slides chap3

ELEC2400 Signals & SystemsChapter 3. Modelling, Differential Equations and

System Properties

Brett Ninnes

[email protected].

School of Electrical Engineering and Computer Science

The University of Newcastle

Slides by Juan I. Yuz ([email protected]) - July 24, 2003 – p.1/100

3. Modelling, Differential Equations and System Properties

In most engineering scenarios, signals are related bythe laws of physics.

The equations involved can be very complicated,although the essential characteristics can be capturedby a simplified analysis.

Physical models can be very often expressed asdifferential equations.

This provide insight into the general principles toderive, describe and analyse these models.

Chapter 3. Modelling, Differential Equations and System Properties – p.2/100

Electrical Circuits Modelling

Kirchoff Current and Voltage conservation laws:

(KCL)∑

k

ik(t) = 0

(KVL)∑

n

vn(t) = 0

Fundamental passive components:

Resistor (Ohm’s Law) vn(t) = R ik(t)

Inductor vn(t) = Ld

dtik(t)

Capacitor ik(t) = Cd

dtvn(t)


Example: RC circuit

+−

+

+

−

−

vR(t)

i(t)

vC(t)v(t)

We have that

by Kirchoff’s Voltage Law:v(t) = vR(t) + vC(t)

by Ohm’s Law:vR(t) = R i(t)

by the capacitive action:

i(t) = Cd

dtvC(t).

Replacing in KVL, this yields:

v(t) = RCd

dtvC(t) + vC(t) ⇐⇒

d

dtvC(t) +

1

RCvC(t) =

1

RCv(t)


Mechanical System Modelling

For a given a body, we have:

position u(t)

velocity v(t) = ddt

u(t)

acceleration a(t) = ddt

v(t) = d2

dt2u(t)

The sum of forces on a body of mass m, moving onlyin one axis, makes it accelerate:

(Newton’s 2nd Law)∑

k

Fk(t) = ma(t)


Mechanical System Modelling

As well, many physical mechanical components obeylaws expressible via differential equation.

For a spring:

Fk(t) = −ksxk(t)

where ks is the spring constant, and positive xk(t)corresponds to spring stretching.For a damper (or dash-pot):

Fk(t) = −kd

d

dtxk(t)

where kd is the damper viscosity, and positive xk(t)corresponds to extension (of the damper).


Example 1: Car Shock Absorber

Spring Damper

F1(t) = mg

Road

CAR CHASSIS

F3(t) = −kd

d

dt[y(t) − u(t)]

u(t)

y(t)

F2(t) = −ks [y(t) − u(t)]

Three forces:(gravity) F1(t) = mg

(the spring) F2(t) = −ks[y(t) − u(t)]

(the damper) F3(t) = −kdddt

[y(t) − u(t)]


Example 1: Car Shock Absorber

By the Newton 2nd Law:

F1(t) + F2(t) + F3(t) = ma(t)

mg − ks[y(t) − u(t)] − kdd

dt[y(t) − u(t)] = m

d2

dt2y(t)

Which (rearranging) yields:

d2

dt2y(t) +

kd

m

d

dty(t) +

ks

my(t) = g +

kd

m

d

dtu(t) +

ks

mu(t)

And, making y(t) = y(t) − mg/ks :

d2

dt2y(t) +

kd

m

d

dty(t) +

ks

my(t) =

kd

m

d

dtu(t) +

ks

mu(t)


Example 2: Car Velocity

x(t)

F1(t) F2(t)

Consider a car subject to the engine force F1(t), and thewind resistance F2(t) proportional to velocity:

F2(t) = kv(t) , v(t) ,d

dtx(t)

And the acceleration a(t) is the time rate of change ofvelocity v(t):

a(t) =d

dtv(t)


Example 2: Car Velocity

x(t)

F1(t) F2(t)

Using Newton’s 2nd law, we get a relationship betweenthe motor induced force F1(t) = f(t) and the car velocityv(t):

F1(t) − F2(t) = f(t) − kv(t) = md

dtv(t)

Which may be re-written as:

d

dtv(t) +

k

mv(t) =

1

mf(t)


Example 3: Water Tank

Suface Area = A m2

y(t)

u(t)

h(t)

Consider a circular water tank of surface area A [m2], subject to aninflow u(t) [m3/s] and a resulting outflow y(t) [m3/s].The total volume V (t) of water in the tank at any time is:

V (t) = Ah(t)

Thus:d

dtV (t) = A

d

dth(t) = u(t) − y(t)


Example 3: Water Tank

Suface Area = A m2

y(t)

u(t)

h(t)

The output flow rate y(t) is proportional to the pressure at thebottom of the tank, and in turn this pressure is proportional to theheight h(t):

y(t) = Kh(t) for some K ∈ R+

Thus, the final differential equation model is:

A

K·

d

dty(t) + y(t) = u(t)


Example 4: Coupled Water Tanks

u(t)

y(t)

Suface Area = A2 m2

h2(t)

h1(t)

Suface Area = A1 m2

x(t)

Consider now that the outflow x(t) of one tank of surface area A1

becomes the inflow of a second tank of surface area A2.

As per the previous example, we have that:

A1d

dth1(t) = u(t) − x(t)

A2d

dth2(t) = x(t) − y(t)



u(t)

y(t)

Suface Area = A2 m2

h2(t)

h1(t)

Suface Area = A1 m2

x(t)

The flow rate x(t) is proportional to the pressure difference, which isproportional to h1(t) − h2(t). That is:

x(t) = K1[h1(t) − h2(t)] for some K1 ∈ R+

(Note that x(t) can be negative)Finally, as in the previous example:

y(t) = K2h2(t) for some K2 ∈ R+



Substituting the expressions for x(t) and h2(t), we have:

A1d

dth1(t) = u(t) − K1

[

h1(t) −1

K2y(t)

]

A2

K2

d

dty(t) = K1

[

h1(t) −1

K2y(t)

]

− y(t) (∗)

Which can be rewritten as:

d

dth1(t) = −

K1

A1h1(t) +

1

A1

[

u(t) +1

K2y(t)

]

h1(t) =A2

K1K2

d

dty(t) +

(K1 + K2

K1K2

)

y(t)

Differentiating (∗) and replacing h1(t), we obtain the final differentialequation model relating input flow rate u(t) to output flow rate y(t):

(A1A2

K1K2

)d2

dt2y(t) +

(A1 + A2

K2+

A1

K1

)d

dty(t) + y(t) = u(t)


General Differential Equation Models

For a wide class of physical phenomena, two signals y(t)and u(t) are related by a differential equation:

dn

dtny(t) + an−1

dn−1

dtn−1y(t) + · · · + a1

d

dty(t) + a0y(t) =

bm

dm

dtmu(t) + · · · + b1

d

dtu(t) + b0u(t)

which is linear, time-invariant and finite dimensional.

