n.ethz.chwhen transferring a system equation from the form of newton’s law of motion to the one...

Signals and Systems II – John Lygeros Stefan Rickli http://blogs.ethz.ch/ricklis

Signals and Systems II

1 ExamChecklist:

ScriptSeries w/ solutionsNyquist summary (by assistant)Laplace transform table (NuS style)Linear algebra: summary & bookAnalysis I/II summaryAnalysis III summaryComplex Analysis summaryElectronic Circuits summary (bode plots)NuS summaryPhysics I summary

Remarks: no bode plots by hand (probably)

2 Modeling2.1 Differentiator circuitsThey cannot be put into state space form as the output depends on the derivative of the system’s input!

This is the case when the transfer function of the system turns out to be non-proper, i.e.

deg (denominator)<deg (numerator)(Exam Summer 2010, Ex 3)

2.2 Reverse engineering transfer function

When one is given a transfer function for a certain element, e.g. a non-ideal op amp, and it depends on s → introduce an additional state!

Example: if the transfer function of a non-ideal op amp is given by

Y (s )= Ks+1⏟G (s )

U ( s) (1 )

then go back to the time domain by calculating its inverse Laplace transform

¿ (1 )⇔∧¿sY ( s )+Y ( s )=KU ( s)¿∧L−1{}⇒∧¿ y ( t )+ y (t )=Ku (t )¿∧⇔∧¿ y (t )=− y (t )+Ku (t )

Now use the controllable or observable canonical form.

2.3 Common pitfall: pendulum & gravity

Assumption: small angle θ ⇒ spring force always horizontal

F s∧¿−( l ⋅sinθ )⋅ k⏟magnitude

⋅ cosθ⏟w/ respect to eθ

FG∧¿−m⋅ g ⋅ sinθ⏟w/ respect to eθ

2.4 Common pitfall: forgetting to divide all the terms

When transferring a system equation from the form of newton’s law of motion to the one needed for the state space model, one easily forgets to divide all the terms with the mass/moment of inertia term.

Example:

M x1∧¿b ( x2− x1 )+k (x2−x1 )+u

⇔ x1∧¿− kMx1−

bMx1+

kMx2+

bMx2+

1Mu

3 Existence and uniqueness of solutions

Definition of Lipschitz continuity:

A function f :Rn→Rn is called Lipschitz if

∃ λ>0 :∀ x , x∈Rn:‖f ( x )−f ( x )‖≤ λ‖x− x‖

Criteria to decide whether a function is lipschitz: Continuous everywhere. Differentiable almost everywhere (except discrete points, e.g. abs(x)

is OK). The derivative is bounded where it exists.

Examples:function globally lipschitz locally lipschitz

x yes yes

x2 no yes

√ x no yes

√ x2+k , x∈R ,k∈R≥ 0 yes yes

|x| yes yes

sgn x no yes

sin x yes yes

3.1 Requirements3.1.1 Non-autonomous (general)

systemsFor a system, given by

x (t )= f (x ( t ) ,u (t ) ,t ) , x (t 0 )=x0with

f (⋅ ) :Rn×Rn×R≥0→Rn , u (⋅) : [ t0 , t 1 ]→Rm

the solution is a function of the form

x (⋅ ): [t 0 t1 ]→Rn

Strictly mathematically speaking, we need f to be continuous in u and t

and u (⋅) continuous in t.

More loose definitions need

f ( x ,u ,t ) Lipschitz in x, continuous in u and t u ( t ) continuous in „almost all“ t

more precisely: there can be discontinuities at given single points

3.1.2 Autonomous systems (corner case)

A system is autonomous if it’s system dynamics doesn’t explicitly depend on an independent variable.

The above definition makes it clear that time-variant system can never be autonomous (because t is an independent variable)!

Existence and uniqueness of solutions:

If f is Lipschitz then the differential equation x (t )=f (x ( t ) ) with

initial condition x0∈ Rn has a unique solution

x (⋅ ): [0 , T ]→Rn∀T ≥0 , x0∈Rn

Continous dependence on initial condition:

If f is Lipschitz then the solutions starting at x0 , x0∈Rn are such

that ∀ t ≥0 :‖x ( t )− x ( t )‖≤eλt‖x0− x0‖

Zuletzt gespeichert: 20.02.2017, 13:23, Version 12 1 / 16


4 Continous LTI in time domain

4.1 System representationState Space Model:

x (t )=(x1 (t )⋮

xn (t )) ,u (t )=(u1 (t )⋮

um (t ))x ( t )∧¿A x (t )+Bu (t )y ( t )∧¿C x (t )+Du (t )

4.2 Solution of state space model

x ( t )=¿Φ ( t ) x0⏟ZIT

+¿∧∫0

t

Φ (t−τ )Bu ( τ )⋅ dτ⏟

ZST

y (t )=¿CΦ (t ) x0⏟ZIR

+¿∧∫0

t

CΦ ( t−τ )Bu ( τ )⋅ dτ+Du ( t )⏟

ZSRZIT: „Zero Input Transition“ZST: „Zero State Transition“ZIR: „Zero Input Response“ZSR: „Zero State Response“

where

Φ ( t )=e At=I+At+ A2 t2

2!+…+ A

k tk

k !+…∈Rn×n

Properties of the state transition matrix:

1. Φ (0 )=I

2.∂∂ t

(Φ ) ( t )=A ⋅Φ ( t )

3. Φ (−t )=[Φ (t ) ]−1

4. Φ (t 1+t2 )=Φ (t 1 )⋅Φ (t2 )

4.2.1 computation of e At

4.2.1.1 A diagonizableIf A has distinct Eigenvalues: A is diagonizable and so the matrixexponential can be computed as follows:

A=W ΛW−1

with Λ containing exp(eigenvalues) of A on its diagonal and W the corresponding eigenvectors.So:

W=(w1 … wn ) Λ=(eλ1 t … 0⋮ ⋱ ⋮0 … eλn t)

and e At=W ⋅ eΛ t ⋅W−1

Watch out! In the heat of the moment one easily writes e−0 t wrongly

as 0!

e−0 t=1

Example:Let

A=(−2 0−1 −1), B=(21), x0=0 , u (t )=σ (t )

