Theory of Linear Systems Lecture Notes - disi.unitn.itdisi.unitn.it/~palopoli/courses/TS/TSchap3.pdf · Theory of Linear Systems Lecture Notes Luigi Palopoli Dipartimento di Ingegneria

Theory of Linear Systems

Lecture Notes

Luigi Palopoli

Dipartimento di Ingegneria e Scienza dell’Informazione

Universita di Trento

Contents

List of symbols and notations vii

Chapter 1. Introduction 1

Chapter 2. Definition of system 32.1. Signals 52.2. Systems 82.3. I/O Representation 112.4. State Space representation 152.5. Linear Systems 202.6. Time Invariance 222.7. Hybrid Systems 23

Chapter 3. Impulse response and convolutions for linear and timeinvariant IO Representations 25

3.1. Forced Evolution of discrete–time systems 253.2. Forced Evolution of continuous–time systems 313.3. Properties of the impulse response 35

Chapter 4. Laplace and z-Transform 394.1. Complex exponentials 394.2. Complex Exponential functions 404.3. Definition of the Laplace Transform 404.4. Existence and Uniqueness of the Laplace transform 424.5. Properties of the Laplace transform 444.6. Inversion of the Laplace Transform 504.7. BIBO Stability for CT systems 604.8. Use of Laplace Transform for Control Design 664.9. The z-Transform 704.10. Existence and Uniqueness of z-Transform 714.11. z-Transform of sampled data signals 72

Chapter 5. State space Realisation 75

Chapter 6. Structural Properties of Systems 77

Chapter 7. Linear Systems Control 79

Appendix A. Some mathematical definition of interest 81

v

vi CONTENTS

Bibliography 83

List of symbols and notations

• R: set of real numbers• N: set of natural numbers• C: set of complex numbers• Z (:) set of integer numbers• T : time space• U : class of input signals• Y : class of output signal• S : system• 1(:) step function defined as

1(=)

{0 if t < 0

1 if t ≥ 0

• L (.): Laplace transform• L−1 (.): inverse Laplace transform• Real (z): real part of a complex number• Imag (z): imaginary part of a complex number• |z|: modulus of a complex number• ∠z: phase of a complex number• z: complex conjugate.

vii

CHAPTER 1

Introduction

This lecture notes are related to the course of “Theory of Linear Sys-tems£ taught in the undergraduate class of Telecommunications and Com-puting Engineering at the University of Trento. The presentation of LinearSystems proposed here is necessarily synthetic and introductory. The readerintested in additional details is referred to one of the books that exist on thethe topic. Two excellent examples are offered by the book written by JoaHespanha on Linear Systems [Hes09] or by the book writtern by Philippset al. [PPR95] on linear systems.

1

CHAPTER 2

Definition of system

This course revolves around the notion of timed systems. For our pur-poses, a system is a physical or artificial entity (e.g., a computer program),where a number of meaningful quantities evolve in a way that can be de-scribed using a mathematical formalism. The evolution of our quantities ofinterest is described by a mathematical function that associates a value of avariable defined “time” with a value taken by the quantity which is definedin an appropriate set (e.g., voltage, position, velocity etc.). Such functionswill be defined “signals”.

In some cases we will have a convenience in defining a system as a relationbetween input signals and output signals. In other cases, a system is bestthought as a relation between the evolution of some quantities. An exampleis offered by an economic system or by a human body. In this case, we couldhave a convenience in identifying input and output quantities or not (as anexample, the system could evolve autonomously and have no input).

Our illustration of this notion is best given using examples.

Example 2.1. Consider the RC circuit shown in Figure 2.1. In thissystem, we can easily identify several quantities of interest, for which it ispossible to take measurements. The most evident ones are currents andvoltages across the resistor and the capacitor. The laws of physics allow usto describe the evolution of the different quantities in time.

In this example, by time we mean exactly the “physical time”, whichflows with continuity and allows us to observe the evolution of physicalphenomena.

Let us move to a completely different example.

Figure 1. A simple RC network

3

4 2. DEFINITION OF SYSTEM

0 1

23

0/00

1/01

0/01

1/10

0/10

1/11

0/11

1/00

Figure 2. State machine implementing a modulo 4 counter

Example 2.2. Consider a digital circuit that counts the number of timesthat a certain event occurs (e.g., input = 1) modulo 4 and that prints thisnumber as an output on the occurrence of any event. Recalling basic notionsof digital circuits, it is easy to see that a circuit like this can be modelledas the state machine shown in Figure 2.2. What we have just shown in theFigure is commonly referred to as a “state transition diagram”, which is agraph where circles represent states and arcs are associated to transitions. Atransition is labelled by an input/output pair, meaning that the transitionis taken for the specified input and results in the emission of the specifiedoutput. In our example, if the system receives a 1 as an input when instate 0, it moves into state 1 producing as output 01. The label associatedwith the transitions is therefore “1/01”, where the first number representsthe input that triggers the transition and in the second number representsthe output generated when the transition is taken. In this example, we areimplicitly using a binary representation for numbers.

The system in the example could be implemented by a sequential elec-tronic circuit or by a software programme and is different from the analoguecircuit discussed in Example 2.1 in many respects. The most importantone is that we are interested in describing the input/output relation (i.e.,what each input sequence produces) without being necessarily interested inthe point in time where each “reaction” of the machine takes place. Thenumber of events that the system can respond to is clearly infinite, butgiven the amount of time that the system needs to produce a reaction wecannot generate events continuously in time. One possible way to formalisethis is that the number of events that intervene between any two events isfinite [LSV98] is always finite.

2.1. SIGNALS 5

As known from elementary courses of computer systems, it is possible toreduce our freedom and require that the system reacts only at precise pointsin time. Indeed, most sequential networks are implemented in a “synchro-nous” way, with reaction to events forced to take place upon the rising orfalling edge of a clock signal1. In this case, the minimum time intervalbetween two different reactions is bound to the clock period. However le-gitimate and reasonable, this is just one of the possible implementation ofa state machine. In a different asynchronous implementations, the machinereacts to an event whenever it occurs. We could say that in Figure 2.2 wehave offered an abstract specification of the state machine, which is open todifferent implementation options for the designer.

There are cases where the system is inherently synchronous. Considerthe following example.

Example 2.3. Consider a banking account where losses or gains arecapitalised every T time units (where T is a sub-multiple of one year). Thecustomer is allowed to deposit or withdraw money every time an intervalexpires. If at the beginning of an interval the capital is positive, at the endthe credit grows with an annual interest rate I+. If it is negative, at the endthe debt grows with an annual interest rate I−.

Let C(kT ) be the capital at the beginning of the k−th observation in-terval and let C((k + 1)T ) be the capital at the end. Finally, let S(kT ) bethe amount of money (positive or negative) that the customer deposits orwithdraws. The evolution of the capital is given by:(2.1)

C((k + 1)T ) =

{(1 + I+/12)T (C(kT ) + S(kT )) Se C(kT ) + S(kT ) ≥ 0

(1 + I−/12)T (C(kT ) + S(kT )) Se C(kT ) + S(kT ) < 0

At a fist sight, this system looks very similar to the state machine de-scribed earlier. Once again, it is important to know the exact sequence ofdeposits and withdrawals. Therefore, it is required that the time space Tbe an ordered set. However, it is not enough. This time, we need to assumethat the time between intervals between events are of the same durationand we need to know how large this interval is (otherwise the computationof the interest is not possible). Contrary to the modulo counter in Exam-ple 2.2, the banking account example is not only synchronous in one of itspossible implementation, but it is inherently so since its evolution is tiedto the evolution of physical time. However, contrary to the RC network inExample 2.1, it is not required (nor possible) to have arbitrarily close events.

2.1. Signals

A common presence in all the systems mentioned above are “signals”.A signal is simply defined as a function from a time space T to a set U. At

1When no event occurs on a clock, the network does not react


this point, we need not make any particular assumption on the set U (wewill have to later on).

We will frequently use calligraphic letter to denote classes of signals. Forexample,

U = {u(·) : T → U} .

2.1.1. The notion of time. A first problem to face is to define theexact meaning for the notion of time, a set that we will henceforth denoteby T .

2.1.1.1. Continuous Time. Continuous time (CT) signals are based ona time space that has to be:

(1) totally ordered,(2) metric,(3) a continuum (in the mathematical sense).

The exact meaning of these concepts is recalled in Appendix A. At thispoint, it is more useful to elaborate on the intuition underneath and on whywe need them to describe CT signals (and systems).

Let us start with an example.

Example 2.4. Consider the RC circuit in Example 2.1, VR(t) can beused to denote the voltage drop across the resistor R as a function of timeand is a CT signal. The same applies to the voltage VC(t) measured overthe capacitor C, for Vin(t) and for I(t):

CT signals are used to represent the evolution of physical quantities. Inthis setting, if a quantity takes on two different values at different time itis relevant to know which one comes first and which one follows, and howfar away the two events are. By the adjective “relevant” we mean that theevolution of the system will be quite different if the order of two events isinverted or if their distance is changed. To understand this, consider anelectrical RC circuit as in Example 2.1. Suppose that a 5V power sourceis connected to the terminals at time t1 and switched off at time t2. Theevolution of the system is understandably different if t2 − t1 = 10−3s ort2− t1 = 105s. Moreover, the value of a quantity like Vr can me measured atany time t and is potentially affected by the evolution of a different quantityat all possible times t′ ∈ T . These considerations explain why we need aset that is a continuum to mode CT signals. Therefore, the most obviouschoice for the time space is to choose it as the real set: T = R.

2.1.1.2. Discrete Events. The opposite case to CT signals is when theconnection between the time space T and the physical time is very shallow.Consider the system shown in Example 2.2. If we are only interested in theevolution of the output sequence given the input sequence, all we need toknow on the timing of the event is condensed in their order. For instance,the input sequence 011 will certainly produce a different sequence of statesand of output than the sequence 101 which only differs from the first onefor the order of the first two symbols.

2.1. SIGNALS 7

An important point is that an input sequence (e.g., 0 · 1 · 1) will producethe same output sequence whatever the spacing between any two events(be it one millisecond or one century). So if our goal is to specify theoutput sequence given an input sequence, we do not need continuity ormetric structure for our time space T .

Example 2.5. For the modulo counter in Example 2.2, a sequence ofinput such as 0 ·1 ·0 ·1 ·1 · . . . is an example of a DE signal. So is the sequenceof output 00 · 01 · 01 · 10 · . . ..

Discrete event (DE) signals are defined over time spaces that are totallyordered but are not required to be a continuum and to be metric. Some au-thors [LSV98] suggest that a in order to define a DE signal the time spacehas to satisfy the additional property that between any two events we canhave a finite number of intervening events. This requirement is formally cap-tured by imposing that events are order-isomorphic to the natural numbers.This subtle but important point is beyond the scope of these notes.

For our purposes, we can simply say that a DE signal consists of asequence of totally ordered events. Each of them can be associated with anincreasing natural number that reflects its position in the sequence. So, inthe sequence 011, the first 0 will be associated with the event 0, the first 1with event 1 and the third 1 with event 3.

2.1.1.3. Discrete Time. As mentioned above, DE signals are ordered se-quences of events each one associated with a time instant. For an importantsubclass of these signals, events have to be synchronous, meaning that theyoccur/are registered at specified time instants. This means that the timevariable the signal is defined over can take on only specific values. A typical(but not mandatory choice) is that such instants be multiple of a specifiedperiod T .

Generally speaking, we define a discrete–time (DT) signal a DE signalwhere events are constrained to be synchronous.

Example 2.6. In the banking account described in Example 2.3, theevolution of the capital C(kT ) and the sum S(kT ) deposited or withdrawnare examples of DT signal.

As in the case of DE signals, we can order the different time instantsusing the natural numbers. But, for DT signals the algebraic structure ofthe time space (i.e., its being an abelian group) is used in full for sums anddifferences. hence, we will use the set Z (t)o represent DT systems.

2.1.2. Sampled Data. Finally, it is useful to mention the presence ofa different type of systems, compounded of a collection of heterogeneoussubsystems, each one associated with a different time space. A classic ex-ample is offered by sample data systems, which are DT systems obtainedfrom CT systems restricting the points in time where certain quantities canbe measured (sampled) or certain input variables be changed.


Figure 3. Example of a sampled data system

An example of this type is shown in Figure 2.1.2. A given physicalquantity that evolves in continuous time is sampled at some points in timethat are spaced by a fixed amount of time (sampling interval). The outcomeis a sequence of numbers that is a DT system. We can have this signalprocessed through a DT system that generates a new DT signal (sequence ofsamples), which can in its turn be converted back into a CT signal througha digital to analogue converter. This is typically done by sampling thesignal value at time kT and holding it constant throughout the interval[kT, (k + 1)T ] (Zero Order Hold).

2.2. Systems

In this course, we will concentrate on systems defined on CT signalsand DT signals. We start by giving an abstract definition general enough toencompass most systems of interest. Let U taking value in the set Urepresenta class of input signals and Y represent a class of output signals taking valuein the set Y . We could define the system as a binary relation between U andY : S ⊆ U × Y .

We recall that a binary relation is a set of pairs. In this case one ele-ment of the pair is the input signal and the other one is the output signal.Importantly, the relation can associate more than one output signal to thesame input.

We remark that this definition of system is “oriented”, i.e., it assumes thedistinction between an input (cause) and of an output (effect). In a moregeneral setting, we could simply define it as a relation between tuples ofsignals, no-matter the role each one plays. This is a “speculative” viewpointwhich is of interest for such areas as theoretical physics or biology. In thiscourse, on the contrary, we will take an “engineering perspective” where adistinction is made between quantities that can be acted upon and otherquantities that evolve as a result of these actions.

Example 2.7. Consider the system in Example 2.1 and suppose thatthe input is given by Vin(t) and the output is given by Vc(t). Suppose weapply a step input Vf1(t) where the step 1(t) is defined as

1(t) =

{0 if t < 0

1 if t ≥ 0.

2.2. SYSTEMS 9

It can be seen (we will) that VC(t) = Vc(0)e−t/RC+(1−et/RC)Vf . Therefore,depending on the initial value Vc(0) we obtain a different output for the sameinput.

Example 2.8. Consider the system in Example 2.3. Suppose that attime t = 0 we start with a capital C0 and that we deposit a constant amountof money S at the beginning of every capitalisation period. It is possible tosee that the capital at time kT is given by

c(kT ) = (1 + I+)kC0 + S(1 + I+)k+1 − 1

I+.

Depending on the initial capital, the evolution can be much different.

As we have just seen, we could have a different output associated to thesame input both for the case of CT systems and of DT systems.

In the examples above, we also underscore the choice of an initial pointin time to start the observation of the evolution of the system. Generallyspeaking, we can choose an instant t0. Denote by T (t0) the subset of thetime space containing instants

T (t0) = {t ∈ T : t ≥ t0} .

