Introduction to Computer Control Systems · Introduction to Computer Control Systems Lecture 7: LTI system stability Dave Zachariah Div. Systems and Control, Dept. Information Technology,

Introduction to Computer Control SystemsLecture 7: LTI system stability

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 / 1

Today’s lecture: What and why?

BIBO stabilityWhy: If system is unstable, controller becomes critical

Nyquist stability criterionWhy: Will the controller stabilize or destabilize closed-loop system?Local stability of nonlinear systemsWhy: Many real systems are really nonlinear



BIBO stabilityWhy: If system is unstable, controller becomes criticalNyquist stability criterionWhy: Will the controller stabilize or destabilize closed-loop system?

Local stability of nonlinear systemsWhy: Many real systems are really nonlinear



BIBO stabilityWhy: If system is unstable, controller becomes criticalNyquist stability criterionWhy: Will the controller stabilize or destabilize closed-loop system?Local stability of nonlinear systemsWhy: Many real systems are really nonlinear


Bounded input-bounded output (BIBO) stability

System G is stable in BIBO sense if bounded input

|u(t)| < K

yields bounded output|y(t)| < K ′,

for some constants K and K ′. That is, system will not ‘explode’.Example: Water tank vs. spring

0 50 100

−50

0

50

t [s]

inpu

t u(t

), o

utpu

t y(t

)

u(t)y(t)

0 50 100−6

−4

−2

0

2

4

t [s]

inpu

t u(t

), o

utpu

t y(t

)

u(t)y(t)


Bounded input-bounded output (BIBO) stability

System G is stable in BIBO sense if bounded input

|u(t)| < K

yields bounded output|y(t)| < K ′,

for some constants K and K ′. That is, system will not ‘explode’.Example: Water tank vs. spring

0 50 100

−50

0

50

t [s]

inpu

t u(t

), o

utpu

t y(t

)

u(t)y(t)

0 50 100−6

−4

−2

0

2

4

t [s]

inpu

t u(t

), o

utpu

t y(t

)

u(t)y(t)


Bounded input-bounded output (BIBO) stability, cont’d

For LTI systems we have y(t) =∫ tτ=0 g(τ)u(t− τ)dτ .

Then

|y(t)| ≤∫ t

τ=0|g(τ)||u(t− τ)|dτ ≤

∫ t

τ=0|g(τ)|Kdτ

Using partial fraction of transfer function

G(s) =n∑i=1

αisi − pi

⇒ g(τ) = L−1{G(s)} =n∑i=1

α′iepiτ

where pi are the poles.



For LTI systems we have y(t) =∫ tτ=0 g(τ)u(t− τ)dτ . Then

|y(t)| ≤∫ t

τ=0|g(τ)||u(t− τ)|dτ ≤

∫ t

τ=0|g(τ)|Kdτ


G(s) =n∑i=1

αisi − pi

⇒ g(τ) = L−1{G(s)} =n∑i=1

α′iepiτ




For LTI systems we have y(t) =∫ tτ=0 g(τ)u(t− τ)dτ . Then

|y(t)| ≤∫ t

τ=0|g(τ)||u(t− τ)|dτ ≤

∫ t

τ=0|g(τ)|Kdτ


G(s) =

n∑i=1

αisi − pi

⇒ g(τ) = L−1{G(s)} =

n∑i=1

α′iepiτ




When real part of pi is negative, then epiτ → 0 quickly as τ →∞and

∫ tτ=0 |g(τ)|dτ ≤ K ′′ is bounded.

⇒ |y(t)| ≤∫ t

τ=0|g(τ)|Kdτ ≤KK ′′ = K ′

So, all Re{pi} < 0 ⇒ System G is BIBO stable!

Re{s}

Im{s}

p3

p1

p2

Recall: poles pi are also the eigenvalues of A





⇒ |y(t)| ≤∫ t

τ=0|g(τ)|Kdτ ≤KK ′′ = K ′


Re{s}

Im{s}

p3

p1

p2






⇒ |y(t)| ≤∫ t

τ=0|g(τ)|Kdτ ≤KK ′′ = K ′


Re{s}

Im{s}

p3

p1

p2






⇒ |y(t)| ≤∫ t

τ=0|g(τ)|Kdτ ≤KK ′′ = K ′


Re{s}

Im{s}

p3

p1

p2



Nyquist stability criterion

Q: Is our closed-loop system (controller H and system G) stable?

yur+

−G

eH


Nyquist stability criterion, cont’d

A: If poles of closed-loop system on left half of complex plane, then itis stable!

yur+

−G

eH

Gtot

Recall:

Y (s) = Gtot(s)R(s) =G(s)H(s)

