Plot the equation y=cos(A*x) + i*sin(A*x) in MATLAB

  • MATLAB
  • Thread starter mtirmize
  • Start date
  • #1
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Main Question or Discussion Point

Hi Everyone,

I am a new Matlab User, I am want to plot the equation y=cos(A*x) + i*sin(A*x)

where A = 400*pi/340 and want to plot y and vary x from pi:0.1:pi

clear all;
close all;

A = 400*pi/340;
x = -180:0,1:180;


a=cos(A*x);
b=i*sin(A*x);

i= imag(b);

hold on;
grid on;

plot(a,i, 'r-*','MarkerEdgeColor','b');

xlabel('Distance');
ylabel('Primary Source');
title('Simulation Results');
 

Answers and Replies

  • #2
kreil
Insights Author
Gold Member
668
67
You defined x in degrees, then used the radian trig functions. There are degree counterparts, like sind(x), that calculate the trig values in degrees. So choose one or the other, but make it consistent.

It's a best practice to use elementwise operations for your functions. Of course, the operators * and .* are the same when a scalar is involved, but it will save you headaches in the future.

Your call to plot confuses me. You are plotting i against a? I think this is why your current plot looks strange. You should be plotting a + i against x.


Code:
clear
clc

A = 400*pi/340;
x = -pi:0.1:pi;

y = cos(A.*x) + imag(i.*sin(A.*x));

hold on; grid on;
plot(x,y,'r-*','MarkerEdgeColor','b')
xlabel('Distance');
ylabel('Primary Source');
title('Simulation Results');
 

Attachments

  • #3
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thanks you so much for your quick reply

can you let me know if this is also correct ?

clear all;
close all;

A=400*pi/340;

theta= -2*pi:0.03:2*pi;

y=exp(1i*A*theta);


grid on;
hold on;

plot(theta,y, 'r-*','MarkerEdgeColor','b');


xlabel('Distance');
ylabel('Source');
title('Simulation Results');
 
  • #4
kreil
Insights Author
Gold Member
668
67
The only thing is that y = exp(1i*A*theta); yields a vector of complex numbers. If you care only about the real part, you should plot real(y), if you care only about the imaginary part, you should plot imag(y).

When you use plot(theta, y) it ignores all imaginary components.

HOWEVER, when you supply only a single imaginary argument, MATLAB plots the real part vs the imaginary part. For your second example, it produces the unit circle

Code:
clear all; close all;

A=400*pi/340;
theta= -2*pi:0.03:2*pi;

y=exp(1i*A*theta);

grid on; hold on;
plot(y, 'r-*','MarkerEdgeColor','b');
axis equal;
xlabel('Distance');
ylabel('Source');
title('Simulation Results');
Edit: I should have used different axis names
 

Attachments

  • #5
9
0
okay i understand that now. I really appreciate your help and your time. God Bless you
 
  • #6
33,507
5,192
thanks you so much for your quick reply

can you let me know if this is also correct ?

Code:
clear all; 
close all;

A=400*pi/340;

theta= -2*pi:0.03:2*pi;

y=exp(1i*A*theta);
The line above bothers me. I'm not an expert in matlab, and don't have it to try things out, but no programming language that I'm familiar with allows a variable whose name starts with a decimal digit. I would be surprised to find that matlab allows a variable whose name is 1i or 2i or 3i, etc.

If your intent was 1 * i, that would be OK, but what's the point?
Code:
grid on;
hold on;

plot(theta,y, 'r-*','MarkerEdgeColor','b');


xlabel('Distance');
ylabel('Source');
title('Simulation Results');
PS - instead of color tags, please use [code] tags around your code.
 
  • #7
kreil
Insights Author
Gold Member
668
67
Yes, 1i is "a thing" in MATLAB

Since "i" and "j" are functions in MATLAB for doing complex maths, when you use "i" or "j" in a loop, for example, you may be silently breaking code that relied on using them as complex. So one solution is to always use "1i" or "1j" for complex maths, leaving "i" and "j" open for use elsewhere.

To top it off, 1i and 1j are actually recommended in the documentation for improved speed with complex calculations.
 
Last edited:

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