Physiological Modeling - Introduction to Biomedical Engineering - page 854

.

F ant

B ag x 2

K lt x 3

F ag

K se ( x 4 - x 3 )

K se ( x 2 - x 1 )

B ant x .

K lt x 2

.

.

F ag = B ag x 2 + K se ( x 2 - x 1 ) + K lt x 2

K se ( x 4 - x 3 ) = F ant + K lt x 3 + B ant x 3

..

rK se ( x 2 - x 1 )

J p q

.

B p q

K p q
rK se ( x 1 - x 3 )

.. .

rK se ( x 2 - x 1 ) = J q + B q + K q + rK se ( x 1 - x 3 )

FIGURE 13.31 Free body diagrams for the system in Figure 13.30.

To solve for y, we use the Laplace transform analysis (details omitted here; see Enderle,
1984, or Enderle, 2010a).

...:

3 ... 1 :

2 ‥

þ B ant F ag B ag F ant d K st F ag F ant ¼ þ C þ C þ C þ C

0 y

ð 13 : 35 Þ

where

J ¼ J p B ¼ B p K ¼ K p 10 3 , 10 3 , 10 3 ,

r

5 : 2087

r

5 : 2087

r

5 : 2087

180
p ¼

57 : 296

¼ x

K se
rJB ant B ag

10 3

y

r

5 : 2087 x , and d ¼

þ BB ant B ag
JB ant B ag

C 3 ¼ JK st B ag þ B ant

2

¼ JK

st þ BK st B ag þ B ant

þ B ag B ant K þ ð 2 K se Þ

C

2

JB ant B ag

KK st þ

2 2

se

C 1 ¼ BK

st þ B ag þ B ant

2 K se K st K

JB ant B ag

C 0 ¼ KK

2

st þ

K se K st K lt
JB ant B ag

2