In the previous examples:

ddt

vC(t) + 1RC

vC(t) = 1RC

v(t)

d2

dt2y(t) + kd

m

ddt

y(t) + ks

my(t) = kd

m

ddt

u(t) + ks

mu(t)


First Order Differential Equations

Given a differential equation which describes therelationship between y(t) and u(t), can an explicitformula for y(t) be given?

We consider a simple case, where n = 1 and m = 0:

d

dty(t) + a0y(t) = b0u(t).

It is called a first order differential equation (onlyfirst derivatives present).We are interested in obtaining a solution y(t) interms of u(t) and the initial condition y(t0) = y0.


Solution of First Order Differential Equations

To solve the equation we use an integrating factor, ea0t :

ea0td

dty(t) + ea0ta0y(t) = ea0tb0u(t)

d

dt

[ea0t · y(t)

]= ea0tb0u(t)

And, using the fact that:

∫ t

t0

[d

dσf(σ)

]

dσ = f(t) − f(t0)

we can integrate both sides to give:

y(t) · ea0t = y(t0) ea0t0 +

∫ t

t0

ea0σb0 u(σ) dσ


Solution of First Order Differential Equations

In summary, the first order differential equation:

d

dty(t) + a0y(t) = b0u(t)

has solution y(t) which is given by:

y(t) = y0ea0(t0−t) +

∫ t

t0

ea0(σ−t)b0 u(σ) dσ ; t ≥ t0

This solution, at time t = t0 passes through the point:

y(t0) = y0


Example: First Order System with Step Input

Consider u(t) = 1(t) and y(0) = y0, for the equation:d

dty(t) + a0y(t) = b0u(t).

The solution is given by y(t) = y0e−a0t +

b0

a0

[1 − e−a0t

].

b0

a0

y0

0Time (seconds) 5/a0

y(t) = y0e−a0t +b0

a0[1 − e−a0t]


State Space Descriptions

More generally, n,m > 1 in:

dn

dtny(t) + an−1

dn−1

dtn−1y(t) + · · · + a1

d

dty(t) + a0y(t) =

bmdm

dtmu(t) + · · · + b1

d

dtu(t) + b0u(t)

State-space descriptions reduce high order scalardifferential equations to vector first order ones.

The associated vector quantity is known as thestate-vector, and its dimension is equal to n.

The advantage of reduction to first-order vectordifferential equation is ability to provide solutions.


State Space Descriptions: The Second Order Case

Consider a second order differential equation (n = 2):

d2

dt2y(t) + a1

d

dty(t) + a0y(t) = b1

d

dtu(t) + b0u(t)

It may be re-written as:

d

dt

[d

dty(t) + a1y(t) − b1u(t)

]

+ a0y(t) = b0u(t)

Introduce the new signals:

x1(t) , y(t)

x2(t) ,d

dty(t) + a1y(t) − b1u(t)



Replacing, we have:

d

dtx1(t) + a1x1(t) − b1 = x2(t) (definition of x2(t))

d

dtx2(t) + a0x1(t) = b0u(t) (original 2nd order eq.)

These two coupled first-order differential equations may becompactly expressed in matrix-vector form as:

d

dt

[

x1(t)

x2(t)

]

=

[

−a1 1

−a0 0

][

x1(t)

x2(t)

]

+

[

b1

b0

]

u(t)

y(t) =[

1 0][

x1(t)

x2(t)

]



We then have a state-space description for the 2nd orderdifferential equation:

d

dtx(t) = Ax(t) + Bu(t)

y(t) = Cx(t).

where:

x(t) ,

[

x1(t)

x2(t)

]

, A ,

[

−a1 1

−a0 0

]

, B ,

[

b1

b0

]

, C ,[

1 0]


Example: Shock Absorber (Revisited)

For a car shock absorber, with y(t) = y(t) − mg/ks, weobtained a 2nd order model of the form:

d2

dt2y(t) +

kd

m

d

dty(t) +

ks

my(t) =

kd

m

d

dtu(t) +

ks

mu(t)

The corresponding state-space description is given by:

d

dtx(t) =

−kd

m1

−ks

m0

x(t) +

kd

m

ks

m

u(t)

y(t) =[

1 0]

x(t)


State Space Descriptions: The General Case

Consider the general differential equation, with m ≤ n − 1:n∑

k=1

ak

dk

dtky(t) + a0y(t) =

n−1∑

`=1

b`

d`

dt`u(t) + b0u(t)

It may be re-written as:

d

dt

[n∑

k=1

ak

dk−1

dtk−1y(t) −

n−1∑

`=1

b`

d`−1

dt`−1u(t)

]

︸︷︷︸

xn(t)

= −a0 y(t)︸︷︷︸

x1(t)

+b0u(t).

Which implies a first-order relationship:

d

dtxn(t) = −a0x1(t) + b0u(t).



This re-parameterisation can be repeated:

xn(t) ,

n∑

k=1

ak

dk−1

dtk−1y(t) −

n−1∑

`=1

b`

d`−1

dt`−1u(t)

=d

dt

[n∑

k=2

ak

dk−2

dtk−2y(t) −

n−1∑

`=2

b`

d`−2

dt`−2u(t)

]

︸︷︷︸

xn−1(t)

+a1 y(t)︸︷︷︸

x1(t)

−b1u(t)

so that:

d

dtxn−1(t) = xn(t) − a1x1(t) + b1u(t)



Repeating this process n − 1 times yields n coupled firstorder differential equations:

d

dtxn(t) = −a0x1(t) + b0u(t)

d

dtxn−1(t) = xn(t) − a1x1(t) + b1u(t)

d

dtxn−2(t) = xn−1(t) − a2x1(t) + b2u(t)

......

d

dtx1(t) = x2(t) − an−1x1(t) + bn−1u(t)

and where y(t) = x1(t)



Putting in matrix-vector form:

d

dt

x1(t)

x2(t)...

xn−1(t)

xn(t)

=

−an−1 1 0 · · · 0

−an−2 0 1 · · · 0... . . . . . . ...

