Control Theory

Control Theory

Session 6 – Transfer Functions

Standard form of first order TF

1)()()(

c)()()(

,,

te,

sK

sUsZsT

tuKtzdttdz

upup

up

Step response:

t

up eKtz 1)( ,

We will consider changeswith respect to a given initialequilibrium

Step response of 264s

A

Δz(t)

t

B

Definition of step response: Δz(t) if Δu(t) is a step of size 1

Second order processesTypical example: mass-spring-damper

)()()()(2

2

tutkzdttdzc

dttzdm

u(t)z(t)

(set-up in a horizontal plane, spring in rest position when x=0, initial velocity v=0: equilibrium!)

Standard form of second order TF

22

2,

,

te,

22

2

2)()()(

c)()()(2)(

nn

nupup

upnn

ssK

sUsZsT

tuKtzdttdz

dttzd

Step respones of 2nd order processes

+

>1: Overdamped

212

21

12

11)()1)(1(

)(21

,

tt eeKtz

ssKsT stepup

=1: Critically damped = fastest without oscillations

<1: Underdamped: Oscillations!

tnt

stepnn teeKtz 1)(

)(tzstep

Overshoot in 2nd order systems

Overshoot in 2nd order systems

21

22

100..

11

2

eOP

TTn

peak

n

osc

Group Task

m=1 [kg]k=1 [N/m]

Find the TF andplot the step response for1) c= 4 [Ns/m]2) c=2 [Ns/m]3) c=1 [Ns/m]

Group Task 2

m=1 [kg]k=1 [N/m]

Can we now add a P controller and calculate thetransfer function of the closed loop?

(by the way, what’s the transfer function of a P controller?)