Set up a function corresponding to this graph

hi, the following sketch was given to me and i need to set up a corresponding function. One problem, i have absolutely no clue, can somebody help me?

Answer 1

[Adam Danz]

In addition to the square() function (requires signal processing toolbox) and the heaviside() function (symbolic math toolbox), here's a much lower level, custom function that produces one step as is depicted in your image.

% Define step function
% * x is a vector, monotonically increasing
% * start: where the step should rise (scalar, member of x)
% * width: the width of the step (scalar, positive)
% * height: the height of the step (scalar, positive or negative)
% * base: the minimum base of the step (scalar)
stepFunc = @(x, start, width, height, base) ...
    [zeros(1,sum(x < start)), 0, ones(1, sum(x >= start & x <= start+width)), 0, zeros(1, sum(x > start+width))] ...
    * height + base; 

% Calculate step function
% Define parameters
x = 0:1000; 
start = 200; 
width = 250; 
height = 4; 
base = 1.0; 
y = stepFunc(x, start, width, height, base); 

% Now we have to adjust x so that there are duplicate values where the step rises and falls
% in order to make it vertical
stepIdx = find(abs(diff(y)) == abs(height)); 
xIdx = [1:stepIdx(1), stepIdx(1):stepIdx(2)-1, stepIdx(2)-1:length(x)]; 
newX = x(xIdx); 

% Plot results
figure
plot(newX, y)
ylim(ylim + [-1.1, 1.1]*range(ylim)) %incease y lim by 10%

[Danny Helwegen]

Thanks, this was exactly what i needed

Answer 2

[Mark Sherstan]

Use the square function. The following should get you started:

t = linspace(0,3*pi)';
x = square(t);

plot(t,x)
ylim([-2,2])

[Danny Helwegen]

thx for the answer, it was not exactly what i meant but still thanks