Biomedical Engineering Reference
In-Depth Information
metabolism. If the level of thyroxine is too low, TSH is released by the pituitary gland,
which increases the release of thyroxine by the thyroid gland. If the level of thyroxine is
too high, then less TSH is released, which reduces the production of thyroxine by the
thyroid gland. To examine thyroid disorders, radioactive iodine (
131
) is given to the patient,
and its uptake by the thyroid gland is measured. Simply put, too much iodine uptake
is indicative of hyperthyroidism (too much thyroxine is produced) or too little uptake is
indicative of hypothyroidism (too little thyroxine is produced). Removal of
I
131
I
from the
plasma is via the urine by the kidneys or taken up by the thyroid gland.
EXAMPLE PROBLEM 7.5
Consider the removal of
131
I
for the plasma compartment shown in Figure 7.9. Assume that a
131
then moves from the plasma into the thyroid gland or
is excreted into the urine. Assume the injected quantity of
131
is injected into the plasma.
bolus of
I
I
131
I
in the plasma is
q
1
(0). Find the
response
q
1
(
t
).
Solution
Our solution is easily considered by modeling the bolus as a change in initial condition at
0,
and a zero input. Using conservation of mass, the differential equation describing the rate of
change of the quantity of
t
¼
131
I
in the plasma compartment is
q
1
¼
K
1
T
þ
K
1
U
ð
Þ
q
1
The solution is
ðÞ
e
K
1
T
þ
K
1
U
ð
Þ
t
q
1
¼
q
1
0
u
ð
t
Þ
131
Notice that the effect of the removal of
I
from the plasma is easily carried out with a single
transfer rate,
K
10
¼
K
1
T
þ
K
1
U
, with no change in the outcome. However, if we are interested in
tracking the uptake of
131
in the thyroid gland, separating the removals with two different trans-
fer rates allow us to track it.
This example is also solved by applying an input of
I
q
1
(0)d(
t
) with a zero initial condition. We
start by solving for the response to
u(t)
,
q
1
u
ð
t
Þ
, differentiating the unit step response to yield the
d(
t
) response,
).
Conservation of mass yields the following differential equation,
q
1
¼
u
ð
t
Þ
K
1
T
þ
K
1
U
q
1
d
ð
t
Þ
, and scale by
q
1
(0) to give
q
1
(
t
ð
Þ
q
1
which has the solution
u
ð
t
Þ
e
K
1
T
þ
K
1
U
ð
Þ
t
1
q
1
u
¼
K
1
T
þ
K
1
U
The impulse response is
q
1
d
¼
q
1
u
¼
e
K
1
T
þ
K
1
U
ð
Þ
t
u
ð
t
Þ
ðÞ
e
K
1
T
þ
K
1
U
ð
Þ
t
and scaled by
q
1
(0), gives the response
q
1
¼
q
1
0
u
ð
t
Þ:
Note that treating a bolus as a
change in initial conditions yields a much easier solution.