We can denote by W T (t0) the set of functions defined over T (t0) that takevalue in W :

W T (t0) = {w0(·) : ∀t ≥ t0, t→ w0(t) ∈W} .

By w0|T (t1) we will denote the truncation of the function, which is evaluatedfrom t1 > t0 onward.

Now we can formally define a system as follows:

Definition 2.9. An abstract dynamic system is a 3-tuple {T ,U ×Y ,Σ}where

• T is the time space• U is the set of input functions• Y is the set of output functions

and

Σ ={

Σ(t0) ⊂ U T (t0) × Y T (t0) : t0 ∈ T and CRT is satisfied},

where CRT stands for closure with respect to truncation: i.e., ∀t1 ≥ t0

(u0, y0) ∈ Σ(t0) =⇒(u0|T (t1), y0|T (t1)

)∈ Σ(t1).

In plain words, the CRT property means that if a couple of functionsbelong to the system from t0, so will their truncation from t1 on.


2.2.1. Parametric representation. The binary relation R = U × Yis made of ordered pairs (u, y), we denote by D(R) the domain of the relation(i.e., U ) and by R(R) its range (i.e., Y ).

A general result applicable to binary relations is the following [RI79]

Lemma 2.10. Given a binary relation R, it is possible to define a set Pand a function π : P ×D(R)→ R(R) such that

(a, b) ∈ R =⇒ ∃p : b = π(p, a)(2.2)

p ∈ P, a ∈ D(R) =⇒ (a, π(p, a)) ∈ R(2.3)

π is said parametric representation and (P, π) is said parametrisation ofthe relation.

This essentially means that it is possible to represent a relation as theunion of a set of function graphs parametrised by a suitable choice of pa-rameters. The main interest of this result lies in the following:

Theorem 2.11. Consider a system defined as in Definition 2.9. It ispossible to identify a parametrisation (Xt0 , π) such that

(2.4) π = {πt0 : Xt0 ×D(Σ(t0))→ R(Σ(t0))/t0 ∈ T }satisfying the following properties:

(u0, y0) ∈ Σ(t0) =⇒ ∃x0 : y0 = πt0(x0, y0)(2.5)

x0 ∈ Xt0 , u0 ∈ D(Σ(t0)) =⇒ (u0, πt0(x0, u0)) ∈ Σ(t0).(2.6)

This result is easily understood looking at Example 2.7 and 2.8, wherethe Vc(0) and C0 are used as parameters.

2.2.2. Causality. We are now in condition to state the notion of causal-ity.

Definition 2.12. Let u|[t0,t] be the restriction of the function u to the

closed interval [t0, t]. A system is causal if it has a representation (Xt0 , π)such that

∀t0 ∈ T , ∀x0 ∈ Xt0 , ∀t ∈ T(2.7)

u[t0, t]= u′[t0, t] =⇒ [πt0(x0, u)](t) = [πt0(x0, u

′)](t).(2.8)

A parametric representation of this type is said causal. If instead of theclosed interval [t0, t] we use the semi-open interval [t0, t), the parametricrepresentation and the system is said strictly causal.

Observe that π is a functional, i.e., a function defined over space offunctions. So πt0(x0, u) is the output function associated to the parameterx0 and to the function u, and [πt0(x0, u)](t) is the value it takes at time t.In plain words, the definition just introduced means that the values of ubeyond t do not affect the value of the output at time t. A simple way toput it is that a causal system does not foresee the future. If the system isstrictly the output at time t is only affected by the input at time strictlysmaller than t.

2.3. I/O REPRESENTATION 11

2.2.3. Number of input and output signals. In the discussionabove, we have not made specific assumptions on the range U of the in-put functions U and on the range Y of the output functions Y . In ourcourse, we will have cases where these quantities are scalars, and other casesin which they are vectors.

The first case is when both input and output functions are scalar. Wecall this type of systems Single Input Single Output (SISO) systems.

If the input is not a scalar function we will refer to the system as MultipleInput (MI); likewise, if the output is not scalar we will use the definition Mul-tiple Output (MO). Clearly, we can have all different combinations: SingleInput Single Output (SISO), Single Input Multiple Output (SIMO), Multi-ple Input Single Output (MISO), Multiple Input Multiple Output (MIMO).

As an example, the RC network in Example 2.1 is a SISO system becauseits input is given by the the input voltage Vin and the output Vc are bothscalar functions. However, if our output functions are both Vin and I thesystem should be considered as SIMO. As for the banking account describedin Example 2.3, the amount of money s(kT ) deposited or withdrawn is thescalar input and the capital c(kT ) is the scalar output. So the system isSISO. However, if the interest rates I+ and I− are allowed to change intime, the system will be MISO.

2.3. I/O Representation

In the previous sections, we have seen that a system is essentially arelation between spaces of functions defined over a common time space. Inview of a general results any relation can be described as the union of thegraph of a set of functions parametrised by a suitable set. This lead us tothe parametric representation of system that applies to any type of system.We are now left with the problem of understanding how such a parametricdescription can be found.

We are now going to delve more deeply inside the subclass of continuoustime (T = R) and of discrete time systems (T =Z ()).

One possible way to describe a system is by expressing a mathematicalproperty that relates input to output. We will now turn our attention tosystems that can be described by differential equations (CT systems) ordifference equations (DT systems) of an appropriate order n:

(2.9) F (y(t), Dy(t), D2y(t), . . . , Dny(t), u(t), Du(t), . . . , Dpu(t), t) = 0,

where the operator D, when applied to a generic function f , is defined as:

Dkf =

{dkfdtk

for CT systems

y(t+ k) for DT systems.

For well posed systems, it is possible to write the equation in a normal form(2.10)

Dny(t) = F (y(t), Dy(t), D2y(t), . . . , Dny(t), u,Du(t), . . . , Dpu(t), t),


Example 2.13. Consider the differential equation:

(2.11) v(t) = −γv(t) + g,

which could describe a rocket moving in the atmosphere under the action ofgravity. Let us It is possible to solve the differential equation as follows:

dv

dt= −γv − g

dv

γv + g= −dt∫

dx

γv + g=

∫−t

1

γlog |γv + g| = −t+ c

γv + g = e−γt+cγ

v(t) = −g/γ +Ae−γt

The last equation tells us that the solution is determined modulo a constant(A), which can be found imposing the initial condition. As we can see wehave found a parametrisation of the system (although applicable only toconstant input function).

Figure 4. Example of mass-spring-damper system

Example 2.14. Consider the mass-spring-damper system in Figure 4.Assume that the x-axis is perpendicular to the ground and points upward.The Newton equation leads us to following differential equation

(2.12) mx = F (t)−mg −K(x− xrest)−Rxwhere the term −K(x−xrest) accounts for the elastic reaction of the springand −Rx accounts for Coulomb friction. This is an instance of the moregeneral expression in Equation (2.10). For this example, the solution of theequation is not so straightforward with standard means (although perfectlyfeasible). We will see generic ways to find a parametrisation of this systemfor any input function F (t) (if they are analytically “nice” enough).

2.3. I/O REPRESENTATION 13

Example 2.15. Let us go back to our differential equation

(2.13) v(t) = −γv(t) + F (t),

which could describe a rocket moving in the atmosphere under the action ofits thrusters (F (t)). Suppose we want to integrate the equation numericallyusing a fixed integration interval. We can integrate both sides of the equation∫ (k+1)T

kTvdt =

∫ (k+1)T

kT(−γv(t) + F (t))dt.

If the integration interval is small enough, we can assume that both x(t) andF (t) remain constant throughout the interval. We can also use the Eulerapproximation for the derivative:

v(t) ≈ v((k + 1)T )− v(kT )

T.

Therefore the integral above can be written as:

v((k + 1)T )− v(kT ) ≈ −γv(kT )T + F (kT )T.

As we can see, we have obtained a DT time system in the normal formin (2.10). Given an initial velocity v0, the equation above allows us toiteratively construct the parametrisation of the system.

In the examples before we have seen that CT systems are described bydifferential equations, DT systems are described by difference equations. Wehave also seen that for the latter that the difference equation is in fact analgorithmic way to obtain a parametrisation of the system. In this paramet-ric representation we need to initialise the input u and the output y up tothe for a number of steps equal than the maximum forward step with whichthey appear in the equation minus 1. For instance for the equation

y(k + 2) = y(k + 1) + y(k) + u(k + 1) + u(k)

in order to start the algorithm, we need y(0), y(1), u(0). We will also seehow to derive explicit forms for this representation.

For CT systems the same result is obtained solving a differential equa-tion (a task that is not always easy). A couple of very well known results(Peano and Lipschitz Theorems) specify conditions (e.g., Lipschitz continu-ity) under which a differential equation has a unique solution. This allowsus to construct a parametrisation, where we have to consider the initialconditions for the input and output function and for their derivatives withan order lower than the maximum order with which they appear in theequation. For instance for the equation:

y = −y + u+ 3y − uwe will need y(0), y(0), u(0), u(0). The complex discussion on existence anduniqueness of solutions will be left out of these notes, and we will assumethat the differential equation is well behaved and that it admits a uniquesolutions for the input function of our interest.


Figure 5. Inverted Pendulum

2.3.1. Causality. In order for the system to be causal, we have to haven ≥ p. Indeed, if we consider a simple DT example like:

y(k + 1) = y(k)− 3u(k) + 4u(k + 2),

in which n = 1 and p = 2, we easily see that the output at a given timedepends on the future value of the input.

Consider now an even simpler and apparently inoffensive example of CTsystem like

y(t) = u.

In order to define a derivative, we have to compute the limit of the leftand of the right increment as see that they are equal. But knowing theright increment is tantamount to peering into the future (although by aninfinitesimal amount).

2.3.2. MIMO systems. The definition in Equation (2.10) refers toSISO systems. It is easy to generalise to a MIMO systems as follows:

Dnyj(t) = F (y1(t), . . . , Dny1(t), . . . , ym(t), . . . , Dnym(t), u1, Du1(t), . . . ,(2.14)

. . . , Dpu1(t), . . . uh, Dum(t), . . . , Dpuh(t), t), j = 1, . . .m

for a system with m output functions and h input functions.

Example 2.16. Consider the inverted pendulum in Figure 2.16. Let Mbe the mass of the cart, m the mass of the pendulum, l the length of thependulum, y1 the position x of the centre of mass of the cart, y2 the angle θof the pendulum with the vertical line, and u be the force F applied to thecart. The equation of motion can be found as follows:

y1 =ml

M +m

(y2 cos y2 − y2

2 sin y2

)+

1

M +mu

y2 =1

l(g sin y2 + y1 cos y2)

2.4. STATE SPACE REPRESENTATION 15

2.4. State Space representation

In the sections above we have seen that: 1. for a system the sameinput function can be associated with multiple output functions, 2. for eacht0, we can introduce a parametric representation such that, given an inputfunction, the associated output function can be “disambiguated” using aparameter x0, 3. for a class of systems (causal systems), this parametricrepresentation is causal meaning that, for a given value of the parameter x0,the output at any given time t′ > t0 only depends on the values taken bythe input function in the interval [t0, t

′) (or [t0, t′] if the causality is strict).

Given that t0 can be chosen freely, this means that: 1. a causal systemcannot “foresee” the future, 2. the parameter x0 summarises all the storyof the system previous to t0. These are qualitative arguments. In order toturn them into a consistent and mathematically sound definition of state weneed a few more steps.

In particular, by changing the initial instant t0 we will find a differentvalue for the parameters x0. Our first assumption is that whatever thechoice of t0 such parameters live in the same set X: ∀t0 x0 ∈ X. Thisis obviously the case if we adopt the IO representation of the system; asdiscussed in Section 2.3, if a CT system is described by means of a differentialequations, and if this is mathematically well behaved (such as to admit aunique solution), the parameters are conveniently chosen as the set of allinitial conditions for the derivatives of input and output function up to acertain order. The same reasoning can be applied to DT system where theparameters are the initial values. Therefore, whatever the initial time t0,the parameters x0 live in the space given by the initial conditions.

In order to have a complete identification between x0 at time t0 andthe past history of the system (up to t0) we need some form of consistencybetween the parameter x0 at time t0 and the parameter x1 at time any othertime t1. We will assume the existence of a function φ, such that

x1 = φ(t1, t0, x0, u),

where u is restricted to its values in the set [t0, t1). Roughly speaking, φ tellsus how to transform the story up until t0 into the story up until t1, giventhe knowledge of the input in between t0 and t1. Lt (T × T )∗ be defined as:

(T × T )∗ = {(t, t0) : t, t0 ∈ T and t ≥ t0} .

The function φ is defined as

φ : (T × T ) ∗ ×X × U → X(2.15)

x(t) = φ(t, t0, x0, u).

We will require that it respects the following three properties:

Consistency:

∀t ∈ T, ∀u ∈ U , φ(t, t, x, u) = x,


Causality:

∀t, t0 ∈ T ,∀x0 ∈ X u|[t0,t) = u′∣∣[t0,t)

=⇒ φ(t, t0, x0, u) = φ(t, t0, x0, u′)

Separation:

∀(t, t0),∀x0 ∈ X,∀u ∈ U(2.16)

t > t1 > t0 =⇒ φ(t, t0, x0, u) = φ(t, t1, φ(t1, t0, x0, u), u).(2.17)

The first two properties are obvious; the third one formalises the idea thatthe state captures the story of the system.

The notion of state we are introducing is rooted in that of a parametricrepresentation introduced earlier. The same notion leads to the link betweeninput state and output. Indeed, assuming a causal system, the function InEquation (2.4) leads us to:

∀t0, y0(t) = [πt0(x0, u0)] (t)∀t ≥ t0,where u0 is restricted to [t0, t]. Assuming t0 = t we can introduce thedefinition of a function η as follows:

y(t) = [πt(x, u)] (t) = η(t, x(t), u(t)).

The function η is described as

η : T ×X × U → Y(2.18)

y(t) = η(t, x(t), u(t)).

2.4.1. Input signals, output signals and states. The variable x(t)is called state of the system, and the set X it lives in is called state space.While systems with finite state space exist and are relevant (e.g., see Exam-ple 2.2), we will focus on systems with dense state space.

Generally speaking, a system has a vector of input signals, part of whichare directly controllable and can be used as command variables to drive thesystem toward a desired behaviour. Other input signals are not under directcontrol of the user and act as external disturbance or noise.

As an example, if we model a car the throttle and the steering wheelsare command variables, whereas the inclination of the road can be seen asan external disturbance.

Output signals are measurable quantities and can be used to formulatedesign specifications. For instance, in the model of a car, one can formulate aspecification like if the driver open the throttle completely, the speed reaches100km/h in 11 seconds, in which the speed plays the role of our controlledvariable.