−a1 0 0 1

−a0 0 · · · 0 0

x1(t)

x2(t)...

xn−1(t)

xn(t)

+

bn−1

bn−2...b1

b0

u(t)

y(t) =[

1 0 · · · 0 0]

x1(t)

x2(t)...

xn−1(t)

xn(t)



In summary, the n’th order differential equation:

dn

dtny(t) + an−1

dn−1

dtn−1y(t) + · · · + a1

d

dty(t) + a0y(t) =

bn−1dn−1

dtn−1u(t) + · · · + b0u(t).

may be re-expressed as a state-space representation, interms of the n dimensional state vector x(t):

d


y(t) = Cx(t)



Where:

A ,

−an−1 1 0 · · · 0

−an−2 0 1 · · · 0... . . . . . . ...

−a1 0 0 1

−a0 0 · · · 0 0

, B ,

bn−1

bn−2...b1

b0

,

C ,[

1 0 · · · 0 0]

This formulation is known as the observer canonical form.


State Transformations and Canonical Forms

The state space representation of a differentialequation is not unique.

Specifically, for any invertible matrix M ∈ Rn×n, thestate vector x(t) ∈ Rn×1 can be transformed to a newstate vector z(t), such that:

z(t) = Mx(t) ⇐⇒ x(t) = M−1z(t)

In this case:d

dtz(t) =

d

dtMx(t) = MAx(t) + MBu(t)

= MAM−1z(t) + MBu(t)

and y(t) = Cx(t) = CM−1z(t)



That is, the state space representation:

d

dtz(t) = A′z(t) + B′u(t)

y(t) = C′z(t)

where:

A′ , MAM−1, B′ , MB, C′ , CM−1

describes exactly the same relationship between thesignals y(t) and u(t)



For example, controller canonical form:

A ,

−an−1 −an−2 · · · −a1 −a0

1 0 · · · 0 0

0 1 · · · 0 0...

... . . . ......

0 0 · · · 1 0

B ,

1

0...0

0

C ,[

bn−1 bn−2 · · · b1 b0

]

There are also observability and controllabilitycanonical forms, similar to the preceding.


State Space Descriptions: The Feed-through Term

We have considered the general differential equation:

dn

dtny(t) + · · · + a2

d2

dt2y(t) + a1

d

dty(t) + a0y(t) =

bm

dm

dtmu(t) + · · · + b0u(t)

where we restricted m < n.

We now consider the case m = n, and hence there is aterm dn

dtn u(t) on the right hand side:

n∑

k=0

ak

dk

dtky(t) = bn

dn

dtnu(t) +

n−1∑

k=0

bk

dk

dtku(t).



We can rewrite the equation, remembering an = 1:

n∑

k=0

akdk

dtky(t) − bn

dn

dtnu(t) =

n−1∑

k=0

bkdk

dtku(t)

n∑

k=0

akdk

dtky(t) −

n∑

k=0

akbndk

dtku(t) =

n−1∑

k=0

bkdk

dtku(t) −

n−1∑

k=0

akbndk

dtku(t)

n∑

k=0

akdk

dtk(y(t) − bnu(t))︸︷︷︸

z(t)

=n−1∑

k=0

(bk − akbn)︸︷︷︸

bn

dk

dtku(t)

Considering z(t) as output variable, this can now be written instate-space form.



Then we have:

d


z(t) = Cx(t)

and using z(t) = y(t) − bnu(t), it is equivalent to:

d


y(t) = Cx(t) + D u(t)

where D = bn, and it is called the feed-through term:it is the proportion of u(t) directly fed-through to y(t).



In summary, the differential equation relationship:

dn

dtny(t)+an−1

dn−1

dtn−1y(t)+ · · ·+a0y(t) = bn

dn

dtnu(t)+ · · ·+b0u(t)

may be expressed as a state-space representation:d


y(t) = Cx(t) + Du(t)

In observer canonical form A,B,C, D are given as:

A �

��

�

−an−1 1 0 · · · 0

−an−2 0 1 · · · 0

......

. . ....

−a1 0 1

−a0 0 · · · · · · 0

��

�

B �

��

�

bn−1 − bnan−1

bn−2 − bnan−2

...

b1 − bna1

b0 − bna0

��

�

C � �1 0 0 · · · 0 � D � bn



While, in controller canonical form the matricesA,B,C, D are given as:

A �

��

�

−an−1 −an−2 · · · −a1 −a0

1 0 · · · 0 0

0 1 · · · 0 0

......

. . ....

...

0 0 · · · 1 0

��

�

B �

��

�

1

0

...

0

0

��

�

C �

��

�

bn−1 − bnan−1

bn−2 − bnan−2

...

b1 − bna1

b0 − bna0

��

�

T

D � bn


Direct State-Space Modelling

Sometimes, in physical modelling, it may be natural toprogress directly to a state space form:

+

−

+

−

+ −

vL(t)

vC(t) R vR(t)C

L

i(t)

Kirchoff’s laws:

iC(t) = Cd

dtvC(t) = i(t) − iL(t)

vL(t) = Ld

dtiL(t) = vC(t) − RiL(t)

Which are expressible as:

d

dtvC(t) = −

1

CiL(t) +

1

Ci(t)

d

dtiL(t) =

1

LvC(t) −

R

LiL(t)



Writing these two differential equations in matrix-vectorform gives:

+

−

+

−

+ −

vL(t)

vC(t) R vR(t)C

L

i(t)

State-space model:

d

dt

vC(t)

iL(t)

︸︷︷︸

x(t)

=

0 −1/C

1/L −R/L

︸︷︷︸

A

vC(t)

iL(t)

