em blocos, no domínio de Laplace, contendo um bloco de retardo, um somador e um </div><div class="t m1 x5c h5b ye9 ff1e fs2b fc3 sc0 lsd ws6">integrador<span class="blank _3"></span>. <span class="blank _3"></span>Aplicando um impulso unitário <span class="blank _5"> </span><span class="ff1f ls79">δ</span>(t) na entrada, a forma de onda da saída h(t) é </div><div class="t m1 x5c h5b yea ff1e fs2b fc3 sc0 lsd ws89">(A) <span class="blank _31"></span> (B)</div><div class="t m1 x5c h5b yeb ff1e fs2b fc3 sc0 lsd ws89">(C) <span class="blank _32"></span> (D)</div><div class="t m1 x5c h5b yec ff1e fs2b fc3 sc0 lsd">(E)</div><div class="t m1 x5c h5a yed ff1d fs2a fc3 sc0 lsd">60</div><div class="t m1 x5c h5b yee ff1e fs2b fc3 sc0 lsd ws6">Um sistema linear<span class="blank _3"></span>, causal e de segunda ordem é representado pela seguinte função de T<span class="blank _3"></span>ransferência:</div><div class="t m1 x5c h5b yef ff1e fs2b fc3 sc0 lsd ws8a">Esse sistema opera com razão de amortecimento 0,7 e frequência natural não amortecida de 15 rad/s. Quando alimentado </div><div class="t m1 x5c h5c yf0 ff1e fs2b fc3 sc0 lsd ws8b">por um degrau unitário em sua entrada, a saída, em regime permanente, atinge o valor 0,4. Os valores de <span class="ff1d ls7a">a</span> e <span class="ff1d ls7b">K</span><span class="ws6">, </span></div><div class="t m1 x5c h5b yf1 ff1e fs2b fc3 sc0 lsd ws6">respectivamente, são</div><div class="t m1 x5c h5b yf2 ff1e fs2b fc3 sc0 lsd ws6">(A) <span class="blank _6"> </span>42 e 180</div><div class="t m1 x5c h5b yf3 ff1e fs2b fc3 sc0 lsd ws6">(B) <span class="blank _6"> </span>21 e 90 </div><div class="t m1 x5c h5b yf4 ff1e fs2b fc3 sc0 lsd ws6">(C) <span class="blank _5"> </span>21 e 15</div><div class="t m1 x5c h5b yf5 ff1e fs2b fc3 sc0 lsd ws6">(D) <span class="blank _5"> </span>10,5 e 90</div><div class="t m1 x5c h5b yf6 ff1e fs2b fc3 sc0 lsd ws6">(E) <span class="blank _6"> </span>10,5 e 45</div></div><div class="t m1 x3d hc yf7 ff3 fs5 fc1 sc0 lsd">Resolução:</div><div class="t m1 x3d h6 yf8 ff1 fs5 fc1 sc0 lsd ws7c">A e<span class="blank _3"></span>xpressão no domínio de Laplace do diagrama de blocos dado é f<span class="blank _3"></span>acil-</div><div class="t m1 x41 h6 yf9 ff1 fs5 fc1 sc0 lsd ws6">mente deduzida:</div><div class="t m1 x5d h3e yfa fff fs5 fc1 sc0 ls7c">H<span class="ff10 lsd ws3c">(</span><span class="ls7d">s<span class="ff10 lsd ws7d">) = </span><span class="ls7e">U<span class="ff10 lsd ws3c">(</span></span>s<span class="ff10 lsd ws7e">)[ <span class="vf">1</span></span></span></div><div class="t m1 x5e h5d yfb fff fs5 fc1 sc0 ls7f">s<span class="ff15 ls80 vf">−</span><span class="ls23 vd">e</span><span class="ff13 fs23 lsd ws3f v1b">−<span class="ff14 ws7f">τ s</span></span></div><div class="t m1 x5f h41 yfb fff fs5 fc1 sc0 ls81">s<span class="ff10 lsd vf">]</span></div><div class="t m1 x41 h6 yfc ff1 fs5 fc1 sc0 lsd ws80">P<span class="blank _4"></span>orém,<span class="blank _10"> </span>sabemos que a entrada é um impulso unitário<span class="blank _3"></span>,<span class="blank _10"> </span>ou seja,<span class="blank _c"> </span><span class="fff ws3e">u<span class="ff10 ls22">(</span>t<span class="ff10 ws81">) = </span><span class="ls82">δ</span><span class="ff10 ws3c">(</span>t<span class="ff10 ws3c">)</span></span><span class="ws82">, logo</span></div><div class="t m1 x41 h6 yfd fff fs5 fc1 sc0 ls7e">U<span class="ff10 ls22">(</span><span class="ls7d">s<span class="ff10 ls83">)<span class="ff1 lsd ws6">é 1.</span></span></span></div><div class="t m1 x4c h3e yfe fff fs5 fc1 sc0 ls7c">H<span class="ff10 lsd ws3c">(</span><span class="ls7d">s<span class="ff10 ls84 ws83">)=1<span class="blank _4"></span><span class="ff15 ls54">×<span class="ff10 ls85">[<span class="lsd vf">1</span></span></span></span></span></div><div class="t m1 x45 h5d yff fff fs5 fc1 sc0 ls7f">s<span class="ff15 ls86 vf">−</span><span class="lsd ws3e vd">e</span><span class="ff13 fs23 ls20 v1b">−<span class="ff14 lsd ws84">τ s</span></span></div><div class="t m1 x60 h5e yff fff fs5 fc1 sc0 ls87">s<span class="ff10 lsd ws5e vf">] =<span class="blank _c"> </span><span class="vf">1</span></span></div><div class="t m1 x61 h5e yff fff fs5 fc1 sc0 ls88">s<span class="ff15 ls86 vf">−</span><span class="ff10 lsd vd">1</span></div><div class="t m1 x62 h5f yff fff fs5 fc1 sc0 ls7f">s<span class="ff15 ls54 vf">×</span><span class="ls23 vf">e</span><span class="ff13 fs23 lsd ws3f ve">−<span class="ff14 ws7f">τ s</span></span></div><div class="t m1 x3d h6 y100 ff1 fs5 fc1 sc0 lsd ws85">Sabemos que uma translação no tempo igual a <span class="fff ls89">τ</span>segundos equiv<span class="blank _3"></span>ale a uma</div><div class="t m1 x41 h6 y101 ff1 fs5 fc1 sc0 lsd ws86">multiplicação por <span class="fff ls23">e</span><span class="ff13 fs23 ws3f v12">−<span class="ff14 ws84">τ s<span class="blank _33"> </span></span></span>no domínio de Laplace, ou seja:<span class="blank _d"> </span><span class="fff ls23">e</span><span class="ff13 fs23 ws3f v12">−<span class="ff14 ws7f">τ s<span class="blank _b"> </span></span></span><span class="ff15 ls8a">→<span class="fff ls82">δ<span class="ff10 ls22">(</span><span class="ls8b">t</span></span><span class="ls8c">−<span class="fff ls8d">τ</span></span></span><span class="ff10 ws3c">)</span><span class="ws87">. Logo<span class="blank _3"></span>,</span></div><div class="t m1 x41 h6 y102 ff1 fs5 fc1 sc0 lsd ws6">sabendo que<span class="blank _c"> </span><span class="ff12 fs23 v9">1</span></div><div class="t m1 x1 h60 y103 ff14 fs23 fc1 sc0 ls8e">s<span class="ff1 fs5 lsd ws6 v12">no domínio de Laplace equiv<span class="blank _0"></span>ale a um degrau (<span