In other cases, an output signal conveys useful elements of informationthat allow reconstructing the system state when the latter is not directlymeasurable. For instance, the speed of a vehicle is typically a state variablethat cannot be measured directly. It is possible to mount encoder on thewheels that measure the number of rotations n and from this derive anestimate of the velocity. The most obvious way is to compute a numeric


“derivative” using the left increments. This strategy typically produce noisyestimates. We will see in the course that better results can be obtained insystematic ways using the so called “observer”, a system that builds on theknowledge of the system model and of the story of the output signal toreconstruct an estimate of the state.

2.4.2. Existence of a state. The functions φ and η that we havequalitatively introduced above allow us to generate a system:

Σ(t0) ={

(u0, y0) ∈ U T (t0) × Y T (t0),(2.19)

u0 = u|T (t0) , y0 : y0(t) = η(t, φ(t, t0, x0, u), u(t)), with u0 ∈ U and x0 ∈ X} .

An important point is when is the converse true? In other words, whenis it possible to construct a state representation as outlined above for genericabstract system? It turns out that this is indeed possible for an abstractsystem under mild conditions. Essentially, what is required is that inputfunctions have the same range and that they be a closed set under concate-nation. The existence and uniqueness of a state representation for abstractsystems is beyond the scope of this course and we refer the interested readerto specialised text books [RI79] or to the scientific literature. In this course,we will always assume that a representation of this kind exists for our sys-tems of interest.

2.4.3. Implicit Representation. The η and φ functions that we haveintroduced in Equation (2.15) and (2.18) compound the so-called explicitrepresentation of a system. By “explicit” we mean that given an inputfunction and an initial state the two function enable to compute immediatelythe evolution of the state and of the output.

In contrast, an implicit representation is one in which the computationthe output is not immediate but it requires the solution of a system ofdifference or of differential equations.

For DT systems, an implicit representation can be found by by simplyevaluating φ across a single step:

x(t+ 1) = φ(t+ 1, t, x(t), u(t)) = f(t, x(t), u(t))(2.20)

The function f(. . .) is said generator function. The representation (X, f, η)is said implicit representation. Clearly the function φ can be recovered fromf through a simple iteration.

Example 2.17. Consider a DT system with scalar state (i.e., state withdimension 1) and suppose that its generator function is given by the linearfunction:

x(t+ 1) = a(t)x(t) + b(t)u(t).


If we want to compute the evolution from a state x0 at time t0 to a state x1

at time t1, we can do it by unrolling the following iteration:

x(t0 + 1) = a(t0)x0 + b(t0)u(t0)

x(t0 + 2) = a(t0 + 1)x(t0 + 1) + b(t0 + 1)u(t0 + 1) =

= a(t0 + 1)a(t0)x0 + a(t0 + 1)b(t0)u(t0) + b(t0 + 1)u(t0 + 1)

x(t0 + 3) = a(t0 + 2)x(t0 + 2) + b(t0 + 2)u(t0 + 2) =

= a(t0 + 2)a(t0 + 1)a(t0)x0 + a(t0 + 2)a(t0 + 1)b(t0)u(t0)+

+ a(t0 + 2)b(t0 + 1)u(t0 + 1) + b(t0 + 2)u(t0 + 2)

. . . . . .

We can prove by induction for a general H that:

x(t0 +H) = φ(t0 +H, t0, x0, u) =

= ψ(t0 +H, t0)x0 +

t0+H−1∑t=t0

G(t0 +H − 1, t)u(t)

where

ψ(t, τ) =

{∏t−1t′=τ a(t′) if t > τ

1 if t ≤ τG(t, τ) = ψ(t+ 1, τ + 1)b(τ)

When we move to the CT domain, things are not so immediate. It isreasonable to think of the evolution of the state in real time as describedby means of a differential equation. This requires a sufficient regularity (atthe very least differentiability) of the function φ. If we admit that φ is thesolution to the equation:

∂

∂tφ = f(t, φ, u(t)),

then our implicit representation is given by:

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

Generally speaking the state is a vector and f is a vector values function.This is easy to understand looking at the example below.

Example 2.18. Consider the Mass Spring system in Figure 4. Let xrepresent the vertical position of the mass. If we choose the state as

x(t) =

[x1(t)x2(t)

]=

[x− xrestd(x−xrest)

dt

]the use of standard laws of mechanics leads us to

x(t) =˙[

x1(t)x2(t)

]=

[x2

F (t)−Kx1 −Rx2

].


2.4.4. Link between state and I/O representation. Looking atExample 2.14 and at Example 2.18, the reader can infer an easy way to derivea state representation from an I/O representation. Suppose our system isdescribed by an a differential (or by a difference) equation in Equation (2.10).Consider the simplified case in which p = 1. Suppose we set xi = Di−1y,with the obvious implication that

Dxi = DDiy = Di+1y = xi+1

. We can write Equation (2.10) as:

D

x1

x2

. . .xn

=

x2

x3

. . .F (x1, x2, . . . , xn, u, t)

(2.21)

y = x1.(2.22)

This way, we have found the expressions for the vector valued functionf(x, u, t) and for the function h(x, u, t):

f(x, u, t) =

x2

x3

. . .F (x1, x2, . . . , xn, u, t)

h(x, u, t) = x1,

where x is defined as x = [x1, x2, . . . , xn]T . If n ≥ p > 1, the constructioncan be generalised by setting:

xi = Di−1y for i ∈ [1, n](2.23)

xi+n = Di−1u for i ∈ [1, p].

This strategy leads to our results but is often inefficient, i.e., it introducesmore states than strictly needed. Consider the following example:

Example 2.19. Consider a system whose I/O representation is given by

y = y + 5u− u.

If we want to find a state space representation and approach the problem asdiscussed before, we would have to introduce two state x1 = y and x2 = u.Suppose, instead, that we set x1 = y − 5u. We have,

x1 = y − 5u

= y − u =

= y − 5u+ 4u

= x1 + 4u

y = x1 + 5u


The example above reveals that: 1) there are more than one state spacerepresentation for the same I/O description, 2) some systems (that we willlearn to qualify as linear) allow for a much more compact state space thanthe one using the canonical set of variables in Equation (2.23)

2.5. Linear Systems

As we discussed above, an abstract system Σ(t0) is defined as a set of

pairs (u0(t), y0(t)) with u0(t) ∈ U T (t0) and y0(t) ∈ Y T (t0). Suppose that thesets U and Y are both vector spaces defined over the real set R. Clearlythe range sets U and Y where the functions take value have to possess thesame algebraic structure. In plain words this means that if u1(t) ∈ U andu2(t) ∈ U , then αu1(t) + βu2(t) ∈ U for any α ∈ R and β ∈ R (and thesame for any pair y1(t) and y2(t) belonging to Y .

This being said, we can introduce the following definition

Definition 2.20. The system Σ(t0) is linear if given any two pairs(u1(t), y1(t)), (u2(t), y2(t)), if they both belong to the system Σ(t0), so willtheir linear combinationa: α (u1(t), y1(t)) + β (u2(t), y2(t)) ∈ Σ(t0), whereα, β ∈ R.

Consider now a parametric description of the system with a parameterspace Xt0 and a function π. For a given choice of the parameter x0 thesystem responds to an input u1 with the output y1 = πt0(x0, u1) and to theinput u2 with the output y2 = πt0(x0, u2). The intuitive meaning of thelinearity is captured by the following two properties:

• Superimposition: the response to the combination u1(t)+u2(t) willbe the superimposition of the effect of u1(t) and u2(t): y1(t)+y2(t)• Scaling : the response to αu1(t) is αy2(t).

Example 2.21. Consider the mass spring system in Example 2.14. Sup-pose that starting from an initial condition x1(t is the evolution of the systemfor F1(t) and x2(t) is the solution to corresponding to the force F2(t). Wecan write

mx1 = F1(t)−mg −K(x1 − xrest)−Rx1

mx2 = F2(t)−mg −K(x2 − xrest)−Rx2

If we multiply the first equation by α, the second equation by β and sumup the result we will see that x1(t) + x2(t) is a solution to F1(t) + F2(t).

The considerations in the example before apply because the differen-tial equation describing the system is linear and the derivative is a linearoperator: D(αf(t) + βg(t)) = αD(f(t)) + βD(g(t)).

If we repeat the same line of reasoning with the system in Example 2.16,we will fail because the equation is not linear. More simply, if we considerthe mass spring example but we impose a maximum value for the force F (t)the scaling principle will not apply and the linearity will be lost.

2.5. LINEAR SYSTEMS 21

2.5.1. Linear I/O representations. The same argument of Exam-ple 2.21 can be repeated for any system described by a linear difference ordifferential equation of:

(2.24) Dny(t) =

n−1∑i=0

αi(t)Dhy(t) +

p∑j=0

βj(t)Dju(t).

Generally speaking, when this equation has a unique solution (which isgenerally true for “well behaved” functions α(t) and β(t)) and of causal sys-tems (p ≤ n), such a solution is a function of u and of the initial conditions:

y(t) = φ(t, t0, y(0), Dy(0), . . . , Dn−1y(0), u(0), . . . , Dpu(0), u|[t0,t]

).

There are two known facts stated in the following:

Lemma 2.22. Given the Equation (2.24), define its associated homoge-neous equation as follows:

(2.25) Dny(t) =n−1∑i=0

αh(t)Diy(t)

The solution to this equation are a vector space and the dimension of thisspace is n. This applies to both differential and difference equations.

Lemma 2.23. Suppose that y is a particular solution of Equation (2.24)related to the input u. If we take solution of Equation (2.25) y and we sumit to y we still obtain a solution of Equation (2.24).

By choosing any basis {y1, . . . , yn} of the vector space of the solutionsof Equation (2.25) , we can write a general solution of of Equation (2.24) as

y(t) = y(t) +H1y1(t) + . . .+Hny2n(t)

where the coefficients H1 can be found by imposing n conditions derivedfrom the initial conditions. In essence, the linearity of the system leads usto a break down of the solution in two pieces. The first one, y, is calledforced evolution because it depends on the input term u. The second one is+H1y1(t) + . . .+Hnyn(t) and is called free evolution because it depends onthe initial conditions.

An obvious but fundamental fact is that a linear representation as de-scribed above generates a linear system as in Definition 2.20.

2.5.2. Linear State Space Representations. A state space explicitrepresentation is given by a couple of functions φ and η, defined as:

φ : T × T ×X × U → X x(t) = φ(t, t0, x0, u)

η : T ×X × U → X y(t) = η(t, x, u)


A representation of this type is linear if both functions are linear with respectto their arguments X × U . In plain words, if we consider φ we have:{φ(t, t0, x0, u1|[t0,t)) = x1

φ(t, t0, x0, u2|[t0,t)) = x2=⇒ φ(t, t0, x0, αu1|[t0,t)+βu2|[t0,t)) = αx1 + βx2{

φ(t, t0, x0,1, u|[t0,t)) = x1

φ(t, t0, x0,2, u|[t0,t)) = x2=⇒ φ(t, t0, αx0,1 + βx0,2, u) = αx1 + βx2.

We can write similar equations for η,{η(t, x0, u1) = y1

η(t, x0, u2) = y2=⇒ η(t, x0, αu1 + βu2) = αy1 + βy2{

η(t, x0,1, u) = y1

η(t, x0,2, u) = y2=⇒ η(t, αx0,1 + βx0,2, u) = αy1 + βy2.

The implicit representation can be obtained from the explicit one bydifferentiation (assuming that φ is differentiable) or by considering one stepof evolution.

f : T ×X × U → X Dx(t) = f(t, x, u)

η : T ×X × U → X y(t) = η(t, x, u)

In this case f has to be linear with respect to X×U . Observe the difference.For φ we require linearity with respect to X × U (where U is a space ofsignals), while for f and for η we require linearity with respect to X × U(where U is the range of the input signals U ).

The linearity of f and η can be expressed in matrix notation as:

Dx(t) = A(t)x(t) +B(t)u(t)

y(t) = C(t)x(t) +D(t)u(t).(2.26)

2.6. Time Invariance

We can introduce a general definition of time invariance for an abstractsystem Σ(t0) defined as a set of pairs (u0(t), y0(t)).

Definition 2.24. A system Σ(t0) is time invariant iff

(u0(t), y0(t)) ∈ Σ(t0) =⇒ (u0(t− t0), y0(t− t0)) ∈ Σ(t0 − t0).

In plain world, what we require is that if a behaviour (u0(t), y0(t)) isgenerated by the system, so will the same behaviour with u0 and y0 shiftedforward or backward in time of the same amount.

The definition of time invariance applies also to the representation ofthe system.

2.7. HYBRID SYSTEMS 23

2.6.1. IO representation. An IO representation of a system is timeinvariant if the differential or the difference equation do not explicitly dependon t.

In order to understand this definition, consider Example 2.3. If we allowthe interest rates I+, I− to change in time the system is not obviously timeinvariant. If they are kept constant, the system is time invariant. Likewise,for the mass-spring system in Example 2.14, we can say that the system istime invariant. But if one of the parameters (e.g., the Hooke constant of thespring) changes with time, then the system is time-varying.

Generally speaking, if we consider the linear IO representation in Equa-tion (2.24), we can say that the system is time–invariant if the parametersαi(t) and βj(t) are constant over time. In this case, the equation becomes

(2.27) Dny(t) =n−1∑i=0

αiDhy(t) +

p∑j=0

βjDju(t).

2.6.2. State Representation. If we consider a state representationin explicit form, it is time invariant if and only if phi does not dependssingularly on t and t0, but it depends on their difference, and if η does notdepend on t

φ : T × T ×X × U → X x(t) = φ(t, t0, x0, u) = φ(t− t0, x0, u)

η : T ×X × U → X y(t) = η(t, x, u) = η(x, u)

If we derive the implicit form, it is easy to see that this implies that f andη do not depend on t.

f : T ×X × U → X Dx(t) = f(t, x, u) = f(x, u)

η : T ×X × U → X y(t) = η(t, x, u) = η(x, u)

In the case of linear system, time invariance is translated into a set ofmatrices A,B,C,D that do not depend on time:

Dx(t) = Ax(t) +Bu(t)

y(t) = Cx(t) +Du(t).(2.28)

It is possible to see that linear IO and state space representations areassociated with linear systems

2.7. Hybrid Systems

Different hybrid systems are obtained combining a DES event that de-termines changes in the “physical laws” that govern the evolution of a CTsystem. Consider the example in Figure 2.7. A tank is filled with a fluid,which flows out with a rate Q0 from an outlet in the bottom of the tank.


Figure 6. Example of an Hybrid System

When the level of the fluid reaches a threshold, a sensor detects the eventand opens a valve which produces an incoming flow Q1 > Q2 until the tankis full. In this example, we have a state machine that determines switchesbetween two different evolutions of the system, each of them governed byCT dynamics. This type of systems are beyond the scope of this course.