Then

x (t )=Φ ( t ) ⋅ x0⏟¿ 0

+∫0

t

Φ ( t−τ )Bu (τ )⏟¿ 1

⋅dτ

The eigenvalues are λ1=−2 , λ2=−1

The corresponding eigenvectors are w1=(11) ,w2=(01)

Φ ( t )∧¿e At=W ⋅e Λt ⋅W−1

¿=(1 01 1)⋅(e

−2 t 00 e−t)⋅( 1 0

−1 1)=( e−2t 0e−2t−e−t e−t)

Φ ( t−τ )=( e−2 ( t−τ ) 0e−2 (t−τ )−e−(t−τ ) e−(t−τ ))

So the final solution is

x (t )∧¿∫0

t

Φ (t−τ )B ⋅dτ=∫0

t

( 2e−2t+2 τ

2e−2 t+2 τ−e−t+ τ)⋅dτ¿=…=( 1−e−2 t

e−t−e−2t) ,∀ t∈ [ 0 , T ]

4.2.1.2 A nilpotent

Φ ( t )=I+At+ 12 !

A2t 2+…⏟

¿eAt

+ 1k !

A k tk+…⏟

¿0

4.2.1.3 A non-diagonizable, obviously sum of a diagonal and a nilpotent matrix

If A is non-diagonizable, but can be written as sum of a diagonal matrix M

and a nilpotent matrix N (with N q=0 ,∀ q>q0), the

matrixexponential can be computed as follows:

e At=e(M +N ) t=eMt ⋅eNt , if M and N commute:

MN=NM

eNt=I ⋅ t+Nt+ N2 t2

2!+…+ N

q tq

q !Example:

Let A=(−1 1

0 −1)=(−1 00 −1)⏟

D

+(0 10 0)⏟

N

1. D ⋅N=? N ⋅Dhere

(−1 ) ⋅ I⏟D

⋅N=…=N ⋅ I ⋅ (−1 )⏟D

2.

Φ (t )=e At=e (D+N )t=eDt⋅ eNt=(e−t t ⋅e−t

0 e−t )Note: The factor in front of the exponential term stem from eNt .

4.2.1.4 A non-diagonizable, most general case

Theorem:Every square matrix can be written as the product of a matrix in Jordan normal form, J, and a transformation matrix, P, consisting of generalized eigenvectors.The matrix operations can be performed on every Jordan block on the diagonals. For further information refer to the Wikipedia articles „Jordan normal form”, „Matrix operation”, ...

4.2.1.5 Finally:Always check that

1. Φ (0 )=e A ⋅0=I2.ddtΦ (t )= d

dteA ⋅t=A ⋅eA ⋅t

4.2.2 Evolution of the solutionEvery starting condition can be written as a linear combination of the system’s eigenvectors:

x0=a1 v1+a2 v2+…+an vnWhen we choose to use only one scaled eigenvector, a i vi , as the initial

state with no input, then

x (t )=eAt x0=eλi t ai v iwhich can be proven by using the definition of e At as a series and

translates to „if I start on an eigenvector, I’ll stay on an eigenvector“.(Exam Summer 2009, Ex 3)



For an arbitrary initial state we can deduce the system’s ZIT as

x (t )=eAt x0∧¿e At ⋅ (a1 v1+a2v2+…+an vn )¿=e At a1 v1+e

At a2 v2+…+e At anvn¿=eλ1 t a1v1+e

λ2t a2v2+…+e λn t anvn4.2.3 Matrix properties ↔ System

prop. The nullspace of A is the set of states that force a system to have

zero dynamics if left autonomous. A invertible ⇒ null(A) = {0} ⇒ origin is the only equilibrium point

The eigenvalues of a Matrix ( A+αI ) are ~λ i=λi+α

(series 2)

4.3 Coordinate changeLast page hand results series 2 + result from series 4.1, Slide 2.26

WARNING: in the lecture and slides coordinate changes are handled as

x=Tx whereas the LinAlg theory uses x=T−1 x .

x=Tx⇒ A=TAT−1 , B=TB ,C=CT−1 , D=D

x=T−1 x⇒ A=T−1 AT , B=T−1B , C=CT , D=DIn FS16, Ex 4.1 one even uses

x∧¿T 1−1 x

y∧¿T 2−1 y

u∧¿T 3−1u

⇒ A=T 1−1 AT 1 , B=T 1

−1BT 3 ,C=T 2−1CT1 , D=T 2

−1DT 3

Under any coordinate change the transfer function doesn’t change, i.e.

G (s )=G (s )(proof: PVK 15.6.15-4)

If x=Tx , what is the relationship between the ZIR and ZSR in both

bases?

Ce At x0∧¿ C e A t x0∧¿∀ t ≥0

C∫0

t

e A ( t−τ )Bu (τ ) ⋅dτ∧¿C∫0

t

e A (t−τ ) Bu ( τ )⋅ dτ∧¿ ∀ t ≥0

(proof: PVK 15.6.15-4)

4.4 Possible problem typesQ: show that some matrix A has the matrix-exponential e At.A: remember the Highlander! There can only be THE ONE.Explicitly: there can be only be one solution (state transition matrix) to the following differential equation:

{¿ x ( t )=A ⋅ x ( t )¿ x (0 )=x0

The (only) solution is the state transition matrix Φ (t )=e A ⋅ t.Check these conditions:

1. Φ (0 )=e A ⋅0=I2.ddtΦ ( t )= d

dteA ⋅t=A ⋅eA ⋅t

Q: Draw the locus diagram / phase plane plot of a system.

A: First, we need expressions for the evolutions of the state variables. I.e.

{x1 (t )=…x2 (t )=…

Then try to replace a term in x1 (t ) or x2 ( t ) by an expression

derived by the other function and solve for x1 (t ) resp. x2 (t ).

(Series 3, Ex1.5)

Q: Calculate a system’s state space representation in new coordinates. The transformation is supplied.

