Answer:
a) [tex]\int\limits^T_0 {(k_{T}i(t)-x(t)) } \, dt=I(w(T)-w(0))=Iw(T)[/tex]
b) The block-diagram/model that explain the equation [tex]w(T)=\frac{1}{I}\int\limits^T_0 {(k_{T}i-x(t)) } \, dt[/tex] is attached to the image
Explanation:
a) Re-write the equation. We have to assume that the motor is at rest, in this way we have to:
w(t=0)=0
Integrate the equation:
[tex]\int\limits^T_0 {(k_{T}i(t-x(t))) } \, dt=I\int\limits^T_0 {} \, dw(t)[/tex]
[tex]\int\limits^T_0 {(k_{T}i(t)-x(t)) } \, dt=I(w(T)-w(0))=Iw(T)[/tex]
b) The expression for w(T) is:
[tex]w(T)=\frac{1}{I}\int\limits^T_0 {(k_{T}i-x(t)) } \, dt[/tex]