CHAPTER 3

Impulse response and convolutions for linear andtime invariant IO Representations

In this chapter, we restrict our focus on IO representation of SISO linearand time invariant systems. Generally speaking, the evolution of the systemis described by the following equation

(3.1) Dny(t) =n−1∑i=0

αiDiy(t) +

p∑j=0

βjDju(t),

where time invariance requires that αi and βi be constant. The system iscausal if p ≤ n and strictly causal if p < n. We have seen that the initialconditions:

y(0), Dy(0), . . . , Dn−1y(0), . . . , Dpu(0), . . . , Du(0)

and of the input u|[t0,t]. We have finally seen that the linearity of the systemallows us to split the evolution depending on the initial conditions and theevolution depending on the input in two separate terms:

y(t) = yfree(t) + yforced(t)

with

yfree(t) = Ffree(y(0), Dy(0), . . . , Dn−1y(0), . . . , Dpu(0), . . . , Du(0))

yforced(t) = Fforced(u|[t0, t]).

Let us start with the computation of the forced evolution.

3.1. Forced Evolution of discrete–time systems

It is useful to introduce a special function, which we will call impulsefunction.

3.1.1. Kronecker δ. For discrete time system the impulse function isalso known as Kronecker Delta and is defined as follows:

(3.2) δ(t) =

{1 t = 0

0 t 6= 0.

25

263. IMPULSE RESPONSE AND CONVOLUTIONS FOR LINEAR AND TIME INVARIANT IO REPRESENTATIONS

If we consider any discrete–time function we have the following property:

(3.3) f(t)δ(t− t0) =

{f(t0) t = t0

0 t 6= t0

In plain words we will say that multiplying a Kronecker δ shifted to t0 bya function generates a new function that is zero at all times except for t0(where it is f(t0)). This is equivalent to saying that multiplying by δ hasthe effect of extracting a sample (“sampling”) of the function in the pointwhere δ is centred. Another property is:

(3.4)

τ=∞∑τ=−∞

f(τ)δ(t− τ) = f(t)

In other words we can express any function as a sum of infinite δ.

3.1.2. Impulse Response. Consider the DT system in Equation (3.1).Our goal is to compute the forced response to a generic signal u(t).

Let us define h(t) the forced response to the signal δ(t).

h(t) = Fforced(δ).

As we discussed above we can write u(t) as

u(t) =

τ=∞∑τ=−∞

u(τ)δ(t− τ)

Thanks to the system’s linearity we can write:

Fforced(u) = Fforced(τ=∞∑τ=−∞

u(τ)δ(t− τ))

=

τ=∞∑τ=−∞

u(τ)Fforced(δ(t− τ)),

and using time–invariance we get.

Fforced(u) =

τ=∞∑τ=−∞

u(τ)h(t− τ)

The summation above is called discrete–time convolution (also denoted asu(t) ∗ h(t). What we just said is that for a DT system it is sufficient toknow the impulse response h(t) to be able to reconstruct the response toany signal.

The meaning of the convolution sum is quite simple. Suppose we wantto find y(t) = u(t) ∗ h(t). What we need to do is the following:

(1) Compute the sequence h(−τ), which is the reflection of h(τ) throught = 0;

(2) Shift the obtained sequence by t and compute the product element–wise;

(3) Sum up the products, and this produces y(t).

3.1. FORCED EVOLUTION OF DISCRETE–TIME SYSTEMS 27

This is shown in the two examples below.

Example 3.1. Suppose that:

h(t) =

1 t = 0

2 t = 1

−2 t = 2

0 Otherwise

and

u(t) =

5 t = 0

−3 t = 1

0 Otherwise

compute the system’s response. We can see

y(t) =τ=∞∑τ=−∞

u(τ)h(t− τ)

= 5h(t)− 3h(t− 1) = =

5 t = 0

10− 3 t = 1

−10− 6 t = 2

6 t = 3.

From this example we can clearly see that if the support of h(t) is [0,M ]and the support of u(t) is [0, N ], the support of h(t)∗u(t) will be [0, N+M ].

Example 3.2. Suppose we know that the impulse response to a systemis given by 1(t)at, let us compute the forced response to the input functionu(t) = 1(t)bt, where

1(t) =

{1 t > 0

0 t ≤ 0.

We have:

Fforced(u) =τ=∞∑τ=−∞

1(τ)bτ1(t− τ)at−τ =

=

τ=t−1∑τ=1

bτat−τ =

= atτ=t−1∑τ=1

(b

a

)τ=

= at(b/a)− (b/a)t

1− (b/a)=

=b

a− bat − a

a− bbt.


System:h(t)

u(t) y(t) = h(t) * u(t)

Figure 1. Example block scheme

Our previous discussion can be summarised in the block scheme in Fig-ure 3.1.2. The box represent the system and the arrows the input and theoutput signals.

3.1.3. Properties of the convolution sum. Three important prop-erties of the convolution sum are stated in the following.

Theorem 3.3. The convolution sum enjoys the following properties:

(1) Commutative Property: h(t) ∗ u(t) = h(t) ∗ u(t).(2) Distributive Property: (h1(t) + h2(t)) ∗ u(t) = h1(t) ∗ u(t) + h2(t) ∗

u(t)(3) Associative Property: h1(t) ∗ (h2(t) ∗ u(t)) = (h1(t) ∗ h2(t)) ∗ u(t).

Proof. (1) Commutative Property, considering that h(t) ∗ u(t) =∑+∞τ=−∞ h(τ)u(t − τ). By changing the variable as t − τ = τ1 we

can write :

+∞∑τ=−∞

h(τ)u(t− τ) =−∞∑τ1=∞

h(t− τ1)u(τ1)

= u(t) ∗ h(t).

(2) Distributive Property: this is a straightforward implication of thelinearity of the convolution operator:

(h1(t) + h2(t)) ∗ u(t) =

+∞∑τ=−∞

(h1(τ) + h2(τ))u(t− τ) =

=−∞∑τ1=∞

h1(τ)u(t− τ) +−∞∑τ1=∞

h2(τ)u(t− τ) =

= h1(t) ∗ u(t) + h2(t) ∗ u(t)

3.1. FORCED EVOLUTION OF DISCRETE–TIME SYSTEMS 29

+

h1

h2

uh1+h2

Figure 2. Parallel composition

h2 h1*h2

Figure 3. Series composition

(3) Associative Property:

(h1(t) ∗ h2(t)) ∗ u(t) =∞∑

τ2=−∞u(t− τ2)

∞∑τ1=−∞

h1(τ1)h2(τ2 − τ1) =

=

∞∑τ2=−∞

∞∑τ1=−∞

u(t− τ2)h1(τ1)h2(τ2 − τ1) =

=∞∑

τ ′2=−∞

h1(τ1)∞∑

τ1=−∞u(τ ′2)h2(t− τ ′2 − τ1) =

=∞∑

τ ′2=−∞

h1(τ1)(u(t) ∗ h2(t))|t−τ1

= h1(t) ∗ (u(t) ∗ h2(t))

�

The first property has merely an operational value: if we want to processu(t) through a system h(t) we can compute the most convenient betweenh(t) ∗ u(t) and u(t) ∗ h(t).

The second and the third property have a more substantial meaningrelated to the composition of two systems. The distributive property revealsthat the parallel composition of two systems produces the same result ashaving one system whose impulse response is given by the sum of the twoimpulse responses (see Figure 2).

The associative property is related to the series composition of two sys-tems, which is equivalent to one system having an impulse response givenby the convolution of the two impulse responses (see Figure 3).


3.1.4. Eigenfunctions. There is particular type of functions that aretreated in special way by LTI system. Consider function u(t) = zt andcompute the forced evolution of a LTI system. The response to this inputis given by:

+∞∑τ=−∞

h(τ)u(t− τ) =

∞∑τ=−∞

u(t− τ)h(τ)

=∞∑

τ=−∞zt−τh(τ)

= zt∞∑

τ=−∞z−τh(τ)

= ztH(z)

where H(z) =∞∑

τ=−∞z−τh(τ).

The property just shown can be stated by saying that whenever a signalzt (i.e., a power of t) gets processed by a DT LTI system, the outcome isthe same signal scaled by a constant H(z) which is the limit of the series∑∞

τ=−∞ z−τh(τ), as far as the series converge. For this reason zt is called

an eigenfunction. An eigenfunction is essentially similar to an eigenvector ofa standard linear application, with H(z) being its eigenvalue. This appliesboth to real and complex z.

3.1.4.1. A key property of eigenvalues. We should recall that eigenvec-tors enjoy an extremely useful property. Consider linear application definedfrom Rn to Rn. Suppose that it is associated to a matrix A:

y = Ax.

Suppose that we can identify n independent eigenvectors: {u1, u2, . . . , un}related to the eigenvalues {λ1, ldots, λn}. Let M = [u1u2 . . . un] be thematrix composed using these vectors and assume that these vectors form abasis. Let x be the coordinates in this basis of a generic vector x expressedin the canonical basis. We have:

x = M−1x.

Similarly the transformed version of y is given by:

y = M−1y.

By combining the two conditions we find:

y = M−1y =

= M−1Ax =

= M−1AMx.(3.5)

3.2. FORCED EVOLUTION OF CONTINUOUS–TIME SYSTEMS 31

It can easily be seen that M−1AM is a diagonal matrix. In simple words,this means that if we express a vector using a basis of eigenvectors, thesystem operates on each component in a decoupled way.

The same property holds if we express any signal as a linear combinationof Eigenfunctions. The convenience of this choice will become apparent inthe next few pages.

3.2. Forced Evolution of continuous–time systems

The line of arguments used for DT systems can be applied to CT systemsas well. The key point is the definition of an appropriate impulse function.

3.2.1. Dirac δ. The notion of impulse is not as obvious for CT systemsas it is for DT systems. Indeed, the type of functions we normally dealwith are at the very least continuous and often differentiable. The impulsefunction does not fall within this category, but it is useful to think of it asthe limit case of a relatively well behaved signal. The simplest choice thatcan be made is to define:

δ(t) = lim∆→0

δ∆(t)(3.6)

δ∆(t) =

{0 t 6∈

[−∆

2 ,∆2

]1∆ t ∈

[−∆

2 ,∆2

](3.7)

A few properties that are easy to show are the following:

Property 1: If we compute the integral of Dirac δ on any intervalenclosing the origin, we get 1.0:

∀a > 0, b > 0

∫ b

−aδ(τ)dτ = 1.

Property 2: Multiplying a δ(t− τ) by any function has the effect of“sampling” the value of the function in τ , meaning that we obtainan impulse function multiplied by τ :

f(t)δ(t− τ) = f(τ)δ(t− τ).

Property 3: Any function can be expressed as an integral of impulsefunctions.

∀ε > 0,

∫ t+ε

t−εf(τ)δ(t− τ)dτ.

Indeed,∫ t+ε

t−εf(τ)δ(t− τ)dτ =

∫ t+ε

t−εf(t)δ(t− τ)dτ

= f(t)

∫ t+ε

t−εδ(t− τ)dτ

= f(t).


Property 4: The integral from any negative number to a genericinstant t produces a step function:

∀ε > 0,

∫ t

−εδ(τ)dτ = 1(t),

where 1(t) is defined as

1(t)

{1 t > 1

0 t ≤ 0

Property 5: We could define

δ(t) =d

dt1(t).

This is an obvious abuse of notation because we know that the stepfunction is not differentiable in t = 0.

3.2.2. Impulse Response. We can repeat the same line of argumentsfollowed in Section 3.1.2 to compute the forced response of system (3.1)starting from the impulse response (i.e., the forced response to δ(t)):

h(t) = Fforced(δ).

Indeed, we can write

u(t) =

∫ τ=∞

τ=−∞u(τ)δ(t− τ)dτ.

Using linearity and time–invariance:

y(t) = Fforced(u)

= Fforced(

∫ τ=∞

τ=−∞u(τ)δ(t− τ)dτ)

=

∫ τ=∞

τ=−∞u(τ)Fforced(δ(t− τ))dτ, =

∫ τ=∞

τ=−∞u(τ)h(t− τ)dτ.

This allows us to conclude that once we know h(t), we can compute theresponse to any signal. This is called “convolution integral” and is denotedby the same symbol as the convolution sum: u(t) ∗ h(t). The computationof y(t) requires the following steps:

(1) Compute the “reflection” of h(τ) through τ = 0,(2) Translate the result to the right of t (to the left if t is negative).(3) Compute the product by u(τ) and then the integral of the function

thus obtained.

These steps can be appreciated through an example.

3.2. FORCED EVOLUTION OF CONTINUOUS–TIME SYSTEMS 33

Example 3.4. Suppose that h(t) = 1(t)e−3t. Compute the response ofthe system to u(t) = 1(t). The result can be found as follows:

y(t) = Fforced(u)

=

∫ τ=∞

τ=−∞u(τ)h(t− τ)dτ

=

∫ τ=∞

τ=−∞1(t)1(t− τ)e−3t+3τdτ

=

∫ τ=t

0e−3t+3τdτ

= e−3t 1

3e3τ |tτ=0

=1− e−3t

3.

3.2.3. Properties of the convolution integral. The convolution in-tegral has the same properties Three important properties of the convolutionsum are stated in the following.

Theorem 3.5. The convolution sum enjoys the following properties:

(1) Commutative Property: h(t) ∗ u(t) = h(t) ∗ u(t).(2) Distributive Property: (h1(t) + h2(t)) ∗ u(t) = h1(t) ∗ u(t) + h2(t) ∗

u(t)(3) Associative Property: h1(t) ∗ (h2(t) ∗ u(t)) = (h1(t) ∗ h2(t)) ∗ u(t).

The proof is very similar to one we have given for the convolution sum.

3.2.4. Eigenfunctions. Suppose that we use as input an exponentialfunction u(t) = est.

y(t) =

∫ +∞

τ=−∞h(τ)u(t− τ)dτ

=

∫ +∞

τ=−∞h(τ)es(t−τ)dτ

= est∫ +∞

τ=−∞h(τ)e−sτdτ.

Let H(s) be the integral∫ +∞τ=−∞ h(τ)e−sτdτ , assuming that it converges. We

can summarise the computation above by saying that est is an Eigenfunc-tions related to the eigenvalue H(s).

The above results apples to both real and complex exponentials. Thisleads us to an interesting fact if we analyse the output to an harmonicfunction: u(t) = cosωt. Using Euler’s expression, we have

cosωt =ejωt + e−jωt

2.


