Blood Pressure and Sound (2)
Dept. of Biomedical Engineering
2003200449
YOUNHO HONG
IBP ( Invasive BP ) measurement
diaphragm
Blood
vessel
catheter
Pi
cable
Po
pill up with
some liquid
(ex.saline)
strain gages
Pi
Po
t
t
If we choose sticky and dense liquid,
We can’t get the signal of Pi just like the graph.
To get the signal, We should concern with
distance, diameter of a catheter and liquid, airbubble inside a catheter.
IBP ( Invasive BP ) measurement
# Equivalent Circuit Model Of Catheter-Sensor System
(1) Resistance
L
A

Electrical resistance : V  Ri , R 
V


L
A
resistivity
F  ( P1  P2)
P1
R
P2
Liquid resistance : op  R  F
R  
L
A
viscosity
IBP ( Invasive BP ) measurement
(2) Capacitance or Compliance
iC
dv
A
, C 
dt
x
dp
f C
dt
(3) Inductance or Inertance
di
V L
dt
df
PL ,
dt
m
L 2
A

C  Y  Young's modulus
IBP ( Invasive BP ) measurement
multi
physics
Electric Circuit
Fluid Mechanics
voltage
current
charge
V
L
[ ] (   )
I
A
V
L
dI
dt
I
C
dV
dt
R
pressure
flow
volume
 Pa  s
R [
F
8L
)
3 ] (
4
m
r
L
( 2 )
r
P
dF
dt
C  Young's modulus
L
Equivalent Circuit Model of IBP
catheter
Pi
Rc
Po diaphragm
cable
Vi
liquid
strain gages
Lc
i
+
Vo
-
Cd
Compliance of
diaphragm
dV0
di
Vi  Rc  i  Lc  V0
i  Cd
dt
dt
dV0
d 2V0
Vi  V0  Rc Cd
 Lc Cd
: 2nd order ODE
2
dt
dt
 d
Rc Cd
D
: operator
K  1,
 
dt
2 Lc
damping
D 2 2D
1
ratio
[ 2
 1] Vo (t )  KVi (t )
Wn 
Wn
Wn
Lc Cd
natural
frequency
Equivalent Circuit Model of IBP
(1) Frequency Transfer Function
H ( jw) 
Vo ( jw)
1
1


Vi ( jw) ( jw ) 2  2 jw  1 1  ( w ) 2  j 2 w
Wn
Wn
Wn
Wn
w
2
Wn
1
 H ( jw) 
  tan1 (
)
w
2
w 2 2
2 w 2
1

(
)
[1  ( ) ]  4 ( )
Wn
Wn
Wn
1

[1  (
w 2 2
w
) ]  4 2 ( ) 2
Wn
Wn
1
 tan (
2
w Wn

Wn w
)
Equivalent Circuit Model of IBP
  0.5 (underdamped)
|H|
  1 (criticallydamped)
  2 (overdamped)
Wn
∠H
w
w
  0.5 (underdamped)
-π/2
  1 (criticallydamped)
  2 (overdamped)
-π
Equivalent Circuit Model of IBP
d 2V0
dV
LC 2  RC 0  Vo  Vi
dt
dt
Methods to solve 2nd order ODE
(1) LCD 2Vo  RCDVo  Vo  Vi
( LCD  RCD  1)Vo  Vi
2
d 
D

dt
Vo
1
H ( D) 

Vi LCD 2  RCD  1
: operational transferfunction
(2) LC ( jw) 2 Vo  RCjwVo  Vo  Vi
H ( jw) 
Vo
1

Vi LC ( jw) 2  RC( jw)  1
 H ( D) D  jw
Steady State Freq. Response
|H|
∠H
K
-4/π
0.5
f1
f2
-1.8π
f1
f2
f
Vi (t )  A sin(2f1t )

Vo (t )  KA sin(2f1t  )
4
Vi (t )  A sin(2f 2t )
Vo (t )  0.5 A sin(2f 2t  1.8 )
Unit Step Response
In reality, We need a unit step function for a starting point.
For example, Vi (t )  A sin(2f1t ) should be Vi (t )  A sin(2f1t )  u(t )
underdamping
input signal
overdamping
critical damping
Transient Step Response
Po
balloon
P
bulb
saline
overdamping
critical damping
underdamping
Example
(7.1) A 5mm-long air bubble has formed in the rigid-walled catheter to
a Statham P23Dd sensor. The catheter is 1m long, 6 French diameter,
and filled with water at 20 ℃. Plot the frequency-response curve of the
system with and without the bubble.
r 1 P 12
fn  (
)  91Hz
2 L L
4 L V 12
  3(
)  0.033
r  P
f n,bub  22Hz
 bub  0.137
1.34
1.95
log f
Example
(7.2) By changing only the radius of the catheter, redesign the (no-bubble)
catheter of Figure 7.9 to achieve the damping ratio ζ=1.
Calculate the resulting natiral frequency fn.
r 3 r0    0
3
r 3  0.0032
r  0.147
f n f n 0  r r0
f n  29Hz
1.46
log f
Thank you.