︸︷︷︸

x(t)

+

1/C

0

︸︷︷︸

B

i(t)



u(t)

y(t)

Suface Area = A2 m2

h2(t)

h1(t)

Suface Area = A1 m2

x(t)

Recall the coupled tank system, where:

A1d

dth1(t) = u(t) − x(t)

A2d

dth2(t) = x(t) − y(t)

The flow between the tanks is x(t) = K1[h1(t) − h2(t)].The output flow is y(t) = K2h2(t).



u(t)

y(t)

Suface Area = A2 m2

h2(t)

h1(t)

Suface Area = A1 m2

x(t)

This yields the state-space model:

d

dt

h1(t)

h2(t)

=

−K1/A1 K1/A1

K1/A2 −(K1 + K2)/A2

h1(t)

h2(t)

+

1/A1

0

u(t)

y(t) =[

0 K2

]

h1(t)

h2(t)


Differential equation solution via state-space

Recall that the 1st order scalar differential equation:

d

dty(t) = −a0y(t) + b0u(t), y(t0) = y0

is solved, for a given signal u(t), by:

y(t) = e−a0(t−t0)y0 +

∫ t

t0

ea0(σ−t)b0u(σ) dσ.

What is solution the 1st order vector state-spacedifferential equation:

d

dtx(t) = Ax(t) + Bu(t), x(t0) = x0 ?



Using the previous methods, the solution of:

d

dtx(t) = Ax(t) + Bu(t), x(t0) = x0

is: x(t) = eA(t−t0)x0 +

∫ t

t0

eA(t−σ)Bu(σ) dσ

where the Matrix-Exponential eAt is defined as:

exp(At) = eAt = I + At +A2t2

2!+ . . . =

∞∑

k=0

Aktk

k!

which satisfiesd

dteAt = AeAt = eAtA



Matrix Exponential: Example

Consider the matrix A =

1 2

3 4

.

Then according to the definition:

eA =

1 0

0 1

+

1 2

3 4

+1

2

1 2

3 4

2

+1

6

1 2

3 4

3

+ · · ·

=

51.9690 74.7366

112.1048 164.0738

Note that this is quite different to the matrix:

e1 e2

e3 e4

=

2.7183 7.3891

20.0855 54.5982



In summary, the solution y(t) of the equation:

dn

dtny(t) + an−1

dn−1

dtn−1y(t) + . . . + a0y(t) = bn

dn

dtnu(t) + . . . + b0u(t)

is given by:

y(t) = CeA(t−t0)x0 +

∫ t

t0

CeA(t−σ)Bu(σ) dσ + Du(t)

where: d



The solution y(t) satisfies y(t0) = Cx0 + Du(t0).


Differential Equation Solution via State-Space: Example

Consider the 2nd order equation (α, β ∈ R):

d2

dt2y(t) + (α + β)

d

dty(t) + αβy(t) = b1

d

dtu(t) + b0u(t).

In state-space form:

d


y(t) = Cx(t)

where:

A ,

[

−(α + β) 1

−αβ 0

]

, B ,

[

b1

b0

]

, C ,[

1 0]



The solution is:

y(t) = Cx(t) = CeA(t−t0)x0 +

∫ t

t0

g(t − σ)u(σ) dσ

where g(t) , CeAtB. To obtain eAt, we diagonalize A:

A = S−1ΛS

⇒ Ak = S−1ΛSS−1ΛS · · ·S−1ΛS = S−1ΛkS

⇒ eAt =

∞∑

k=0

Aktk

k!= S−1

(∞∑

k=0

Λktk

k!

)

S



Therefore:

g(t) = CeAtB

=1

(α − β)[1, 0]

αe−αt − βe−βt e−βt − e−αt

αβ[e−αt − e−βt

]αe−βt − βe−αt

b1

b0

= κ1e−αt + κ2e

−βt

And, if we consider u(t) = 1(t), x0 = 0, and t0 = 0:

y(t) =

∫ t

t0

g(t − σ)1(σ) dσ =

∫ t

0κ1e

−α(t−σ) + κ2e−β(t−σ) dσ

=κ1

α(1 − e−αt) +

κ2

β(1 − e−βt)



In the example, the assumptions

a1 = α + β

a0 = αβ

for some α, β ∈ R, are not completely general.

We didn’t consider the cases α = β or α, β ∈ C.

The difficulty with which the matrix exponential eAt wasobtained, is an indication that state-space formulationsmay not be the best method.

This will motivate our study of Laplace-Transformmethods (Chapter 4).


Initial Conditions

An important point is that the solution:

y(t) = CeA(t−t0)x0 +

∫ t

t0


is unique only up to the choice of x(t0) = x0.

The uniqueness of the solution depends upon theinitial conditions at time t = t0 for

y(t), dy(t)/dt, . . . , dn−1y(t)/dtn−1 and

u(t), du(t)/dt, . . . , dn−1u(t)/dtn−1.


Initial Conditions

Rewriting the solution y(t) as:

y(t) = CeA(t−t0)x0 + CeAt

∫ t

t0

e−AσBu(σ) dσ + Du(t)

then, differentiating with respect to t, yields:

d

dty(t) = CAeA(t−t0)x0+

CAeAt

∫ t

t0

e−AσBu(σ) dσ + CBu(t) + Dd

dtu(t)

Therefore, at time t = t0:

d

dty(t0) = CAx0 + CBu(t0) + D

d

dtu(t0)


Initial Conditions

Repeating this process, differentiating up to the n − 1’stderivative and setting t = t0, a pattern clearly emerges:

y(t0)

ddty(t0)

d2

dt2 y(t0)...

dn−1

dtn−1 y(t0)

︸︷︷︸

y0

=

C

CA

CA2

...

CAn−1

︸︷︷︸

O

x0+

D 0 · · · · · · 0

CB D 0 0

CAB CB D...

.... . .

. . ....

CAn−2B · · · · · · CB D

︸︷︷︸

Γ

u(t0)

ddtu(t0)

d2

dt2 u(t0)...

dn−1

dtn−1 u(t0)

︸︷︷︸

u0

which is expressible as y0 = Ox0 + Γu0.


