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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!

Image Analyst
on 17 Apr 2021 at 16:57

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]);

Image Analyst
on 19 Apr 2021 at 13:28

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
on 17 Apr 2021 at 17:01

Edited: Star Strider
on 17 Apr 2021 at 17:12

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 —

.

Star Strider
on 19 Apr 2021 at 11:09

O.K.

I still do not completely understand it, however I am happy you got it to work.

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