matlab words no help needed

I need to plot the sum of 2 sine curves. I can successfully plot (x,y1) and plot(x,y2), they are smooth curves, however when i try to plot(x,y) where y = y1+y2 the curve comes out like this? how do i fix this i dont know what i am doing wrong? Thanks

[Stephen]

"...however when i try to plot... the curve comes out like this"

You are plotting lots of zeros (to within floating point precision), because that is exactly what you calculated.

"how do i fix this i dont know what i am doing wrong?"

You need to consider the B factor. Note that for a B which is an integer multiple of 2*pi (and you defined B as 4*pi, which is clearly a multiple of 2*pi), the following identity holds true:

sin(B*X) = -sin(B*(1-X))

For example:

B = 2*pi;
X = rand(1,9);
sin(B*X)

ans = 1×9

-0.6044 -0.8824 -0.4910 -0.9397 -0.7891 0.9338 0.8413 0.9918 -0.9736

sin(B*(1-X))

ans = 1×9

0.6044 0.8824 0.4910 0.9397 0.7891 -0.9338 -0.8413 -0.9918 0.9736

That is why you are plotting zeros (or at least values very very very close to zero).

[Stephen]

Original question by Tyler Daines retrieved from Google Cache (unfortunately their deleted comments are not included):

https://web.archive.org/web/20210516175443/https://webcache.googleusercontent.com/search?q=cache%3A6xOVMwrjOj8J%3Ahttps%3A%2F%2Fwww.mathworks.com%2Fmatlabcentral%2Fanswers%2F827265-trying-to-plot-a-sine-sine-function-but-not-getting-a-smooth-curve

https://web.archive.org/web/20210516175823/https://www.mathworks.com/matlabcentral/answers/827265-matlab-words-no-help-needed

Trying to plot a sine+sine function but not getting a smooth curve

I need to plot the sum of 2 sine curves. I can successfully plot (x,y1) and plot(x,y2), they are smooth curves, however when i try to plot(x,y) where y = y1+y2 the curve comes out like this? how do i fix this i dont know what i am doing wrong? Thanks

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[DGM]

What you're looking at isn't meaningful. It's all just rounding error. You're essentially doing this:

y = sin(x) - sin(x);

I'm guessing there's some phase component that's not defined right yet, but I can't really guess at what you need it to be.

f = 2;
A = 1;
v = 1;
L = 1;
l = v/f;
B = (2*pi)/l;
w = 2*pi*f;
t = 0;
x = 0:0.0001:L;
z1 = A*sin(B*x - w*t);
z2 = A*sin(B*(L-x) - w*t);
y = z1+z2;

plot(x,z2); hold on
plot(x,z1);

For the future, please just paste your code using the code formatting tools. I had to retype all that and deal with the sub-pixel ambiguity differentiating the character '1' and 'l'.

[Tyler Daines]

Sorry mate i havent used this forum before will do in the future.

[DGM]

No worries. It was short enough.

Answer 2

[VBBV]

%if true
 f = 2

f = 2

V = 1; A= 1; L = 1;
LL = V/f

LL = 0.5000

B = 2*pi/LL

B = 12.5664

t = 0; W = 2*pi*f

W = 12.5664

x = 0:0.0001:L;
z1 = A*sin(B*x-W*t)

z1 = 1×10001

0 0.0013 0.0025 0.0038 0.0050 0.0063 0.0075 0.0088 0.0101 0.0113 0.0126 0.0138 0.0151 0.0163 0.0176 0.0188 0.0201 0.0214 0.0226 0.0239 0.0251 0.0264 0.0276 0.0289 0.0302 0.0314 0.0327 0.0339 0.0352 0.0364

z2 = A*sin(B*(L-x)-W*t);
z = z1-z2

z = 1×10001

0.0000 0.0025 0.0050 0.0075 0.0101 0.0126 0.0151 0.0176 0.0201 0.0226 0.0251 0.0276 0.0302 0.0327 0.0352 0.0377 0.0402 0.0427 0.0452 0.0477 0.0503 0.0528 0.0553 0.0578 0.0603 0.0628 0.0653 0.0678 0.0704 0.0729

plot(x,z)

Use z1-z2 instead for smooth curve

[DGM]

If

z2 = -z1

Then

z1+z2 = 0 % what OP was getting

and

z1-z2 = 2*z1

Hence the amplitude of 2 instead of 1.

[Stephen]

"yields a smooth graph that looks like the one attached"

It might look smooth, but have you actually looked at the Y-values? The peak amplitude is 5e-16. Miniscule!

So far all you have shown is:

1. plotting either version of that identity leads to Y-values which are zero + some floating point noise.
2. different numeric operations in different sequences accumulate floating point error differently (thus the "smooth" and "non-smooth" plots). This is expected.

There is probably no significance in the "smoothness" of that floating point noise, because most likely those curves are beyond any meaningful precision of the data type that you are using, i.e. they do not represent data. It is incorrect to assume that any non-zero floating point value must represent some meaningful data. But you are certainly free to check this yourself.

As everyone on this thread has already commented, you are plotting various combinations of X+(-X).

What do you expect to be plotted for X+(-X) ?

[Stephen]

"Does this mean that there is no way to plot y = ysrc + yref?"

I am not sure why you have reached that conclusion.

"I do see why my plot is meaningless."

The plot itself is not meaningless: you have correctly plotted lots of tiny values, because that is what you told MATLAB to plot. The values are numeric noise resulting from operations which are basically equivalent to X+(-X). MATLAB has correctly plotted that numeric noise (because it was all your numeric values are).

Important question: What do you expect to be plotted for X+(-X) ?

Perhaps you could scale the Y limits to suitable values then the numeric noise would not be noticeable at all.

Answer 3

[RAHUL MAURYA]

clear all;

clc;

f = 2;

A = 1;

v = 1;

L = 1*(180/(pi));

l = v/f;

B = ((2*pi)/l)*(180/pi);

w = 2*pi*f;

t = 0;

x=0:0.0001:L;

z1 = (A*sind(B*x - w*t));

z2 = A*sind(B*(L-x) - w*t);

y=(z1+z2);

subplot(3,1,1);

plot(x,z1)

title('X,Z1')

axis([0 10 -1 1])

subplot(3,1,2);

plot(x,z2)

title('X,Z2')

axis([0 10 -1 1])

subplot(3,1,3);

plot(x,y)

title('X,Z1+Z2')

axis([0 10 -2 2])

Convert B and L in radian to degree.

[RAHUL MAURYA]

Convert B and L in degree from Radian