Initial Conditions

A natural question is: given the initial conditions for u(t)and its derivatives (in u0), how should x0 be chosen tosatisfy the initial condition on the solution y(t) and itsderivatives (in y0) ?

The answer is simple:

x0 = O−1 [y0 − Γu0] , O ,

C

CA

CA2

...

CAn−1

The matrix O is called the observability matrix.


Initial Conditions: Example

Consider 2nd order system:

d2

dt2y(t) + 3

d

dty(t) + 2y(t) = 3

d

dtu(t) + 2u(t).

The state space description (observer canonical form) is:

d

dtx(t) =

−3 1

−2 0

x(t) +

3

2

u(t)

y(t) =[

1 0]

x(t)

Then y(t) = CeAtx0 +

∫ t

0

g(t − σ)u(σ) dσ

where g(t) = CeAtB = 4e−2t − e−t


Initial Conditions: Example

The observability matrix is given by:

O =

C

CA

=

1 0

−3 1

Now, suppose y0 = [−0.5,−1]T and u0 = [0, 0]T , then:

x0 = O−1y0 =

1 0

−3 1

−1

−0.5

−1

=

−0.5

−2.5

CeAtx0 =[

1 0]

2e−2t − e−t e−t − e−2t

2[e−2t − e−t

]2e−t − e−2t

−0.5

−2.5

= 1.5e−2t − 2e−t

so the complete solution is:

y(t) = 1.5e−2t − 2e−t +

∫ t

0

[

4e−2(t−σ) − e−(t−σ)]

u(σ) dσ.


Initial Conditions: Observability

For some state space realizations, the observabilitymatrix O, could be non invertible. For example:

d2

dt2y(t) + 3

d

dty(t) + 2y(t) =

d

dtu(t) + u(t)

has observer canonical form matrices:

A =

−3 1

−2 0

,C =[

1 0]

⇒ O =

C

CA

=

1 0

−3 1

But, the controller canonical form matrices are:

A =

−3 −2

1 0

,C =[

1 1]

⇒ O =

C

CA

=

1 1

−2 −2


Initial Conditions: Observability

This means that the state space model:

d

dtx(t) = Ax(t), x(t0) = x0

y(t) = Cx(t)

describes a richer class of systems than:

dn

dtny(t) + an−1

dn−1

dtn−1y(t) + · · · + a1

d

dty(t) + a0y(t) = 0

which only describes one input-output relationship.

If for a pair {A,C}, the observability matrix O isnon-singular, then the realisation is called observable.


Initial Conditions: Controllability

The concept of controllability involves the question ofwhether the state can be driven to an arbitrary valuex(t) from an arbitrary initial point x0.

To address this issue, we can assume D = 0 andx0 = 0. The state trajectory then is:

x(t) =

∫ t

t0

eA(t−σ)Bu(σ) dσ

where, the matrix exponential term is:

exp(At) = eAt = I + At +A2t2

2!+

A3t3

3!+ · · · =

∞∑

k=0

Aktk

k!



The Cayley–Hamilton theorem establishes that if A isn × n, then for some constant α0, · · · , αn−1:

An = α0I + α1A + α2A2 + · · · + αn−1A

n−1

As a consequence, every power of A can beexpressed in terms of {I,A, . . . ,An−1}.

Applying the theorem above, we have that:

eAt = I + At +A2t2

2!+

A3t3

3!+ · · · =

∞∑

k=0

Aktk

k!

= β0(t)I + β1(t)A + β2(t)A2 + · · · + βn−1(t)A

n−1 =n−1∑

k=0

βk(t)Ak



Substituting into the state trajectory:

x(t) =

∫ t

t0

n−1∑

k=0

βk(t − σ)AkBu(σ) dσ =n−1∑

k=0

AkB

∫ t

t0

βk(t − σ)u(σ) dσ

=[B,AB, · · · ,An−1B

]

︸︷︷︸

C

z0(t)

z1(t)...

zn−1(t)

= Cz(t)

where zk(t) ,

∫ t

t0

βk(t − σ)u(σ) dσ

The matrix C is known as the controllability matrix:

Any x(t) that are achievable by manipulation of u(t)must be a linear combination of the columns of C.



In summary, given an n-dimensional linear timeinvariant system with state space description:

d



the n × n controllability matrix C is defined as:

C ,[B,AB,A2B, · · · ,An−1B

]

The system is termed controllable if and only if C is aninvertible matrix (full rank or det C 6= 0).

Otherwise the system is termed uncontrollable.


Forced and Natural Response

The solution of any differential equation:dn

dtny(t) + an−1

dn−1

dtn−1y(t) + · · · + a0y(t) = bn

dn

dtnu(t) + · · · + b0u(t).

is expressible as:

y(t) = CeA(t−t0)x0︸︷︷︸

Natural Response

+

∫ t

t0


︸︷︷︸

Forced Response

.

The Natural Response dependes on thecoefficients {an−1, . . . , a0, bn, . . . , b0} and the initialconditions x0.The Forced Response dependes on thecoefficients {an−1, . . . , a0, bn, . . . , b0} and the inputsignal u(t).


Forced and Natural Response: Example

Recall the shock absorber model,

d2

dt2y(t) +

kd

m

d

dty(t) +

ks

my(t) =

kd

m

d

dtu(t) +

ks

mu(t)

with m = 1000 kg, ks = 2000 N/m, kd = 3000 Ns/m,andinitial conditions y(0) = −0.5 and dy(0)/dt = −1.

The response of the shock absorber is:

y(t) = 1.5e−2t − 2e−t

︸︷︷︸

Natural Response

+

∫ t

0

[

4e−2(t−σ) − e−(t−σ)]

u(σ) dσ

︸︷︷︸

Forced Response

.


Forced and Natural Response: Example

Suppose that u(t) = 1(t − 1), then for t > 1:

∫ t

0

[

4e−2(t−σ) − e−(t−σ)]

1(σ−1) dσ = 1−2e−2(t−1)+e−(t−1)

t2 4 6 8

Forced Response

Total Response

Natural Response

1

0

−0.5


Impulse Response

Suppose that:Initial time t0 → −∞, i.e., a long time ago,The initial conditions, given by the vector quantitiesy0,u0 and, hence x0, are all zero at t0 = −∞.

