why wont my code show my line in the plot?

2 次查看(过去 30 天)
r=1.1;c=1.2;po=80.;qo=700.;
rqo=r*qo;rc=r*c;
t1=linspace(0.,0.2,10);
t2=linspace(0.2,0.8,10);
p1=(qo/c*((rc*exp(t1/rc)-rc)-5.0.*...
(t1*rc.*exp(t1/rc)-rc^2.0*(exp(t1/rc-1.)))+po).*exp(-t1/rc))
p2=(qo/c*((rc*exp(0.2/rc)-rc)-5.0.*...
(0.2*rc.*exp(0.2/rc)-rc^2.0*(exp(0.2/rc-1.)))+po).*exp(-t2/rc))
figure(1);
plot(t1,p1,t2,p2);
ylabel('Pressure (mmHg)'); xlabel('Time (s)');
axis([0 0.8 1 130]);

回答(2 个)

Adam Danz
Adam Danz 2020-11-17
编辑:Adam Danz 2020-11-17
The code works well if you remove the last line. Note that y-limit of your plot is 25000-50000. The last line of your code is setting the ylim to [1,130] which doesn't show any of the data.
Perhaps you're looking for
axis tight
r=1.1;
c=1.2;
po=80.;
qo=700.;
rqo=r*qo;
rc=r*c;
t1=linspace(0.,0.2,10);
t2=linspace(0.2,0.8,10);
p1=(qo/c*((rc*exp(t1/rc)-rc)-5.0.*...
(t1*rc.*exp(t1/rc)-rc^2.0*(exp(t1/rc-1.)))+po).*exp(-t1/rc));
p2=(qo/c*((rc*exp(0.2/rc)-rc)-5.0.*...
(0.2*rc.*exp(0.2/rc)-rc^2.0*(exp(0.2/rc-1.)))+po).*exp(-t2/rc));
figure(1);
plot(t1,p1,t2,p2);
ylabel('Pressure (mmHg)'); xlabel('Time (s)');
ylim(gca)
ans = 1×2
25000 50000
  2 个评论
ALEX MC QUAID
ALEX MC QUAID 2020-11-17
the pressure values are supposed to be between 1 and 130 does this mean my calculations are wrong?
Adam Danz
Adam Danz 2020-11-17
编辑:Adam Danz 2020-11-17
Yes, unless the difference can be explained by a unit-conversion.

请先登录,再进行评论。


Setsuna Yuuki.
Setsuna Yuuki. 2020-11-17
编辑:Setsuna Yuuki. 2020-11-17
you should change the last line
r=1.1;c=1.2;po=80.;qo=700.;
rqo=r*qo;rc=r*c;
t1=linspace(0.,0.2,10);
t2=linspace(0.2,0.8,10);
p1=(qo/c*((rc*exp(t1/rc)-rc)-5.0.*...
(t1*rc.*exp(t1/rc)-rc^2.0*(exp(t1/rc-1.)))+po).*exp(-t1/rc))
p2=(qo/c*((rc*exp(0.2/rc)-rc)-5.0.*...
(0.2*rc.*exp(0.2/rc)-rc^2.0*(exp(0.2/rc-1.)))+po).*exp(-t2/rc))
figure
plot(t1,p1,t2,p2);
ylabel('Pressure (mmHg)'); xlabel('Time (s)');
axis([0 0.8 1 50000]); % before: axis([0 0.8 1 130]);

类别

Help CenterFile Exchange 中查找有关 Contour Plots 的更多信息

标签

产品


版本

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by