class="blank _0"></span><span class="fff ls8f">D<span class="ff10 lsd ws3c">(</span><span class="ls6b">t<span class="ff10 lsd ws3c">)<span class="ff1 ws6">) no domínio do</span></span></span></span></span></div><div class="t m1 x41 h6 y104 ff1 fs5 fc1 sc0 lsd ws6">tempo<span class="blank _3"></span>, a e<span class="blank _0"></span>xpressão:</div><div class="t m1 x63 h3e y105 fff fs5 fc1 sc0 ls7c">H<span class="ff10 lsd ws3c">(</span><span class="ls7d">s<span class="ff10 lsd ws5e">) =<span class="blank _c"> </span><span class="vf">1</span></span></span></div><div class="t m1 x64 h61 y106 fff fs5 fc1 sc0 ls7f">s<span class="ff15 ls80 vf">−</span><span class="ff10 lsd vd">1</span></div><div class="t m1 x5e h5f y106 fff fs5 fc1 sc0 ls88">s<span class="ff15 ls54 vf">×</span><span class="ls23 vf">e</span><span class="ff13 fs23 lsd ws3f ve">−<span class="ff14 ws7f">τ s</span></span></div><div class="t m1 x0 h44 ya2 ff1 fs23 fc1 sc0 lsd ws6">Material de uso exclusiv<span class="blank _3"></span>o do Comprador Cód.<span class="blank _a"> </span>T34TRJ59YNKS<span class="blank _3"></span>. Sendo vedada, por quaisquer meios e a qualquer título<span class="blank _0"></span>, a sua</div><div class="t m1 x0 h44 ya3 ff1 fs23 fc1 sc0 lsd ws6">reprodução<span class="blank _3"></span>, cópia, divulgação e distr<span class="blank _5"> </span>ibuição<span class="blank _3"></span>.<span class="blank _a"> </span>Sujeitando-se o infrator à responsabilização civil e criminal.</div><a class="l"><div class="d m2" style="border-width:1.000000px;border-style:solid;border-color:rgb(0,255,255);position:absolute;left:229.264000px;bottom:779.309000px;width:163.094000px;height:12.948000px;background-color:rgba(255,255,255,0.000001);"></div></a></div><div class="pi" data-data="{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}"></div></div> <div id="pf8" class="pf w0 h0" data-page-no="8"><div class="pc pc8 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 y107 w2 h62" alt src="https://files.passeidireto.com/49cd2de9-9ce3-46e1-a6f7-51bd167dd6b3/bg8.png" alt="Pré-visualização de imagem de arquivo"><div class="t m3 xb hb y1e ff1 fs9 fc2 sc0 lsd ws19">T34TRJ59YNKS T34TRJ59YNKS T34TRJ59YNKS</div><div class="t m1 x0 h9 y90 ff1 fs7 fc1 sc0 lsd ws6">ANÁLISE DE SINAIS<span class="blank _26"> </span><span class="ff2 ws58">www.concursopetrobraseng.com.br </span>6</div><div class="t m1 x3d h6 y108 ff1 fs5 fc1 sc0 lsd ws6">No domínio do tempo fica:</div><div class="t m1 x65 h42 y109 fff fs5 fc1 sc0 lsd ws3e">h<span class="ff10 ls22">(</span>t<span class="ff10 ws3d">) = </span><span class="ls90">D</span><span class="ff10 ws3c">(</span>t<span class="ff10 ls91">)<span class="ff15 ls6a">−</span></span><span class="ls8f">D</span><span class="ff10 ws3c">(</span>t<span class="ff10 ls22">)</span><span class="ls82">δ</span><span class="ff10 ws3c">(</span><span class="ls69">t<span class="ff15 ls54">−</span><span class="ls8d">τ</span></span><span class="ff10">)</span></div><div class="t m1 x65 h42 y10a fff fs5 fc1 sc0 lsd ws3e">h<span class="ff10 ls22">(</span>t<span class="ff10 ws3d">) = </span><span class="ls90">D</span><span class="ff10 ws3c">(</span>t<span class="ff10 ls91">)<span class="ff15 ls6a">−</span></span><span class="ls8f">D</span><span class="ff10 ws3c">(</span><span class="ls69">t<span class="ff15 ls54">−</span><span class="ls8d">τ</span></span><span class="ff10">)</span></div><div class="t m1 x3d h6 y10b ff1 fs5 fc1 sc0 lsd ws8c">Ou seja,<span class="blank _33"> </span><span class="fff ws3e">h<span class="ff10 ws3c">(</span>t<span class="ff10 ls92">)</span></span>será um degr<span class="blank _3"></span>au unitár<span class="blank _5"> </span>io de <span class="fff ls93">t</span><span class="ff10 ws8d">=<span class="blank _1"> </span>0 </span><span class="ws8e">até <span class="fff ls93">t<span class="ff10 ls94">=</span><span class="ls8d">τ</span></span></span>,<span class="blank _33"> </span>quando então</div><div class="t m1 x41 h6 y10c ff1 fs5 fc1 sc0 lsd ws8f">subtrai-se um degr<span class="blank _3"></span>au também unitár<span class="blank _5"> </span>io<span class="blank _3"></span>,<span class="blank _8"> </span>zerando a saída.<span class="blank _33"> </span>A alternativa que mostr<span class="blank _0"></span>a</div><div class="t m1 x41 h6 y10d ff1 fs5 fc1 sc0 lsd ws6">este compor<span class="blank _5"> </span>tamento da saída é a alter<span class="blank _5"> </span>nativa (A).</div><div class="t m1 x47 h43 y10e ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x47 h43 y10f ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 y10e ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 y10f ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x49 hc y110 ff3 fs5 fc1 sc0 lsd ws6">Alternativa (A)</div><div class="t m1 x0 h44 ya2 ff1 fs23 fc1 sc0 lsd ws6">Material de uso exclusiv<span class="blank _3"></span>o do Comprador Cód.<span class="blank _a"> </span>T34TRJ59YNKS<span class="blank _3"></span>. Sendo vedada, por quaisquer meios e a qualquer título<span class="blank _0"></span>, a sua</div><div class="t m1 x0 h44 ya3 ff1 fs23 fc1 sc0 lsd ws6">reprodução<span class="blank _3"></span>, cópia, divulgação e distr<span class="blank _5"> </span>ibuição<span class="blank _3"></span>.<span class="blank _a"> </span>Sujeitando-se o infrator à responsabilização civil e criminal.