How to plot the magnitude and the phase of this function f(1/z)=(0.5 + 000i)z^(−1 )+ (0.2500 − 0.4330i)z^(−2) + (−0.2500 − 0.4330i)*z^(−3)?

Question

Aisha Mohamed 2022-5-18

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1722005-how-to-plot-the-magnitude-and-the-phase-of-this-function-f-1-z-0-5-000i-z-1-0-2500-0-43

回答： David Goodmanson 2022-5-21

Thank you for every one helped me to understand how to plot comples function ,Torsten and David xplain ed different ways to plot this function which finally gave the same result.(Many thanks)

In my question, when this function is in(1/z) .

How can we plot this function

f(1/z)=(0.5 + 000i)*z^(−1 )+ (0.2500 − 0.4330i)*z^(−2) + (−0.2500 − 0.4330i)*z^(−3)?

I appreciate any help.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

David Goodmanson 2022-5-21

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1722005-how-to-plot-the-magnitude-and-the-phase-of-this-function-f-1-z-0-5-000i-z-1-0-2500-0-43#answer_968520

在 MATLAB Online 中打开

Hello Aisha,

technically this equation should read

f(z)=(0.5 + 000i)*z.^(−1 )+ (0.2500 − 0.4330i)*z.^(−2) + (−0.2500 − 0.4330i)*z.^(−3) 

i.e. f(z) on the left hand side. Also, using .^ instead of ^ is necessary to make progress.

From the previous answer that I posted, you can plot this in the same way. The only difference is that, since z.^(-1) goes infinite for z = 0, you have to pick a grid that does not include the point (0,0) such as

xx = -6.005:.01:60.05;
yy = -6.005:.01:6.005;
[x y] =meshgrid(xx,yy);
z = x+i*y; 

Of course the closer you get to the origin with a mesh point, the larger the result is. But if you plot log(abs(z)) rather than abs(z) this is not really a problem.