Introduction to Computer Control Systems€¦ · Introduction to Computer Control Systems Lecture...

Introduction to Computer Control SystemsLecture 6: LTI system response

Dave Zachariah

Div. Systems and Control, Dept. Information Technology,Uppsala University

December 9, 2014

(UU/Info Technology/SysCon) Intro. Computer Control Sys. December 9, 2014 1 / 12

Today’s lecture: What and why?

LTI system responsesWhy: Characterize what your system does to a well-defined inputsignal. Control design criteria are often defined by using specificationsof the step response.

Observability and controllabilityWhy: Is the system model such that we can observe all state changesthrough the output signal? Can we affect all the states using our inputsignal?



LTI system responsesWhy: Characterize what your system does to a well-defined inputsignal. Control design criteria are often defined by using specificationsof the step response.Observability and controllabilityWhy: Is the system model such that we can observe all state changesthrough the output signal? Can we affect all the states using our inputsignal?


System example: Spring with input force

LTI system in state-space form:

x = Ax + Bu

y = Cx +Du

y

u

States and output:

x =

[x1x2

]=

[yy

]y =

[1 0

]x



Input u(t) is an impulse δ(t)

0 20 40 60 80 100−0.2

0

0.2

0.4

0.6

0.8

1

1.2

t [s]

inp

ut

u(t

), o

utp

ut

y(t

)

u(t)

y(t)

Temporal perspective




−1 −0.5 0 0.5 10

0.02

0.04

0.06

0.08

0.1

x1 (position) [m]

x2 (

ve

locity)

[m/s

]

State-space perspective x(t) at t = 0+.




0 0.05 0.1 0.15 0.2 0.25 0.3−0.02

0

0.02

0.04

0.06

0.08

0.1

x1 (position) [m]

x2 (

ve

locity)

[m/s

]

State-space perspective x(t) at t = 5.




−0.2 −0.1 0 0.1 0.2 0.3

−0.05

0

0.05

x1 (position) [m]

x2 (

ve

locity)

[m/s

]



LTI system response: transient and steady-state

y(t)u(t)G

LTI system in state-space form

x = Ax + Bu

y = Cx +Du

Assume x(0) = 0 then

y(t) =

∫ t

τ=0CeA(t−τ)Bu(τ)dτ +Du(t)

= dinput is a step u(t) = 1 for t ≥ 0e= CA−1eAtB︸︷︷︸

transient response

+ −CA−1B +D︸︷︷︸steady-state response

, t ≥ 0



Input u(t) is a unit step (u(t) = 1 for t ≥ 0.)

0 20 40 60 800

5

10

15

t [s]

inp

ut

u(t

), o

utp

ut

y(t

)

u(t)

y(t)

Temporal perspective




0 20 40 60 800

5

10

15

t [s]

inp

ut

u(t

), o

utp

ut

y(t

)

u(t)

y(t)

Steady-state yss (recall final value theorem) and overshoot M .

Rise time Tr: time it take for y(t) to go from 0.1yss to 0.9yss.

Settling time T ps : time it takes for y(t) to stay within (1± p)yss.(UU/Info Technology/SysCon) Intro. Computer Control Sys. December 9, 2014 6 / 12



0 5 10 15−2

−1

0

1

2

3

x1 (position) [m]

x2 (

ve

locity)

[m/s

]




Input u(t)= cos(ωt) for t ≥ 0.

0 20 40 60 80 100−6

−4

−2

0

2

4

t [s]

inp

ut

u(t

), o

utp

ut

y(t

)

u(t)

y(t)

Temporal: Note transient vs. stationary/steady-state of y(t)



Input u(t)= cos(ωt) for t ≥ 0.

−6 −4 −2 0 2 4−3

−2

−1

0

1

2

3

x1 (position) [m]

x2 (

ve

locity)

[m/s

]



Controllability of LTI systems

LTI system of order n in state-space form

x = Ax + Bu

y = Cx +Du.

A particular state x? is controllable if we can apply an input u(t)that takes the system from x(0) = 0 to x? in finite time T .On the board: Illustrate

Recall solution of state

x(T ) =

∫ T

t=0eA(T−τ)Bu(τ)dτ

= dusing Cayley-Hamilton’s theorem we get following forme= Bγ0 + ABγ1 + · · ·+ An−1Bγn−1

⇒ a given state x(T ) is linear combination of B,AB, · · · ,An−1B.


Controllability of LTI systems, cont’d


x = Ax + Bu

y = Cx +Du.

⇒ a given state x(T ) is linear combination of B,AB, · · · ,An−1B.

All states x? are controllable if and only if matrix

S(A,B) =[B AB · · · An−1B

]has n independent columns. That is, S has full rank.

System G is controllable ⇔ all x? are controllable ⇔ rank(S) = n.

Important property for designing controllers.


Observability of LTI systems


x = Ax + Bu

y = Cx +Du.

Suppose we have zero input u(t) ≡ 0 and initialize system at somestate x(0) = x? 6= 0. Then x? is unobservable if output isunchanged y(t) ≡ 0.On the board: Illustrate

If output signal is constant y(t) = 0, then all derivatives at t = 0are

dk

dtky(t)|t=0 = C

dkx(t)

dtk|t=0=CAkx?= 0


Observability of LTI systems


x = Ax + Bu

y = Cx +Du.

Constant y(t) = 0, means that for observable x? 6= 0 we have

dk

dtky(t)|t=0=CAkx? 6= 0

⇒ Cx? 6= 0, CAx? 6= 0, · · · , CAn−1x? 6= 0

That is, x? 6= 0 is observable if

O(C,A) =

CCA

...CAn−1

x? 6= 0

System G is observable ⇔ all x? 6= 0 are observable ⇔rank(O) = n. (So, Ox? = 0 impossible for x? 6= 0)



LTI system responsesWhy: Characterize what your system does to a well-defined inputsignal. Control design criteria are often defined by using specificationsof the step response.Observability and controllabilityWhy: Is the system model such that we can observe all state changesthrough the output signal? Can we affect all the states using our inputsignal?


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