i want to draw a half staircase sinusoidal curve

I want to draw a half sinusoidal curve. In other means a pattern insert in a picture. I used for or if loop but not get success.

x-axis increases 20 sec and against it y-axis 1 sec.

I am highly appreciated for your answers.

Answer 1

[Walter Roberson]

Use the Computer Vision insertShape function requesting a 'line' .

The coordinates for the line would need to duplicate every middle x value -- once for the bottom of the stair and once for the top of the stair.

[Walter Roberson]

t = linspace(0,350,17);
y = 8 * sin(pi*t/350) + 7;
stairs(t, y); ylim([6 16])

tt = reshape([t;t], 1, []);
tt([1 end]) = [];
yy = reshape([y(1:end-1);y(1:end-1)], 1, []);
tt(1:6)

ans = 1×6

0 21.8750 21.8750 43.7500 43.7500 65.6250

yy(1:6)

ans = 1×6

7.0000 7.0000 8.5607 8.5607 10.0615 10.0615

plot(tt, yy); ylim([6 16])

So you can use tt and yy as your coordinates.

[Muhammad Asad]

Thanks for your resposne. I highly appreciate your answer. In your coding the y-axis length decrease but it is not constant. Actually it is a input signal that i put to wind turbine. That's why i am very curious.

Is it is not possible using if or for loop conditions. or anyother way which give expact replica.

[Walter Roberson]

For a sinusoidal curve, either Δx or Δy can be created as constant, but not both. Your original plot is not sinusoidal.

[Muhammad Asad]

Yes you are right. I just add word "sinusoidal" because it looks that. In above figure, it looks that x-axis increase 20 sec and then y-axis rise to 1 sec then x-axis steady for 20 sec and so on. it is just steps. i try to plot it using if and else condition but no results on graph.

That's why i posted it here. Is it is possible to draw half figure like steps rise then another half it is replica but inverse.

[Walter Roberson]

t =  0:5:350;
t(end) = [];

y = floor(t/20);
mask = t > 170;
y(mask) = fliplr(y(~mask));
t(end+1) = 350;
y(end+1) = 0;
y = y + 7;


stairs(t, y); ylim([6 16])

tt = reshape([t;t], 1, []);
tt([1 end]) = [];
yy = reshape([y(1:end-1);y(1:end-1)], 1, []);
plot(tt, yy); ylim([6 16])

You can now use tt and yy in your insertShape()

[Muhammad Asad]

Thanks waltor. Delay in acceptance of an answer is due to health issue. Your answer is very helpful.

Only one more assitance in your above code the values are stored in "fliplr". My purpose to ask you this question is that the above image will be the input of my turbine. I try to apply it directly but it is not working. I want to apply the above siganl as a input to my turbine. any suggestions?

[Walter Roberson]

The original question was about drawing that particular shape into an image.

You now talk about using the signal as an input to your turbine. You would never use an image of a signal as the input to your turbine; you would want the value of the points. And you would want the value to be computable based on the current clock, rather than having to pre-compute the value.

The below function generate_hump() accepts a time value and returns the appropriate signal level.

The function would need to be modified slightly if you wanted a periodic signal; in such a case we would need to know the exact point at which you wanted the signal to start to rise again. At the moment, your output for 350 is an extra 0; if your waveform were completely symmetric it would end at 340 rather than at 350.

t =  0:5:350;

num_t = length(t);
y = zeros(1, num_t);

for K = 1 : num_t
    y(K) = generate_hump(t(K));
end
stairs(t, y); ylim([6 16])

function y = generate_hump(t)
    if t >= 350
        y = 0;
    elseif t > 170
        y = floor((350-t)/20);
    else
        y = floor(t/20);
    end
    y = y + 7;
 end

[Muhammad Asad]

Thanks. your answers are valueable.