ESCUELA SUPERIOR POLITECNICA DE CHIMBORAZO
FACULTAD DE MECANICA
ESCUELA DE ING. MECANICA
“CONTROL AUTOMATICO”
TRABAJO 2
TEMA: DIAGRAMA DE BLOQUES
INTEGRANTES: CLAUDIA SOLORZANO
JOSE LUIS ULLOA
TANIA YANCHALIQUIN
ING.: JORGE LEMA
RIOBAMBA - ECUADOR
Encontrar las ecuaciones diferenciales
NODO 1
f=M 1
d x1dt
+k1 x1+B1( d x1dt −d x3dt )+k 2( x1−x2)
NODO 2
0=M 2
d x2dt
+B2d x2dt
+k2(x2−x1)
NODO 3
fa=M3
d x3dt
+B1( d x3dt −d x1dt )+B2( d x3dt −
d x2dt )
D= ddt→operador diferencial
f=M 1D x1+k1 x1+B1D ( x₁−x₂ )+k2(x1−x2)
0=M 2D ² x₂+B2Dx₂+k2(x2−x1)
fa=M3D ² x₃+B1D ( x₃−x₁ )+B2D ( x₃−x₁ )
Encontrar las ecuaciones diferenciales
NODO 1
0=k1 x1+M 1
d ² x1dt ²
+B1d x1dt
+B2ddt (x1−x2 )+B3
d x1dt
NODO 2
f=M 2
d ² x2dt ²
+B3d x2dt
+B2ddt (x2−x1 )
D= ddt→operador diferencial
0=k1 x1+M 1D ² x ₁+B1Dx ₁+B2D (x1−x2 )+B3Dx₁
f=M 2D ² x₂+B3Dx₂+B2D (x2−x1)
1. Encontrar la función de transferencia2. De qué orden es el modelo matemático3. Encontrar x₃(t) si la señal es de tipo escalón F= 7N
Nodo 1
0=k2 X1+M 1d2 X1dt
+k3 (X1−X2 )+β2( d X1dt −d X2dt )+β1( d X1dt −
d X3dt )
D= ddt
0=k2 X1+M 1D2 X1+k3 (X1−X2 )+β2 (D X1−D X 2)+ β1 (D X1−D X3 )
D= S0=k2 X1+M 1S
2 X1+k3 (X1−X2 )+β2 (S X1−S X2 )+β1 (S X1−S X 3)
0=X1 (k2+M 1S2+k3+β2S+β1S)−X2 (k3+ β2S )−β1S X3
0=X1 (2+2S+S2 )−X2 (1+S )−X3S (1)
Nodo 2
f (s )=M 2d2 X2dt
+k 3 ( X2−X1 )+β2( d X2dt −d X1dt )+β1( d X2dt −
d X3dt )
D= ddt
F=M 2D2 X2+k3 (X2−X1 )+β2 (D X 2−D X1 )+β1 (D X2−D X3 )
D= S
F=M 2S2 X2+k3 (X 2−X1 )+β2 (S X2−S X1 )+β1 (S X2−S X3 )
F=−X1 (1+S )+X2 (2+2S+S2 )−X3S (2)
Nodo 3
0=k1 X3+M 3d2 X3dt
+β1(d X3dt −d X1dt )+β3( d X3dt −
d X2dt )
D= ddt
0=k1 X3+M 3D2 X3+β1 (D X3−D X1 )+β3 (D X3−D X2 )
D= S
0=k1 X3+M 3S2 X3+β1 (S X3−S X1 )+β3 (S X 3−S X2 )
F=X1S−X2S+(2+2 S+S2 )−X3S (3)
0=X1 (2+2S+S2 )−X2 (1+S )−X3S (1)
F=−X1 (1+S )+X2 (2+2S+S2 )−X3S (2)
F=X1S−X2S+(2+2 S+S2 )−X3S (3)
Resolviendo el sistema se tiene
X3F
= 3 s+3 s2+s3
4+18 s+32 s2+32 s3+20 s4+7 s5+s6
F (t )=172sinh (t )+ 1
2(1+((e−2t )−(2e−t )))
3 s+3 s2+s3
4+18 s+32 s2+32 s3+20 s4+7 s5+s6X3F