Keep y value below certain value in a loop
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Hey,
I am trying to run a function, where y can never be below zero and never above Capacity.
The function looks like that:
remainingCharge = Capacity - usage + ChargingPower
So remainingCharge = y and is not allowed to go below 0 or beyond Capacity.
I've tried:
for x = 1:1:tablesize
t.remainingCharge = Capacity - t.usage(x) + t.ChargingPower(x)
if t.remainingCharge(x) > Capacity
remainingCharge(x) == Capacity
end
end
This is not working, as Capacity needs to be as long as the tablerows - currently I am adding Capacity to the table - hopfeully this works.
In the meantime if anyone has a suggestion on how to solve this I am happily taking hints!
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回答(2 个)
Image Analyst
2021-4-17
t.remainingCharge(x) = Capacity - t.usage(x) + t.ChargingPower(x) % x is the row number I think.
% Make it never be above Capacity
t.remainingCharge(x) = min([t.remainingCharge(x), Capacity]);
% Make it never be below 0
t.remainingCharge(x) = max([t.remainingCharge(x), 0]);
1 个评论
Image Analyst
2021-4-19
Lukas, did min() and max() not work to clip the signal for some reason? Why not?
Also note I indexed t.remainingCharge so that you're not overwriting it each time.
Star Strider
2021-4-17
编辑:Star Strider
2021-4-17
Try something like this:
x = linspace(0, 5*pi, 250);
Capacity = 0.7;
limit_y = @(y,Capacity) (y<0).*0 + (y>Capacity).*Capacity + ((y>=0) & (y<Capacity)).*y;
y = sin(x);
figure
plot(x, y, 'LineWidth',1)
hold on
plot(x, limit_y(y,Capacity), '--', 'LineWidth',1.5)
hold off
grid
xlabel('x')
ylabel('y')
legend('Original','Limited', 'Location','best')
Note that no explicit loops are necessary, since the anonymous functrion does everything in one call.
The anonymous function is designed to work on vectors, however it is certainly possible to call the anonymous function in a loop with individual scalar values as the ‘y’ argument. It will work the same way, regardless.
Experiment with it to get the result you want.
EDIT — (17 Apr 2021 at 17:12)
Added plot image to illustrate function effect —
.
7 个评论
Star Strider
2021-4-19
O.K.
I still do not completely understand it, however I am happy you got it to work.
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