Applying linearity we find:

y(t) =

∫ +∞

τ=−∞h(τ)u(t− τ)dτ

=

∫ +∞

τ=−∞h(τ)

ejω(t−τ) + e−jω(t−τ)

2dτ

=1

2

∫ +∞

τ=−∞h(τ)ejω(t−τ)dτ +

1

2

∫ +∞

τ=−∞e−jω(t−τ)dτ

=1

2ejωtH(jω) +

1

2e−jωtH(−jω),

where H(jω) =∫ +∞τ=−∞ h(τ)e−jωτdτ . Let z represent the complex conjugate

of a complex number z. If z1 and z2 are two complex numbers and α is areal, we can easily show that:

(1) z1 + z2 = z1 + z2

(2) z1z2 = z1z2

(3) αz1 = αz1.

Applying these properties, we can see:

H(jω) =

∫ +∞

τ=−∞h(τ)e−jωτdτ

=

∫ +∞


=

∫ +∞


= H(−jω)

Therefore,

y(t) =1

2ejωtH(jω) +

1

2e−jωtH(−jω)

=1

2ejωtH(jω) +

1

2e−jωtH(jω).

If we use for H(jω) its modulus/phase representation:

H(jω) = |H(jω)| ej∠H(jω),

3.3. PROPERTIES OF THE IMPULSE RESPONSE 35

we find:

y(t) =1

2ejωtH(jω) +

1

2e−jωtH(jω)

=1

2|H(jω)| ej∠H(jω)ejωt +

1

2e−jωt |H(jω)| e−j∠H(jω)

=|H(jω)|

2

(ej(∠H(jω)+ωt) + e−j(∠H(jω)+ωt)

)=|H(jω)|

2cos (ωt+ ∠H(jω)) .

The result above can be summarised in the following:

Theorem 3.6. Consider a TC LTI system. If∫ +∞τ=−∞ h(τ)e−jωτdτ con-

verges to a value H(jω), then the system responds to an harmonic inputfunction cosωt with an harmonic output function having the same fre-quency.

3.3. Properties of the impulse response

Many important properties of the system can be inferred by looking atthe analytical features of h(t), both for DT and CT systems.

3.3.1. Causality. The first important problem that can be read fromh(t) is causality, as specified in the following Theorem.

Theorem 3.7. Let h(t) be the impulse response of a system Σ. Thesystem is causal if and only if h(t) = 0 for t < 0.

Proof. Let us focus for simplicity on CT systems. Necessity derivesfrom the observation that if we choose u(t) = δ(t), causality requires thath(t) = 0 for t < 0.

Sufficiency is shown considering that if h(t) = 0 then∫ ∞−∞

h(t− τ)u(τ)dτ =

∫ t

−∞h(t− τ)u(τ)dτ.

As shown in the formula above, the computation of y at time t is onlyaffected by u(·) until t. The proof for DT systems is absolutely similar. �

3.3.2. BIBO Stability. Key to this course is the idea of stability.Roughly speaking, a stable system is one for which small perturbationsof various kind on the evolution of the output signal. We will see severalpossible way in which this intuitive notions can be formalised in mathemat-ical terms. The first one applies to IO representations and is the so-calledBounded-Input-Bounded-Output (BIBO) stability formalised below.

Definition 3.8 (BIBO stability). A system is BIBO stable iff for allε > 0 there exists a positive real δ > 0 such that

|u(t)| ≤ ε =⇒ |y(t)| < δ.


Figure 4. Scheme of a unicycle mobile robot.

The idea is that if we apply “small” input signals produce “small” outputsignals.

Example 3.9. Consider a cylindrical robot that moves along a path.Suppose we are able to control the vehicle forward speed v(t) and the angularspeed ω(t). We can easily see that the differential equations describing thekinematics of the system are

y = v sin θ

θ = ω.

Suppose that our output is ω We can see that if apply the following signal

ω =

{ε t ∈ [0, 0.1s]

0 t 6∈ [0, 0.1s]

the system changes slightly its orientation, but from that moment onward ystarts to grow unbounded even if ε is very small. This is the perfect exampleof a BIBO unstable system.

The BIBO stability property is particularly easy to study for LTI system,for which we can prove the following.

Theorem 3.10. Consider a LTI system Σ with impulse response h(t). Ifthe system is DT then it is BIBO stable if and only if there exists a constantS such that

∑∞infty |h(t)| = S < ∞. If the system is CT then it is BIBO

stable if and only if there exist a constant S such that∫∞infty |h(τ)|dτ = S <

∞.

Proof. We give the proof for the case of DT systems. The output ofthe system is given by

y(t) =

∞∑τ=−∞

h(τ)u(t− τ)

3.3. PROPERTIES OF THE IMPULSE RESPONSE 37

To prove sufficiency, assume that for some ε > 0 we have u(t) ≤ ε,∀t.We have:

y(t) =∞∑

τ=−∞h(τ)u(t− τ)

≤∞∑

τ=−∞|h(τ)u(t− τ)|

≤∞∑

τ=−∞|h(τ)||u(t− τ)|

≤∞∑

τ=−∞|h(τ)|ε

≤ ε∞∑

τ=−∞|h(τ)|

≤ εSε.Hence,

|u(t)| ≤ ε =⇒ |y(t)| ≤ δ = Sε,

which proves sufficiency.To prove necessity, it is sufficient to consider the input signal u(t) =

εsign(−t). If we compute y(0), we have:

y(0) =∞∑

τ=−∞h(τ)u(−τ)

≤∞∑

τ=−∞εh(τ)sign(h(τ))

≤ ε∞∑

τ=−∞|h(τ)|

Therefore, if∑∞

τ=−∞ |h(τ)| diverges, so will y(0) in the face of a boundedsignal u(t). �

CHAPTER 4

Laplace and z-Transform

In the past chapter we have seen that LTI system respond to exponentialsignals (est for CT systems and zt for DT systems) in a special way. Suchfunctions are “eigenfunctions”, meaning that the corresponding output isthe same function multiplied by an eigenvalue, which is given by:

H(s) =

∫ ∞0

h(τ)e−sτdτ For CT systems

H(z) =

∞∑0

h(τ)z−τ For DT systems.

This is the basis for a solution strategy based on the so-called Laplace andZ transform. Before delving into this, it is useful to re-cap some basic factsabout complex exponentials.

4.1. Complex exponentials

An exponential function has a fundamental property:

eaeb = ea+b

ea/eb = ea−b

(ea)n = ena.

The very same properties are also enjoyed by a strange complex valuedfunction:

ejθ = cos θ + j sin θ,

which we call the Euler exponential. We can see this for the first property:

ejθejψ = (cos θ + j sin θ)(cosψ + j sinψ)

= (cos θ cosψ − sin θ sinψ) + j(sin θ cosψ + sinψ cos θ)

= (cos(θ + ψ) + j(sin θ + ψ)

= ej(θ+ψ).

It is possible define exponential with real and imaginary part:

eσ+jθ = eσejθ

= eσ(cos θ + j sin θ).

Another important property has to do with the computation of a powerof a complex number z. If we express z = ρejθ, we have zn = ρnejnθ.

39

40 4. LAPLACE AND Z-TRANSFORM

4.2. Complex Exponential functions

Based on the definition of complex exponential, we can introduce com-plex exponential functions.

In the CT domain, given any complex number s = σ + jω, with σ ∈ R,ω ∈ R, we define CT complex exponential the function:

est : R ∈ = e(σ+jω)t =

= eσt (cosωt+ j sinωt) .

Both the real and the imaginary part of this function are sinusoidal oscilla-tions whose amplitude is modulated by an exponential function (increasingif σ > 0 and decreasing if σ < 0).

Likewise, if we consider any complex variable z, we can express z = ρejθ.In this setting zt is given by:

zt = ρtejtθ.

4.3. Definition of the Laplace Transform

The Laplace transform is an integral operator defined as

L (f(t)) = F (s) =

∫ ∞0

f(τ)e−sτdτ.

The idea is to associate each signal f (which is a function of t) with a newfunction F (s), which is a function of the complex variable s.

Example 4.1. Let us compute the Laplace transform of 1(t).

L (1(t)) =

∫ ∞0

1(τ)e−sτdτ =

=

∫ ∞0

e−sτdτ =

=1

s

(1− lim

t→∞e−st

).

We can easily see that

limt→∞e−st =

{0 Real (s) > 0

Does not converge Real (s) ≤ 0.

Therefore,

L (1(t)) =

{1s if Real (s) > 0

undefined otherwise

4.3. DEFINITION OF THE LAPLACE TRANSFORM 41

Example 4.2. Let us compute the Laplace transform of 1(t)eat, witha ∈ R.

L (1(t)) =

∫ ∞0

1(t)eaτe−sτdτ =

=

∫ ∞0

e−(s−a)τdτ =

=1

s− a

(1− lim

t→∞e−(s−a)t

).

We can easily see that

limt→∞e−(s−a)t =

{0 Real (s) = σ > a

Does not converge Real (s) = σ ≤ a.Hence, the Laplace transform is given by:

L(1(t)eat

)=

{1s−a if Real (s) > a

undefined otherwise

The same line of reasoning also applies to exponential functions with com-plex exponent: e(σ+ω)t with σ ∈ R and ω ∈ R. In this case the result is1/(s− σ − jω) and the convergence condition is Real (s) ≥ σ.

Example 4.3. Let us compute the Laplace transform of the DIRACδ(t).

L (δ(t)) =

∫ ∞0

δ(t)e−sτdτ =

=

∫ ∞0

δ(t)e−s0dτ =

=

∫ ∞0

δ(t)dτ =

1

Contrary to the examples before this transform is defined for all possible s.

Looking at the examples above, we can derive the following conclusions:

• The Laplace transform of a CT signal is a function of a complexvariable s,• The Transform is not defined for all possible values. In general, we

have a region of convergence (ROC) where the Transform makessense. For the step function 1(t) the ROC is Real (s) ≥ 0. For anexponential signal 1(t)eat the ROC is Real (s) > a.

There are several points that remain to be addressed. The first one is ifthe relation between a function and its Laplace transform is a bijection.In the affirmative case, another open point is how to invert the Laplace


transform. The third point is: what is the meaning and the practical use ofthis function.

4.4. Existence and Uniqueness of the Laplace transform

For the discussion on existence and uniqueness of the Laplace transform,we need the following definitions.

Definition 4.4. A function f(t) is said of exponential order γ if thereexist a constant A such that

f(t) ≤ Aeγt.

It is easy to show the following.

Theorem 4.5. Consider a function f(t) and assume that: 1) f(t) iscontinuous, 2) f(t) is of exponential order γ. Then the Laplace transform:

L (f(t)) =

∫ ∞0

f(τ)e−sτdτ,

exists and the ROC contains the half-space Real (s) > γ.

This theorem gives us conditions for the existence of the Laplace trans-form for some values of s.

The condition on invertibility requires the same definition.

Definition 4.6. Two functions f(t) and g(t) are said almost equal,f(t) ≈ g(t), if f(t) and g(t) are equal for all t except for a set of points ofnull measure.

A simple way to think of two functions that are almost equal is in thefollowing example.

Example 4.7. Consider the two functions f(t) = e3t and

g(t) =

{0 if t = 2, 4, 6, 8 . . .

e3t otherwise.

The two functions are almost equal because the set of isolated points givenby t = 2k, with k ∈ N has null measure.

We are now able to state the following:

Theorem 4.8 (Lerch’s Theorem). Suppose f(t) and g(t) are continuousexcept for a countable number of isolated points, and that they are of expo-nential order γ. Then if L (f(t)) = L (g(t)) for all s > γ the two functionsare almost equal: f(t) ≈ g(t).

From an engineering viewpoint saying that two functions are almostequal is tantamount to saying that they are the same, therefore with aslight abuse of notation we will write f(t) = g(t).

4.4. EXISTENCE AND UNIQUENESS OF THE LAPLACE TRANSFORM 43

4.4.1. Inverse Laplace Transform. The importance of Lerch’s theo-rem is lies in the fact that under reasonable and mild conditions the Laplacetransform is actually invertible. The inverse is given by:

f(t) = L−1 (F (s)) =1

2πj

∫ σ+j∞

σ−j∞F (s)estds,

where Real (s) = σ belongs to the ROC. This complex integral is generallydifficult to compute and we will find more manageable means to computethe inverse transform. However, the expression is quite insightful, becauseit reveals that the Laplace transform is actually a decomposition of a vectorf(t) using a basis given by the family of exponential functions, with F (s)being in some sense the coordinate related to the basis element est.

4.4.2. A first application of the Laplace transform. As we dis-cussed above, the Laplace transform H(s) of the impulse response has ac-tually a very precise meaning: it is the eigenvalue of the eigenfunctions est.The function H(s) is commonly called transfer function. This allows us toimmediate answer the query in the following example.

Example 4.9. Suppose that a system has impulse response 1(t)et. Findthe forced response to eαt. For the reasons that we have seen H(s) =1/(s − 1) with ROC Real (s) > 1. The eigenvalue is given by H(α) andis is defined only if α > 1. In this case the forced response is given byy(t) = eαt/(α− 1).

With a moderate effort, we can also answer the query in the followingexample.

Example 4.10. Suppose that a system has impulse response 1(t)e−3t.Find the forced response to cos 4t. As discussed in Chapter 2 (Theorem 3.6),the response is well defined if H(j4) converges, and in this case it is givenby:

y(t) = |H(j4)| cos(4t+ ∠H(j4).

H(s) is given by 1/(s+ 3) and its ROC is Real (s) > −3. Since j4 is on theimaginary axis H(j4) converges and the response above is well defined. Wecan find:

|H(j4)| =∣∣∣∣ 1

j4 + 3

∣∣∣∣ =

=|1|

|3 + j4|=

=1√

9 + 16=

=1

5,


and

∠H(j4) = ∠1

j4 + 3=

= ∠1− ∠3 + 4j

= − arctan 4/3.

4.5. Properties of the Laplace transform

We have seen expressions for both the direct and the inverse Laplacetransform. The easiest way to accomplish these tasks is not the applica-tion of these formulas, but the use of a number of properties that we willdiscuss below and the knowledge of the expressions of a few “elementary”transforms. In this section we will quickly review some of the properties.

4.5.1. Linearity. Given the linearity of the integral operator, it isstraightforward to prove the following.

Theorem 4.11. Let F1(s) = L (f1(t)) and F2(s) = L (f2(t)). Then,

L (h1f1(t) + h2f2(t)) = h1F1(s) + h2F2(s).

We immediately see an application of this result.

Example 4.12. Let us compute the Laplace transform of cos 3t.

L (cos(3t)) = L(ejωt + e−jωt

2

)=

1

2

(L(ejωt + e−jωt

))=

1

2

(1

s− jω+

1

s+ jω

)=

1

2

(2s

s2 − (jω)2

)=

s

s2 + ω)2.

4.5.2. Time Shifting. This property applies when we delay or antici-pate a causal function.

Theorem 4.13. Let F (s) = L (1(t)f(t)). Then F (s)e−st0 = L (1t− t0f(t− t0)).