This yields only the forced response part:

y(t) =

∫ t

−∞

g(t − σ)u(σ) dσ + Du(t)

where g(t) , CeAtB.

Clearly, given u(t), the response y(t) depends on thefunction g(t) and the feed-through term D.


Impulse Response

In particular, if u(t) = δ(t) (Dirac delta), for t > 0:

y(t) =

∫ t

−∞

g(t − σ)δ(σ) dσ + Dδ(t) = g(t) + Dδ(t)

Thus, the system impulse response is defined as:

h(t) ,

{

g(t) + Dδ(t) ; t ≥ 0

0 ; t < 0

For t < 0, h(t) is zero because of the zero initialconditions.

From physical modelling, commonly D = bn = 0, soh(t) = g(t).


Impulse Response: Example 1

For an RC-circuit, we have seen that:

d

dtvC(t)︸︷︷︸

x(t)

= −1

RC︸︷︷︸

A

vC(t)︸︷︷︸

x(t)

+1

RC︸︷︷︸

B

v(t)

Therefore:

g(t) = CeAtB =1

RCe−t/RC ⇒ h(t) =

1

RCe−t/RC ; t ≥ 0

0 ; t < 0

t

1

RC

0

1

RCe−t/RC


Impulse Response: Example 2

For a car shock absorver, we obtained:

d2

dt2y(t) +

kd

m

d

dty(t) +

ks

my(t) =

kd

m

d

dtu(t) +

ks

mu(t)

If m = 1000 kg, ks = 1000 N/m, kd = 500 Ns/m, then:

h(t) = CeAtB = 1.0328e−0.25t cos(0.9682t − 1.0654), t > 0

t

1.0328e−0.25t cos(0.9682t − 1.0654)

0


Convolution

A key result we have is that for any input u(t), and zeroinitial conditions:

y(t) =

∫ t

−∞

g(t − σ)u(σ) dσ + Du(t), g(t) , CeAtB

=

∫ t

−∞

(g(t − σ) + Dδ(t − σ)) u(σ) dσ

=

∫ ∞

−∞

h(t − σ)u(σ) dσ

Noting: σ = [t,∞) ⇒ (t − σ) ∈ (−∞, 0] ⇒ h(t − σ) = 0.

We say that y(t) is the convolution of h(t) with u(t).

The new limits give a more general expression, oftennecessary (for example, for filter design).


Convolution

y(t) =

∫∞

−∞

h(t − σ)u(σ) dσ = [h ~ u] (t)

Intuitively, it means that y(t) is the sum of the effects ofu(t), weighted by h(t).

Graphically, h(t) characterizes the system:

u(t)

Linear System

h(t)

y(t)u(t)

y(t) =

∫ ∞

−∞



Properties of Convolution

Commutative Property:

[h ~ u](t) =

∫ ∞

−∞

h(t − σ)u(σ) dσ = −

∫ −∞

∞

h(x)u(t − x) dx

=

∫ ∞

−∞

u(t − x)h(x) dx = [u ~ h](t).

Linearity:

[h ~ (αu + βw)] (t) =

∫ ∞

−∞

h(t − σ) (αu(σ) + βw(σ)) dσ

= α

∫ ∞

−∞

h(t − σ)u(σ) dσ + β

∫ ∞

−∞

h(t − σ)w(σ) dσ

= α[h ~ u](t) + β[h ~ w](t).



Graphically, linearity means that linear combination ofsignals can be done before or after a convolution operation,leading to the same final result.

+

+ +

+

y(t)u(t)

y(t)

Linear Systemh(t)

Gain α

w(t)

w(t)Gain β

y(t)

Linear Systemh(t)

Linear Systemh(t)

w(t)

Gain α

Gain β

u(t)

w(t)

[h � (αu + βw)](t) = α[h � u](t) + β[h � w](t)



Convolution and Dirac δ functions:

[h ~ δ](t) =

∫ ∞

−∞

h(t − σ)δ(σ) dσ = h(t)

Convolution and Time Shifted Signals:

[h ~ g](t) =

∫ ∞

−∞

h(t − σ)g(σ) dσ

=

∫ ∞

−∞

h(t − σ)u(σ − T ) dσ

and changing variable x = σ − T :

[h ~ g](t) =

∫ ∞

−∞

h((t − T ) − x)u(x) dx = [h ~ u](t − T )



Associative property:

[h ~ (u ~ f)](t) =

∫ ∞

−∞

h(t − σ)[u ~ f ](σ) dσ

=

∫ ∞

−∞

h(t − σ)

[∫ ∞

−∞

u(σ − x)f(x) dx

]

dσ

=

∫ ∞

−∞

[∫ ∞

−∞

h(t − σ)u(σ − x) dσ

]

f(x) dx

changing variable τ = σ − x:

[h ~ (u ~ f)](t) =

∫ ∞

−∞

[∫ ∞

−∞

h(t − x − τ)u(τ) dτ

]

f(x) dx

=

∫ ∞

−∞

[h ~ u](t − x)f(x) dx

= [(h ~ u) ~ f ](t).



Combining commutativity and associativity, we have,for example:

[h~(u~f)](t) = [h~(f~u)](t) = [(h~f)~u](t) = [(f~h)~u](t)

Linear System

h(t)

y(t)

u(t)

u(t)

Linear System

f(t)

y(t)

u(t)

u(t)

Linear SystemLinear System

f(t) h(t)

SameInput

SameOutput

OrderReversed


Convolution: Example

To apply the convolution:

y(t) =

∫∞

−∞


in practice, we have to consider:the effect of changing h(σ) to h(t − σ) involves theflipping and sliding of that signal,the integral is then the area obtained for differentintervals, andconmutativity ensures that we can also flip andslide u(t), obtaining the same final result.