</div><a class="l"><div class="d m2" style="border-width:1.000000px;border-style:solid;border-color:rgb(0,255,255);position:absolute;left:229.264000px;bottom:779.309000px;width:163.094000px;height:12.948000px;background-color:rgba(255,255,255,0.000001);"></div></a></div><div class="pi" data-data="{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}"></div></div> <div id="pf9" class="pf w0 h0" data-page-no="9"><div class="pc pc9 w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 y1d w2 h56" alt src="https://files.passeidireto.com/49cd2de9-9ce3-46e1-a6f7-51bd167dd6b3/bg9.png" alt="Pré-visualização de imagem de arquivo"><div class="t m3 xb hb y1e ff1 fs9 fc2 sc0 lsd ws19">T34TRJ59YNKS T34TRJ59YNKS T34TRJ59YNKS</div><div class="t m1 x0 h9 y90 ff1 fs7 fc1 sc0 lsd ws6">ANÁLISE DE SINAIS<span class="blank _26"> </span><span class="ff2 ws58">www.concursopetrobraseng.com.br </span>7</div><div class="t m1 xc hc ya5 ff3 fs5 fc1 sc0 lsd ws6">Questão 5<span class="blank _12"> </span><span class="ff1 fs7 v1">(Eng.<span class="blank _11"> </span>de Equipamentos Jr Eletrônica - P<span class="blank _3"></span>etrobras 2010/2)</span></div><div class="c x41 y111 w8 h63"><div class="t m1 x51 h58 y112 ff20 fs28 fc3 sc0 lsd ws6">ENGENHEIRO(A) DE EQUIP<span class="blank _3"></span>AMENTOS JÚNIOR</div><div class="t m1 x51 h58 y113 ff20 fs28 fc3 sc0 lsd">ELETRÔNICA</div><div class="t m1 x4c h59 y114 ff21 fs29 fc3 sc0 lsd">6</div><div class="t m1 x66 h59 y115 ff21 fs29 fc3 sc0 lsd ws6">CONHECIMENTOS ESPECÍFICOS</div><div class="t m1 x50 h59 y116 ff21 fs29 fc3 sc0 lsd ws6">BLOCO 1</div><div class="t m1 x30 h5a y117 ff21 fs2a fc3 sc0 lsd">21</div><div class="t m1 x30 h5b y118 ff22 fs2b fc3 sc0 ls95 ws9d">A<span class="blank _3"></span> figura acima mostra dois sinais, na forma de pulsos limitados no tempo. Considere que a transformada de Fourier de </div><div class="t m1 x30 h64 y119 ff22 fs2b fc3 sc0 ls95 ws9e">v(t) é dada pela expressão, na forma polar<span class="blank _3"></span>, V(<span class="ff23 ls96">ω</span><span class="lsd ws90">)=|V<span class="ff23 ws91">(ω</span>)|e<span class="fs2c ls97 ws6 v0"> <span class="fs2d ls98 v1c">j<span class="ff23 lsd ws92">φ<span class="ff22 ls99">(</span>ω<span class="ff22 ls9a">)</span></span></span></span><span class="ws9e v0">. Com base nas propriedades da transformada de Fourier e </span></span></div><div class="t m1 x30 h5b y11a ff22 fs2b fc3 sc0 lsb8 ws9f">considerando as semelhanças e <span class="lsb9 ws6">simetrias entre os dois pulsos, a expressão da transformada <span class="lsd">de w(t) é</span></span></div><div class="t m1 x30 h5b y11b ff22 fs2b fc3 sc0 lsd ws6">(A) <span class="blank _34"></span>W(<span class="ff23 ls96">ω</span>) = 2|V(<span class="ff23 ls96">ω</span><span class="wsa0">)| (B) <span class="blank _35"></span>W(<span class="ff23 ls9b">ω</span><span class="ws6">) = 2|V(<span class="ff23 ls9c">ω</span><span class="ws90">)|cos[<span class="ff23 ls9d">φ</span>(<span class="ff23 ls9c">ω</span>)]</span></span></span></div><div class="t m1 x30 h5b y11c ff22 fs2b fc3 sc0 lsd wsa1">(C) W(<span class="ff23 ls96">ω</span><span class="ws6">) = j2|V(<span class="ff23 ls96">ω</span><span class="ws90">)|sen[<span class="ff23 ws91">φ</span>(<span class="ff23 ws91">ω</span><span class="wsa2">)] (D) <span class="blank _36"></span>W(<span class="ff23 ls9b">ω</span><span class="ws6">) = j2|V(<span class="ff23 ls9e">ω</span><span class="ws90">)|cos[<span class="ff23 ws91">φ</span><span class="ls9d">(</span><span class="ff23 ws91">ω</span>)]</span></span></span></span></span></div><div class="t m1 x30 h5b y11d ff22 fs2b fc3 sc0 lsd wsa3">(E) W(<span class="ff23 ls96">ω</span><span class="ws6">) = 2|V(<span class="ff23 ls96">ω</span><span class="ws90">)|sen[<span class="ff23 ls96">φ</span>(<span class="ff23 ls9f">ω</span>)]</span></span></div><div class="t m1 x30 h5a y11e ff21 fs2a fc3 sc0 lsd">22</div><div class="t m1 x30 h65 y11f ff22 fs2b fc3 sc0 lsd wsa4">Um sistema discreto de 2<span class="fs2d lsa0 v1c">a</span><span class="wsa5"> ordem é composto por dois polos complexos, conjugados, que estão representados no diagra-</span></div><div class="t m1 x30 h5b y120 ff22 fs2b fc3 sc0 lsd wsa6">ma de polos e zeros da figura acima. O círculo unitário está traçado com linha pontilhada. <span class="blank _3"></span>A<span class="blank _3"></span> resposta ao impulso desse </div><div class="t m1 x30 h5b y121 ff22 fs2b fc3 sc0 lsd wsa7">sistema gera um sinal, discreto, senoidal amortecido e que oscila na frequência de 25<span class="ff23 ws6">π </span>rad/s. Nessas condições, o período </div><div class="t m1 x30 h5b y122 ff22 fs2b fc3 sc0 lsd ws6">de amostragem, em ms, usado na discretização desse sistema, é</div><div class="t m1 x30 h5b y123 ff22 fs2b fc3 sc0 lsd ws6">(A) <span class="blank _6"> </span> 5,0</div><div class="t m1 x30 h5b y124 ff22 fs2b fc3 sc0 lsd wsa3">(B) 10,0</div><div class="t m1 x30 h5b y125 ff22 fs2b fc3 sc0 lsd wsa1">(C) 12,0</div><div class="t m1 x30 h5b y126 ff22 fs2b fc3 sc0 lsd wsa1">(D) 15,5</div><div class="t m1 x30 h5b y127 ff22 fs2b fc3 sc0 lsd wsa3">(E) 20,2</div></div><div class="t m1 x3d hc y128 ff3 fs5 fc1 sc0 lsd">Resolução:</div><div class="t m1 x3d h6 y129 ff1 fs5 fc1 sc0 lsd ws93">P<span class="blank _3"></span>ara resolv<span class="blank _3"></span>er esta questão<span class="blank _0"></span>,<span class="blank _1"> </span>primeiro vamos relembrar a <span class="ff3 ws94">pr<span class="blank _3"></span>opriedade de</span></div><div class="t m1 x41 h6 y12a ff3 fs5 fc1 sc0 lsd ws95">escalonamento <span class="ff1 ws6">de um sinal:</span></div><div class="t m1 x67 h40 y12b fff fs5 fc1 sc0 ls5a">x<span class="ff10 lsd ws3c">(</span><span class="lsd ws3e">at<span class="ff10 ls5b">)<span class="ff15 lsa1">⇔</span><span class="lsd vf">1</span></span></span></div><div class="t m1 x57 h61 y12c ff15 fs5 fc1 sc0 lsd ws5f">|<span class="fff lsa2">a</span><span class="lsa3">|<span class="fff lsa4 vf">X<span class="ff10 lsa5">(</span></span></span><span class="fff vd">ω</span></div><div