Example 1
A derivative controller has a derivative constant Kd of 0.4 s. What
will be the controller output when the error (a) changes at 2%/s,
(b) is constant at 4%?
Answer 1
(a) Using the equation given above, i.e. controller output = Kd x rate
of change of error, then we have:
controller output = 0.4 x 2 = 0.8%
This is a constant output.
(b) With a constant error there is no change of error with time and
thus the controller output is zero
Example 2
What will the controller output be for a proportional plus derivative
controller (a) initially and (b) 2 s after the error begins to change
from the zero error at the rate of 2%/s. The controller
has Kp = 4 and Td = 0.4 s.
Answer 2
(a) Initially the error is zero and so there is no controller output due
to proportional action. There will, however, be an output due to
derivative action since the error is changing at 2%/s. Since the
output of the controller, even when giving a response due to
derivative action alone, is multiplied by the proportional gain, we
have:
controller output = Td x rate of change of error
= 4 x 0.4 x 2% = 3.2%
(b) Because the rate of change is constant, after 2 s the error will
have become 4%. Hence, then the controller output due to the
proportional mode will be given by:
controller output = Kp x error
and so that part of the output is:
controller output = 4 x 4% = 16%
The error is still changing and so there will still be an output due to
the derivative mode. This will be given by:
controller output = Kp Td x rate of change of error
and so:
controller output = 4 x 0.4 x 2% = 3.2%
Hence the total controller output due to both modes is the sum of
these two outputs and 16% + 3.2% = 19.2%.
PID control
Example
Determine the controller output of a three-mode controller having
Kp as 4, Ti as 0.2 s, Td as 0.5 s at time (a) t = 0 and (b) t = 2 s when
there is an error input which starts at 0 at time t = 0 and increases at
1%/s.
Answer
(a) Using the equation:
controller output = Kp(error + (1/Ti) x integral of error
+ Td x rate of change of error)
we have for time t = 0 an error of 0, a rate of change of error with
time of 1/s and an area between this value of t and t = 0 of 0.
Thus:
controller output = 4(0 + 0 + 0.5 x 1) = 2.0%
(b) When t = 2 s, the error has become 1%, the rate of change of the
error with time is 1%/s and the area under between t = 2 and t = 0 is
1%s. Thus:
controller output = 4(1 + (1/0.2) x 1 + 0.5 x 1) = 26%
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