Respuesta :
Answer:The required transfer function is ,
[tex]\frac{Kt}{La*Jm*s{2} +(Ra*Jm+Bm*La)*s+(Kt*Ke+Ra*Bm)]}[/tex]
Step-by-step explanation:
Given:
Consider a DC motor
To Find :
Transfer function of load shaft and Input voltage.
Solution:
(Refer the attachment because this website dont allows the some words and so written on paper )
[tex]Ea(t)=Ra*Ia(t)+L*dIa(t)/dt+Eb(t)[/tex]............Equation(1)
[tex]T(t)=Jm*dWm(t)/dt+B*Wm(t)[/tex]............................Equation(2)
i.e. Electrical -mechanical equations we get,
E=K*Wm(t)
T(t)=Kt*Ia(t)...................... [ t is used since function are in Time domain]
Taking Laplace transform equation (1) and (2) we get ,
For Electrical,
[tex]Ia(s)=[1/L*s+Ra][Ea(s)-Eb(s)].[/tex].......................(Frequency domain)
For mechanical ,
[tex]T(s)=[Jm*s+Bm]Wm(s)[/tex]
And we know that
[tex]T(s)=Kt*Ia(s)[/tex]
And,
[tex]Wm(s)=[1/(Jm*s+Bm)]*T(s)[/tex]
And also
[tex]Eb(s)=Ke*Wm(s)[/tex]
By definition of transfer function:
T.F.= G(s)/[(1+H(s)*G(s)]
[tex]Here is G(s)=Kt*[1/L*s+Ra]*[1/(Jm*s+Bm)][/tex]
Using G(s) value in transfer function we get as ,
Wm(s)/Ea(s)=[[tex]\frac{Kt}{La*Jm*s{2} +(Ra*Jm+Bm*La)*s+(Kt*Ke+Ra*Bm)]}[/tex]
This is required transfer function
