PICCOLO AIUTINO

85 Risposte - Pagina 4

• G

  giug

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  641

  Re: PICCOLO AIUTINO

  01 feb 2012, 18:01

  Sì, le righe... ma di A, non di B.
  Diciamo che la risposta 15x15=225 sarebbe corretta se A avesse tutti gli elementi di B...
• J

  Joseph101

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  46

  Re: PICCOLO AIUTINO

  01 feb 2012, 18:22

  L'indicie "i" deve generare le righe di B ? vero? dai un aiutino finale ?
• J

  Joseph101

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  46

  Re: PICCOLO AIUTINO

  01 feb 2012, 18:30

  > for i=1:225
  B(A(i,1),A(i,2))=A(i,3)
  end
  questo non va bene ....sono quasi arrivato
• G

  giug

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  641

  Re: PICCOLO AIUTINO

  01 feb 2012, 18:31

  Dai che ci siamo.
  Non va bene nel tuo caso perché le righe di A non sono 225.
• J

  Joseph101

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  46

  Re: PICCOLO AIUTINO

  01 feb 2012, 18:46

  Forse sono sulla strada giusta , ma manca qualcosa
  for i=1:15
  B(C(i,1),C(i,2))=C(i,3)
  end
  
  B =
  
  1.0e+005 *
  
  Columns 1 through 11
  
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  B =
  
  1.0e+005 *
  
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  B =
  
  1.0e+005 *
  
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  0.2739 0 0 0 0 0 0 0 0 0 0
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  B =
  
  1.0e+005 *
  
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  0.2739 0 0 0 0 0 0 0 0 0 0
  -0.1250 0 0 0 0 0 0 0 0 0 0
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  B =
  
  1.0e+005 *
  
  Columns 1 through 11
  
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  0.2739 0 0 0 0 0 0 0 0 0 0
  -0.1250 0 0 0 0 0 0 0 0 0 0
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  B =
  
  1.0e+005 *
  
  Columns 1 through 11
  
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  0.2739 0 0 0 0 0 0 0 0 0 0
  -0.1250 0 0 0 0 0 0 0 0 0 0
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  B =
  
  1.0e+005 *
  
  Columns 1 through 11
  
  7.5336 0 0 0 0 0 0 0 0 0 0
  0.2739 0 0 0 0 0 0 0 0 0 0
  -0.1250 0 0 0 0 0 0 0 0 0 0
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  0 0 0 0
  0 0 0 0
  0 0 0 0
  
  
  B =
  
  1.0e+005 *
  
  Columns 1 through 11
  
  7.5336 0 0 0 0 0 0 0 0 0 0
  0.2739 0 0 0 0 0 0 0 0 0 0
  -0.1250 0 0 0 0 0 0 0 0 0 0
  0 0 0 0 0 0 0 0 0 0 0
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  B =
  
  1.0e+005 *
  
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  7.5336 0 0 0 0 0 0 0 0 0 0
  0.2739 0 0 0 0 0 0 0 0 0 0
  -0.1250 0 0 0 0 0 0 0 0 0 0
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  B =
  
  1.0e+005 *
  
  Columns 1 through 11
  
  7.5336 0 0 0 0 0 0 0 0 0 0
  0.2739 0 0 0 0 0 0 0 0 0 0
  -0.1250 0 0 0 0 0 0 0 0 0 0
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  B =
  
  1.0e+005 *
  
  Columns 1 through 11
  
  7.5336 0 0 0 0 0 0 0 0 0 0
  0.2739 0 0 0 0 0 0 0 0 0 0
  -0.1250 0 0 0 0 0 0 0 0 0 0
  0 0 0 0 0 0 0 0 0 0 0
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  B =
  
  1.0e+005 *
  
  Columns 1 through 11
  
  7.5336 0 0 0 0 0 0 0 0 0 0
  0.2739 0 0 0 0 0 0 0 0 0 0
  -0.1250 0 0 0 0 0 0 0 0 0 0
  0 0 0 0 0 0 0 0 0 0 0
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  B =
  
  1.0e+005 *
  
  Columns 1 through 11
  
  7.5336 0 0 0 0 0 0 0 0 0 0
  0.2739 0.9994 0 0 0 0 0 0 0 0 0
  -0.1250 0 0 0 0 0 0 0 0 0 0
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  B =
  
  1.0e+005 *
  
  Columns 1 through 11
  
  7.5336 0 0 0 0 0 0 0 0 0 0
  0.2739 0.9994 0 0 0 0 0 0 0 0 0
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  B =
  
