ploting a specific function.

hi, anyone knows how can I plot this function in Matlab?

thanks...

[Walter Roberson]

Are the vertical parts intended to be sudden jumps ("the value reached 1 and jumped to 0") or are they intended to be lines?

Do you just need to plot the values, or do you need all of the intermediate values?

Answer 1

[Walter Roberson]

v0 = 0; v2 = 0.2; v3 = 0.3; v7 = 0.7; v8 = 0.8; v1=1;
xr = [v0, v2*(1-eps), v2, v3*(1-eps), v3, v7*(1-eps), v7, v8*(1-eps), v8, v1];
yr = [0,           1,  0,          0,  1,          -1, 0,          0, -1,  0];
x = linspace(v0, v1, 500);
y = interp1(xr, yr, x);
plot(x, y);

[Walter Roberson]

Note: the assigning to variables such as v2 is there so that you can be sure that you get bitwise identical meanings of literal constants. You could also write,

xr = [0, 0.2*(1-eps), 0.2, 0.3*(1-eps), 0.3, 0.7*(1-eps), 0.7, 0.8*(1-eps), 0.8, 1];
yr = [0, 1, 0, 0, 1, -1, 0, 0, -1, 0];
x = linspace(0, 1, 500);
y = interp1(xr, yr, x);
plot(x, y);

but then you have the worry about whether the 0.2*(1-eps) as a literal constant will definitely evaluate to a different value than 0.2 as a literal constant -- because if it happens to round to the same value due to some quirk of the parser, then interp1() will complain about the values not being monotonically increasing.

Answer 2

[Kenny Kim]

t = linspace(0,1,10001);
x = nan(size(t));
for i =1:numel(t)
    if t(i) <=0.2
        x(i) = 5*t(i);
    elseif t(i) >0.2 && t(i) <= 0.3
        x(i) = 0;
    elseif t(i) > 0.3 && t(i) <= 0.7
        x(i) = 1 - 5*(t(i) - 0.3);
    elseif t(i) > 0.7 && t(i) <= 0.8
        x(i) = 0;
    else
        x(i) = -1 + 5*(t(i) - 0.8);
    end
end

plot(t,x); xlabel('Time (s)'); ylabel('x(t)'); title('Giris Isareti');

Answer 3

[ahmet cakar]

First of all, thanks for the answers these really works for me.(Btw sorry for late return) So, I need to learn just one more thing. How can I do amplitude modulation to this function. so, I need to multiply x(t) by cos2*pi*fc*t (fc=100Hz , t=as I give in figure). Thanks again..

[Kenny Kim]

Hopefully this is what you meant.

If you continue from where I left:

fc = 100; % cosine freq

% Next multiply by 100 hz signal
x = x.*cos(2*pi*fc*t);

plot(t,x);

you get:

If you are using Walter's method then replace x with y and t with x.

[ahmet cakar]

Yeah, I think that's what I want :) thanks so much. but I want to ask something. here: x = x. <What does this "."(dot) do? (btw sorry for my bad English :) )

[Walter Roberson]

• is the mtimes operator, which does algebraic matrix multiplication

.* is the times operator, which does element-by-element multiplication.

[ahmet cakar]

Hmm okey. Thanks again.