Transcript Control Theory Closed Loop Transfer Functions Group Task 2 m=1 [kg] c=2 [Ns/m] k=1 [N/m] Can we now add a P controller and calculate the transfer.

Control Theory
Closed Loop Transfer Functions
Group Task 2
m=1 [kg]
c=2 [Ns/m]
k=1 [N/m]
Can we now add a P controller and calculate the
transfer function of the closed loop?
(by the way, what’s the transfer function of a P controller?)
General formulas
Second order + P
8
Tp,u ( s )  2
s  8s  4
Tp,d (s)  0, TA (s)  TS (s)  1
Tc (s)  Kc
Which Kc makes the closed loop (servo) critically damped?
Draw the closed loop step response.
Second order + P
8
Tp,u ( s )  2
s  8s  4
Tp,d (s)  0, TA (s)  TS (s)  1
Tc (s)  Kc
Choose Kc such that the 63% rise time equals 0.1875 seconds. How big is the
overshoot? Draw the closed loop step response.
Second order + P
2
Tp,u ( s)  2
s  s 1
Tp,d (s)  0, TA (s)  TS (s)  1
Tc (s)  Kc
Which Kc makes the closed loop (servo) critically damped?
Draw the step response.
BTW – Representation with static
characteristics (here for Kc = -7/16)
Process
u = OP
Control 2
b
Control 1
b and r0
chosen
randomly
SS=8
z1
r0
r1
y = z = PV
Group Task 1
1
Tp,u ( s ) 
( s  3)
Tp,d (s)  0
Tc (s)  Kc
1
TA ( s ) 
s 1
1
TS ( s ) 
s5
Find Kc > 0 that makes this feedback system (servo) marginally stable.
Draw the closed loop step response (sketch + Matlab)
Integration in the loop
2
Tp,u ( s ) 
s
Tp,d (s)  0, TA (s)  TS (s)  1
Tc ( s )  K c
How big is the closed loop steady state error (servo)?
Draw the closed loop step response for Kc = 4.
Steady state error of the servo
problem to a step of size 1 (r =z =0)?
0
A.
B.
C.
D.
0
2Kc
+
None of the above
0
Integral Action
When an integrator is present in the loop: steady state error = 0?
Idea! We add an integral in the control action.
The PI controller now looks like (in the time domain):
t
Kc
u (t )  K c e(t ) 
e(t )dt

TI 0
TI is the ‘reset time’: it’s a measure for the time over which this
integral is ‘smeared out’. A smaller reset time gives a stronger
influence over a shorter period of time. (So notice: to put the I
action off, the reset time should be infinity!)
Integral Action
The PI controller now looks like (in the Laplace domain):
K c K c (TI s  1)
Tc ( s)  K c 

TI s
TI s
First order + PI
2
Tp,u ( s ) 
s 1
Tp,d (s)  0, TA (s)  TS (s)  1
K c K c (TI s  1)
Tc ( s)  K c 

TI s
TI s
Calculate the closed loop TF and choose Kc, TI such that the P.O. is less than 20% and the 2%
settling time is about 1s. Draw the step response (sketch+Matlab)