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Current approach (re-written to be more concise, although possibly less clear): I = I + deltaT * Ki * mean_error; I = (I>limImax) ? limImax : ((I<limImin) ? limiImin : I); /* Calculate out from P, I, and D, then clamp */ out = P + I + D; out = (out>limMax) ? limMax : ((out<limMin) ? limiMin : out); Example where this fails, or at least performs poorly: - Assume P = 90%; D = 0%; I = 20%; limImax=50%; limMax = 100%; /* I < limImax, so no anti-windup is in effect; however ... */ out = P + I + D; /* => 110%, i.e. out is over-saturated */ out - (out>limMax) ? limMax : ...; /* => 100%, out is clamped, but I is not */ New approach: I = I + deltaT * Ki * mean_error; /* Same */ /* Remove static limiting of I */ /* Same */ out = P + I + D; out = (out>limMax) ? limMax : ((out<limMin) ? limiMin : out); /* New - solve for I from [out=P+I+D] above */ I = out - (P + D); /* [I] will not change unless out is being clamped
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[Just a suggestion]
Current approach (re-written to be more concise, although possibly less clear):
I = I + deltaT * Ki * mean_error;
I = (I>limImax) ? limImax : ((I<limImin) ? limiImin : I);
/* Calculate out from P, I, and D, then clamp */
out = P + I + D;
out = (out>limMax) ? limMax : ((out<limMin) ? limiMin : out);
Example where this fails, or at least performs poorly:
Assume
P = 90%; D = 0%; I = 20%; limImax=50%; limMax = 100%;
/* I < limImax, so no anti-windup is in effect; however ... */
out = P + I + D; /* => 110%, i.e. out is over-saturated /
out - (out>limMax) ? limMax : ...; / => 100%, out is clamped, but I is not */
New approach:
I = I + deltaT * Ki * mean_error; /* Same /
/ Remove static limiting of I */
/* Same */
out = P + I + D;
out = (out>limMax) ? limMax : ((out<limMin) ? limiMin : out);
/* New - solve for I from [out=P+I+D] above /
I = out - (P + D); / [I] will not change unless out is being clamped