Biomedical Engineering Reference
In-Depth Information
Thus,
q
1
¼
f
ð
t
Þ
K
ð
7
:
29
Þ
q
10
1
The solution of Eq. (7.29) consists of the natural and forced responses. The natural
response has the form
e
K
10
t
, where the root
K
10
. The forced response
takes the form of the input. For instance, if the input is a constant continuous infusion,
q
n
¼
B
is
1
1
1
K
10
:
u(t)
, the forced response is
q
1
f
¼
Thus, the complete unit step response is
q
1
u
¼
q
1
n
þ
1
K
e
K
10
t
þ
q
f
¼
B
u
ð
t
Þ
, where
B
1
is determined from the initial condition of the system.
1
1
10
u
ð
t
Þ:
1
K
e
K
10
t
If the initial condition is zero, then
q
u
¼
1
If the initial condition is not
1
10
1
K
zero, then
B
1
¼
q
1
ð
0
Þ
10
:
If the magnitude of the unit step input is z, then the response
u
ð
t
Þ
z
K
10
e
K
10
t
with zero initial conditions is
q
1
u
¼
1
, and with initial condition
q
1
(0) is
z
K
z
K
e
K
10
t
q
u
¼
10
10
q
ð
0
Þ
u
ð
t
Þ:
1
1
), the solution method begins the same as the
unit step response with zero initial conditions, and then the bolus response is
q
1
d
¼
dq
If the input is a bolus injection, d(
t
1
u
dt
¼
e
K
10
t
u
ð
t
Þ:
e
K
10
t
u
ð
t
Þ:
If the input is zd(
t
), then the response is
q
1
d
¼
z
If the input is hypodermic needle injection,
u
(
t
)
u
(
t
t
1
) with zero initial conditions, then the
u
ð
t
Þ
1
K
1
K
e
K
10
t
e
K
10
ð
t
t
1
Þ
response based on superposition is
q
1
p
¼
1
1
u
ð
t
t
Þ:
1
10
10
If the input is z(
(
)
u
(
t
t
1
)), then the response is
u
t
u
ð
t
t
u
ð
t
Þ
z
K
z
K
e
K
10
t
e
K
10
ð
t
t
1
Þ
q
p
¼
1
1
Þ:
1
1
10
10
7.5.1 Half-Life
When tracking a solute in the body, the half-life of the concentration is an important met-
ric, where the half-life is the time required to reduce the concentration by 50 percent from
maximum. Consider the case in which a bolus of zd(
) is injected into the system in
t
e
K
10
t
u
ð
t
Þ:
Figure 7.8, with solution
q
1
d
¼
z
To find the half-life,
t
, we convert to concentra-
1
2
¼
c
1
¼
q
1
d
z
V
1
e
K
10
t
u
ð
t
Þ
z
tion by letting
V
1
¼
, then set
c
1
t
2
, and solve for
t
2
. Thus, we have
2
V
1
K
10
t
1
2
z
z
V
1
¼
1
e
2
V
Taking the natural logarithm gives
¼
K
10
t
2
1
2
ln