A: Use one of the following approaches:1. Either perform the transformation in LinAlg style, i.e. like in 4.3.2. Or directly use the supplied transformation equations and factor out

the new terms for the state variables in the new coordinates.Example in PVK 15.6.15-1, Exam Spring 2013



5 Discrete time LTI systems

5.1 SamplingIn the beginning there was sampling…

Slides 6.5-6.7, FS16 series 8 prep

If we denote A ,B ,C ,D as the matrices of the continuous time

system then we get after sampling

¿ A=eA T ,∧¿B=∫0

T

eA ⋅ (T−τ )⋅B ⋅dτ

¿C=C ,∧¿D=D

5.2 State space representation

xk+1∧¿A xk+Buky k∧¿C xk+Duk

xk=A k x0⏟ZIT

+∑i=0

k−1

A k−i−1Bu i⏟ZST

The system’s impulse response is

hk=A k−1B

5.3 Computation of Ak

If A is diagonalizable, then it can be written as

A=W ΛW−1

So

Ak=W ΛkW−1

with

Λ k=[ λ1k ¿0

¿⋱ ¿¿λn

k¿]5.4 Deadbeat responseFor a deadbeat response we need λ i=0∀ i .

Then AN=0 for some 0<N ≤n where n=dim (A ).

So

xk=A k x0=0∀ k ≥ N

5.4.1 Deadbeat controlThere always (?) exists an input uk=F xk such that

( A+BF ) shows a deadbeat behavior.

This is very desirable, but the corresponding input doesn’t care about energy!

5.5 Coordinate changeLet xk=T xk T∈Rn×n

invertible

Then

xk+1∧¿ A xk+ B uky k∧¿ C xk+Duk

with

A=TAT−1 , B=TB , C=CT−1 , D=D(slide 6.14)

6 Nonlinear Systems6.1 linear affine systemsA system of the form

{x (t )∧¿ Ax ( t )+Bu ( t )+xconsty (t )∧¿Cx (t )+Du (t )

is called linear affine and is the simplest (to me known) case of a nonlinear system.

Its linearized system matrix is equal to A, the only difference between the

linear affine and the linearized system is the absence of xconst in the

latter.



7 Stability of systems7.1 Stability of continuous LTI

systemsTheorem 3.1:Let A be diagonalizable

The system is { ¿ stable∧⇔ℜ [ λi ]≤0 ∀ i¿asymptotically stable∧⇔ℜ [ λ i ]<0∀ i

¿unstable∧⇔∃ i : ℜ [ λ i ]>0

Theorem 3.2:Let A be non-diagonalizable

The system is {¿asymptotically stable∧⇔ℜ [ λ i ]<0∀ i¿unstable∧⇐∃i : ℜ [ λi ]>0

In the case of a non-diagonalizable system with ℜ[λ i]≤0 but

∃i :ℜ [ λi ]=0 stability cannot be inferred from the eigenvalues

of A alone. We need to have a look on limt→∞

Φ ( t ) .

Remark about asymptotical stability:

We need stability first, only then do the limit calculation for t→∞ .

Consider something like the function x (t )=| 1t−1|. It goes to

zero eventually, but has a pole at t=1. So we definitely need stability

first.

7.1.1 Energy-like / Lypunov functions

Consider V ( t )≔ 12xT ( t )Qx ( t ). It’s known as the

Lyapunov function.Its derivative is

V (t )∧¿ xTQx+xTQ x¿=(Ax )TQx+xTQ (Ax )¿=xT ATQx+xTQAx¿=xT (ATQ+QA ) x

¿=xT Rx

If we now require R to be some negative definite matrix then the

components of V (t ) are all negative and so V (t ) t→∞→

0

which means that the system is asymptotically stable.

The last conclusion only works if we find a unique and positive definite solution for P !

In short

Q=QT>0 and unique∧⇒system is asymptotically stableno sol., multiple sol.non-positive definite sol.}∧⇒ system is not asymptotically stable

See theorem 4.1 on slide 4.9 for the exact formulation.

Remark from last page in series 4, 4.3

7.1.2 Example

Let A=( λ1 10 λ2

). Compute Φ ( t ) and discuss stability.

Distinguish two cases: λ1≠ λ2 and λ=λ1=λ2.

λ1≠ λ2

Λ=( λ1 00 λ2) ,W=(1 −1

λ1−λ2

0 1 )⇒A is

diagonalizable

Φ ( t )∧¿e At=W ⋅ e Λt ⋅W−1=(1 −1λ1−λ2

0 1 )⋅(λ1 00 λ2)⋅(1 1

λ1−λ2

0 1 )¿=…=(eλ 1 t eλ 1 t−eλ2 t

λ1− λ2

0 eλ2 t )x (t )=Φ ( t ) ⋅ x0:{ ¿ λ1>0 and/or λ2>0:∧¿ unstable

¿ λ1<0 , i=1,2 :∧¿asymptotically stable¿ λ1=0 , λ2<0 or λ2=0 , λ1<0 :∧¿ stable

λ=λ1=λ2

In this case A=( λ 10 λ).

Λ=( λ 00 λ) ,W=(10)⇒A is not diagonalizable

But A can be decomposed into a diagonal and nilpotent part:

A=( λ 00 λ)⏟

D

+(0 10 0)⏟

NDo D and N commute?

D ⋅N=λ ⋅ I ⋅N=(0 λ0 0) ,N ⋅D=N ⋅ λ ⋅ I=λ ⋅N ⋅ I=(0 λ

0 0)⇒ Yes, they commute. This allows us to express the matrix exponential as follows:

e At=Φ (t )=e (D+N )t=eDt⋅ eNtwhereas

eNt=I+Nt=(1 t0 1), eDt=(eλt 0

0 e λt)and

e At=Φ (t )=(eλt t ⋅eλt

0 eλt )Conclusion:

The system is { ¿unstable ,∧¿ λ>0¿ unstable ,∧¿ λ=0 t ⋅eλt t→∞

→∞

¿asymptotically stable ,∧¿ λ<0 t ⋅e λt t→∞→

0

7.2 Stability of continuous time-variant systems

We cannot infer stability from the eigenvalues of A (t )! Therefore we

need an ε-δ proof.

Example (midterm 2016):

Let x (t )= −11+t

x (t ). Discuss stability.

Use δ=|x ( t )|≤ε . Then photos 2.5.16



7.3 Stability of discrete time LTI systems

The eigenvalues can be written as λ i=σ i+i ωi and

therefore |λ i|=√σ i2+ωi2

.

So

|λ i|<1⇒|λi|k k→∞

→0

|λi|=1⇒|λi|k=1 ∀ k

|λ i|>1⇒|λi|kk→∞

→∞

Beware! We’re talking about the eigenvalues’ absolute value, not their real part!