  1.0e+005 *
  
  Columns 1 through 11
  
  7.5336 0 0 0 0 0 0 0 0 0 0
  0.2739 0.9994 0 0 0 0 0 0 0 0 0
  -0.1250 0 0 0 0 0 0 0 0 0 0
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  0 0 0 0
  La prima colonna sembra ok , ma non riempie il resto
• G

  giug

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  641

  Re: PICCOLO AIUTINO

  01 feb 2012, 19:00

  Allora, stai considerando solo le prime 15 righe di A (non hai capito a che valore deve arrivare il ciclo), quindi queste:
  1 1 753362,3173
  2 1 27394,91824
  3 1 -12498,27972
  7 1 0,00E+00
  8 1 0,00E+00
  9 1 0,00E+00
  10 1 -4166,09324
  11 1 0,00E+00
  12 1 -12498,27972
  13 1 -13697,45912
  14 1 27394,91824
  15 1 0,00E+00
  2 2 99935,4343
  3 2 0,00E+00
  7 2 0,00E+00
  
  Ora, matlab fa il ciclo e controlla una a una tutte le righe della matrice
  per i=1 dice "devo mettere nella 1° riga e 1°colonna di B il numero 753362,3173
  per i=2 dice "devo mettere nella 2° riga e 1°colonna di B il numero 27394,91824
  per i=3 dice "devo mettere nella 3° riga e 1°colonna di B il numero -12498,27972
  Per i=4 cosa dice?
• J

  Joseph101

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  Re: PICCOLO AIUTINO

  01 feb 2012, 19:20

  Ok,
  for i=1:102
  B(C(i,1),C(i,2))=C(i,3)
  end
  Mi da la matrice (almeno spero sia quella ) ...
  7.5336 0 0 0 0 0 0 0 0 0 0
  0.2739 0.9994 0 0 0 0 0 0 0 0 0
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  0.2739 0.3564 0 0 -0.0443 0 0 0.1070 -0.3210 0 0
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  0 7.5336 0 0
  0 -0.2739 0.9994 0
  0 -0.1250 0 0.7847
  
  Adesso come riempio tutta la matrice sapendo che è simmetrica rispetto alla diagonale ?
• G

  giug

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  Re: PICCOLO AIUTINO

  01 feb 2012, 19:34

  Sicuramente ci sono diversi modi... adesso mi viene in mente questo.
  Trasponi la matrice C e metti a zero gli elementi sulla diagonale, poi sommi la matrice ottenuta a C.
  Esempio:
  C =[1 0 0;
  2 3 0;
  4 5 6]
  Trasponi
  D=
  1 2 4
  0 3 5
  0 0 6
  Cancelli gli elementi sulla diagonale
  E=
  0 2 4
  0 0 5
  0 0 0
  Sommi E+C
  1 2 4
  2 3 5
  4 5 6
  
  Ti torna? La trasposta si fa usando l'apice, per mettere a zero gli elementi sulla diagonale usi la funzione diga (cercala sull'help).
• J

  Joseph101

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  Re: PICCOLO AIUTINO

  01 feb 2012, 19:43

  Ok , una volta fatta questa operazione devo raggiungere il risultato finale
  |C-omega^2*M|=0 ; adesso conosco sia C che M come ricavo omega?
  T
• G

  giug

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  Re: PICCOLO AIUTINO

  02 feb 2012, 10:33

  Svolgi a mano l'equazione in modo che venga omega=...
  e poi dai la nuova equazione in pasto a matlab che ti darà il risultato.
• J

  Joseph101

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  Re: PICCOLO AIUTINO

  02 feb 2012, 10:53

  Ciao ,
  secondo te questa procedura di ricare la matrice simmetrica potrebbe andare bene ?
  >> A=[1 0 0; 2 3 0; 4 5 6]
  
  A =
  
  1 0 0
  2 3 0
  4 5 6
  
  >> B=A'
  
  B =
  
  1 2 4
  0 3 5
  0 0 6
  
  >> C=B-diag(diag(B))
  
  C =
  
  0 2 4
  0 0 5
  0 0 0
  
  >> D=A+C
  
  D =
  
  1 2 4
  2 3 5
  4 5 6
  Oppure esiste un metodo più veloce ?
  quanto alla seconda risposta basta fare allora :
  Omega=sqrt(K/M)
• G

  giug

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  Re: PICCOLO AIUTINO

  02 feb 2012, 11:04

  Sì, a parte che c'è il modulo e quindi il risultato potrà essere positivo o negativo.
  Il metodo per trovare l'altra parte della matrice simmetrica va bene. Al momento non mi viene in mente altro di più immediato, anche se probabilmente esiste.
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  Joseph101

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  Re: PICCOLO AIUTINO

  02 feb 2012, 11:21

  Penso ci sia qualche problema, nel senso che Omega rappresenta la frequenza del sistema e per valutare essa ho bisogno di risolvere l'equazione agli autovalori
  |K-Omega^2*M|=0 , pertanto prima devo ricavare questa matrice in cui Omega è incognito , successivamente ricavare gli autovalori .
  Ho provato a fare tale differenza come K-O^2*M=0 ma mi da errore ....
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  giug

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  Re: PICCOLO AIUTINO

  02 feb 2012, 11:22

  Prova con solve('abs(K-O^2*M)')
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  Joseph101

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  Re: PICCOLO AIUTINO

  02 feb 2012, 11:39

  No... mi da questo
  >> ('abs(K-O^2*M)')
  
  ans =
  
  abs(K-O^2*M)
  
  >>