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if you assume tau = 0, e[n] - e[n-1] = 0, then you'll get: D[n] = D[n-1], so it keeps always the previous value (and doesn't go to 0).
if you assume tau >>T, e[n] - e[n-1] = 0, then you'll get: D[n] = -1*D[n-1], so it keeps oscillating (and doesn't go to zero).
After reviewing other sources, I suggest using Backward-Euler transform s→(z-1)/z, then you'll get:
Code now (bi-linear transform):
if you assume
tau = 0, e[n] - e[n-1] = 0
, then you'll get:D[n] = D[n-1]
, so it keeps always the previous value (and doesn't go to 0).if you assume
tau >>T, e[n] - e[n-1] = 0
, then you'll get:D[n] = -1*D[n-1]
, so it keeps oscillating (and doesn't go to zero).After reviewing other sources, I suggest using Backward-Euler transform
s→(z-1)/z
, then you'll get:in this case it works as expected, for both "test cases" from above, if
e[n] - e[n-1] = 0
thenD[n] = 0
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