A diagonalizable discrete time LTI system is

{ ¿ stable∧¿⇔∀ i :|λi|≤1¿asymptotically stable∧¿⇔∀ i :|λi|<1

¿unstable∧¿⇔∃ i :|λi|>1

A non-diagonalizable discrete time LTI system is

{¿asymptotically stable∧¿⇔∀ i :|λi|<1¿unstable∧¿⇔∃ i :|λi|>1

Otherwise, when we have an eigenvalue with |λ i|=1, we need to

look at the repetition pattern of the eigenvalues.

7.3.1 Energy-like / Lypunov functions

Slide 6.15

7.3.2 Minimum energy input / control

Above slide 6.16Slides 6.16-6.18



8 Stability of equilibria8.1.1 DefinitionsThe set of vectors in the nullspace of a system dynamics matrix A form the set of equilibrium points E of that LTI system.

E={v∈Rn|v∈null ( A ) }If A has full rank, then the only equilibrium point is the origin.

8.2 Stability of equilibria in linear systems

As a linear system’s set of equilibria is one connected subset of Rn×n it

doesn’t make any difference if one is talking about the stability of the whole system or the stability of that set of equilibria.

So

Equilibria of E are {as. stable &⇔ system is as. stablestable&⇔ system is stableunstable &⇔ system is unstable

8.3 Stability of equilibria in nonlinear systems

FS15 series 8 prep

Say we have a set of equilibria points x i. What kind of equilibria are

they?Use Lyapunovs direct or indirect method to prove whether they are stable.

8.3.1 Lyapunovs indirect methodLet x (t )=f (x (t ) ) and let x be an equilibrium of the system,

i.e. f ( x )=0.

Taylor about some arbitrary point x0 provides us with

f ( x )=f (x0 )+A (x0 )⋅ (x−x0 )+higher order termswhere

f ( x )=[ f 1 ( x )⋮

f n ( x )] , A ( x0 )=[∂ f 1

∂ x1( x0 ) …

∂f 1

∂ xn(x0 )

⋮ ⋱ ⋮∂ f n∂ x1

( x0 ) …∂f n∂ xn

(x0 )](A is also known as the Jacobian matrix)

To determine the stability of an equilibrium point x , one has to look at

the eigenvalues of the linearization of f ( x ) around x ,

i.e. the eigenvalues of A ( x )

So one calculates det (λI−A ( x ))=! 0⇔λi=…Theorem 7.1:

We can’t make any statement for the case ∃i :ℜ {λi }=0

8.3.2 Lyapunovs direct methodTheorem 7.2:

To show 3. one needs the chain rule!

Examples:

a (b (c (x , y ) ) )y=abbc c y

a (b (x , y ) ,c ( x , y ) )x=abbx+ac cx

a(b (x− y⏟c ( x, y ) ))xy∧¿ (abbc 1 )y

¿=bc (abbbc c y⏟−1)+ab (bcc c y )

especially

V (a (t ) ,b (t ) )t=abb t+acc tV ( x )t=∂xV ⋅ x1+∂yV ⋅ x2

see notes PVK 16.6.15-4

If one finds an energy-function V ( x ) of a system then every state that

appears only with negative sign in ∂V∂x

( x ) will dissipate energy from

the system.


An equilibrium x is

locally asymptotically stable if the eigenvalues of the linearization all have negative real part

unstable if the linearization has at least one eigenvalue with positive real part

Assume there exists an open set S⊆Rn with x∈S and a

differentiable function V (⋅) :Rn→R which satisfies

1. V ( x )=02. V ( x )>0∀ x∈S ∖ {x }


9 Controllability9.1 Controllability in

continuous timeEquivalent definitions:

¿ ∀ x0 , x1∧¿∃u (⋅) : [ 0 , t ]→Rm : x1=eAt x0+∫

0

t

eA (t−τ )Bu (τ ) ⋅dτ∧¿ (1 )

¿⇔∀ x1∧¿∃u (⋅ ): [ 0 , t ]→Rm: x1=∫0

t

e A ( t−τ )Bu (τ ) ⋅dτ∧¿ (2 )

¿⇔∀ x0∧¿∃u (⋅) : [ 0 ,t ]→Rm :e At x0+∫0

t

e A ( t−τ )Bu (τ ) ⋅dτ=0∧¿ (3 )(1) is „controllability: go everywhere from an arbitrary starting point“.(2) is „reachability from origin“.(3) is „ability to go to the origin from anywhere“.

Controllability Gramian WC (t ):

WC ( t )=∫0

t

e AτBBT e AT τ ⋅dτ∈Rn× n

Theorem: The system is controllable over [ 0 , t ] ⇔ WC ( t ) invertible

Properties of WC (t ): symmetric positive semi-definite1

Eigenvalues are ≥0 WC (t ) is invertible for some t ≥0⇔ invertible

∀ t>0(Fact 4.4)

Note about WC (t ):(1) WC ( t ) real symmetric → semi-simple → all eigenvalues have

same algebraic multiplicity as geometric multiplicity → WC (t ) is

diagonalizable

(2) If at least one of the eigenvalues is =0 then WC (t ) is not

invertible, but it’s still diagonalizable and semi-positive definite!

1 The controllability Gramian is always positive semi-definite!Proof:

xTW c x=∫ xT (… ) x dτ=…=∫‖v‖2dτ ≥0

(s. 4.17)

(3) If we look at the Lyapunov equation ATQ+QA=−R

and

set R=B BT , then the solution is Q=W C (t →∞ ) .

Controllability matrix:

P= [B AB A2B … An−1B ]∈ Rn×n ⋅m

Theorem: The system is controllable ⇔ Rank [P ]=nControllability Gramian ↔ Controllability matrix:

It turns out that Range [P ]=Range [W C ( t ) ], that is

the set of reachable states, which is, of course, nothing else than

span {columns of P }.

This gives us a nice equivalence chain:

WC ( t ) invertible⇔Range [W C ( t ) ]=Rn

⇔ system is controllable⇔Range [P ]=Rn∧⇔Rank [P ]=n

For m=1 (single input) we get a square matrix and can check if P is

invertible (i.e. det (P )≠0).

Also it turns out that the set of uncontrollable states is null (P ).