class="t m1 x61 h41 y12c fff fs5 fc1 sc0 lsa6">a<span class="ff10 lsd vf">)</span></div><div class="t m1 x41 h6 y12d ff1 fs5 fc1 sc0 lsd ws68">Se <span class="fff lsa7">a<span class="ff10 lsa8">=</span></span><span class="ff15 ws5f">−<span class="ff10 lsa9">1</span></span>temos:</div><div class="t m1 x68 h42 y12e fff fs5 fc1 sc0 ls5a">x<span class="ff10 ls22">(<span class="ff15 lsd ws5f">−</span></span><span class="lsd ws3e">t<span class="ff10 ls5b">)<span class="ff15 ls5c">⇔</span></span><span class="lsaa">X</span><span class="ff10 ws3c">(<span class="ff15 ws5f">−</span></span><span class="ls1e">ω</span><span class="ff10">)</span></span></div><div class="t m1 x3d h6 y12f ff1 fs5 fc1 sc0 lsd ws96">Baseando-nos nos gráficos<span class="blank _0"></span>, podemos expressar o sinal <span class="fff lsab">w</span><span class="ff10 ws3c">(<span class="fff ws3e">t</span><span class="lsac">)</span></span>em função do</div><div class="t m1 x41 h6 y130 ff1 fs5 fc1 sc0 lsd ws97">sinal <span class="fff lsad">v<span class="ff10 ls22">(</span><span class="lsd ws3e">t<span class="ff10 ws3c">)</span></span></span>:</div><div class="t m1 x2f h42 y131 fff fs5 fc1 sc0 lsab">w<span class="ff10 lsd ws3c">(</span><span class="lsd ws3e">t<span class="ff10 ws7d">) = </span><span class="lsad">v</span><span class="ff10 ws3c">(</span>t<span class="ff10 ls91">)<span class="ff15 ls6a">−</span></span><span class="lsad">v</span><span class="ff10 ws3c">(<span class="ff15 ls65">−</span></span>t<span class="ff10">)</span></span></div><div class="t m1 x3d h6 y132 ff1 fs5 fc1 sc0 lsd ws98">Agora,<span class="blank _37"> </span>aplicando a T<span class="blank _34"></span>ransf<span class="blank _3"></span>or<span class="blank _5"> </span>mada de Fourier<span class="blank _4"></span>,<span class="blank _38"> </span>utilizando a f<span class="blank _3"></span>or<span class="blank _5"> </span>ma polar</div><div class="t m1 x41 h6 y133 ff1 fs5 fc1 sc0 lsd ws6">mostrada no enunciado e lembr<span class="blank _3"></span>ando da propr<span class="blank _5"> </span>iedade, temos:</div><div class="t m1 x66 h42 y134 fff fs5 fc1 sc0 lsae">W<span class="ff10 lsd ws3c">(</span><span class="lsd ws3e">t<span class="ff10 ws7d">) = </span><span class="lsaf">V<span class="ff10 ls22">(</span><span class="ls53">ω<span class="ff10 ls91">)<span class="ff15 ls54">−</span></span><span class="lsb0">V</span></span></span><span class="ff10 ws3c">(<span class="ff15 ls65">−</span></span><span class="ls53">ω</span><span class="ff10">)</span></span></div><div class="t m1 x66 h3d y135 fff fs5 fc1 sc0 lsae">W<span class="ff10 lsd ws3c">(</span><span class="lsd ws3e">t<span class="ff10 ws7d">) = <span class="ff15 ws5f">|</span></span><span class="lsaf">V<span class="ff10 ls22">(</span><span class="ls53">ω</span></span><span class="ff10 ws3c">)<span class="ff15 ws5f">|</span></span><span class="ls23">e</span><span class="ff14 fs23 ws99 v9">j φ<span class="ff12 ws59">(<span class="ff14 lsb1">ω</span><span class="lsb2">)</span></span></span><span class="ff15 ws9a">− |</span><span class="lsb0">V</span><span class="ff10 ws3c">(<span class="ff15 ls65">−</span></span><span class="ls53">ω</span><span class="ff10 ws3c">)<span class="ff15 ws5f">|</span></span><span class="ls23">e</span><span class="ff13 fs23 ws3f v9">−<span class="ff14 ws99">j φ<span class="ff12 ws59">(</span><span class="lsb3">ω</span><span class="ff12">)</span></span></span></span></div><div class="t m1 x66 h66 y136 fff fs5 fc1 sc0 lsae">W<span class="ff10 lsd ws3c">(</span><span class="lsd ws3e">t<span class="ff10 ws7d">) = <span class="ff15 ws5f">|</span></span><span class="lsaf">V<span class="ff10 ls22">(</span><span class="ls53">ω</span></span><span class="ff10 ws3c">)<span class="ff15 lsb4">|</span><span class="ff11 ws62 v1d"></span></span>e<span class="ff14 fs23 ws99 v9">j φ<span class="ff12 ls4d">(<span class="ff14 lsb1">ω</span><span class="lsb2">)</span></span></span><span class="ff15 ls54">−</span><span class="ls23">e</span><span class="ff13 fs23 ws3f v9">−<span class="ff14 ws99">j φ<span class="ff12 ws59">(</span><span class="lsb3">ω<span class="ff12 lsb5">)</span></span></span></span><span class="ff11 v1d"></span></span></div><div class="t m1 x66 h42 y137 fff fs5 fc1 sc0 lsae">W<span class="ff10 lsd ws3c">(</span><span class="lsd ws3e">t<span class="ff10 ws7d">) = <span class="ff15 ws5f">|</span></span><span class="lsaf">V<span class="ff10 ls22">(</span><span class="ls53">ω</span></span><span class="ff10 ws3c">)<span class="ff15 lsb4">|</span>[(</span>cos<span class="ff10 ls73">[</span>φ<span class="ff10 ws3c">(</span><span class="ls53">ω</span><span class="ff10 ws77">)] + </span><span class="ws60">j sen<span class="ff10 ws3c">[</span><span class="lsb6">φ</span><span class="ff10 ws3c">(</span><span class="ls53">ω</span><span class="ff10 ws9b">)]) <span class="ff15 ls54">−</span><span class="ws3c">(</span></span></span>cos<span class="ff10 ls73">[</span>φ<span class="ff10 ws3c">(</span><span class="ls53">ω</span><span class="ff10 ws9c">)] <span class="ff15 ls54">−</span></span><span class="ws60">j sen<span class="ff10 ws3c">[</span></span>φ<span class="ff10 ls22">(</span><span class="ls53">ω</span><span class="ff10">)])]</span></span></div><div class="t m1 x66 h42 y138 fff fs5 fc1 sc0 lsae">W<span class="ff10 lsd ws3c">(</span><span class="lsd ws3e">t<span class="ff10 ws7d">) = </span><span class="lsb7">j</span><span class="ff10 ws3c">2<span class="ff15 ws5f">|</span></span><span class="lsb0">V</span><span class="ff10 ws3c">(</span><span class="ls53">ω<span class="ff10 ls22">)</span></span><span class="ff15 ws5f">|</span>sen<span class="ff10 ws3c">[</span>φ<span class="ff10 ls22">(</span><span class="ls53">ω</span><span class="ff10">)]</span></span></div><div class="t m1 x47 h43 y139 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x47 h43 y13a ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 y139 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 y13a ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x49 hc y13b ff3 fs5 fc1 sc0 lsd ws6">Alternativa (C)</div><div class="t m1 x0 h44 ya2 ff1 fs23 fc1 sc0 lsd ws6">Material de uso exclusiv<span class="blank _3"></span>o do Comprador Cód.