Any other ways to program tan(x) between [-3*pi/2 to 3*pi/2])?

Hello!
I write down the code for the tan(x) between [-3*pi/2 to 3*pi/2]), but I want to reduce it, or doing it in another way. Any ideas?
x1=-(3*pi/2)+0.01:0.01:-(pi/2)-0.01;
plot(x1,tan(x1),'r','linewidth',2.5)
hold on
x2=-(pi/2)+0.01:0.01:(pi/2)-0.01;
plot(x2,tan(x2),'r','linewidth',2.5)
x3=(pi/2)+0.01:0.01:(3*pi/2)-0.01;
plot(x3,tan(x3),'r','linewidth',2.5)
hold off

 채택된 답변

Image Analyst
Image Analyst 2013년 7월 13일
x = -3*pi/2 + .1 : 0.1 : 3*pi/2 - 0.1;
y = tan(x);
plot(x, tan(x), 'r','linewidth',2);
grid on;

댓글 수: 5

Thanks but I know that! The problem is tan(x) doesn't converge to [-3*pi/2 -pi/2 pi/2 3*pi/2]...That was my reason to program it like that!
It works. I don't know what you want to do. Do you just want to display it but not have the axes go from -inf to +inf? If so, just use ylim() to set it to whatever you want. MATLAB does handle infinity so you just need to define exactly what you want to plot.
x = -3*pi/2 : 0.1 : 3*pi/2;
y = tan(x);
plot(x, y, 'r','linewidth',2);
ylim([-10, 10]);
grid on;
Now it is correct, What about if I want to remove the vertical lines??
You can cover them up with black lines using the line() function, or you can set values more than 10 or 1000 or whatever to nan:
x = -3*pi/2 : 0.01 : 3*pi/2;
y = tan(x);
maxYtoPlot = 10;
minYtoPlot = -10;
y(y > maxYtoPlot) = nan;
y(y < minYtoPlot) = nan;
plot(x, y, 'r','linewidth',2);
ylim([minYtoPlot, maxYtoPlot])
grid on;
Good JOB!

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Azzi Abdelmalek
Azzi Abdelmalek 2013년 7월 13일
x=-3*pi/2+0.01:0.01:-(pi/2)-0.01
y=tan(x)
y=[y nan y nan y]
x=linspace(x(1),pi/2-0.01,numel(y))
plot(x,y)

댓글 수: 3

I am sorry to say it is not correct because the tan(x) must pass through [-pi 0 and pi] which yours doesn't!!!
Look at x, you will see that x takes values near 0, if you want to be more closer to 0, you have to choose a very small sample time
Thanks, I need to try it to see!

댓글을 달려면 로그인하십시오.

카테고리

도움말 센터File Exchange에서 Images에 대해 자세히 알아보기

태그

Community Treasure Hunt

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

Start Hunting!

Translated by