|
|
 | | From: | chilly charly | | Subject: | Re: calculating lung volume from airflow signal | | Date: | Sat, 22 Jan 2005 20:40:59 -0600 (CST) |
|
|
 | Hi Burcu !
Still stuck with peaks ? ;)
There is nothing strange in yo=
ur integration signal : You are integrating over the whole signal range, an=
d your data present a small offset. That gives a continuously increasing re=
sult.
Remember : the integral of a constant is a straight line.
And t=
he integral of a sine function is a cosine function.
This second stateme=
nt explains why you do not obtain the lung volume : you are integrating ove=
r the whole signal, instead of integrating on only a breath duration!
I =
suppose that your signal is the air flow rate (BTW, what is the signal unit=
? ml/s, l/min ?). You are apparently recording both positive and negative =
air flow : air entering and leaving the lungs. Of course, the sum of the tw=
o operations should be near zero, since respiration produces an amount of C=
O2 (as volume) nearly equal to the consumed O2. Otherwise the lungs would c=
ollapse (out>in) or explode (in>out)!
What you have to do is to detect t=
he instants when the air flow crosses the zero value (ie going from breath =
in to breathe out, or the reverse), and integrate the signal only between t=
hese two instants. You will get result which sign will depend on the breath=
direction.
I'm sure that with Randall help's you will rapidly find the =
solution now.
CC
|
|
|
