ploting a specific function.
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hi, anyone knows how can I plot this function in Matlab?
thanks...
1 个评论
Walter Roberson
2016-12-12
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?
采纳的回答
Walter Roberson
2016-12-12
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);
1 个评论
Walter Roberson
2016-12-12
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.
更多回答（2 个）
Kenny Kim
2016-12-12
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');
ahmet cakar
2016-12-13
0 个投票
4 个评论
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
2016-12-13
Walter Roberson
2016-12-13
ahmet cakar
2016-12-14
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