9.1.1 Minimum energy input / control

Theorem 4.2:

Assume that the system is controllable. Given x1∈Rn and T>0, the input that drives the system from x (0 )=0 to x (T )=x1

and has the minimum energy is given by

um ( t )=BT e AT (T−t )WC (T )−1 x1t∈ [ 0 ,T ]

How to deal with uncontrollable system?⇒ Series 5.1 FS16

9.2 Controllability in discrete time

Use the controllability matrix

P= [B AB A2B … An−1B ] like in the

continuous case.

9.3 Is a certain eigenvalue controllable?

Let λ i be the eigenvalue in question.

Simply compute

Pi≔ [ (λi I−A ) B ]Then

λ i is controllable ⇔ Pi has full rank

9.4 Controllers / Feedback

x=Ax+BKx= (A+BK )⏟~A

x

After applying of feedback the system is autonomous again.

One can now compare the characteristic polynomial of ~A with the

characteristic polynomial of some desired dynamics and calculate the Feedback coefficients from that.

10 Observability10.1 Observability in

continuous timeObservability matrix:

Q=[ CCA⋮

C An−1]∈Rp ⋅ n×n

How to determine whether a system ( A ,C ) is observable:

A system ( A ,C ) is observable ⇔ rank (Q )=n (full rank)

Unobservable states:

Given u (⋅) : [ 0 , t ] and y (⋅) : [ 0 ,t ], can we uniquely

determine x (⋅ ): [0 , t ]?An unobservable state is an

x∈ Rn:Ce Aτ x=0∀ τ∈ [0 , t ]Alternative check for observability: Compute Q for different C’s, i.e.

[ 1 0 0 … 0 ] , [ 0 1 0 … 0 ] ,… and compute the set of unobservable states.Series 5, ex 3.2

As soon as a system is observable for a small t ⇒ the system is observable for all t!

10.2 Observers in continuous time

We measure y ( t ) , u ( t )˙x (t )=A x (t )+Bu (t )⏟

mimic evolution of x ( t )

+L ⋅ ( y (t )−C x (t )−Du (t ) )⏟correction term

The error dynamics is

e ( t )∧¿ x (t )− x ( t )e ( t )∧¿ (A−LC )⋅e (t )



10.3 Observability in discrete time

Use the observability matrix Q=[ CCA⋮

C An−1] like in the

continuous case.

10.4 Observers in discrete timeExam Summer 2008

10.5 Is a certain eigenvalue observable?

Let λ i be the eigenvalue in question.

Simply compute

Qi≔[ (λ i I−A )C ]

Then

λ i is controllable ⇔ Qi has full rankThis has not been proofed but is a deduction from a remark that „a similar check like the controllability check [to the left] can be done for its dual system“.


unstable

stable


11 System representationsSeries 6 FS15 prep

State space

Causal signals

g (t )=0∀ t<0

Laplace domain

Fourier space (or steady state)

Structure diagram

11.1 Transition of spaces11.1.1 State space → Laplace

domainIn the process of this transition we lose information that can’t be undone in the reverse process!

G (s )=C ⋅ (sI−A )−1B+DNote that

{poles of G (s ) }⊆ {Eigenvalues of A }.

Example:Let

x∧¿ (−2 00 1)⋅ x+(11)⋅u

y∧¿ (1 0 )⋅ xThis system is Unstable, because one eigenvalue has positive real part Controllable, because its states are decoupled and both states are

influencable Not observable, because its states are decoupled and one state

can’t be seen

After applying above formula for the transfer function G (s ) we get

G (s )=(1 0 )⋅(1s+2

0

0 1s−1

)⋅(11)= 1s+2

Performing the inverse Laplace transformation yields

g ( t )=e−2 t

which looks stable on the paper.

So we just lost all information about the state x2 (t )!

After calculation of the controllable canonical form we get

x∧¿−2x+uy∧¿ x

This resulting „original“ system – again – looks stable.

11.1.2 Laplace space → state spaceThe lost information from the transition of state space → Laplace space can’t be undone!

For SISO (single-input-single-output) systems there are several approaches to perform this transition.

11.1.2.1 Strictly proper systems deg ( num )<deg (den )

Let G (s ) be the strictly proper rational transfer function consisting of

a nominator and denominator polynomial:

G (s )=∑k=1

n

bk ⋅sn−k

sn+∑k=1

n

ak ⋅ sn− k

, n>1

Then the controllable canonical form of the above system is simply:

x=(0∈R( n−1)×1 I ∈R(n−1 )×(n−1 )

−an −an−1…−a1 )⋅ x+(0⋮01)⋅u

y=(bn … b1 )⋅ xSome examples for quick reference:SISO system with one dimension

G (s )=b1

s+a1

controllable canonical form:

¿ x ( t )=−a1 x (t )+u ( t )¿ y ( t )=b1 x ( t )

observable canonical form:

¿ x ( t )=−a1 x (t )+b1u (t )¿ y (t )=x (t )

SISO system with two dimensions

G (s )=b1 s+b2

s2+a1 s+a2


¿ x ( t )=[ 0 1−a2 −a1] x ( t )+[01 ]u ( t )

¿∧ y ( t )=[b2 b1 ] x ( t )observable canonical form:

¿ x ( t )=[0 −a2

1 −a1] x (t )+[b2

b1]u (t )

¿∧ y (t )=[0 1 ] x (t )

SISO system with four dimensions

G (s )=b1 s

3+b2 s2+b3 s+b4

s4+a1 s3+a2 s

2+a3 s+a4


x (t )∧¿ [ 0 1 0 00 0 1 00 0 0 1

−a4 −a3 −a2 −a1]⋅ x (t )+[0001 ]⋅ u (t )

¿ y (t )∧¿ [b4 b3 b2 b1 ] ⋅ x ( t )observable canonical form:

x (t )∧¿ [0 0 0 −a4

1 0 0 −a3

0 1 0 −a2

0 0 1 −a1]⋅ x ( t )+[b4

b3

b2

b1]⋅u ( t )

¿ y (t )∧¿ [ 0 0 0 1 ] ⋅ x (t )(diagonal canonical form link2)

11.1.2.2 Systems with deg ( num )=deg (den )

Example with given transfer function G (s ):

G (s )= Y (s )U (s )

∧¿ s+4s+2

=( s+2 )+2s+2

=1+ 2s+2

¿⇔Y (s )∧¿U (s ) ⋅(1+ 2s+2 )

We can draw this function as the following block diagram

Calculating the inverse Laplace transformation leads to

y ( t )=w (t )+u ( t )=2 z (t )+u (t )with

{z (t )∧¿−¿∧2 z (t )+u (t )w (t )∧¿∧¿2 z (t )

that comes from the controllable canonical form of

the path U (s )→W ( s) .

