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f’(0): Value of the first derivation of f(t) at time t = 0
f’’(0): Value of the second derivation of f(t) at time t = 0
With the initial values at t = 0, this results the following Laplace transform:
x
a
+ a
1
sx
a
+ a
2
s
2
x
a
= k * x
e
This yields the transfer function: G(s) = x
a
/ x
e
= k / (1+a
1
s+a
2
s
2
).
So, instead of solving closed-loop control problems with differential equations, the Laplace
transform is used to calculate the transfer function. With this transfer function it is possible to
obtain information about the stability of the system using suitable methods, for example, the
root locus method (see the relevant literature).
In drive control loops, frequency response analyses are often the route to success that
requires the least effort. The transfer function can be obtained at any time by substituting
p = j by the complex expression s.
Below, the transfer function is dealt with as a response to a step change in input for several
transfer elements.
Proportional transfer element
There is no time delay between input and output, only the amplitude of the output is less than
or greater than the input.
Kp is the gain of the P element.
First-order lag element
Example: Series connection of the resistance and inductance, e.g. in the armature circuit of
the motor
This element results in a sluggish time response of the control loop, large time constants
having a greater effect. The aim is therefore to compensate for this response with a suitable
controller.
x
x
a
= x
e
* Kp
x
e
x
a
U
R * i(t)
t
T
63% U
R
L
i
u: voltage, i: current, R: resistance, L: inductance
u(t) = R * i(t) + L * di(t) / dt; time constant T = L / R
R * i(t) = U * (1-e
-t/T
); V
P
= U / R
Time equation: f(t) = x
a
(t) / X
e
= V
P
* (1-e
-t/T
)
Frequency response: F(p) = V
P
/ (1+pT)
U: step change in voltage at the input
Step response: R * i(t)
Transfer function,
Time equation:
F (t) = x
a
(t) / x
e
(t) = Kp
Frequency response:
F(p) = x
a
(p) / x
e
(p) = Kp
