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Zbirka rijesenih zadataka - PBF - Sveučilište u Zagrebu

Zbirka rijesenih zadataka - PBF - Sveučilište u Zagrebu

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Ž. Kurtanjek: <strong>PBF</strong> Mjerenja i automatizacija 2007/2008 39<br />

Zadatak 15<br />

Izračunajte prijenosne funkcije (matricu) sustava sa dvije ulazne i dvije<br />

izlazne veličine (prikazanog na slici)<br />

X 1<br />

Y 1<br />

W(s)<br />

X 2 Y 2<br />

zadanog sa sustavom od dvije linearne jednadžbe:<br />

Rješenje<br />

dy<br />

2 ⋅<br />

dt<br />

1<br />

dy<br />

− 3⋅<br />

dt<br />

2<br />

+ y<br />

t<br />

1<br />

2<br />

− ⋅ + 2 ⋅ + y2<br />

+<br />

dt dt<br />

∫<br />

0<br />

1<br />

= 4 ⋅ x<br />

1<br />

1<br />

+<br />

1<br />

2<br />

⋅<br />

t<br />

∫<br />

0<br />

x<br />

1<br />

2<br />

⋅ dt<br />

dy dy<br />

3 y ⋅ dt = 2 ⋅ x + 5⋅<br />

x<br />

2<br />

Primijenimo Laplaceove transformacije za sustav jednadžbi:<br />

1 1<br />

2 ⋅ s ⋅Y1<br />

− 3⋅<br />

s ⋅Y2<br />

+ Y1<br />

= 4 ⋅ X<br />

1<br />

+ ⋅ ⋅ X<br />

2<br />

2 s<br />

1<br />

− 3⋅<br />

s ⋅Y1 + 2 ⋅ s ⋅Y2<br />

+ Y2<br />

+ ⋅Y1<br />

= 2 ⋅ X<br />

1<br />

+ 5 ⋅ X<br />

s<br />

Izlazne zavisne veličine, koje su nepoznanice, izdvojimo na lijevoj strani, a ulazne<br />

nezavisne veličine izdvojimo na desnoj strani sustava jednadžbi:<br />

( 2 ⋅ s + 1)<br />

⋅Y<br />

− 3⋅<br />

s ⋅Y<br />

⎛ 1 ⎞<br />

⎜−<br />

3⋅<br />

s + ⎟ ⋅Y<br />

⎝ s ⎠<br />

Isto napisano u matričnom obliku je:<br />

1<br />

1<br />

2<br />

+ 2 ⋅ s ⋅Y<br />

= 4 ⋅ X<br />

2<br />

1<br />

+<br />

= 2 ⋅ X<br />

1<br />

2 ⋅ s<br />

1<br />

⋅ X<br />

2<br />

+ 5⋅<br />

X<br />

2<br />

2<br />

⎛ 2 ⋅ s + 1<br />

⎜ 1<br />

⎜−<br />

3⋅<br />

s +<br />

⎝ s<br />

− 3⋅<br />

s⎞<br />

⎟ ⎛Y<br />

⋅⎜<br />

2 ⋅ s ⎟<br />

⎠ ⎝Y<br />

1<br />

2<br />

⎞<br />

⎛<br />

⎜4<br />

⎟ =<br />

⎠<br />

⎜<br />

⎝2<br />

1 ⎞<br />

⎟ ⎛ X<br />

2 ⋅ s ⋅⎜<br />

⎟<br />

5 ⎠<br />

⎝ X<br />

1<br />

2<br />

⎞<br />

⎟<br />

⎠<br />

Riješimo inverzijom matrice:<br />

⎛Y<br />

⎜<br />

⎝Y<br />

1<br />

2<br />

⎞ ⎛ 2 ⋅ s + 1<br />

⎜<br />

⎟ = 1<br />

⎠ ⎜−<br />

3⋅<br />

s +<br />

⎝ s<br />

− 3⋅<br />

s⎞<br />

⎟<br />

2 ⋅ s ⎟<br />

⎠<br />

−1<br />

⎛<br />

⎜4<br />

⋅<br />

⎜<br />

⎝2<br />

1 ⎞<br />

⎟ ⎛ X<br />

2 ⋅ s ⋅<br />

⎜<br />

⎟<br />

5 ⎠<br />

⎝ X<br />

1<br />

2<br />

⎞<br />

⎟<br />

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