why the figure of cos function does not symmetric around y axis?

조회 수: 2 (최근 30일)
Aisha Mohamed . 2022년 8월 29일
편집: Torsten . 2022년 8월 30일
I am tring to describe the figurs of these function .
fk(z) = (0.1000 + 0.3000i) + (0.4243 + 0.0017i)z + (0.9000- 0.0010i)z2
fb(z) = (0.1000 0.3000i)z1 + (0.2121 0.0008i)z2 + (0.9000 +0.0010i)z3
but from the following figurs I faced many question:
1- Why cos(phase(f_k(z))) and cos(phase(f_b(1/z))) not symetric around the y axis?
2 - why there is overlab between the positiv and negative values of cos(phase(f_b(1/z))) in the origin?
3- what is the best points I have to describe them to cover these figures?
I will appriciate any help
댓글 수: 4이전 댓글 3개 표시이전 댓글 3개 숨기기
Aisha Mohamed 2022년 8월 30일
When I plot these figures in plane I got the previous figures.
Is this wrong?
and as we know the cos figure is symmetric around y axis, why these figures do not?
I appreciate any help

댓글을 달려면 로그인하십시오.

답변 (1개)

Chunru 2022년 8월 30일
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);
figure
z = re_z + 1i*im_z;
f_of_z_result = polyval(p,z);
f_of_1_over_z_result = polyval(p1,1./z);
figure();
subplot(1,2,1)
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
view(2);
axis tight
subplot(1,2,2)
surf(re_z,im_z, cos(angle(f_of_1_over_z_result)),'EdgeColor','none')
colorbar
title(('phase of f(1/z)'))
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
axis tight
view(2);
1- Why cos(phase(f_k(z))) and cos(phase(f_b(1/z))) not symetric around the y axis?
==> It is a general polynomial and cos(angle(f_x(x+iy))) is in general not equals to cos(angle(f_x(-x+iy))
2 - why there is overlab between the positiv and negative values of cos(phase(f_b(1/z))) in the origin?
==> What do you mean overlap? This is quite a general polynomial and it may not be straightforward to explain how the very nonlinear function cos(phase(f_b(1/z))) behavors.
3- what is the best points I have to describe them to cover these figures?
==> Without criteria given, it is not possible to tell what is the best.
댓글 수: 3이전 댓글 2개 표시이전 댓글 2개 숨기기
Torsten 2022년 8월 30일
편집: Torsten 님. 2022년 8월 30일
Yes, as I suspected: you wrote something, but meant something else.
If you want to plot what is shown in your original graphics for cos(angle(fb(1/z))), you must define fb as
fb(z) = (0.1000 0.3000i)*z + (0.2121 0.0008i)*z^2+ (0.9000 +0.0010i)*z^3,
not as
fb(z) = (0.1000 0.3000i)z1 + (0.2121 0.0008i)z2 + (0.9000 +0.0010i)z3
It's up to you to decide which of the definitions is correct. Depending on your decision, you get different plots.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Axis Labels에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by