Convolution Example: RC Circuit

For an RC Circuit we already obtained:

h(t) =

1

RCe−t/RC ; t ≥ 0

0 ; t < 0

σ0

1RC h(σ) =

1

RCe−σ/RC

Assume that the voltage is:

v(t) = 1(t) − 1(t − 1)

1 σ

1

0

v(σ)

Note that,flippingand shifting, weobtain h(t − σ):

h(t − σ)

σ0 t



Note that:

y(t) = 0, t ≤ 0

σ0 1

v(σ)h(t − σ), t = 0

y(t) =

∫ t

0

1

RCe(σ−t)/RC dσ

= 1 − e−t/RC

0 < t ≤ 1 � ��

� ��

∞

−∞


h(t − σ), t = 0.5 v(σ)

10.50

y(t) =

∫ 1

0

1

RCe(σ−t)/RC dσ

= e−t/RC[

e1/RC − 1]

t > 1 � � ��

σ0 1 t

h(t − σ), t > 1v(σ)

�

∞

−∞




Combining the 3 previous results:

vC(t) =

0 ; t ≤ 0

1 − e−t/RC ; t ∈ (0, 1][e1/RC − 1

]e−t/RC ; t > 1

t10

1 − e−1/RC

1 − e−t/RC

vC(t) =

∫ ∞

−∞

h(σ)v(t − σ) dσ

[e1/RC − 1]e−t/RC


Convolution Example: RC Circuit (again)

The previous example can be repeated, but using thecommutative property:

∫ ∞

−∞

h(t − σ)v(σ) dσ =

∫ ∞

−∞

v(t − σ)h(σ) dσ

In this case, the impulse response remains static for allt as:

σ0

1RC h(σ) =

1

RCe−σ/RC

However, now v is flipped and shifted.



Then we have:

y(t) = 0, t ≤ 0

σ01

v(t − σ), t = 0

h(σ)

y(t) =

∫ t

0

1

RCe−σ/RC dσ

= 1 − e−t/RC

0 < t ≤ 1

� ��

� ��

σ0

h(σ)

v(t − σ), t = 0.5

tt − 1 = −0.5

y(t) =

∫ t

t−1

1

RCe−σ/RC dσ

= e−t/RC[

e1/RC − 1]

t > 1 � � ��

σ

h(σ)v(t − σ), t > 1

t0 t − 1Chapter 3. Modelling, Differential Equations and System Properties – p.83/100


Combining the 3 results, we obtain the same previoussolution:

vC(t) =

0 ; t ≤ 0

1 − e−t/RC ; t ∈ (0, 1][

e1/RC − 1]

e−t/RC ; t > 1

��

��

t10

1 − e−1/RC

1 − e−t/RC

vC(t) =

∫ ∞

−∞

h(σ)v(t − σ) dσ

[e1/RC − 1]e−t/RC

This verifies that it does not matter which of the twofunction in a convolution is flipped and shifted.


Convolution Example: Rectangular Signals

As a final more abstract example of convolution,suppose that u(σ) and h(σ) are rectangular signals:

u(σ)

1 σ

1

0

h(σ)

σ

1

0 0.5

In this case, if we elect to flip and slide h as h(t − σ):

h(t − σ)

t − 0.5 σ0 t



Then we have:

y(t) = 0, t ≥ 1.5

σ

u(σ)

0 1 1.5

h(t − σ), t = 1.5

y(t) =

∫ t−0.5

1

dσ

= σ∣∣∣

σ=t−0.5

σ=1= 1.5 − t

1 ≤ t < 1.5 �� ∞−∞


σ

u(σ)

1.250.75

h(t − σ), t = 1.25

0

y(t) =

∫ t

t−0.5

dσ

= σ∣∣∣

σ=t−0.5

σ=t= 0.5

0.5 ≤ t < 1 � � ��

σ0

�∞−∞

h(t − σ)u(σ) dσu(σ)

h(t − σ), t = 0.75

0.750.25 1



Using similar arguments, if 0 ≤ t < 0.5:

y(t) =

∫ t

0

dσ = σ∣∣∣

σ=t

σ=0= t

And finally, if t ≤ 0, the product h(t − σ)u(σ) is zero forall σ and hence:

y(t) = 0 ; t < 0.

Combining all the results, we have the full solution:

y(t) =

��

0 ; t ≥ 1.5

1.5 − t ; t ∈ [1, 1.5)

0.5 ; t ∈ [0.5, 1)

t ; t ∈ [0, 0.5)

0 ; t < 0 t0.5 1.510

0.5

y(t) = �∞−∞

h(σ)u(t − σ) dσ


System Properties

Via electrical, mechanical and differential equationmodelling examples we have gradually introduced theidea of a system.

In a more abstract and general sense we have:

A system S is a governing influence whichregulates the relative behaviour of signals. Thenotation used to denote this regulation is thatthe nature of a signal y(t) relative to anothersignal u(t) is given as

y(t) = S[u](t).


System Properties

Linearity/Non-linearity:A system S is linear if, for any signals u(t), x(t) and

any constant numbers α, β:

S[αu + βx](t) = αS[u](t) + βS[x](t)

The systems expressible as convolution with animpulse response, S[u](t) = [h ~ u](t), are linear:

S[αu + βx](t) = [h ~ (αu + βx)](t) = α[h ~ u](t) + β[h ~ x](t)

= αS[u](t) + βS[x](t)

The system S[u](t) = u2(t) is non-linear since:

S[αu+βx](t) = (αu(t) + βx(t))2

= α2u2(t)+β2x2(t)+2αβu(t)x(t)

6= αu2(t) + βx2(t) = αS[u](t) + βS[x](t)


System Properties

Causality:A system is causal or non-anticipative if y(t?) = S[u](t?)

does not depend on future values of u(t) (i.e. t > t?).

All physical systems are causal (except in certainquantum phenomena).The linear system:

y(t) = [h ~ u](t) is causal ⇐⇒ h(t) = 0 for t < 0.

To see this, in

y(t?) = [h ~ u](t?) =

∫ ∞

−∞

h(σ)u(t? − σ) dσ.

note that if h(σ) 6= 0 for σ < 0, then y(t) depends onfuture values of u(t), because t? − σ > t?.


System Properties

Memory:A causal system S is memory-less if y(t?) = S[u](t?)

depends only on u(t) at t = t? (otherwise it has memory,since y(t?) is affected by prior values of u).