<span class="blank _a"> </span>T34TRJ59YNKS<span class="blank _3"></span>. Sendo vedada, por quaisquer meios e a qualquer título<span class="blank _0"></span>, a sua</div><div class="t m1 x0 h44 ya3 ff1 fs23 fc1 sc0 lsd ws6">reprodução<span class="blank _3"></span>, cópia, divulgação e distr<span class="blank _5"> </span>ibuição<span class="blank _3"></span>.<span class="blank _a"> </span>Sujeitando-se o infrator à responsabilização civil e criminal.</div><a class="l"><div class="d m2" style="border-width:1.000000px;border-style:solid;border-color:rgb(0,255,255);position:absolute;left:229.264000px;bottom:779.309000px;width:163.094000px;height:12.948000px;background-color:rgba(255,255,255,0.000001);"></div></a></div><div class="pi" data-data="{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}"></div></div> <div id="pfa" class="pf w0 h0" data-page-no="a"><div class="pc pca w0 h0"><img fetchpriority="low" loading="lazy" class="bi x0 y13c w2 h67" alt src="https://files.passeidireto.com/49cd2de9-9ce3-46e1-a6f7-51bd167dd6b3/bga.png" alt="Pré-visualização de imagem de arquivo"><div class="t m3 xb hb y1e ff1 fs9 fc2 sc0 lsd ws19">T34TRJ59YNKS T34TRJ59YNKS T34TRJ59YNKS</div><div class="t m1 x0 h9 y90 ff1 fs7 fc1 sc0 lsd ws6">ANÁLISE DE SINAIS<span class="blank _26"> </span><span class="ff2 ws58">www.concursopetrobraseng.com.br </span>8</div><div class="t m1 xc hc ya5 ff3 fs5 fc1 sc0 lsd ws6">Questão 6<span class="blank _12"> </span><span class="ff1 fs7 v1">(Eng.<span class="blank _11"> </span>de Equipamentos Jr Eletrônica - P<span class="blank _3"></span>etrobras 2010/2)</span></div><div class="c x69 y13d w9 h68"><div class="t m1 x6a h69 y13e ff24 fs2e fc3 sc0 lsd ws6">ENGENHEIRO(A) DE EQUIP<span class="blank _4"></span>AMENTOS JÚNIOR</div><div class="t m1 x6a h69 y13f ff24 fs2e fc3 sc0 lsd">ELETRÔNICA</div><div class="t m1 x6b h6a y140 ff25 fs22 fc3 sc0 lsd">10</div><div class="t m1 x6c h6b y141 ff25 fs2f fc3 sc0 lsd">33</div><div class="t m1 x6d h47 y142 ff26 fs24 fc3 sc0 lsba wsb9">A<span class="blank _3"></span> resposta de um sistema linear à aplicação de um im-</div><div class="t m1 x6d h47 y143 ff26 fs24 fc3 sc0 lsba ws6">pulso <span class="blank _39"> </span><span class="ff27 lsbb">δ</span><span class="wsba">(t) (delta de Dirac) é dada por h(t) = <span class="blank _3"></span>A<span class="ff27 lsbc">δ</span><span class="ws6">(t <span class="blank _39"> </span><span class="ff27 lsbd">−</span><span class="lsdb"> t</span></span></span></div><div class="t m1 x9 h6c y144 ff26 fs27 fc3 sc0 lsbe">0<span class="fs24 lsba ws6 v7">), </span></div><div class="t m1 x6d h47 y145 ff26 fs24 fc3 sc0 lsba wsbb">onde <span class="blank _3"></span>A<span class="blank _3"></span> e t<span class="fs27 lsbf v5">0</span> são constantes positivas. <span class="blank _3"></span>Admitindo-se que </div><div class="t m1 x6d h47 y146 ff26 fs24 fc3 sc0 lsba wsbc">este sistema tenha como entrada um sinal senoidal defi-</div><div class="t m1 x6d h6d y147 ff26 fs24 fc3 sc0 lsba wsbd">nido por <span class="lsd">x(t) = B cos(2<span class="blank _39"> </span><span class="ff27 lsc0">π<span class="ff28 lsc1">f</span></span><span class="fs27 lsc2 v5">0</span><span class="v0">t<span class="blank _39"> </span>), o espectro do sinal de saída, </span></span></div><div class="t m1 x6d h47 y148 ff26 fs24 fc3 sc0 lsd ws6">correspondente a essa entrada, é dado pela expressão</div><div class="t m1 x5c h6e y149 ff26 fs24 fc3 sc0 lsd ws6">(A) <span class="blank _3a"> </span><span class="wsa8 v0">(e<span class="fs27 wsa9 v7">j2</span><span class="ff27 fs28 wsaa v7">π</span></span><span class="ff28 fs27 v7">f <span class="ff26 wsa9">t<span class="fs30 lsc3 v1e">0</span></span></span><span class="v0"> + e</span></div><div class="t md x25 h6f y14a ff26 fs27 fc3 sc0 lsd">-</div><div class="t m1 x6e h70 y14a ff26 fs27 fc3 sc0 lsd wsa9">j2<span class="ff27 fs28 wsaa v0">π</span><span class="ff28 ws6">f </span>t<span class="fs30 wsab v1e">0</span><span class="fs24 v5">)</span></div><div class="t m1 x5c h6d y14b ff26 fs24 fc3 sc0 lsd ws70">(B) ABcos(2<span class="ff27 ws72">π<span class="ff28 ws6">f <span class="blank _3"></span><span class="ff26 wsa8">t<span class="fs27 lsc4 v5">0</span>)</span></span></span></div><div class="t m1 x5c h6d y14c ff26 fs24 fc3 sc0 lsd ws6">(C) <span class="blank _3b"> </span><span class="ff27 ws72">δ(</span><span class="ff28">f <span class="ff27 ws72">)</span></span><span class="wsa8">cos(2<span class="ff27 ws72">π</span></span><span class="ff28">f <span class="blank _3"></span><span class="ff26 wsa8">t<span class="fs27 lsc5 v5">0</span>)</span></span></div><div class="t m1 x6f h71 y14d ff26 fs24 fc3 sc0 lsd ws6">(D) <span class="blank _3b"> </span><span class="wsa8 v0">[<span class="ff27 ws72">δ(<span class="ff28 ws6">f </span></span></span></div><div class="t md x70 h47 y14e ff26 fs24 fc3 sc0 lsd">-</div><div class="t m1 x2a h6d y14e ff28 fs24 fc3 sc0 lsd ws72">f<span class="ff26 fs27 lsc6 v5">0</span><span class="ff27 ws6">) + δ(<span class="ff28">f <span class="ff26 wsa8">+</span></span></span>f<span class="ff26 fs27 lsc7 v5">0</span><span class="ff27">)]<span class="ff26">e</span></span></div><div class="t md x71 h6f