2 http://www.engr.mun.ca/~millan/Eng6825/canonicals.pdf



11.1.2.3 Systems with deg ( num )>deg (den )

These systems are differentiators and thus can’t be put into state space form!

This is because the output of the system depends on the derivative of the input. (Exam Summer 2010, Ex 3)

11.1.3 Structure diagram → Laplace domain

T ( s )= forward path1+ loop

Because (?)

YR

= G1+G ⋅C

ER

= 11+G ⋅C



12 LTI systems in the frequency domain

12.1 Laplace TransformGiven: LTI system of the form

x ( t )∧¿Ax (t )+Bu ( t )y ( t )∧¿Cx ( t )+Du (t )

After Laplace transformation we have

X ( s )∧¿ (sI−A )−1 x0+( sI−A )−1BU ( s )Y (s )∧¿CX (s )+DU ( s )

We can show that L {e At }=L {Φ ( t ) }=( sI−A )−1

So

L−1 {( sI−A )−1 }=Φ (t )See slides 5.8,5.9

The „Transfer function“ of an LTI system is:

G (s )=C ⋅ (sI−A )−1B+DIt contains only information about states that are controllable and observable. Modes that are either uncontrollable or unobservable don’t show up in the transfer function.

(As there is a bijection between transfer functions and the controllable / observable canonical form we can say that when we encounter a system that is in the controllable canonical form then it’s also observable and vice versa.)Wrong. This only applies if the resulting transfer function is minimal, i.e. without Pole/Zero cancellations.For a counter example see https://blogs.ethz.ch/ricklis/Archive/306The controllable and observable canonical still don’t describe the same physical system! They are merely different realizations of the same transfer function.

Initial value theorem:

limt →0

f ( t )=lims→∞

( s ⋅F ( s) )

Final value theorem:

limt→∞

f (t )=lims→0

( s ⋅F (s ) )

12.2 Block diagramsIf all involved systems are LTI, then I can use superposition to get a composed transfer function. I.e. series 6, 4.1

How to calculate Y (s ):

1. Set R (s )=0 , calculate Gd ( s ) :d (t )→y ( t )2. Set D (s )=0 , calculate Gr ( s) :r ( t )→y ( t )

3. Superpose:

Y (s )∧¿Gr ( s) ⋅R (s )+Gd (s ) ⋅D ( s)

¿=G (s )K ( s)

1+G (s ) K (s )R (s )+ G ( s )

1+G (s )K ( s)D ( s )

12.3 Nyquist stability criterion

N: # encirclements (clockwise) of −1k

P: # open loop poles with positive real partZ: # closed loop poles with positive real part

N=Z−P=(¿ )−P

(*): we want Z=0 for stability

This ensures that F ( s )=1+KG (s ) has no poles with

positive real part.

The task is now to choose −1k

such that N=−P holds.

12.4 Z-Transform: discrete analog to Laplace transform

6.19-6.22



Physical Aspects

1 EnergyThe longer I had to cope with „Energy“, the more abstract this concept became to me. It helps to remember the following definitions and derivations.

Law of conservation of energy:The change in internal energy of a closed system is expressed in a general form by

dE=δQ−δWwhere δQ is the heat supplied to the system and δW is the work

applied to the system.

1.1 WorkWork is the result of a force on a point that moves through a distance. As the point moves, it follows a curve Γ, with a velocity v, at each instant. The small amount of work δW that occurs over an instant of time dt is calculated as

δW=F ⋅d s=F ⋅ v ⋅dtwhere the F ⋅ v is the power over the instant dt. The sum of these

small amounts of work over the trajectory of the point yields the work,

W=∫t1

t2

F ⋅v ⋅dt=∫t 1

t 2

F ⋅ d sdtdt=∫

Γ

F ⋅ d s

where Γ is the trajectory from x ( t1 ) to x ( t2 ).

1.1.1 Work and energyWork on a free (no fields), rigid (no internal degrees of freedom) body, is equal to the change in kinetic energy of the velocity and rotation of that body,

W=ΔEkinThe work of forces generated by a potential function is known as potential energy and the forces are said to be conservative. Therefore, work on an object that is merely displaced in a conservative force field, without change in velocity or rotation, is equal to minus the change of potential energy of the object,

W=−ΔEpot

If the work for an applied force is independent of the path, then the work done by the force is, by the gradient theorem, the potential function evaluated at the start and end of the trajectory of the point of application. It is tradition to define this function with a negative sign so that positive work is a reduction in the potential, that is

W=∫ΓF ⋅d x=∫

x (t 1)

x (t 2)

F ⋅ d x=U ( x (t 1 ))−U ( x (t 2 ))The function U ( x ) is called the potential energy associated with the

applied force.

Work by gravity:

Fg=−GMmr3 r

W=−∫r ( t1)

r ( t2)GMmr3 r ⋅dr=…=GMm

r (t2 )−GMmr (t1 )

with

U (r )=−GMmr

On the surface of the earth, GMr

is approximately constant:

W=Fg ( y2− y1 )=Fg Δ y=−mg Δ yWork by a spring:

W=∫0

t

F ⋅v ⋅dt=−∫0

t

kx vx dt=−12k x2

2 NuS stuffcomponent resistor inductivity capacitance

impedance ZR=R ZL= jωL ZC=1jωC

admittance Y R=1R Y L=

1jωL

Y C= jωC

time domain iR=1Ru iL=

1L∫u tⅆ iC=C

uⅆtⅆ

uR=Ri uL=L iⅆtⅆ

uC=1C∫ i tⅆ

Laplace domain U=RI U=sL I−Li(+0) I= 1

sLU+

i (+0 )s

I=U /R U= 1sC

I+ u (+0 )s

I=sCU−Cu(+0)



