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Dainez_Lista_4 - UDESC Universidade do Estado de Santa...

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UDESC – Universidade do Estado de Santa Catarina Centro de Ciências Tecnológicas Mestrado em Engenharia Elétrica Aluno: Paulo Sérgio Dainez Lista de Exercicio no. 4 Problema B-3-4 Obtenha X(z): Método 1: ) 2 ( ) 1 ( ) ( ) 2 )( 1 ( 3 ) ( + + + = + + + = s b s a s X s s s s X Onde: 1 ) 2 )( 1 ( ) 2 )( 3 ( 2 ) 2 )( 1 ( ) 1 )( 3 ( 2 1 - = + + + + = = + + + + = - = - = s s s s s s b s s s s a Logo: ) 2 ( 1 ) 1 ( 2 ) ( + - + = s s s X Aplicando a Transformada Z: ) 1 )( 1 ( ) 1 ( 1 ) ( ) 1 )( 1 ( 1 2 2 ) ( ) 1 ( 1 ) 1 ( 2 ) ( 1 2 1 1 1 2 1 1 1 2 1 2 1 - - - - - - - - - - - - - - - - - - - - - - + = - - + - - = - - - = z e z e z e e z X z e z e z e z e z X z e z e z X T T T T T T T T T T Método 2: ) 2 )( 1 ( 3 ) ( + + + = s s s s X ) 2 ( 1 ) 1 ( 2 ) ( + - + = s s s X Paulo Sérgio Dainez 1 Lista n 3 - ASL
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Aplicando a transforma L inversa: t t e e t x 2 2 ) ( - - - = e: < = - = - = - - - - 0 0 ,... 3 , 2 , 1 , 0 2 ) ( 2 ) ( 2 2 k k e e kT x e e kT x kT kT kT kT Aplicando agora a definição de transformada Z: = - - = - - = - - - = - - = - = = 0 2 0 0 2 0 2 ) ( ) 2 ( ) ( ) ( ) ( k k kT k k kT k k kT kT k k z e z e z X z e e z X z kT x z X ) 1 )( 1 ( ) 1 ( 1 ) ( ) 1 )( 1 ( 1 2 2 ) ( ) 1 ( 1 ) 1 ( 2 ) ( 1 2 1 1 1 2 1 1 1 2 1 2 1 - - - - - - - - - - - - - - - - - - - - - - + = - - + - - = - - - = z e z e z e e z X z e z e z e z e z X z e z e z X T T T T T T T T T T Problema B-3-5 Obtenha a X(z) atravez do método dos resíduos de: ) )( ( ) ( b s a s K s X + + = Logo: - - - - = - - - - - = - + + + + - + + + = - - - - - - ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) )( ( ) ( lim ) ( ) )( ( ) ( lim ) ( bT aT bT aT Ts b s Ts a s e z z e z z a b K z X e z z a b K e z z a b K z X e z z b s a s b s K e z z b s a s a s K z X Paulo Sérgio Dainez 2 Lista n 3 - ASL
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Problema B-3-6 Obtenha a X(z) de: 2 ) ( 1 ) 1 ( ) ( a s s e s X Ts + - = - Mais: - = - = - - s s G Z z z X s G s e s X Ts ) ( ) 1 ( ) ( ) ( ) 1 ( ) ( 1 Logo: s b a s a a s a s s G a s s s s G a s s G + + + + = + = + = ) ( ) ( ) ( ) ( 1 ) ( ) ( 1 ) ( 2 2 1 2 2 Onde: 2 0 2 2 2 2 2 2 2 1 1 ) ( 1 ) ( ) ( 1 ) ( ) ( a a s s s b a a s s a s ds d a a a s s a s a s a s a s = + = - = + + = - = + + = = - = - = Logo: s a a s a a s a s s G 2 2 2 1 ) ( 1 ) ( 1 ) ( + + - + + - = E: ) 1 ( ) ( ) ( ) ( 2 2 2 - + - - + - - = - - - z a z e z a z e z a Te s s G Z aT aT aT Paulo Sérgio Dainez 3 Lista n 3 - ASL
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Então: - + - - + - - - = - - - - ) 1 ( ) ( ) ( ) 1 ( ) ( 2 2 2 1 z a z e z a z e z a Te z z X aT aT aT - - - + - - - - - - = - - - - ) 1 ( ) ( ) ( ) 1 )( ( ) 1 ( ) 1 ( ) ( 2 2 2 z e z a e z z z e z z z aTe z z z X aT aT aT aT 2 2 2 ) ( ) ( ) 1 )( ( ) 1 ( ) ( aT aT aT aT e z z a e z z z e z z z aTe z X - - - - - - + - - - - - = 2 2 2 ) ( ) 1 ( ) 1 ( ) ( aT aT aT aT aT e z z a aTe z e aT e z e z X - - - - - - + + + - - = Problema B-3-7 Obtenha a reposta ao degrau de: ) ( ) ( 5 , 0 ) 1 ( k x k y k y = + + Onde: 0 ) 0 ( = y Aplicando a tranformada Z: ) ( ) ( 5 , 0 ) 0 ( ) ( . z X z Y zy z Y z = + - Logo: ) 5 , 0 1 ( ) ( ) ( ) 5 , 0 ( 1 ) ( ) ( ) ( ) 5 , 0 )( ( 1 1 - - + = + = = + z z z X z Y z z X z Y z X z z Y Para obter o resultado do matlab usamos o seguinte programa: num=[0 1]; den=[1 0.5]; x=[ones(1,21)]; k=0:20; y1=filter(num,den,x); Para compararmos o resultado: y2(1)=0; for i=1:20 y2(i+1)=-0.5*y2(i)+x(i); end Paulo Sérgio Dainez 4 Lista n 3 - ASL
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Por ultimo plotamos o resultado com o comando plot(k,y1, '+' ,k,y2, 'o' ) 0 2 4 6 8 10 12 14 16 18 20 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Problema B-3-8 Obtenha a reposta ao degrau de: ) ( ) ( ) 2 ( k x k y k y = + + Onde: 0 0 ) ( < = k k y K=-2 → 0 ) 2 ( ) 2 ( ) 0 ( = - + - - = x y y K=-1 → 0 ) 1 ( ) 1 ( ) 1 ( = - + - - = x y y Aplicando a tranformada Z: ) ( ) ( ) 1 ( ) 0 ( ) ( . 2 2 z X z Y zy y z z Y z = + - - Logo: ) ( ) 1 )( ( 2 z X z z Y = + Paulo Sérgio Dainez 5 Lista n 3 - ASL
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) 1 ( ) ( ) ( ) 1 ( 1 ) ( ) ( 2 2 2 - - + = + = z z z X z Y z z X z Y Para obter o resultado do matlab usamos o seguinte programa: num=[0 0 1]; den=[1 0 1]; x=[ones(1,21)]; k=0:20; y1=filter(num,den,x); Para compararmos o resultado: y2(1)=0; y2(2)=0;
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