y14f ff26 fs27 fc3 sc0 lsd">-</div><div class="t m1 x72 h70 y14f ff26 fs27 fc3 sc0 lsd wsa9">j2<span class="ff27 fs28 wsaa v0">π</span><span class="ff28 lsdc ws6">f </span>t<span class="fs30 v1e">0</span></div><div class="t m1 x6f h6d y150 ff26 fs24 fc3 sc0 lsd ws6">(E) <span class="blank _3a"> </span><span class="ff27 ws72">δ(</span><span class="ff28">f </span></div><div class="t md x73 h47 y150 ff26 fs24 fc3 sc0 lsd">-</div><div class="t m1 x74 h6d y150 ff28 fs24 fc3 sc0 lsd ws72">f<span class="ff26 fs27 lsc8 v5">0</span><span class="ff27">)<span class="ff26 wsa8">cos(2</span>π</span><span class="ws6">f <span class="blank _3"></span><span class="ff26 wsa8">t<span class="fs27 wsa9 v5">0</span>)</span></span></div><div class="t m1 x6d h6b y151 ff25 fs2f fc3 sc0 lsd">34</div><div class="t m1 x6d h47 y152 ff26 fs24 fc3 sc0 lsd wsbe">Deseja-se transmitir<span class="blank _3"></span>, digitalmente, um sinal de vídeo cujo </div><div class="t m1 x6d h47 y153 ff26 fs24 fc3 sc0 lsd wsbf">espectro é limitado à faixa de 0 a 4 MHz. Na conversão </div><div class="t m1 x6d h47 y154 ff26 fs24 fc3 sc0 lsd wsc0">A/D desse sinal, utilizam-se um amostrador que opera na </div><div class="t m1 x6d h47 y155 ff26 fs24 fc3 sc0 lsd wsc1">taxa de Nyquist e um codificador que gera na saída, para </div><div class="t m1 x6d h47 y156 ff26 fs24 fc3 sc0 lsd wsc2">cada amostra na sua entrada, uma palavra binária de </div><div class="t m1 x6d h72 y157 ff26 fs24 fc3 sc0 lsd wsc3">comprimento fixo igual a 12 <span class="ff29 wsa8">bits</span>. Para que a interferência </div><div class="t m1 x6d h47 y158 ff26 fs24 fc3 sc0 lsd wsc4">entre símbolos (IES) no receptor seja desprezível, admite-</div><div class="t m1 x6d h47 y159 ff26 fs24 fc3 sc0 lsd wsc5">-se que a largura de banda do canal deve ser<span class="blank _3"></span>, no mínimo, </div><div class="t m1 x6d h47 y15a ff26 fs24 fc3 sc0 lsd wsc6">igual a <span class="blank _3c"> </span>, onde T<span class="fs27 wsa9 v5">s</span> é o intervalo de sinalização na saída </div><div class="t m1 x6d h47 y15b ff26 fs24 fc3 sc0 lsd wsc7">do modulador<span class="blank _3"></span>, ou, em outras palavras, o intervalo entre </div><div class="t m1 x6d h47 y15c ff26 fs24 fc3 sc0 lsd ws6">símbolos (sinais) gerados pelo modulador<span class="blank _3"></span>.</div><div class="t m1 x6d h47 y15d ff26 fs24 fc3 sc0 lsdd wsc8">Dispondo-se de um canal com largura de banda de 25 MHz, </div><div class="t m1 x6d h47 y15e ff26 fs24 fc3 sc0 lsd wsc9">o método de modulação que atende à condição para que </div><div class="t m1 x6d h47 y15f ff26 fs24 fc3 sc0 lsd ws6">a IES seja desprezível é o </div><div class="t m1 x6d h47 y160 ff26 fs24 fc3 sc0 lsd ws6">(A) BPSK</div><div class="t m1 x6d h47 y161 ff26 fs24 fc3 sc0 lsd ws6">(B) QPSK</div><div class="t m1 x6d h47 y162 ff26 fs24 fc3 sc0 lsd ws6">(C) FSK-2</div><div class="t m1 x6d h47 y163 ff26 fs24 fc3 sc0 lsd ws6">(D) PSK-8</div><div class="t m1 x6d h47 y164 ff26 fs24 fc3 sc0 lsd ws6">(E) QAM-16</div><div class="t m1 x75 h6b y165 ff25 fs2f fc3 sc0 lsd">35</div><div class="t m1 x75 h47 y166 ff26 fs24 fc3 sc0 lsd wsca">Considere um sistema de segunda ordem com a seguinte </div><div class="t m1 x75 h47 y167 ff26 fs24 fc3 sc0 lsd ws6">função de transferência: </div><div class="t m1 x76 h47 y168 ff26 fs24 fc3 sc0 lsd ws6"> </div><div class="t m1 x75 h47 y169 ff26 fs24 fc3 sc0 lsd wscb">A<span class="blank _3"></span> partir da análise de estabilidade e de desempenho, </div><div class="t m1 x75 h47 y16a ff26 fs24 fc3 sc0 lsd ws6">afirma-se que G(s) é</div><div class="t m1 x75 h47 y16b ff26 fs24 fc3 sc0 lsd wscc">(A) estável, com a frequência natural amortecida igual a </div><div class="t m1 x45 h47 y16c ff26 fs24 fc3 sc0 lsd ws6">6, e o sistema é subamortecido.</div><div class="t m1 x75 h47 y16d ff26 fs24 fc3 sc0 lsd wscd">(B) <span class="blank _30"> </span>estável, com o coeficiente de amortecimento igual a 1, </div><div class="t m1 x45 h47 y16e ff26 fs24 fc3 sc0 lsd ws6">e o sistema é criticamente amortecido. </div><div class="t m1 x75 h47 y16f ff26 fs24 fc3 sc0 lsd wscd">(C) <span class="blank _6"> </span>estável, com o coeficiente de amortecimento igual a 3, </div><div class="t m1 x45 h47 y170 ff26 fs24 fc3 sc0 lsd ws6">e o sistema é superamortecido.</div><div class="t m1 x75 h47 y171 ff26 fs24 fc3 sc0 lsd wsce">(D) <span class="blank _18"></span>instável, com a frequência natural não amortecida </div><div class="t m1 x45 h47 y172 ff26 fs24 fc3 sc0 lsd ws6">igual a 3, e o sistema é subamortecido.</div><div class="t m1 x75 h47 y173 ff26 fs24 fc3 sc0 lsd wscf">(E) <span class="blank _6"> </span>instável, com frequência natural não amortecida igual </div><div class="t m1 x45 h47 y174 ff26 fs24 fc3 sc0 lsd ws6">a 6, e o sistema é criticamente amortecido.</div><div class="t m1 x75 h6b y175 ff25 fs2f fc3 sc0 lsd">36</div><div class="t m1 x75 h47 y176 ff26 fs24 fc3 sc0 lsdd wsd0">Para análise de estabilidade em sistemas lineares, conside-</div><div class="t m1 x75 h47 y177 ff26 fs24 fc3 sc0 lsdd wsd1">re a função de transferência de um sistema em malha fecha-</div><div class="t m1 x75 h47 y178 ff26 fs24 fc3 sc0 lsdd wsd2">da, dada por <span class="blank _3d"> </span><span class="lsd"> , onde </span></div><div class="t m1 x75 h47 y179 ff26 fs24 fc3 sc0 lsd wsd3">a constante <span class="blank _3e"> </span>. Para garantir a estabilidade desse </div><div class="t m1 x75 h47 y17a ff26 fs24 fc3 sc0 lsd ws6">sistema, o intervalo de variação de k deve ser</div><div class="t m1 x75 h47 y17b ff26 fs24 fc3 sc0 lsd ws6">(A) <span class="blank _6"> </span>0 < k < 2 <span class="blank _3f"> </span>(B) <span class="blank _6"> </span>1 < k < 2</div><div class="t m1 x75 h47 y17c