How can I plot this function such as all values of |f(1/z)| be more clear and how can I plot the cosine of the phase of this function ?

When I plot this function (thanks for every help).

f(1/z) = (0.1000 -0.3000i)z^-1 + (0.2121-0.0008i)z^-2 + (-0.9000-0.0010i)z^-3

I want to study time evolution of this functions, but I have got from its figure that:

1-the |f(1/z)| seems to take zero every where exept at z=o when it goes to infiniy, even when I tried to change the values of z , I always get the same graph

My questions are:

1:How can I plot this figure such that all values of z in the plane appeare more clear to enable me to study its movement over time?

2- How can I plot the cos for the phase of f(1/z) to get more cleare plotting of the phase of this function?

p1=[(0.9000+0.0010i) (0.2121-0.0008i) (0.1000 -0.3000i)];

This is the figure of |f(1/z)| and the phase of f(1/z)

I used this link:

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;

xlabel('x')

ylabel('y')

f_of_1_over_z_result = polyval(p1,1./z);

figure();

subplot(2,2,3)

surf(re_z,im_z,abs(f_of_1_over_z_result),'EdgeColor','none')

colorbar

title('|f(1/z)|')

xlabel('Z_R')

ylabel('Z_I')

subplot(2,2,4)

surf(re_z,im_z,angle(f_of_1_over_z_result),'EdgeColor','none')

colorbar

title('phase of f(1/z)')

xlabel('Z_R')

ylabel('Z_I')

grid on

I appriciate any help.

Answer 1

[Benjamin Kraus]

It looks like the nearly infinite value in your data is drowning out the remaining data. You can fix this by changing the z-limits and color limits, using the zlim and caxis commands.

For example:

% Prepare the data for plotting
p1=[(0.9000+0.0010i) (0.2121-0.0008i) (0.1000 -0.3000i)];
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);

% Create first picture
nexttile
surf(re_z,im_z,abs(f_of_1_over_z_result),'EdgeColor','none')
colorbar
title('|f(1/z)|')
xlabel('Z_R')
ylabel('Z_I')
zlim([0 4]) % Adjust this value as needed
caxis([0 4]) % Adjust this value as needed

%%
nexttile
surf(re_z,im_z,angle(f_of_1_over_z_result),'EdgeColor','none')
colorbar
title('phase of f(1/z)')
xlabel('Z_R')
ylabel('Z_I')
grid on

[Aisha Mohamed]

Thanks Benjamin

I am trying to apply your link to study this figure but the program does nor run and I got this message:

Undefined function or variable 'nexttile'.

[Torsten]

The command was introduced in R2019b.

My guess is you have an earlier version.

[Benjamin Kraus]

nexttile is a new version of subplot.

If you don't have nexttile you can just replace those two lines with calls to subplot like in your origial code.

[Aisha Mohamed]

Thanks, Benjamin and Torsten very much, now I can limit the value of z and study the magnitude of the function. But I can not study the phase of the function because it seems unclear and does not give correct information about the function even when I limited the value of z.

My question is :

How can I plot the phase of this function to be clear? how can I plot the cosine of the phase instead?

I appreciate any help

[Aisha Mohamed]

Hi Dear

Why in this question f(1/z) = (0.1000 -0.3000i)z^-1 + (0.2121-0.0008i)z^-2 + (-0.9000-0.0010i)z^-3

the polynomial f(1/z) is written by a Matlab expert here without adding (0 ) as the last term in this function,

p1=[(0.9000+0.0010i) (0.2121-0.0008i) (0.1000 -0.3000i)]; but we have got from some Matlab experts answers that this function must write in this form:

p1=[(0.9000+0.0010i) (0.2121-0.0008i) (0.1000 -0.3000i) (0)];

Does that mean both answers are correct, or did we only forget to add 0 at the end of this function?

I appriciate any help

[Torsten]

I suggest you test it for a simpler polynomial:

z = 2;
p1 = [1, 2, 3, 0];
polyval(p1,1/z)

ans = 2.1250

(1/z)^3 + 2*(1/z)^2 + 3*(1/z)

ans = 2.1250

p2 = [1, 2, 3];
polyval(p2,1/z)

ans = 4.2500

(1/z)^2 + 2*(1/z) + 3

ans = 4.2500

So the version with the 0 as last element of the vector is the correct one.

[Aisha Mohamed]

Thanks Torsten.

[Benjamin Kraus]

To expand on @Torsten's answer: The doc page for polyval explains the input to polyval:

In the first case (without the 0) you have three elements in your vector, which defines a quadratic polynomial (n==2) and the polynomial becomes:

In the second case (with the 0) you have four elements in your vector, which defines a cubic polynomial (n==3) and the polynomial becomes:

Because equals 0, that reduces to

But, you can clearly see the first element of the vector is treated significantly differently with and without the zero.

[Aisha Mohamed]

Thanks Benjamin so much.