Biomedical Engineering Reference
In-Depth Information
1
1
e
st
o
s
:
e
st
d
t
e
st
d
t
L
f
h
ð
t
t
o
Þg ¼
h
ð
t
t
o
Þ
¼
¼
(A.220)
0
t
o
The derivative of the unit step function is the delta function
d
(
t
) which has the
value 0 for all values of
t
except
t
¼
t
o
. The function may be viewed as the limit as
e
tends to zero of a function that has the value 1/
e
between 0 and
e
and is zero
d
(
t
) are then
everywhere else. These properties of
1
0
dð
d
h
ð
t
t
o
Þ
¼ dð
t
t
o
Þ; dð
t
t
o
Þ¼
0 for
t
6¼
t
o
;
and
t
t
o
Þ
d
t
¼
1
:
(A.221)
d
t
If the delta function is multiplied by any other function
f
(
t
), the product is zero
everywhere except at
t
¼
t
o
and it follows that
1
f
ð
t
Þdð
t
t
o
Þ
d
t
¼
f
ð
t
o
Þ:
(A.222)
0
Example A.16.3
Problem
: Use Laplace transforms to solve the ordinary differential equation
representing the standard linear solid (1.8) for
F
~
and vice
versa assuming that
F
(0
) and
x
(0
) are equal to zero. Then determine the creep
and relaxation functions from the result.
Solution: The Laplace transform of the differential equation for the standard linear
solid (1.8) is given by
ð
s
Þ
as a function of
x
ð
s
Þ
F
þ t
x
s F
ð
s
Þ
ð
s
Þ
¼ ~
x
ð
s
Þþt
F
s
x
~
ð
s
Þ:
(A.223)
k
k
The solutions of this transformed equation for
F
ð
s
Þ
, and for
x
~
ð
s
Þ
, are
F
F
ð
k
¼
ð
s
1
þ
s
t
F
Þ
Þ¼
ð
1
þ
s
t
x
Þ
ð
k
;
s
t
x
Þ
~
x
ð
s
Þ
and
~
x
ð
s
(A.224)
ð
1
þ
s
ð
1
þ
s
t
F
Þ
respectively. To obtain the creep function one sets
F
(
t
)
¼
h
(
t
) in the second of the
equations above; thus by partial fractions
Þ¼
ð
1
þ
s
t
x
Þ
1
s
¼
1
t
x
k
x
~
ð
s
t
F
Þ
þ
t
F
Þ
;
(A.225)
ð
1
þ
s
t
F
Þ
s
ð
1
þ
s
ð
1
þ
s
and to obtain the relaxation function one sets
x
(
t
)
¼
h
(
t
) in the first of the equations
above, thus
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