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Controle Direto de Torque do Motor de Indução ... - D.s.c.e. - Unicamp

Controle Direto de Torque do Motor de Indução ... - D.s.c.e. - Unicamp

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76 <strong>Controle</strong> <strong>Direto</strong> <strong>de</strong> <strong>Torque</strong> Fuzzy<br />

KI<br />

s<br />

KIT<br />

=<br />

2<br />

= KIT<br />

2<br />

=<br />

z +1<br />

z −1<br />

2−(1−z −1 )<br />

1−z −1<br />

KI KIT<br />

−<br />

1−z −1 2<br />

= − KIT<br />

2<br />

Uma vez discretizada a parte integral, tem-se:<br />

Consi<strong>de</strong>ran<strong>do</strong> que:<br />

Tem-se:<br />

+ KIT<br />

1−z −1<br />

(4.16)<br />

(4.17)<br />

(4.18)<br />

(4.19)<br />

U(s) = (KP + KI<br />

)E(s) (4.20)<br />

s<br />

Ũ(z) = (KP − KIT<br />

2<br />

+ KIT<br />

1−z −1)˜ E(z) (4.21)<br />

˜KP = KP − KIT<br />

2<br />

(4.22)<br />

˜KI = KIT (4.23)<br />

Ũ(z) = ( ˜ KP +<br />

˜KI<br />

1−z −1)˜ E(z) (4.24)<br />

(1−z −1 ) Ũ(z) = ˜ KP(1−z −1 ) ˜ E(z)+ ˜ KI ˜ E(z) (4.25)<br />

Utilizan<strong>do</strong> a inversa da transformada Z, tem-se:<br />

u(nT)−u(nT −T) = ˜ KP(e(nT)−e(nT −T))+ ˜ KIe(nT) (4.26)<br />

Dividin<strong>do</strong> pelo perío<strong>do</strong>T , tem-se:<br />

u(nT)−u(nT −T)<br />

T<br />

= ˜ KP(e(nT)−e(nT −T))<br />

T<br />

Consi<strong>de</strong>ran<strong>do</strong> que a saída incremental <strong>do</strong> controla<strong>do</strong>r é:<br />

+ ˜ KI<br />

e(nT) (4.27)<br />

T

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