MATLAB code for right handed circular polarization

Hello, I need help to write simple code that preview right handed circular polarization, using the cos and sin fun. also in the plot must see the direction of the polarization (for example with arrows).
the expression of RHCP is : Acos(t)+Asint(t)

3 commentaires

To check: is Acos(t) to mean arccosine of t, or is it to mean A multiplied by cosine of t ? Is Asint intended to mean A times sine (of something?) times (t applied to t) ??
Roman
Roman le 18 Déc 2012
A*cos(t)+A*sin(t) A- is the amplitude of the sinus and cosine waves.
i want both RHCP and LHCP by sin and cos

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Réponses (3)

You mean like this: ???
fontSize = 20;
A = 10; % Amplitude.
t = linspace(0, 2 * pi, 40);
signal = A .* cos(t) + A .* sin(t);
stem(t, signal, 'bo-', 'LineWidth', 2);
xlabel('t', 'FontSize', fontSize);
ylabel('signal', 'FontSize', fontSize);
title('RHCP', 'FontSize', fontSize);
grid on;
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
% Give a name to the title bar.
set(gcf,'name','Demo by ImageAnalyst','numbertitle','off')

1 commentaire

A .* cos(t) + A .* sin(t) could be simplified to A .* (cos(t) + sin(t))

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That it is possible to rewrite A*cos(t) + A*sin(t) to A*(cos(t)+sin(t)) clearly shows that that is a scalar quantity - such has by definition no polarization. Try instead with something that is a vector-valued function:
A*[cos(w*t-kz),-sin(w*t-k*z),0]
AbdulRehman Khan Abkhan
AbdulRehman Khan Abkhan le 16 Nov 2021

0 votes

A = 10; % Amplitude.
t = linspace(0, 2 * pi, 40);
signal = A .* cos(t) + A .* sin(t);
stem(t, signal, 'bo-', 'LineWidth', 2);
x('t');
y('signal');
title('RHCP');
grid on;
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
% Give a name to the title bar.
set(gcf,'name','Demo by ImageAnalyst','numbertitle','off')

2 commentaires

How is your Answer different than mine?
x('t') is a call to an undefined function or variable. It seems likely that xlabel() and ylabel() were intended.

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Question posée :

le 14 Déc 2012

Commenté :

le 13 Jan 2025

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