ff26 fs24 fc3 sc0 lsd ws71">(C) k <span class="blank _0"></span>> <span class="blank _3"></span><span class="ff27 ws72">−<span class="ff26 ws6"> 2 <span class="blank _40"> </span>(D) <span class="blank _5"> </span>k > </span>−<span class="ff26 ws6"> 1</span></span></div><div class="t m1 x75 h47 y17d ff26 fs24 fc3 sc0 lsd ws6">(E) <span class="blank _6"> </span>k > 0</div><div class="t m1 x75 h6b y17e ff25 fs2f fc3 sc0 lsd">37</div><div class="t m1 x75 h47 y17f ff26 fs24 fc3 sc0 lsd wsd4">Em um determinado processo industrial, sabe-se que a </div><div class="t m1 x75 h47 y180 ff26 fs24 fc3 sc0 lsd wsd5">temperatura de uma de suas etapas varia entre 10 ºC e </div><div class="t m1 x75 h47 y181 ff26 fs24 fc3 sc0 lsd wsd6">50 ºC. O <span class="blank _5"> </span><span class="lsc9">instrumento de medição usado para medir </span></div><div class="t m1 x75 h47 y182 ff26 fs24 fc3 sc0 lsc9 wsd7">essa temperatura possui sua faixa de medida de <span class="ff27 lsca">−</span>50 ºC </div><div class="t m1 x75 h47 y183 ff26 fs24 fc3 sc0 lsd wsd8">a 50 ºC, com uma zona morta de 1%. Diante do exposto, </div><div class="t m1 x75 h47 y184 ff26 fs24 fc3 sc0 lsd ws6">afirma-se que o instrumento de medição</div><div class="t m1 x75 h47 y185 ff26 fs24 fc3 sc0 lsd wsd9">(A) <span class="blank _6"> </span>não apresentará variações de temperaturas inferiores </div><div class="t m1 x45 h47 y186 ff26 fs24 fc3 sc0 lsd ws6">ou iguais a 0,5 ºC.</div><div class="t m1 x75 h47 y187 ff26 fs24 fc3 sc0 lsd wsd9">(B) <span class="blank _6"> </span>não apresentará variações de temperaturas inferiores </div><div class="t m1 x45 h47 y188 ff26 fs24 fc3 sc0 lsd ws6">ou iguais a 1 ºC.</div><div class="t m1 x75 h47 y189 ff26 fs24 fc3 sc0 lsde wsda">(C) não <span class="blank _3"></span><span class="lsdf wsdb">é adequado para a medição na qual é empregado, </span></div><div class="t m1 x45 h47 y18a ff26 fs24 fc3 sc0 lse0 wsdc">visto que pode apresentar distorções na medição da </div><div class="t m1 x45 h47 y18b ff26 fs24 fc3 sc0 lse0 ws6">temperatura se a mesma estiver entre 49 ºC e 50 ºC.</div><div class="t m1 x75 h47 y18c ff26 fs24 fc3 sc0 lsd wsdd">(D) <span class="blank _6"> </span>mede, embora sem confiabilidade na precisão, tempe-</div><div class="t m1 x45 h47 y18d ff26 fs24 fc3 sc0 lsd wsde">raturas variando em até 0,5 ºC além de sua faixa de </div><div class="t m1 x45 h47 y18e ff26 fs24 fc3 sc0 lsd ws6">medida nominal.</div><div class="t m1 x75 h47 y18f ff26 fs24 fc3 sc0 lsd wsdf">(E) <span class="blank _3"></span>mede, embora sem confiabilidade na precisão, tem-</div><div class="t m1 x45 h47 y190 ff26 fs24 fc3 sc0 lsd wse0">peraturas variando em até 1 ºC além de sua faixa de </div><div class="t m1 x45 h47 y191 ff26 fs24 fc3 sc0 lsd ws6">medida nominal.</div></div><div class="t m1 x3d hc y192 ff3 fs5 fc1 sc0 lsd">Resolução:</div><div class="t m1 x3d h6 y193 ff1 fs5 fc1 sc0 lsd wsac">Do enunciado sabemos que para uma entr<span class="blank _3"></span>ada<span class="blank _c"> </span><span class="fff ls82">δ<span class="ff10 ls22">(</span><span class="lsd ws3e">t<span class="ff10 lscb">)</span></span></span>a saída será dada por</div><div class="t m1 x41 h6 y194 fff fs5 fc1 sc0 lsd ws3e">h<span class="ff10 ws3c">(</span>t<span class="ff10 wsad">) = </span><span class="wsae">Aδ <span class="ff10 ws3c">(</span><span class="lscc">t<span class="ff15 lscd">−</span></span></span>t<span class="ff12 fs23 ls5d v13">0</span><span class="ff10 ws3c">)<span class="ff1 ws66">, ou seja,<span class="blank _c"> </span>o sistema multiplica a entrada por uma constante </span></span><span class="lsce">A</span><span class="ff1">e</span></div><div class="t m1 x41 h6 y195 ff1 fs5 fc1 sc0 lsd wsaf">também e<span class="blank _3"></span>xecuta uma tr<span class="blank _0"></span>anslação no tempo igual a <span class="fff ls6b">t<span class="ff12 fs23 ls59 v13">0</span></span><span class="wsb0">.<span class="blank _b"> </span>P<span class="blank _3"></span>or<span class="blank _6"> </span>tanto<span class="blank _3"></span>,<span class="blank _c"> </span>quando<span class="blank _7"> </span>tivermos</span></div><div class="t m1 x41 h6 y196 ff1 fs5 fc1 sc0 lsd wsb1">uma entrada <span class="fff ls5a">x</span><span class="ff10 ws3c">(<span class="fff ls6b">t</span><span class="wsb2">) = <span class="fff wsb3">B cos</span></span>(2<span class="fff ws6c">π f<span class="ff12 fs23 ls59 v13">0</span><span class="ls6b">t</span></span>)</span><span class="wsb4">, a saída será a multiplicação deste sinal por <span class="fff">A</span></span></div><div class="t m1 x41 h6 y197 ff1 fs5 fc1 sc0 lsd ws8f">com um deslocamento no tempo igual a <span class="fff ws3e">t<span class="ff12 fs23 ls59 v13">0</span></span>, ou seja, se chamar<span class="blank _5"> </span>mos<span class="blank _8"> </span>esta saída de</div><div class="t m1 x41 h6 y198 fff fs5 fc1 sc0 lscf">y<span class="ff10 lsd ws3c">(</span><span class="lsd ws3e">t<span class="ff10 lsd0">)</span><span class="ff1">teremos:</span></span></div><div class="t m1 x65 h6 y199 fff fs5 fc1 sc0 lsd1">y<span class="ff10 ls22">(</span><span class="lsd ws3e">t<span class="ff10 ws3d">) = </span><span class="wsb3">AB cos<span class="ff10 ws3c">[2</span><span class="ws6c">π f<span class="ff12 fs23 ls5d v13">0</span><span class="ff10 ws3c">(</span><span class="ls69">t<span class="ff15 ls54">−</span></span></span></span>t<span class="ff12 fs23 ls59 v13">0</span><span class="ff10 wsb5">)] <span class="ff1">(1)</span></span></span></div><div class="t m1 x41 h6 y19a ff1 fs5 fc1 sc0 lsd wsb6">P<span class="blank _4"></span>orém,<span class="blank _c"> </span>sabemos que um deslocamento no tempo pode ser representado na fre-</div><div class="t