How to plot cosine of complex function of third degree?
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I plot the phase of this function
fb(z) = (0.1496 − 0.1692i)z^−1 + (0.2472 + 0.0916i)z^−2 + (−0.8196 − 0.3718i)z^−3
by using this command
surf(re_z,im_z,angle(f_of_z_result),'EdgeColor','none'); .
My question is how can me plot the cosine of this function?
I appriciate any help
6 Comments
Chunru
on 12 Jul 2022
do you mean the following?
surf(re_z, im_z, abs(f_of_z_result),'EdgeColor','none');
Aisha Mohamed
on 12 Jul 2022
Torsten
on 12 Jul 2022
The cosine of a complex-valued function is complex-valued. So you will have to tell us what you want to plot:
cos(abs(f_of_z_result)), abs(cos(f_of_z_result)), angle(cos(f_of_z_result)), cos(angle(f_of_z_result)),...
There are many possibilities.
Aisha Mohamed
on 12 Jul 2022
Torsten
on 12 Jul 2022
Just insert the expression you want in the surf command instead of abs(f_of_z_result).
Aisha Mohamed
on 12 Jul 2022
Answers (1)
Nipun
on 8 Nov 2023
Hi Aisha,
I understand that you are trying to plot the cosine of the given symbolic relation in MATLAB using "surf" plot. In this answer I'm make the following assumption based on the information provided:
- Variable "z"is a complex variable represented as re_z + i*im_z
- re_z and im_zare real doubles that represent axes in the returned 3D plot.
- Since the cosine of a complex valued function is complex, I assume that you want to plot the absolute value.
Based on these assumptions, you can use the following to obtain the plot:
surf(re_z, im_z, abs(cos(f_of_z_result)),'EdgeColor','none');
"abs" and "cos"are in-built MATLAB functions that return the absolute and cosine value of the argument respectively.
If re_z andim_z are complex-valued, use the "abs()" function to plot your result:
surf(abs(re_z), abs(im_z), abs(cos(f_of_z_result)),'EdgeColor','none');
MATLAB has nice documentation for both functions. I have provided their links below:
- abs() function - https://in.mathworks.com/help/matlab/ref/abs.html
- cos() function - https://www.mathworks.com/help/symbolic/cos.html
Hope this helps.
Regards,
Nipun
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