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MATLAB code for right handed circular polarization

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Roman
Roman am 14 Dez. 2012
Kommentiert: Image Analyst am 17 Nov. 2021
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)
  2 Kommentare
Walter Roberson
Walter Roberson am 14 Dez. 2012
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 am 18 Dez. 2012
A*cos(t)+A*sin(t) A- is the amplitude of the sinus and cosine waves.

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Antworten (3)

Image Analyst
Image Analyst am 14 Dez. 2012
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 Kommentar
Walter Roberson
Walter Roberson am 14 Dez. 2012
A .* cos(t) + A .* sin(t) could be simplified to A .* (cos(t) + sin(t))

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Bjorn Gustavsson
Bjorn Gustavsson am 18 Dez. 2012
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 am 16 Nov. 2021
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')

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