4.5. PROPERTIES OF THE LAPLACE TRANSFORM 45

Proof. We can prove this by a simple change of variable:

L (1(t− t0)f(t− t0)) =

∫ ∞0

1(t− t0)f(t− t0)e−stdt

=

∫ ∞0

1(t− t0)f(t− t0)e−ste−st0est0dt

= e−st0∫ ∞

01(t− t0)f(t− t0)e−s(t−t0)dt

= e−st0∫ ∞−t0

1(t′)f(t′)e−st′dt′

= e−st0∫ ∞

01(t′)f(t′)e−st

′dt′

= e−st0F (s).

�

This property is shown in the following example.

Example 4.14. Find the Laplace transform of f(t) given by the follow-ing expression:

f(t) =

{1 If t ∈ [0, 1]

0 otherwise.

We can easily see that f(t) = 1(t)− 1(t− 1). Hence,

L (f(t)) = L (1(t))− L (1(t− 1))

=1

s− e−s

s

=1− e−s

s.

4.5.3. Shifting in the Laplace domain. The dual property of timeshifting applies if we apply a shift in the Laplace complex variable, as shownin the following.

Theorem 4.15. Left F (s) = L (f(t)). The F (s− s0) = L(f(t)es0t

).

Proof. Using the formula for Laplace transform, we have:

L(f(t)es0t

)=

∫ ∞0

f(τ)es0τe−sτdτ

=

∫ ∞0

f(τ)e−(s−s0)τdτ

= F (s− s0),

where the last step derives from the application of the definition of Laplacetransform. �


Example 4.16. Suppose we know that L (1(t)) = 1/s, then it is im-mediate to deduct that L

(1(t)eat

)= 1/(s − a). Likewise, if we know that

L (cosωt) = s/(s2 + ω2), it is immediate to deduct that L(eat cosωt

)=

s−a(s−a)2+ω2 .

4.5.4. Time Scaling. It is interesting to see what happens if we ex-pand or shrink the time scale.

Theorem 4.17. Let F (s) = L (f(t)) and let a ∈ R+. Then L (f(at)) =1aF (s/a).

Proof. We can apply the definition:

L (f(at)) =

∫ ∞0

f(at)e−stdt

=

∫ ∞0

f(t′)e−st′/adt

′

a

=1

aF (s/a).

�

Example 4.18. Suppose we know that L (cos t) = s/(s2 + 1) then

L (cosωt) =1

ω

s/ω

s2/ω2 + 1

=s

s2 + ω2.

4.5.5. Convolution. Now we come to one of the most important prop-erties of the Laplace Transform for its implications on Linear Systems.

Theorem 4.19. Let f(t) and h(t)be two causal functions and let L (f(t)) =F (s) and L (h(t)) = H(s). Then,

L (f(t) ∗ h(t)) = F (s)H(s).

Proof. Let g(t) = f(t) ∗ h(t). We have

L (g(t)) =

∫ ∞0

e−stf(t) ∗ h(t)dt

=

∫ ∞0

e−st(∫ t

0h(τ)f(t− τ)τ

)dt

=

∫ ∞0

∫ t

0e−sth(τ)f(t− τ)dτdt

The integration is carried out in the triangle 0 ≤ τ ≤ t. We can change theorder of integration:

L (g(t)) =

∫ ∞τ=0

∫ ∞t=τ

e−sth(τ)f(t− τ)dtdτ.


Let us make the following change of variable t = t− τ .

L (g(t)) =

∫ ∞τ=0

∫ ∞t=0

e−s(t+τ)th(τ)f(t)dtdτ

=

(∫ ∞τ=0

e−sτh(τ)dτ

)(∫ ∞τ=0

e−stf(t)dt

)= F (s)H(s).

�

We have just discovered a key fact: convolution in the time domain be-comes algebraic product in the Laplace domain. This suggests the followingprocedure to find the forced response to any signal u(t).

(1) compute the Transfer Function H(s) = L (h(t)),(2) compute U(s),(3) compute the inverse transform of H(s)U(s).

The practical execution of these steps requires: 1. a method for the com-putation of H(s) given a system described by a differential equation, 2. aneffective strategy to compute the inverse transform. Such problems will beaddressed below.

4.5.6. Differentiation rule. The computation of the Laplace trans-form of the derivative is key to the practical solution of differential equations.

Theorem 4.20. Let F (s) = L (f(t)). Then,

L(df(t)

dt

)= sF (s)− f(0).

Proof. We can apply integration by part to the definition of the Laplacetransform.

L(df(t)

dt

)=

∫ ∞0

df(t)

dte−stdt

= e−stf(t)|∞0 −∫ ∞

0f(t)(−se−st)dt

= limt→∞

e−stf(t)− f(0) + s

∫ ∞0

f(t)e−stdt

= sF (s)− f(0),

the last step is justified because in the ROC we must have limt→∞ e−stf(t) =

0. �

We now have a clear avenue to the solution of any differential equation.Consider the following example.

Example 4.21. Consider the RC network in discussed in Example 2.1.Set u(t) = Vin(t)1(t) and y(t) = Vc(t). As we now its evolution is described


by the differential equation:

y = − y

RC+u(t)

RC.

Let τ = RC. We can apply the properties of the Laplace transform asfollows:

L (y) = L(− y

RC+u(t)

RC

)sY (s)− y(0) = −L

(− y

RC

)+ L

(u(t)

RC

)sY (s)− y(0) = −− Y (s)

τ+ L

(U(s)

τ

)Y (s) =

U(s)

τ(s+ 1τ )

+y(0)

s+ 1τ

=

Y (s) =1

τs(s+ 1τ )

+y(0)

s+ 1τ

.

We can observe a few facts of interest.

• The decomposition between forced evolution ( 1τs(s+ 1

τ)) and free evo-

lution ( y(0)

s+ 1τ

) comes for free.

• Given that the forced response is given by U(s)

τ(s+ 1τ

)and that we know,

from the convolution rule, that Y (s) = H(s)U(s), where H(s) isthe transfer function, we can deduct that H(s) = 1

τ(s+ 1τ

)

• We can further write

1

τs(s+ 1τ )

=1

τ

(τ

s− τ

s+ 1τ

)

=1

s− 1

s+ 1τ

Under our assumptions, the Laplace transform is invertible. Hence, we canwrite:

y(t) = 1(()t)(1− et/τ ) + y(0)e−t/τ .

We can apply the differentiation recursively. For the case of the secondderivative:

L(D2f(t)

)= sL (Df(t))−Df(0) = s2F (s)− sf(0)−Df(0).

In the general case:

L (Dnf(t)) = snF (s)− sn−1f(0)− sn−2Df(0)− . . . Dn−1f(0).


4.5.7. Integration Rule. The Laplace transform of the integral of afunction is easily derived, as shown in the following:

Theorem 4.22. Le F (s) = L (f(t)), then

L(∫ t

0f(τ)dτ

)=F (s)

s.

Proof. We can simply observe that∫ t

0 f(τ)dτ = f(t) ∗ 1(t) and apply

the convolution rule taking into account that L (1(t)) = 1s . �

The convolution, the integration and the differentiation rule simply tellus that integral and differential operations in the time domain become simplealgebraic operations in the Laplace domain.

4.5.8. Final Value and Initial Value. There is much we can readfrom the expression of the Laplace transform without computing the inversetransform. This is shown in the following:

Theorem 4.23. Let F (s) be the known expression of L (f(t)). Then:I) if limt→0 f(t) exists then limt→0 f(t) = lims→∞ sF (s), II) if limt→∞ f(t)exists and is finite, then limt→∞ f(t) = lims→0 sF (s).

Proof. Let us first prove the first claim. From the differentiation rule,we know:

(4.1) L (Df(t)) =

∫ ∞0

d

dtf(t)e−stdt = sF (s)− f(0).

If we compute∫∞

0ddtf(t)e−stdt for s→∞, we find:∫ ∞

0

d

dtf(t)e−s∞dt = 0,

Therefore, we find f(0) = sF (s). Now, let us move to the second claim. Ifwe evaluate it for s→ 0 of

∫∞0

ddtf(t)e−stdt, we have:

lims−>0

∫ ∞0

d

dtf(t)e−stdt =

∫ ∞0

lims−>0

d

dtf(t)e−stdt =

=d

dtf(t)

∣∣∣∣∞0

= f(∞)− f(0).

Consider once more Equation 4.1, we find for s→ 0:

lims→0

sF (s)− f(0) = f(∞)− f(0),

which leads us straight to the claim. �

Example 4.24. Suppose that the Laplace Transform of a function isgiven by F (s) = s

s(s+2) . Then we have f(0) = lims→∞ sF (s) = 1, and

f(∞) = lims→0 sF (s) = 0.


4.5.9. Differentiation in the Laplace domain. The dual propertyfor the differentiation in time domain is shown in the following.

Theorem 4.25. Let L (f(t)) = F (s). Then L (−tf(t)) = dF (s)/ds.

Proof. We can write:

dF (s)

ds=

d

ds

∫ ∞0

f(t)e−stdt

=

∫ ∞0

df(t)e−st

dsdt

=

∫ ∞0

f(t)de−st

dsdt

=

∫ ∞0

(−t) · f(t)e−stdt

= L (−tf(t))

�

We can generalise the result in the following corollary:

Corollary 4.26. Let L (f(t)) = F (s). Then L ((−1)ntnf(t)) = dnF (s)/dsn.

Proof. It descends from an iterative application of Theorem 4.25. �

4.6. Inversion of the Laplace Transform

The application of the properties described in the previous section leadsto the following consideration.

Fact 4.27. Consider a CT LTI system described by the differential equa-tion

(4.2) Dny(t) =n−1∑i=0

αiDiy(t) +

p∑j=0

βjDju(t),

with p ≤ n and initial conditions:

y(0), Dy(0), . . . , Dn−1y(0), . . . , Dpu(0), . . . , Du(0).

Then the application of the differentiation rule leads to

Y (s) = Yforced(s) + Yfree(s) with

Yforced(s) =

∑pj=0 βjs

j

sn −∑n−1

i=0 αisiU(s)

Yfree(s) =N0(s)

sn −∑n−1

i=0 αisiU(s)

where N0(s) is a polynomial of degree n− 1 whose coefficients are functionsfor the initial conditions.

4.6. INVERSION OF THE LAPLACE TRANSFORM 51

Looking at the form of Yforced(s), we also conclude that the expressionof the transfer functions H(s) (i.e., the transform of the impulse response)is given by:

H(s) = L (h(t)) =

∑pj=0 βjs

j

sn −∑n−1

i=0 αisi.

Observing that: 1. for standard functions u(t) its Laplace Transformis given itself by a fraction of polynomials, 2. we operate in conditionswhere the uniqueness of the inverse Laplace Transform is guaranteed (up toa number of isolated points), we conclude that the solution of a differentialequation of a CT LTI system requires the inversion of two Laplace functionsgiven by fractions of polynomial in the s variable.

4.6.1. Inversion of fractions of polynomials. The general expres-sion of fraction of polynomials can be written as:

Asp + ap−1s

p−1 + ap−2sp−2 + . . . a0

sn + bn−1sn−1 + . . .+ b0,

where p ≤ n for the causality of the system. A fundamental result of algebrais the following:

Theorem 4.28. A polynomial of degree n with complex coefficients hasexactly n complex roots.

This allows us to write the fraction of polynomials as

A(s− z1)(s− z2) . . . (s− zp)(s− p1)(s− p2) . . . (s− pn)

.

The roots z1 of the numerator are called zeros while the roots of the denom-inator are called poles.

We now consider different cases.4.6.1.1. The case of real and distinct poles. The first case we consider is

when poles are real and distinct. In this case it is possible to see that ourfunction can be expressed as

F (s) = A(s− z1)(s− z2) . . . (s− zp)(s− p1)(s− p2) . . . (s− pn)

=A1

s− p1+

A2

s− p2+ . . .+

Ans− pn

Each term A1s−p1 is said a partial fraction and this is called partial fraction

expansion. We can show the following:

Proposition 4.29. Consider a function



with distinct and real roots. Then the coefficients of its partial fractionexpansion

F (s) =A1

s− p1+

A2

s− p2+ . . .+

Ans− pn

,

are given by:

Ai = F (s)(s− pi)|s=pi .

Proof. Without loss of generality we show the proposition for A1. Wecan write


=A1

s− p1+

A2

s− p2+ . . .+

Ans− pn

and multiply both sides by (s− p1).

F (s)(s− p1) = A1 +A2(s− p1)

s− p2+ . . .+

An(s− p1)

s− pn.

If we evaluate the result at s = p1, each of the termsAj(s−p1)s−pj is null (being

the roots distinct). Therefore, the right hand side produces A1. The righthand side is finite because the term s − p1 and the numerator cancels theonly s− p1 term at the denominator. �

Example 4.30. Consider the system:

y = 5y − 4y − 4u+ u.

We want to study the system’s evolution for u(t) = 1(t), y(0) = 1, y(0) = 0,u(0) = 2. Computing the Laplace transform we get:

s2Y (s)− sy(0)− y(0) = 5sY (s)− 5y(0)− 4Y (s)− 4sU(s) + 4u(0) + U(s)(s2 + 5s+ 4

)Y (s) = (1− 4s)U(s) + sy(0)− 5y(0) + y(0) + 4u(0)

Y (s) =1− 4s

s2 + 5s+ 4U(s) +

sy(0)− 5y(0) + y(0) + 4u(0)

s2 + 5s+ 4

Y (s) =1− 4s

s2 + 5s+ 4

1

s+

s− 5 + 8

s2 + 5s+ 4

Observe that s2 + 5s+ 4 = (s+ 4)(s+ 1) and

Y (s) =1− 4s

s(s+ 4)(s+ 1)+

s+ 3

(s+ 4)(s+ 1)


If we compute the partial fraction expansion, we find:

Y (s) =A1

s+

A2

s+ 1+

A3

s+ 4+

+B1

s+ 1+

B2

s+ 4where

A1 =1− 4s

(s+ 4)(s+ 1)ss

∣∣∣∣s=0

=1

4

A2 =1− 4s

(s+ 4)(s+ 1)s(s+ 1)

∣∣∣∣s=−1

=−5

3

A3 =1− 4s

(s+ 4)(s+ 1)s(s+ 4)

∣∣∣∣s=−4

=17

12

B1 =s+ 3

(s+ 4)(s+ 1)(s+ 1)

∣∣∣∣s=−1

=2

3

B1 =s+ 3

(s+ 4)(s+ 1)(s+ 4)

∣∣∣∣s=−4

=1

3.

and

y(t) = 1(t)

(1

4− 5

3e−t +

17

12e−3t

)+

+ 1(t)

(2

3e−t +

1

3e−3t

).