The simple amplifier system y(t) = S[u](t) = Ku(t) forsome K ∈ R is a memory-less system.The linear systemy(t) = [h ~ u](t) is memory-less ⇐⇒ h(t) = Kδ(t), K ∈ R.

To see this, in

y(t?) = [h ~ u](t?) =

∫ ∞

−∞

h(σ)u(t? − σ) dσ

note that, if h(σ?) 6= 0 for σ? > 0 then y(t?) depends onh(σ?)u(t? − σ?).


System Properties

Time Invariance:A system S is time-invariant if, a time time shifting in

the input generates a time shifting in the output:

x(t) = u(t − T ) ⇒ z(t) = S[x](t) = S[u](t − T ) = y(t − T )

The simple amplifier system:

y(t) = S[u](t) = Ku(t)

for some K ∈ R is time invariant. But, if the gain istime-varying:

y(t) = S[u](t) = | sinπt|u(t)

would be time-varying.


System Properties

System Dimension: A system described by an n-thorder differential equations can be represented as:

d



where x(t) is a state vector of dimension n. In thiscase, the system is said to have dimension n.

When n < ∞, the system is finite dimensional.For the non-linear case:

d

dtx(t) = f(x(t), u(t))

y(t) = g(x(t), u(t))

Again, the dimension is that of the vector x(t).Chapter 3. Modelling, Differential Equations and System Properties – p.93/100

System Properties

Stability:Intuitively, a system is stable if small perturbations

to it are only capable of producing small responses.

y(t) = CeA(t−t0)x0︸︷︷︸

Natural Response

+

∫ t

t0

g(t − σ)u(σ) dσ + Du(t)

︸︷︷︸

Forced Response

; g(t) , CeAtB.

Asymptotic Stability:It concerns on the natural response only.If we diagonalize the matrix A = S−1ΛS, then:

y(t) = CeA(t−t0)x0 = CS−1eΛ(t−t0)Sx0

where Λ contains the eigenvalues of A.


System Properties

Asymptotic Stability.If all the eigenvalues λ1, . . . , λn have negative real part:

eΛ(t−t0) =

eλ1(t−t0) 0 · · · 0

0 eλ2(t−t0) · · · 0...

. . ....

0 · · · 0 eλn(t−t0)

−−−→t→∞

0

Hence y(t) → 0, for any x0. Then we say that a linear,finite dimensional and time invariant system:

d



is asymptotically stable ⇐⇒ Re{λk} < 0 ; k = 1, . . . , n.Chapter 3. Modelling, Differential Equations and System Properties – p.95/100

System Properties

Bounded-Input Bounded-Output (BIBO) Stability:This involves the idea that the system response y(t) isbounded, for any bounded input u(t), regardless of theinitial conditions.

|u(t)| ≤ M < ∞ ⇒ |y(t)| =

∣∣∣∣

∫ ∞

−∞


∣∣∣∣

≤ M

∫ ∞

−∞

|h(t − σ)|dσ

And ‖h‖1 =

∫ ∞

−∞

|h(σ)|dσ =

∫ ∞

0

∣∣CeAtB + Dδ(t)

∣∣dt

=

∫ ∞

0

∣∣CS−1eΛt SB

∣∣ dt + D

≤n∑

k=1

|αk|

∫ ∞

0

∣∣eλkt

∣∣ dt + D


System Properties

Bounded-Input Bounded-Output (BIBO) Stability:Then, we have that a linear, finite dimensional and timeinvariant system:

d



is bounded-input bounded-output stable if all theeigenvalues λ1, · · · , λn of A, satisfy:

Re{λk} < 0; k = 1, 2, · · · , n.

However, the converse is not true.


System Properties

Stability: ExampleAgain, we consider the general 2nd order linear time invariantsystem:

d2

dt2y(t) + a1

d

dty(t) + a0 y(t) = b1

d

dtu(t) + b0 u(t)

In state-space representation:

A ,

λ1 + λ2 1

−λ1λ2 0

, B ,

b1

b0

, C ,[

1 0]

The eigenvalues of A are λ1,2 = −1

2

(

a1 ±√

a21 − 4a0

)

Thus, the nature of the response y(t) is completely captured by theterms eλ1t and eλ2t.


System Properties

Stability: ExampleWith x0 = 0, the response to u(t) = ε1(t) for t > 0 is:

y(t) =ε

λ1

(λ1b1 + b0

λ1 − λ2

)

(eλ1t − 1) +ε

λ2

(b0 + λ2b1

λ1 − λ2

)

(1 − eλ2t)

If u(t) = 0 and the initial condition is x0 = [x10 , x2

0]T , then:

y(t) =

(λ1x

20 + x1

0

λ2 − λ1

)

eλ1t +

(x1

0 + λ2x20

λ1 − λ2

)

eλ2t, t > 0

If λ1,2 = α ± jω, then eλ1,2t = eαt [cos ωt ± j sinωt].If α = Re{λ1,2} < 0, then eλ1,2t will tend to zero as t grows.If α > 0, then eλ1,2t will explode towards ∞ as t grows.

Therefore, the system is bounded-input bounded-output andasymptotically stable if Re{λ1,2} < 0.


System PropertiesNote that a system can be BIBO stable even if aneigenvalue violates the condition Re{λk} < 0.Example. Consider the case of:

d2

dt2y(t) +

d

dty(t) − 2 y(t) = b1

d

dtu(t) + b0 u(t)

The eigenvalues of the matrix A =[−1 12 0

]are λ1, λ2 = −2, 1, and

hence Re{λ2} = 1 > 0.However, the response to u(t) = ε1(t) (with x0 = 0) is:

y(t) =ε

λ1

(λ1b1 + b0

λ1 − λ2

)

(eλ1t−1)+ε

λ2

(b0 + λ2b1

λ1 − λ2

)

(1−eλ2t) ; t > 0

But if we choose b1 = 1 and b0 = −1, then

y(t) =ε

−2

(−2 + (−1)

−2 − 1

)

(e−2t−1)+ε

1

(−1 + 1

−2 − 1

)

(1−et) =ε

2(1−e−2t)

and thus the system is BIBO stable.


slides chap3

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