m1 x41 h6 y19b ff1 fs5 fc1 sc0 lsd wsb7">quência<span class="blank _11"> </span>por :</div><div class="t m1 x77 h3d y19c fff fs5 fc1 sc0 ls5a">x<span class="ff10 lsd ws3c">(</span><span class="ls69">t<span class="ff15 ls54">−</span><span class="ls6b">t<span class="ff12 fs23 ls59 v13">0</span><span class="ff10 ls5b">)<span class="ff15 ls5c">⇔</span></span><span class="lsa4">X<span class="ff10 lsd ws3c">(</span><span class="ls1e">ω<span class="ff10 lsd ws3c">)</span><span class="lsd ws3e">e<span class="ff13 fs23 ls20 v9">−<span class="ff14 lsd ws40">j ω t<span class="ff2a fs31 lsd2 v1f">0</span></span></span><span class="ff10 lsa8">=</span></span></span>X<span class="ff10 lsd ws3c">(</span><span class="ls1e">ω<span class="ff10 lsd ws3c">)</span><span class="lsd ws3e">e<span class="ff13 fs23 ls20 v9">−<span class="ff14 lsd3">j<span class="ff12 lsd ws59">2<span class="ff14 ws5b">π f<span class="blank _30"> </span>t<span class="ff2a fs31 v1f">0</span></span></span></span></span></span></span></span></span></span></div><div class="t m1 x3d h6 y19d ff1 fs5 fc1 sc0 lsd ws6">E também sabemos que a T<span class="blank _34"></span>ransf<span class="blank _3"></span>or<span class="blank _5"> </span>mada de Fourier de um cosseno é:</div><div class="t m1 x78 h42 y19e fff fs5 fc1 sc0 lsd ws3e">cos<span class="ff10 ws3c">(</span><span class="ls5f">ω<span class="ff12 fs23 ls59 v13">0</span></span>t<span class="ff10 ls5b">)<span class="ff15 ls5c">⇔</span></span><span class="lsd4">π<span class="ff10 ls73">[</span><span class="ls82">δ</span></span><span class="ff10 ws3c">(</span><span class="lsd5">ω<span class="ff15 ls54">−</span><span class="ls5f">ω<span class="ff12 fs23 ls59 v13">0</span></span></span><span class="ff10 ws77">) + </span><span class="ls82">δ</span><span class="ff10 ws3c">(</span><span class="lsd5">ω<span class="ff10 ls50">+</span></span>ω<span class="ff12 fs23 ls59 v13">0</span><span class="ff10">)]</span></div><div class="t m1 x41 h6 y19f ff1 fs5 fc1 sc0 lsd ws6">Ou ainda, em função de <span class="fff ls21">f</span>:</div><div class="t m1 x36 h40 y1a0 fff fs5 fc1 sc0 lsd ws3e">cos<span class="ff10 ws3c">(2</span><span class="ws6c">π f<span class="ff12 fs23 ls59 v13">0</span></span>t<span class="ff10 ls5b">)<span class="ff15 lsd6">⇔</span><span class="lsd vf">1</span></span></div><div class="t m1 x79 h41 y1a1 ff10 fs5 fc1 sc0 lsd7">2<span class="ls73 vf">[<span class="fff ls82">δ<span class="ff10 lsd ws3c">(</span><span class="lsd8">f<span class="ff15 ls6a">−</span><span class="ls61">f<span class="ff12 fs23 ls59 v13">0</span><span class="ff10 lsd ws77">) + </span></span></span>δ<span class="ff10 ls22">(</span><span class="lsd8">f<span class="ff10 ls6f">+</span><span class="ls61">f<span class="ff12 fs23 ls59 v13">0</span><span class="ff10 lsd">)]</span></span></span></span></span></div><div class="t m1 x3d h6 y1a2 ff1 fs5 fc1 sc0 lsd wsb8">Agora podemos finalmente aplicar a T<span class="blank _2"></span>ransf<span class="blank _0"></span>or<span class="blank _5"> </span>mada de F<span class="blank _0"></span>ourier na equação 1:</div><div class="t m1 x7a h3e y1a3 fff fs5 fc1 sc0 lsd9">Y<span class="ff10 ls22">(</span><span class="ls53">ω<span class="ff10 lsd ws5e">) =<span class="blank _c"> </span><span class="fff vf">AB</span></span></span></div><div class="t m1 x5d h73 y1a4 ff10 fs5 fc1 sc0 lsda">2<span class="ls73 vf">[<span class="fff ls82">δ<span class="ff10 lsd ws3c">(</span><span class="lsd8">f<span class="ff15 ls54">−</span><span class="ls61">f<span class="ff12 fs23 ls5d v13">0</span><span class="ff10 lsd ws77">) + </span></span></span>δ<span class="ff10 lsd ws3c">(</span><span class="lsd8">f<span class="ff10 ls50">+</span><span class="ls61">f<span class="ff12 fs23 ls59 v13">0</span><span class="ff10 lsd ws3c">)]</span><span class="lsd ws3e">e<span class="ff13 fs23 ls20 v9">−<span class="ff14 lsd3">j<span class="ff12 lsd ws59">2<span class="ff14 ws5b">π f<span class="blank _30"> </span>t<span class="ff2a fs31 v1f">0</span></span></span></span></span></span></span></span></span></span></div><div class="t m1 x47 h43 y1a5 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x47 h43 y1a6 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 y1a5 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x48 h43 y1a6 ff16 fs7 fc1 sc0 lsd"></div><div class="t m1 x49 hc y1a7 ff3 fs5 fc1 sc0 lsd ws6">Alternativa (D)</div><div class="t m1 x0 h44 ya2 ff1 fs23 fc1 sc0 lsd ws6">Material de uso exclusiv<span class="blank _3"></span>o do Comprador Cód.<span class="blank _a"> </span>T34TRJ59YNKS<span class="blank _3"></span>. Sendo vedada, por quaisquer meios e a qualquer título<span class="blank _0"></span>, a sua</div><div class="t m1 x0 h44 ya3 ff1 fs23 fc1 sc0 lsd ws6">reprodução<span class="blank _3"></span>, cópia, divulgação e distr<span class="blank _5"> </span>ibuição<span class="blank _3"></span>.<span class="blank _a"> </span>Sujeitando-se o infrator à responsabilização civil e criminal.</div><a class="l"><div class="d m2" style="border-width:1.000000px;border-style:solid;border-color:rgb(0,255,255);position:absolute;left:229.264000px;bottom:779.309000px;width:163.094000px;height:12.948000px;background-color:rgba(255,255,255,0.000001);"></div></a><a class="l" data-dest-detail="[10,"XYZ",243.563,389.742,null]"><div class="d m2" style="border-width:1.000000px;border-style:solid;border-color:rgb(255,0,0);position:absolute;left:517.751000px;bottom:150.142000px;width:7.640000px;height:12.445000px;background-color:rgba(255,255,255,0.000001);"></div></a></div><div class="pi" data-data="{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}"></div></div>
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