4.6.1.2. The case of complex conjugate poles. Before considering the caseof complex poles it is useful to state the following results:

Lemma 4.31. Consider a polynomial P (s) in a complex variable s with

real coefficient. Then P (s) = P (s).

Proof. We can write:

P (p) = pn + an−1pn−1 + an−2p

n−2 + . . .+ a0

= pn + an−1pn−1 + an−2p

n−2 + . . .+ a0

= pn + an−1pn−1 + an−2pn−2 + . . .+ a0 = P (p).

where the first step is applicable because the coefficients ai are real and arenot affected by the conjugation and the second one because the conjugateof a sum is the sum of the conjugates. �

Theorem 4.32. Consider a polynomial in a complex variable s with realcoefficient. If s = p is a root of the polynomial, then also its conjugate s = pis.

Proof. Consider the polynomial

P (s) = sn + an−1sn−1 + an−2s

n−2 + . . .+ a0,


with ai ∈ R. Il p is a root, then

P (p) = pn + an−1pn−1 + an−2p

n−2 + . . .+ a0 = 0.

In view of Lemma 4.31 we have: P (p) = P (p). If we now observe that

P (p) = 0 =⇒ P (p) = 0 our claim easily follows. �

With this result in mind let us consider the case of fractions of polyno-mial with real coefficients. This situation always occurs when dealing withordinary differential equations. Without loss of generality we will considera single pair of complex conjugate roots. We can show the following:

Proposition 4.33. Consider the following function given by a ratio ofcomplex polynomials with real coefficients. Assume that p1 and p1 are apair of complex conjugate poles.

F (s) =n(s)

d(s)

=n(s)

(s− p1)(s− p∗1) . . ..

Let

F (s) =A1

s− p1+

A1′

s− p1+ . . .

be its partial fraction expansion. Then A′1 = A1. In other words the coeffi-cient of the partial fraction related to the complex conjugate of the pole isthe complex conjugate of the partial fraction related to the pole.

Proof. Proposition 4.35 holds independently of whether poles are realor complex. Therefore, we can write:

A′1 = lims→p1

n(s)

d(s)(s− p1).

The denominator d(s) can be written as d(s) = (s − p1)(s − p1)d1(s). It iseasy to see that d1(s) is a polynomial with real coefficients. By ApplyingLemma 4.31, we find

A′1 =n(s)

d(s)(s− p1)

∣∣∣∣s=p1

=n(p1)

d1(p1)(p1 − p1)

=n(p1)

d1(p1)(−2jImag (p1))

=n(p1)

d1(p1)(2jImag (p1))

= A1.


�

An important consequence of the above can be seen when we computethe inverse transform of the two partial fractions related to a complex con-jugate pair. If

F (s) =A1

s− p1+

A1′

s− p1+ F1(s),

with p1 = σ1 + jω1, then

L−1 (F (s)) = L−1

(A1

s− p1+

A1′

s− p1+ F1(s)

),

= L−1

(A1

s− p1

)+ L−1

(A1′

s− p1

)+ L−1 (F1(s))

= L−1

(A1

s− p1

)+ L−1

(A1

s− p1

)+ L−1 (F1(s))

= A1ep1t +A1e

p1t + f1(t)

= A1ep1t +A1ep1t + f1(t)

= 2Real(A1e

p1t)

+ f1(t)

= 21(t) |A1| eσ1t cos(ω1t+ ∠A1)) + f1(t).

In other words, a complex conjugate pair gives rise to a sinusoidal oscillationwith amplitude modulate by a real exponential.

Example 4.34. Consider the differential equation:

y = y − y + u(t)

Compute the system’s response to 1(t) for null initial conditions. Neglectingthe initial conditions the differential Laplace transform is given by:

Y (s)(s2 − s+ 1) = U(s)

Y (s) =1

(s2 − s+ 1)s

=1

(s− 1+√−3

2 )(s− 1−√−3

2 )s

=1

(s− 1+j√

32 )(s− 1−j

√3

2 )s

=A1

s+

A2

s− 1+j√

32

+A2

s− 1−j√

32


with

A1 =1

(s2 − s+ 1)

∣∣∣∣s=0

= 1

A2 =1

(s− 1−j√

32 )s

∣∣∣∣∣s= 1+j

√3

2

=1

j√

31+j√

32

=1

−32 + j

√3

2

=−3

2 − j√

32

94 + 3

4

= −1

2− j√

3

6

Therefore we can compute the inverse as:

y(t) = 1(t)

(1 + 2 |A2| e

12t cos(

√3

2t+ ∠A2)

)

|A2| =√

1

4+

3

36

=

√3

3

∠A2 = atan2(−√

3

6,−1

2)

= −0.8571

4.6.1.3. The case of multiple roots. There cases when a pole occurs asa multiple roots of the denominator of the transfer function. For instance,we could have d(s) factored as (s + 3)2(s2 − s + 1). For the moment beinglet us focus on the case of just one multiple and real pole. Without loss ofgenerality, let us assume that the multiple pole is p1 and that its multiplicityis h:

F (s) = An(s)

(s− p1)hd1(s)

where d1(s) does not divide (s− p1).

Proposition 4.35. Consider the function in Equation (4.3), where thepole p1 is real and with multiplicity h. It is possible to come up with the


following partial fraction expansion:

F (s) = An(s)

(s− p1)hd1(s)

=A1,1

s− p1+

A1,2

(s− p1)2+ . . .+

A1, h

(s− p1)h+ F1(s)(4.3)

(4.4)

where F1(s) is found using the same rules that apply to single poles as wediscussed above and A1,i is given by:

A1,h−r =1

(h− r)!dr

dsr

[F (s)(s− pi)h

]∣∣∣∣s=pi

.

Proof. Multiplying both sides of Equation (4.4) by (s− p1)h we get:

n(s)

d1(s)= A1,1(s− p1)h−1 + . . .+A1,h−1(s− p1) +A1,h + F1(s)(s− p1)h.

Since neither d1(s) and any denominator in the partial fraction expansionof F1(s) divide (s− p1), when we evaluate in s = p1, we will have:

n(p1)

d1(p1)= A1,h.

Similarly, if we differentiate r times we get:

dr

dsr

[n(s)

d1(s)

]=A1,1(h− 1)(h− 2) . . . (h− r)(s− p1)h−1−r +A1,2(h− 2) . . . (h− r − 1)(s− p1)h−2−r + . . .+

. . .+ (h− r)(h− r − 1) . . . 1 ·A1,h−r+

+ h(h− 1) . . . (h− r)(s− p1)h−rF1(s) +dr

dsr[F1(s)] (s− p1)h.

If we evaluate in s = p1 we obtain our claim. �

Another important fact is in the following:

Proposition 4.36. Let F (s) = 1(s−p)h . Then L−1 (F (s)) = th−1

h! ept.

Proof. It descends from the application of Corollary 4.26 to L(1(t)eat

)=

1s−a �

This leads us to:

Theorem 4.37. Consider the function in Equation (4.3), where the polep1 is real and with multiplicity h. Let

F (s) =A1,1

s− p1+

A1,2

(s− p1)2+ . . .+

A1, h

(s− p1)h+ F1(s)

then

L−1 (F (s)) = 1(t)

(A1,1e

p1t +A1,2tep1t +A1,3

t2

2ep1t + . . .+A1,h

th−1

(h− 1)!ep1t)

+ L−1 (F1(s))


Example 4.38. Consider the following differential equation

y = 2y − y + u(t).

Let us compute the free evolution for y(0) = −4, y(0) = 2. Neglecting theinput function u(t), the Laplace function becomes:

(s2 − sy(0)− y(0))Y (s)− (2s− 2y(0))Y (s) + Y (s) = 0,

which is easily written as

Y (s) =sy(0) + y(0)− 2y(0)

s2 − 2s+ 1

=−4s+ 10

(s− 1)2

=A1,1

s− 1+

A1,2

(s− 1)2

A1,2 = (−4s+ 10)|s=1 = 6

A1,1 =d

ds[(−4s+ 10)]

∣∣∣∣s=1

= −4

As a consequence,

y(t) = 1(t)et (−4 + 6t) .

The case of complex conjugate poles with multiplicity greater than 1 isnot significantly different. Following the same line of reasoning as in SectionWe can treat each pole in the complex conjugate pair as in Section 4.6.1.2:1. finding the coefficients of the partial fraction expansion, 2. observingthat the coefficient related to the complex conjugate are complex conjugate,3. recombining the pairs related to the partial fraction with the same powerand reducing them to real functions. This is shown in the following example:

Example 4.39. Let us compute the inverse of

F (s) =1

s(s2 − s+ 1)2.


Let p1 = 1+j√

32 be one of the roots of (s2 − s + 1) (the other one being its

conjugate p1. We can proceed as follows:

F (s) =1

s(s− p1)2(s− p1)2

=A1,1

s− p1+

A1,1

s− p1+

+A1,2

(s− p1)2+

A1,2

(s− p1)2+

+A2

sA2 = F (s)s|s=0 = 1

A1,2 = F (s)(s− p1)2∣∣s=p1

=1

p1(p1 − p1)2

A1,1 =d

ds

[F (s)(s− p1)2

]∣∣∣∣s=p1

=−((s− p1)2 + 2s(s− p1)

s2(s− p1)4

∣∣∣∣s=p1

=−(3s− p1)

s2(s− p1)3

∣∣∣∣s=p1

=−3p1 + p1

p21(p1 − p1)3

Observing that p1 = ejπ/3 and p1 − p1 = 2j√

32 = j√

3, we find

A1,2 = − 11+j√

32 3

= − 1

3ejπ/3

=1

3e−jπ/3


and

A1,1 =e−jπ/3 − 3ejπ/3

ej2π/3(−j3√

3)

=e−jπ/3 − 3ejπ/3

ej2π/3e−jπ/23√

3

=e−jπ/3 − 3ejπ/3

ejπ/63√

3

=e−jπ/2 − 3ejπ/6

3√

3

=−j − 3

√3

2 −32j

3√

3

=−3√

32 −

52j

3√

3

= −1

2− 5

6√

3j

=1

3

√13

3ejatan2(−5/(6

√3),−1/2)

=1

3

√13

3e−j2.3754

We can combine the results as follows:

f(t) = A21(t)+

+ 1(t)(A1,1e

p1t +A1,1ep1t)

+

+ 1(t)(A1,2te

p1t +A1,2tep1t)

= 1(t)(1 + 2Real

(A1,1e

p1t)

+ 2Real(A1,2te

p1t))

=

= 1(t)

(1 + 2

1

3

√13

3e1/2t cos (

√3

2t− 2.3754) +

2

3te1/2t cos (

√3

2t− π/3)

)

4.7. BIBO Stability for CT systems

The discussion above has lead us to the conclusion that a pole in a CTsystem gives rise to a complex exponential possibly multiplied by a power oft (if the root is multiple). In view of Theorem 3.10, this can be translatedinto the following.

Theorem 4.40. Consider a CT LTI system with transfer function:

H(s) =n(s)

d(s).

4.7. BIBO STABILITY FOR CT SYSTEMS 61

Assume that no cancellation between poles and zero takes place: 6 ∃p : n(p) =d(p) = 0. The system is BIBO stable if and only if all of its poles havenegative real part.

Proof. In view of Theorem 3.10, the system is stable if and only if∫∞0 |h(t)|dt = L < +∞. We now start by proving sufficiency. If H(s) =

L (h(t)) has all poles with negative real part then h(t) =∑H

h=1 Fhhh(t),where the functions hh(t) can be one of the following:

hh(t) =

epht Real root (single or multiple)

eReal(ph)t cos (Imag (ph) t+ φh) Complex single or multiple root

tnhepht Real multiple root

tnheReal(ph)t cos (Imag (ph) t+ φh) Complex multiple root,

which can be interpreted as

|hh(t)| ≤ Hh(t) =

epht Real root (single or multiple)

eReal(ph)t Complex single or multiple root

tnhepht Real multiple root

tnheReal(ph)t Complex multiple root,

and∫∞0 Hh(t)dt =

limK→∞ephK−1ph

For real roots

limK→∞eReal(ph)K−1

phFor complex roots

limK→∞∑nh

k=0(−1)nh−k nh!

k!pnh−kh

KkephK − (−1)nh 1pnhh

Real multiple root

limK→∞∑nh

k=0(−1)nh−k nh!

k!Real(ph)nh−kKkeReal(ph)K − (−1)nh 1

Real(ph)nhComplex multiple root

If Real (ph) is smaller than 0 for all poles, this leads to

∫ ∞0

Hh(t)dt =

−1ph

For real roots−1

Real(ph) For complex roots

− (−1)nh

pnhh

Real multiple root

− (−1)nh

Real(ph)nhReal multiple root.

In other words we can find a constant F that upper bounds all Hh(t) for allh. ∫ ∞

0|h(t)|dt ≤ |Fh|

∫ ∞0

∑|Hh(t)|dt ≤ FM.

To prove necessity, Consider a single pole pi with non negative real partexists and assume for simplicity that its multiplicity is1. Then in the partialfraction expansion ofH(s) we will have a term Ai

s−pi with non null Ai. Indeed,

Ai =n(pi)

(pi − p1) . . . (pi − pi−1)(pi − pi+1) . . . (pi − pn),


and by our assumption n(pi) 6= 0. This fraction will give rise to an expo-nential function in h(t) whose integral grows in an unbounded way. �

Example 4.41. Let us consider the following transfer function:

H(s) =s− 1

s2 + 3s+ 2.

Its poles are given by

p1,2 =−3±

√9− 8

2=

{−2

−1.

Both roots are negative. Hence, the system is BIBO stable.

This example is easy to analyse because the roots of a second orderequation can be found using a closed-form expression. This is possible alsofor order 3 and 4, although the expressions are way less intuitive to dealwith.

Definition 4.42. The polynomial d(s) at the denominator of the trans-fer function is said characteristic polynomial and the equation d(s) = 0 iscalled characteristic equation.

4.7.1. The Routh-Hurwitz Criterion. If we have a mere interest insystem’s BIBO stability the explicit computation of the roots is not requiredas long as we are able to compute the sign of their real part. This can bedone using the Routh-Hurwitz criterion. Suppose the denominator d(s) ofthe transfer function is:

ansn + an−1s

n−1 + . . . a0.

Suppose that n is even. The first step is to form the Routh table as follows:

sn an an−2 an−4 . . . a2 a0

sn−1 an−1 an−3 an−5 . . . a3 a1

sn−2 b1 b2 b3 . . . bn2 0sn−2 c1 c2 c3 . . . 0 0. . . . . . . . .s0 q


where computation of each row is made using the results of the two rowsimmediately above. For instance:

b1 =an−1an−2 − anan−3

an−1


an−1


an−1

. . .

c1 =b1an−3 − b2an−1

b1

c2 =b1an−5 − b3an−1

b1

c2 =b1an−7 − b4an−1

b1. . . .

