# Where is the wrong in this code??

조회 수: 3 (최근 30일)
Aisha Mohamed 2022년 8월 31일
댓글: Aisha Mohamed 2022년 9월 1일
Because I trust the answers of all the Matlab experts in this group, and I thank all of them. The genius expert Torsten saw that there is some mistake in my code that used to plot cos(phase(f_b(1/z)), I used his notes but still there are some missing points in this work. I think this is because I did not write my question in the correct way.
So I ask again where is the wrong in this code??
I have two functions f_k(z) and f_b(1/z) where,
1-f_k(z) = (0.1000 + 0.3000i) + (0.4243 + 0.0017i)z + (0.9000- 0.0010i)z^2
we define p from f_k(z)as,
p=[ ( 0.9000 - 0.0010i) (0.4243 + 0.0017i) (0.1000 + 0.3000i) ];
2-f_b(1/z) = (0.1000 - 0.3000i)z^-1 + (0.2121 - 0.0008i)z^-2 + (0.9000 +0.0010i)z^-3
we define p1 from f_b(1/z)as,
p1=[ ( 0.9000 + 0.0010i) (0.2121 - 0.0008i) (0.1000 - 0.3000i) (0)];
The function f_b(1/z) is not defined at z = 0 which means the graph of this function must be undefined at z = 0
But when I use the code %f_of_1_over_z_result = polyval(p1, z); I found the graph for | f_b(1/z) | and | f_k (z) | are same
and | f_b(1/z) | defined at z = 0 (which must not happen in the myfunction |f_b(1/z)| because it is not defined at z = 0) , as you can see from the following figure:
But,
% Where as when I use the code % f_of_1_over_z_result = polyval(p1,1./z);
% the function |f_b(1/z)|does not define at z=0 (that what must happen in my case) as you can see in the following figure,
I used this code,
p=[ ( 0.9000 - 0.0010i) (0.4243 + 0.0017i) (0.1000 + 0.3000i) ];
p1=[ ( 0.9000 + 0.0010i) (0.2121 - 0.0008i) (0.1000 - 0.3000i) (0)];
re_z = -6.005:.01:6.005;
im_z= -6.005:.01:6.005;
[re_z,im_z] = meshgrid(re_z,im_z);
z = re_z + 1i*im_z;
f_of_z_result = polyval(p,z);
% %
f_of_1_over_z_result = polyval(p1,1./z); % I used this code in the first figure
% f_of_1_over_z_result = polyval(p1,z); % I used this code in the second figure
figure();
subplot(2,2,1)
surf(re_z,im_z,abs(f_of_z_result),'EdgeColor','none')
colorbar
title('|f_k(z)|')
xlabel('Z_R')
ylabel('Z_I')
zlim([0 15]) %adjust this value as needed
caxis([-6 6]) %adjust this value as needed
% grid on
subplot(2,2,2)
surf(re_z,im_z,cos(angle(f_of_z_result)),'EdgeColor','none')
colorbar
title('cos(phase of f_k(z))')
xlabel('Z_R')
ylabel('Z_I')
zlim([-5 5]) %adjust this value as needed
caxis([-1 1]) %adjust this value as needed
subplot(2,2,3)
surf(re_z,im_z,abs(f_of_1_over_z_result),'EdgeColor','none')
colorbar
title('|f_b(1/z)|')
xlabel('Z_R')
ylabel('Z_I')
zlim([0 15]) %adjust this value as needed
caxis([-0.15 0.15])
subplot(2,2,4)
surf(re_z,im_z, cos(angle(f_of_1_over_z_result)),'EdgeColor','none')
colorbar
title(('cos(phase of f_b(1/z))'))
xlabel('Z_R')
ylabel('Z_I')
caxis([-1 1]) %adjust this value as needed
zlim([-1 1]) %adjust this value as needed
grid on
Where is the wrong in this code??

댓글을 달려면 로그인하십시오.

### 채택된 답변

Torsten 2022년 8월 31일
편집: Torsten 2022년 9월 1일
f_b(1/z) = (0.1000 - 0.3000i)z^-1 + (0.2121 - 0.0008i)z^-2 + (0.9000 +0.0010i)z^-3
I wrote it several times already, but this function is the same as
f(w) = (0.1000 - 0.3000i)w + (0.2121 - 0.0008i)w^2 + (0.9000 +0.0010i)w^3
except for 1/z=w=0.
But if we define f_b(0) = 0, we remove this singularity and the two functions are identical all over the complex plane.
This can be seen in the first 4 graphics where we get a value of 0 for abs(f_b(1/z)) at 1/z=0.
So summarizing: There is nothing wrong in your code, but you must decide whether you want to define
f_b(1/z) = (0.1000 - 0.3000i)z^-1 + (0.2121 - 0.0008i)z^-2 + (0.9000 +0.0010i)z^-3
or
f_b(z) = (0.1000 - 0.3000i)z^-1 + (0.2121 - 0.0008i)z^-2 + (0.9000 +0.0010i)z^-3
The first definition gives the 2 pictures in the first gallery, the second definition gives the 2 pictures in the second gallery.
##### 댓글 수: 3이전 댓글 1개 표시이전 댓글 1개 숨기기
Torsten 2022년 9월 1일
편집: Torsten 2022년 9월 1일
In order not to confuse people, the usual way to ask your question would have been
Given
f_b(z) = (0.1000 - 0.3000i)z + (0.2121 - 0.0008i)z^2 + (0.9000 +0.0010i)z^3
and setting
g(z) = f_b(1/z)
how to plot abs(g(z)) and cos(phase(g(z))) ?
?
p1=[ ( 0.9000 + 0.0010i) (0.2121 - 0.0008i) (0.1000 - 0.3000i) (0)];
re_z = -6.005:.01:6.005;
im_z= -6.005:.01:6.005;
[re_z,im_z] = meshgrid(re_z,im_z);
z = re_z + 1i*im_z;
f_of_1_over_z_result = polyval(p1,1./z); % I used this code in the first figure
figure
subplot(2,2,1)
surf(re_z,im_z,abs(f_of_1_over_z_result),'EdgeColor','none')
colorbar
title('|f_b(1/z)|')
xlabel('Z_R')
ylabel('Z_I')
zlim([0 15]) %adjust this value as needed
caxis([-0.15 0.15])
subplot(2,2,2)
surf(re_z,im_z, cos(angle(f_of_1_over_z_result)),'EdgeColor','none')
colorbar
title(('cos(phase of f_b(1/z))'))
xlabel('Z_R')
ylabel('Z_I')
caxis([-1 1]) %adjust this value as needed
zlim([-1 1]) %adjust this value as needed
grid on
Aisha Mohamed 2022년 9월 1일
Thanks so much Torsten. your ways really helpful.

댓글을 달려면 로그인하십시오.

### 카테고리

Help CenterFile Exchange에서 Surface and Mesh Plots에 대해 자세히 알아보기

### Community Treasure Hunt

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

Start Hunting!

Translated by