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Obtención de modelos matemáticos
Sistemas eléctricos
1. La caída de voltaje a través de un resistor está dada por
ER=Ri
2.La caída de voltaje a través de un inductor está dada por
3.La caída de voltaje a través de un capacitor está dada por
Sabemos que
Entonces:
Ahora veamos un ejemplo
Dada la malla eléctrica de la figura1, determinar las corrientes de las diferentes ramas , si las corrientes iniciales valen cero.
Figura 1
Solución
Aplicando la segunda ley de Kirchhoff a las mallas KLMNK y JKNPJ
Estas dos ecuaciones son el modelos matemáticos, ahora vamos a encontrar las corrientes.
Aplicando la transformada de Laplace del sistema y utilizando las condiciones iniciales I1(0)=I2(0)=0
-5i1 - {si1 - I1(0)}+ 2{si2-I2(0)} + 10i2= 0
{si1 - I1(0)} + 20i1 + 15i2 = 55/s
(s + 5)i1 - (2s + 10)i2 = 0.............Ecuación (1)
(s + 20)i1 + 15 i2 = 55/s............ Ecuación (2)
Si factorizamos la ecuación 1
(s + 5)i1- 2( s + 5)i2=0
(s + 5)(i1-2i2) = 0
I1- 2i2 = 0 entonces i1=2i2 y sustituyendo en la ecuación 2 nos queda
Si aplicamos la transformada inversa de Laplace
Entonces
Ejemplo 2
Obtener función de transferencia de la siguiente diagrama
Solución
El diagrama es como la siguiente figura y a la vez se transforma como muestra en la figura b
De la malla I
E i (s) = Z 1I + Z2I 2 .......(1)
De la malla 2
-Z2I1+Z3 I2 + Z4I2=0 .. ....(2)
De la malla 3
E o (s)= Z4 I2.................(3)
Por división de corriente sabemos que
Ahora sustituimos la ecuación (5) en la ecuación (1) y nos queda
La ecuación (6) la sustituimos en la ecuación (3) y nos queda
Para hallar la función de transferencia dividimos Eo (s)/ Ei (s)
Aquí sustituimos Z2=1/C1s, Z4=1/C2s, Z1=R1 y Z3 = R2, la función de transferencia y nos queda
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Enviado por: | Alejandro López Ham |
Idioma: | castellano |
País: | México |