Calculus

1 Partial integration

∫a

b

f ( x )g ( x )⋅dx=[ f ( x )G ( x ) ]ab−∫

a

b

f ' ( x )G ( x )⋅ dx

2 Hurwitz-testSecond order polynomials:

a x2+bx+c=0

¿a ,b , c same sign∧¿⇔∀ i∈ {1,2 }: ℜ {x i }<0¿a ,b , c not same sign∧¿⇔∃ i∈ {1,2 }: ℜ {x i }>0

¿ some of a ,b , c=0∧¿⇔degenerate caseHigher order polynomials:Springer: höhere Mathematik, S. 146ffFeedback systems: …Gelbes Rechenbuch Bd 2, S. 101ffSummary „control systems“ by Weili Gao

3 Horner-schemeUseful for rapid calculation of polynomial divisions!

1. row 1: write down coefficients2. row 2: the first factor under the leftmost coefficient is 03. row 3: write down the value of the zero you’re factoring out4. sum the coefficient and the factor into row 35. multiply the last result by the zero and write it down in the next free

spot of row 26. repeat 4/57. write down result with the new coefficients of row 3

(step numbers are wrong in the picture below, make it prettier)

Example with multiple zeros:

( x−1 ) ( x−5 ) ( x+2 )2 ( x+3 )¿ x5+ x4−21 x3−49 x2+8 x+60

1 1 -21 -49 8 600 1 2 -19 -68 -60

1 1 2 -19 -68 -60 0 (x4+2 x3−19 x2−68 x−60 ) ( x−1 )

0 -2 0 38 60

-2 1 0 -19 -30 0 (x3−19 x−30 ) ( x−1 ) ( x+2 )

0 -3 9 30

-3 1 -3 -10 0 (x2−3 x−10 ) ( x−1 ) ( x+2 ) ( x+3 )

Linear Algebra

1 Matrix / Vector Operations

1.1 NotationOne can write

(F T−1

GT−1

⋮X T−1)=(

FG⋮X

)⋅T−1

(series 5.3)

1.2 Matrix inverseIn general:

A−1=adj ( A )det ( A )

Use Gauss algorithm on left side of ( A|I ) until you have ( I|A−1 ).

2x2 matrices:

A=(a bc d)⇒A−1= 1

det (A ) ( d −b−c a )⏟

adj (A )

3x3 matrices:

A=(a b cd e fg h i )⇒ A−1= 1

det A ( ei−fh ch−bi bf−cefg−di ai−cg cd−afdh−eg bg−ah ae−bd)

1.3 Eigenvalues)Make less mistakes, calculate them as following

det ( λI−A )=! 0⇔λ i=…

(In LinAlg one would calculate det (A−λI )=…, which

results in more mistakes in SigSys II)

1.3.1 Eigenvalue operationsdet (A )=∏

iλi

for every square matrix A!

spur (A )=∑iλ ik i

if A diagonalizable

[V ,D ]=eig ( A−b ⋅ I ) ,b∈ R vi'=v i , λi

'= λi−b

A ⋅v i=λi v i⇒ (b ⋅ A ) ⋅v i=b λi v i

⇒det (b ⋅ A )=bn ⋅det ( A )

scaling of A by b leads to scaling of every eigenvalue.

refer to p147 in Nipp’s book on Linear Algebra

1.4 Block diagonal matricesLet

M≔(A 00 B)

where A and B are submatrices.

Then

det (M )=det ( A )⋅det (B )also

EW (M )=EW (A)∪EW (B )EW: eigenvalues

(exam spring 2013, PVK 15.6.15-1, LinAlg Zfsg 15.1.3)



1.5 Energy-like functions2D

¿ (x1 x2 )( q12

q12

2q12

2q22 )(x1

x2)¿∧¿q12 x1

2+q22 x22+q12 x1 x2

3D

¿ (x1 x2 x3 )( q12

q12

2q13

2q12

2 q22

q23

2q13

2q23

2q32

)(x1

x2

x3)

¿∧¿q12 x12+q22 x2

2+q32 x32+q12 x1 x2+q13 x1 x3+q23 x2 x3

This pattern easily generalizes to n dimensions.

2 Other useful properties / theorems

2.1 Cayley-Hamilton theoremEvery matrix A is a solution to its own characteristic polynomial

An+a1 An−1+…+an In=0

Example

Q: Let A=( 1 −1 10 2 −1

−1 0 2 ), provide an expression A4

in terms of A2 , A , I S: the char. polynomial of a 3x3 matrix has degree 3 at most.Idea: make a decomposition

A4=n ( A ) ⋅ pA ( A )⏟¿ 0 , C-H

+ r (A )⏟degree≤2

Determine the char. polynomial:

pA ( λ )=det ( λI−A )=…=λ3−5 λ2+9 λ−5 ( ¿ )

Remember that in each division

am

=n+ rm⇔a=n ⋅m+r

Perform a polynomial division:

¿ A4/ (A3−5 A2+9 A−5 )⏞p A ( A )≅ m

=A+5⏞n (A )

+16 A2−40 A+25⏞

r (A )

pA ( A )⏟m

¿−( A4−5 A3+9 A2−5 A )¿=5 A3−9 A2+5 A

¿−(5 A3−25 A2+45 A−25 )¿=(16 A2−40A+25 )

Now

( ¿ )⇒ A4= (A+5 )⋅ pA (A )⏟¿0

+16 A2−40 A+25

If n denotes the dimension of a square matrix A then the Cayley-Hamilton

theorem lets us express the matrix exponential of A by its first (n−1 ) exponentials, e.g. Ak=f ( I , A ,…, An−1 )

Matlabsyntax remark[W,Λ] = eig(A) eigenvectors & eigenvaluessyms tY = expm(A*t)

matrix exponential

sys = ss(A,B,C,D)lsim(sys,u,t[,x0,…])step(sys)impulse(sys)

[X,Y] = meshgrid(xgv,ygv) creates gridmesh(X,Y,Z) draws wireframe



1 Bemerkungen (V3.1)NotationFett gedruckte Symbole sind Vektoren, evtl. auch Matrizen.Gelbe Markierungen bezeichnen i.d.R. Abschnitte, welche eine Überarbeitung nötig haben, weitere Infos von den markierten Quellen nötig hätten oder unklar sind.

