Making ramp and unit step function in MATLAB

I am having trouble writing the code for these step functions. I would appreciate any help. I am using MATLAB version 9.3.0.701794 (R2017b)

[Ramnarayan Krishnamurthy]

The following links may be useful:

https://www.mathworks.com/matlabcentral/answers/194542-how-do-i-write-code-to-plot-ramp-function-is-there-any-command-for-ramp?s_tid=answers_rc1-1_p1_MLT

https://www.mathworks.com/matlabcentral/answers/305444-help-writing-a-unit-step-function-for-a-signal?s_tid=answers_rc1-3_p3_MLT T

https://www.mathworks.com/help/signal/gs/impulse-step-and-ramp-functions.html

Answer 1

[Robert U]

Hello Steven Franzkowiak,

one solution is to write a function that behaves as the plotted function. One can argue about the points at t = 4 and t = 8 since they are not defined uniquely. I suppose f(t=4) = 0 and f(t=8) = -2.

The function can look like the following code. Matlab 2017b supports piecewise definition as can be found in one of the links posted by Ramnarayan Krishnamurthy.

 function [ vec_out ] = hw_piecewise_R2014b( vec_in )

    % support column/row vectors
    if size(vec_in,1) > 1
        vec_in = vec_in';
        transpose_vec = 1;
    else
        transpose_vec = 0;
    end

    % everywhere else function returns zero
    vec_out = zeros(1,size(vec_in,2));

    % defined sections of function as seen from chart
    vec_out(vec_in >= -6 & vec_in <= 0) =  5/6 * vec_in(vec_in >= -6 & vec_in <= 0) + 5;
    vec_out(vec_in >   0 & vec_in <= 4) = -5/4 * vec_in(vec_in > 0 & vec_in <= 4) + 5;
    vec_out(vec_in >   4 & vec_in <= 8) = -2;

    % return vector according to input
    if transpose_vec
        vec_out = vec_out';
    end
 end

Having that function you must define a vector t and call the four versions:

 t = -10:0.01:10;
 f1 = hw_piecewise_R2014b(t);
 figure; plot(t,f1)

 f2 = hw_piecewise_R2014b(-t);
 figure; plot(t,f2)

 f3 = hw_piecewise_R2014b(t-4);
 figure; plot(t,f3)

 f4 = hw_piecewise_R2014b(2*t - 4)
 figure; plot(t,f4)

 f5 = hw_piecewise_R2014b(-t/2 - 4)
 figure; plot(t,f5)

Kind regards,

Robert

[Jan]

Posting full solutions of homework question has the drawback, that the OP cannot deliver this as his own solution anymore. Steven did not show, what he has tried so far and did not ask a specific question. Your solution is fine and according to the quality I would vote for it. But I'm afraid, it is a short-term help only for Steven.

[Robert U]

Thank you for your comment, Jan Simon. I decided to post my solution despite the drawback you describe.

In my opinion, the implicit link to logical indexing and functions are valuable for other readers as well and outweigh the drawback that Steven could accomplish his homework by copying the full solution.

Kind regards,

Robert

Answer 2

[Sajib Biswas Shuvo]

This code does the job too. It uses the Symbolic Math toolbox.

syms t;
f_negative_t = 5*heaviside(-t) -5/6 * (-t) * heaviside(-t ) ...
                + 5/6 * (-t-6) * heaviside(-t -6);
f_positive_t = +5*heaviside(t) - 5/4 *t* heaviside(t) ...
                - 2*heaviside(t-4) + 5/4*(t-4)*heaviside(t-4)...
                + 2*heaviside(t-8);
f = f_positive_t + f_negative_t;
fplot(f,[-7,10]);
xline(0), yline(0);