1 +G(s)H(s)R(s)

⇒Find roots s = pi such that 1 +G(s)H(s) = 1 +G0(s) = 0



A: If poles of closed-loop system on left half of complex plane, then itis stable!

yur+

−G

eH

Gtot

Recall:

Y (s) = Gtot(s)R(s) =G(s)H(s)

1 +G(s)H(s)R(s)

⇒Find roots s = pi such that 1 +G(s)H(s) = 1 +G0(s) = 0



Graphical way to find stability of closed-loop system. The Nyquistcontour plot: Real and imaginary part of G0(s) as a function ofs = jω

Let ω increase from −∞ to +∞: plot Re{G0} and Im{G0}



Graphical way to find stability of closed-loop system. The Nyquistcontour plot: Real and imaginary part of G0(s) as a function ofs = jωLet ω increase from −∞ to +∞: plot Re{G0} and Im{G0}

−1.5 −1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Real{G(jω)}

Imag

{G(jω

)}




−1.5 −1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Real{G(jω)}

Imag

{G(jω

)}




−1.5 −1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Real{G(jω)}

Imag

{G(jω

)}



If G0(s) = G(s)H(s) has no poles on right plane. SimplifiedNyquist stability criterion:If Nyquist contour does not encircle −1 ⇒ closed-loop system isstable!

−1.5 −1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

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0.8

1

Real{G(jω)}

Imag

{G(jω

)}


Example of graphical test

System and controller with tuning parameter K > 0:

G(s) =1

s(s+ 2)H(s) =

K

s+ 1

Changing K moves poles of closed-loop system Gtot! When is itstable? On the board

Note: G0(s) = G(s)H(s) has no poles on right plane ⇒ we can usesimplified Nyquist criterion.




G(s) =1

s(s+ 2)H(s) =

K

s+ 1

Changing K moves poles of closed-loop system Gtot! When is itstable? On the boardNote: G0(s) = G(s)H(s) has no poles on right plane ⇒ we can usesimplified Nyquist criterion.

−3 −2 −1 0 1−1

−0.5

0

0.5

1

Real{G0(jω)}

Imag

{G0(jω

)}

Figure: K = 2 ⇒ closed-loop system stable!




G(s) =1

s(s+ 2)H(s) =

K

s+ 1

Changing K moves poles of closed-loop system Gtot! When is itstable? On the boardNote: G0(s) = G(s)H(s) has no poles on right plane ⇒ we can usesimplified Nyquist criterion.

−3 −2 −1 0 1−1

−0.5

0

0.5

1

Real{G0(jω)}

Ima

g{G

0(j

ω)}

Figure: K = 8 ⇒ closed-loop system not stable!


Stability of nonlinear systems?

Recall: if poles/eigenvalues of A on the left half complex plane

⇒ LTI system

{x = Ax + Bu

y = Cx +Duis stable

Q: What about nonlinear system{x = f(x, u)

y = h(x, u)?

A: Hard in general, but analyze around some stationary point(x0, u0) for which x = f(x0, u0) = 0


Stability of nonlinear systems?

Recall: if poles/eigenvalues of A on the left half complex plane

⇒ LTI system

{x = Ax + Bu

y = Cx +Duis stable

Q: What about nonlinear system{x = f(x, u)

y = h(x, u)?

A: Hard in general, but analyze around some stationary point(x0, u0) for which x = f(x0, u0) = 0


Stability of nonlinear systems? Cont’d

Linearizing around a stationary point (x0, u0) for whichx = f(x0, u0) = 0, we get LTI system{

˙x = Ax + Bu

y = Cx +Du

If poles/eigenvalues of A on the left half complex plane ⇒nonlinear system is locally stable around (x0, u0).

x1

x2

x0

x(t)

Locally stable: state x(t) that starts within attractor domain willconverge to x0.


Stability of nonlinear systems? Cont’d

Linearizing around a stationary point (x0, u0) for whichx = f(x0, u0) = 0, we get LTI system{

˙x = Ax + Bu

y = Cx +Du

If poles/eigenvalues of A on the left half complex plane ⇒nonlinear system is locally stable around (x0, u0).

x1

x2

x0

x(t)

Locally stable: state x(t) that starts within attractor domain willconverge to x0.



BIBO stabilityWhy: If system is unstable, controller becomes criticalNyquist stability criterionWhy: Will the controller stabilize or destabilize closed-loop system?Local stability of nonlinear systemsWhy: Many real systems are really nonlinear


Documents

Introduction to Computer Control Systems · Introduction to Computer Control Systems Lecture 7: LTI system stability Dave Zachariah Div. Systems and Control, Dept. Information Technology,