DisclaimerMeine Formelsammlungen entstehen und wachsen meist über eine längere Zeit. Es besteht immer ein gewisses Risiko, dass sich einige Fehler über zig Iterationen versteckt gehalten haben. Ich freue mich deshalb über jegliche (Fehler-) Verbesserungen, Anmerkungen, Lob, Dank oder auch Kritik.Meine Zusammenfassungen werden fortlaufend korrigiert und aktualisiert veröffentlicht.

Weiterverarbeitung:Weil ich es nicht ausstehen kann, dass ständig das Rad neu erfunden werden muss, habe ich das Originaldokument mit veröffentlicht mit der Einladung, sich hier für die eigene Formelsammlung zu bedienen. Ihr könnt diese Zusammenfassung also gerne weiterverarbeiten und / oder auch in überarbeiteter Form veröffentlichen.Haltet jedoch die Herkunft der kopierten/übernommenen Teile so gut wie möglich nachvollziehbar, falls ihr weiter veröffentlicht.Obiges gilt auch für alle anderen Formelsammlungen von mir, welche diesen Bemerkungstext (noch) nicht enthalten.

Quellenangaben:Aus Platz- und Zeitgründen (blame the 'Prüfungsstress') fehlen natürlich praktisch jegliche Quellenangaben (worüber ich auch schon ab und zu fluchen musste). Ich versuche jedoch in diesem Abschnitt die Arbeiten zu referenzieren, von denen ich wissentlich kopiert habe:

Wesentliche Bestandteile:allgemeines aus dem Skript "Signal- und Systemtheorie II, D-ITET" von Prof. J. Lygeros, FS15Abschnitt „Energy“ Englische Wikipedia Artikel

https://en.wikipedia.org/wiki/Energyhttps://en.wikipedia.org/wiki/Work_(physics)

Revisionsverlauf:1.0 26.08.2016 erste Veröffentlichung Stefan Rickli

To Do: Notation: Matrizen mit runden/eckigen Klammern, fette/nichtfette Vektoren Verweise zu Handnotizen: Scan oder abtippen

2 Ressourcen zu „Word und Formeleditor“Es gibt ein paar Ressourcen, welche mir sehr geholfen haben, den Formeleditor in Word zu meistern:

- Microsoft Word formula editor: https://support.office.com/en-us/article/Linear-format-equations-and-Math-AutoCorrect-in-Word-2e00618d-b1fd-49d8-8cb4-8d17f25754f8?ui=en-US&rs=en-US&ad=US

- Unicode Nearly Plain-Text Encoding of Mathematics: http://www.unicode.org/notes/tn28/ o das ist der Standard, an den sich der Editor (fast vollständig) hält. Ist sehr gut beschrieben und

dokumentiert. Ausnahme sind Umrahmungen, welche nicht die ganze Funktionalität erhalten haben.- Die Zeichenübersicht des Editors selber: wenn man mit der Maus über ein Zeichen fährt, zeigt es einem den Tastatur-

Shortcut an, den man eingeben kann

Nice to know:- Alt + Shift + 0 erstellt eine neue Formel

(durch einen Bug in Office 2016 muss dieser Shortcut neu manuell definiert werden, Stand Mai 2016)- der Leerschlag ist euer Freund! Er veranlasst den Formeleditor, die Syntax bis zum aktuellen Punkt zu überprüfen und

das Zeug fixfertig bis zu dem Punkt, wo ihr seid, darzustellen (ausser es gibt noch offene Klammern).o verhält sich der Editor mal komisch, liegt es zu 95% daran, dass etwas in der Syntax nicht stimmt. Hier hilft

ab und an mal, sich die Formel im linearen Modus anzuschauen, wo alles bis auf Sonderzeichen wieder auseinander genommen wird. Das einzige, mit dem der Editor Mühe hat, sind grosse Eq-Arrays und mehrzeiliges Zeug.

- Wenn die Formel auf einer eigenen Zeile steht, veranlasst ein Leerschlag ausserhalb nach der Formel (also AUSSERHALB des Formeleditorfelds) den Editor, die Formel im Inline-Modus darzustellen (Formel braucht weniger Platz)

o siehe Tabellen in dieser ZF, dort habe ich das konsequent benutzt. Löscht mal das Leerzeichen nach einer Formel, das ein Integral enthält

- benutzt die Backslash (\) Befehle! Wenn man sich die beiden Dokumente oben ausdruckt und zur Referenz hält, geht es nicht lange, bis man alles mit der Tastatur machen kann und nie absetzen muss, um was mit der Maus zu machen

- \ensp und \emsp können helfen, um grössere, gewollte Abstände zu realisieren- \\eqarray ordnet mit jedem & einmal links und dann wieder rechts an- bastelt euch eure eigenen Shortcuts

o zum Beispiel \La für ⇐ \Ra für ⇒ \Lra für ⇔ oder \to für ein → oder eine leere 4x4 Matrix als \4x4 mit (■(&&&@&&&@&&&@&&&)) und einem Leerschlag

am Ende (damit der Ausdruck gleich aufgebaut wird) etc

o dazu einfach das entsprechende Zeichen in die Zwischenablage kopieren und im Formeleditor unter „Formeloptionen“ (in den Tools als kleiner Pfeil unten rechts zu finden) und „Math. Autokorrektur“ einfügen und den entsprechenden Backslashbefehl definieren

o manchmal ist es sinnvoll, noch eigene Funktionsnamen zu definieren, welche der Editor erkennen soll, wenn man sie häufig benutzt. Z.B. Imag()

ansonsten Leerschlag funktionsname\funcapply Leerschlag- die mathematische Autokorrektur ist manchmal auch ausserhalb des Formeleditors nützlich. Ich hab das in den

Optionen auch aktiviert- Wenn Word langsam wird wegen vielen anzuzeigenden Formeln, hilft

o 1. die Entwurfsansicht (anstatt Seitenlayout). Wenn man sich damit abfindet, dass dann ab und zu das Layout (noch) nicht dargestellt oder updated wird (nicht beirren lassen), kann man gut die kritischen Abschnitte bearbeiten. Achtung: Bilder werden NICHT angezeigt, sondern einfach mit einem Leerschlag repräsentiert!

o 2. reinzoomen, bis weniger Formeln sichtbar sind.


n.ethz.chwhen transferring a system equation from the form of newton’s law of motion to the one...

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