After building this table, we can state the following.

Theorem 4.43. Suppose that the Routh table can be built as:

sn an an−2 an−4 . . . a2 a0

sn−1 an−1 an−3 an−5 . . . a3 a1

sn−2 b1 b2 b3 . . . bn2 0sn−2 c1 c2 c3 . . . 0 0. . . . . . . . .s0 q

Assume that the first column (an, an1, b1, c1, . . . , q) does not contain 0 ele-ments. Then the number of sign changes in the first column (an, an1, b1, c1, . . . , qof the Routh table corresponds to the number of poles in the right half of thecomplex plan. Moreover, a necessary and sufficient condition for all rootsto be in the left a half plan is that all coefficients ai be positive and that allthe elements of the first column be positive.

Example 4.44. Consider the classic second order equation:

a2s2 + a1s+ a0.

The Routh table is given by

s2 a2 a0

s a1 0s0 a2.

Hence the requirements that all poles are in the left half plan reduces to thefact that all coefficients have to be positive. This can be seen by consideringthat if the roots are negative, we can write the polynomial as

(s+ r1)(s+ r2) = s2 + (r1 + r2)s+ r1r2


So if r1 and r2 are positive (meaning that the roots are negative) if and onlyif (r1 + r2) and r1r2 are positive in their turn.

Example 4.45. We can now consider the case of polynomial of degreethree.

a3s3 + a2s

2 + a1s+ a0 = 0.

The Routh table is given by:

s3 a3 a1 0s2 a2 a0 0s a2a1−a3a0

a20 0

s0 a0 0 0

Hence, along with ai > 0, BIBO stability will also require a2a1 − a3a0 > 0.

To simplify the construction of the table, we can use the following result:

Proposition 4.46. If we multiply an entire row of the Routh table bya positive constant Theorem 4.43 still holds.

There are two important special cases:

(1) The first element of a row is 0, thus preventing to derive the fol-lowing row (which requires the division by the first element),

(2) An entire row is 0.

Both cases reveal the presence of roots on the imaginary axis or in thepositive half plan and therefore the loss of BIBO stability. Still we can useRouth criterion to identify the number of stable and unstable roots.

The first problem is solved by perturbing the 0 element (setting it toε) and evaluating the number of changes and permanence both for positiveand negative ε.

Example 4.47. Consider the polynomial

s3 + s− 1.

Its associated table iss3 1 1 0s2 0 −1 0s ???s0 ???

In order to compute the missing elements of the Routh table, we can intro-duce ε:

s3 1 1 0s2 ε −1 0s ε+1

ε 0 0s0 −1 0

If we consider ε→ 0+ we can see observe one sign change in the first column.The same is we consider the changes for negative values: ε→ 0−. We inferthe presence of one root in the right half plan and two roots in the negativehalf plan.


When an entire row is null, this means that two rows are proportionaland that the characteristic polynomial d(s) can be divided by the polynomialassociated with the row immediately above the null row. The di visor poly-nomial is called “auxiliary polynomial” and call it p(s). Having a null rowalso indicates that the auxiliary polynomial has roots that are symmetricalwith respect to the imaginary axis. The number of changes above the linewith zeros accounts for the stability of the roots of the “remainder” polyno-mial. One possibility to complete the construction of the Routh table is to

replace the null line with dp(s)ds and then continue. Looking the first column

of the modified Routh table, the number of sign changes is still equal to thenumber of roots with positive real part. From the null row down, each signchange indicates a root with positive real part. Since roots are symmetric,there is be a corresponding root in the negative half plan. All roots notaccounted in this way (i.e., no sign changes) are on the imaginary axis.

Example 4.48. Consider the polynomial

d(s) = (s2 + 1)(s+ 1)(s+ 3)

= s4 + 4s3 + 4s2 + 4s+ 3

In this case we know up-front that the polynomial has two stable roots anda pair of imaginary roots. Let us see how the modified Roth criterion allowsus to find this information. The first two rows of the Routh table are:

s4 1 4 3s3 4 4 0. . .

We can divide the second row by 4 and compute the third row as

s4 1 4 3s3 1 1 0s2 3 3 0. . .

We divide the third row by 3 and find:

s4 1 4 3s3 1 1 0s2 1 1 0s 0 0 0

The three positive elements of the first rows reveal two roots with negativereal part. We can now complete the construction. The auxiliary polynomialis given by s2 + 1 and its derivative is 2s. Therefore we can complete the


C(s) P(s) R(s) U(s) Y(s)

C(s) P(s) R(s) U(s) Y(s) E(s)

(A)

+ -‐

(B)

Figure 1. (A) series connection, (B) feedback connection

construction as:

s4 1 4 3s3 1 1 0s2 1 1 0s 2 0 01 1 0 0

The two new elements are positive. This means no changes and two rootson the imaginary axis.

4.8. Use of Laplace Transform for Control Design

The Laplace transform is a very useful tool for designing systems thatcontrol other systems. This is done taking advantage of some propertiesof the connections of systems. The two simplest ways of connecting twosystems are shown in Figure 1. For series connection (Figure 1.(A)), weknow already that, given the two impulse response c(t) and p(t) the outputis y(t) = p(t) ∗ c(t) ∗ r(t). In the Laplace domain this property can betranslated as:

Y (s) = P (s)C(s)R(s),

because convolutions correspond to products in the Laplace domain. Thisconnection can be used in different ways as in the following examples.

Example 4.49. Consider a system

P (s) =1

s+ 5.

4.8. USE OF LAPLACE TRANSFORM FOR CONTROL DESIGN 67

Suppose we want it to track with 0 error constant reference signals. If ourreference signal is r(t) = A1(t), this requirement could be stated as:

limt→∞

y(t) = A.

If make a series connection of P (s) and C(s), we have:

Y (s) = P (s)C(s)A

s.

Applying the final value theorem

limt→∞

y(t) = lims→0

sP (s)C(s)A

s= AP (0)C(0).

Hence, the specification requires: C(0) = 1P (0) = 5. The simplest way of

having this is by obviously choosing C(s) = 5. Clearly, if the pole is knownwith some error ∆ the steady state error will be affected by an error of thesame amount.

Suppose, now, that we also require a faster response than the one as-sociated with e−5t, setting it, for instance, to e−10t. One possibility is tochoose the controller

C(s) = 10s+ 5

s+ 10.

Notice that gain has been chosen equal to 10 in order to have P (0)C(0) = 1.

As shown in the previous example, using series connection we can ad-just the steady state response (up to a certain precision) and even changethe dynamics of the system cancelling some of the poles with zeros in thecontroller. This is not possible if some of the poles are unstable, as shownnext.

Example 4.50. Consider the system

P (s) =1

s− 5.

Suppose we want it to track with 0 error constant reference signals. Ifour reference signal is r(t) = A1(t). The evident problem here is that thesystem is unstable. So our first problem is to “stabilise” it. We could thinkof a series connection with a controller system like C(s) = p s−5

s+p with p > 0.

The problem becomes evident if the pole is known with some precision.Suppose that instead of having

P (s) =1

s− 5 + ∆.


with ∆ << 1. The overall response to U(s) = 1s is given by:

P (s)C(s)R(s) = ps− 5

s(s+ p)(s− 5 + ∆)

=5

5−∆

1

s− p+ 5

(p+ 5−∆)

1

s+ p− p −∆

(5−∆)(5−∆ + p)

1

s− 5 + ∆

The term

H1

s− 5 + ∆,

with = −p −∆(5−∆)(5−∆+p) . This means that however small a perturbation

∆ on the pole can be, it gives rise to an exponential term e(5−∆)t, whichdiverges compromising the system’s stability.

From the example above, we have seen that controlling a system by onlyusing a cascaded connection with a controller is a solution that successfullysolves only parts of the possible range of problems that a designer mightencounter. In particular, compensating the error of the system by a “gain”(i.e., multiplication by a constant) can be a viable solution but it is affectedadversely by the limited knowledge of the system parameters. More impor-tantly, an unstable system cannot be made stable by simply “cancelling”the unstable poles with zeros in the controlling system. A more useful solu-tion for this cases is the feedback connection shown in Figure 1.(B) For thisconnection we can write:

Y (s) = P (s)U(s)

U(s) = C(s)E(s)

E(s) = R(s)− Y (s).

By combining the three equations we easily find:

(4.5) Y (s) =P (s)C(s)

1 + P (s)C(s)R(s).

The equation above is called “closed–loop equation” and the term P (s)C(s)1+P (s)C(s)

is the “closed–loop transfer function”.If we want to enforce a 0 steady state error to the step response (U(s) =

As ), we can once again apply the final value theorem:

limt→∞

y(t) = lim s→ 0sY (s)

= lim s→ 0sY (s)

= lim s→ 0sP (s)C(s)

1 + P (s)C(s)

A

s

Now, if lims→0 P (s)C(s) =∞ we have that

limt→∞

y(t) = A,

4.8. USE OF LAPLACE TRANSFORM FOR CONTROL DESIGN 69

no matter how imprecise the knowledge of P (0) is. This means that eitherP (0) = 1

sP1(s) or C(0) = 1sC1(s). To put it in other words: “if we want to

use a feedback connection to follow a step function with steady state error0, an integrator 1

s has to be found in P (s) or in C(s)”. This is a particularexample of principle known as “internal model principle”, whereby if we wantto generate a signal it has to be contained in the system to be controlled orwe have to insert it into the loop through the controller. The fact that C(s)appears in the denominator of the closed loop transfer function allows us toshift the position of the pole. Unlike what we did with a simple pole-zerocancellation, this technique is much more reliable in terms of parametricuncertainty. This is shown in the following.

Example 4.51. Consider the system

P (s) =10 + ∆0

s− 5−∆1

where ∆0 and ∆1 are bounded but unknown. Suppose we choose the feed-back controller:

C(s) =10(s+ 5)

s.

Let us see whether or not it fulfils the specified requirements.In order to fulfil the requirement of having zero steady state error, we

need to have an integrator in the loop, which is verified in our case becauseC(s) has a pole in s = 0 as long as the system is stable. The closed loopresponse is given by

P (s)C(s)

1 + P (s)C(s)=

10+∆0s−5−∆1

10(s+5)s

1 + 10+∆0s−5−∆1

10(s+5)s

=(10 + ∆0)(10(s+ 5))

(s− 5−∆1)s+ 10(s+ 5)(10 + ∆0)

=(10 + ∆0)(10(s+ 5))

s2 + (10∆0 −∆1 + 95)s+ 500 + 50∆0

In order to verify when the system is stable, we can apply Routh Criterion,and require that all coefficients of the characteristic equation be positive,which is verified if:

∆0 > −10

∆1 < 10∆0 + 95.

As long as these conditions are met the system is both BIBO stable andresponds with null error to a step input. We emphasise that if we use an“open loop” (series connection) scheme, it is sufficient to have small ∆1 todestroy stability and that having a ∆0 6= 0 determines errors in he steadystate behaviour.


4.9. The z-Transform

The z-transform is the discrete time counter–part of the Laplace tran-form. In the same way as the Laplace transform, the z-Tranform, providesanalytical methods for the solution of a difference equation, it offers directinsight into the transient and steady state behaviour of a DT signal, andcan be used to evaluate the stability of a system.

The definition of the z-Transform is the following:

Z (f(t)) =∞∑0

f(t)z−t.

This time we associate a DT signal with a function of the complex variablez. We will find it convenient to use the polar representation: z = ρejφ.

Example 4.52. Let us compute the z-Transform of 1(t):

Z (1(t)) =

∞∑0

1(t)z−t

=∞∑0

z−t

= limH→∞

H∑0

z−t

= limH→∞

1− z−H

1− z−1.

Setting z = ρejθ, we have z−H = ρ−He−jHθ. We have two cases:

limH→∞

z−H =

0 if ρ = |z| > 1

∞ if ρ = |z| < 1

e−jHθ if ρ = |z| = 1

Z (1(t)) = limH→∞

1− z−H

1− z−1=

{1

1−z−1 if |z| > 1

is not defined otherwise

4.10. EXISTENCE AND UNIQUENESS OF Z-TRANSFORM 71

Example 4.53. Let us compute the z-Transform of 1(t)at:

Z (1(t)) =∞∑0

1(t)atz−t

=

∞∑0

z−t

= limH→∞

H∑0

(az

)t= lim

H→∞

1−(az

)H1− a

z

.

Suppose a > 0. Setting z = ρejθ, we have(az

)H=(aρ

)He−jHθ. If a < 0,

we have(az

)H=(aρ

)Hejπ−jHθ. This allows us to conclude:

Z(1(t)at

)=

{z

z−a if |z| > |a|is not defined otherwise

From this discussion we can conclude the following facts.

• The z-Transform of a DT signal is a function of a complex variablez (in the same way as the Laplace transform is a function of thecomplex variable s).• As for the Laplace transform, it is not typically possibile to define

the z-Transform for all values of z, but only for a subset that wedefine Region of Convergence (ROC).• Whereas for the Laplace transform the ROC is typically an half

space (Real (s) > α) for the z-Transform it is an anulus |z| ≥ ρ.

4.10. Existence and Uniqueness of z-Transform

On existence of the z - Transform, we can show the following result:

Theorem 4.54. Consider a function f(t) and assume that one of thefollowing limits exists:

Rf = limt→∞|f(t)|1/t

Rf = limt→∞

f(t+ 1)

f(t).

Then:

(1) the z-Transform Z (f(t)) exisats and converges for |z| geqRf .(2) the z-Trasform is analytic, i.e., continouus and infinitely differen-

tiable w.r.t. z, for |z| ≥ Rf .


Showing this result is complex and beyond the scope of this course.

We only remark that the existence of a limit Rf = limt→∞ |f(t)|1/t can berephrased as the existence of an exponential upper bound for the function

f(t) ≤ ARtf ,for some A.

We can also show the following inverse result:

Theorem 4.55. If F (z) and G(z) are z-Transofrms of two functionsf(t) and g(t) and if F (z) = G(z) for all |z| > R, for some R > 0 thenf(t) = g(t) for t = 0, 1, 2, . . ..

Proof. f G(z) = F (z) for all |z| > R then

∞∑t=0

f(t)z−t =∞∑t=0

g(t)z−t ↔

∞∑t=0

(f(t)− g(t)) z−t = 0↔∞∑t=0

atwt = 0

where we have set w = 1/z and at = f(t)− g(t). It is known that if we havea power series

∑∞t=0 atw

t then if∑∞

t=0 atwt = 0 for all |w| ≤W then at = 0.

Using this result, we conclude f(t) = g(t). �

4.10.1. Inverse z-Transform. It is possible to show that the inversetransform of the z-Transform is defined by the following line integral.

f(t) = Z−1 (F (z)) =

∮|z|=R

F (z)zt−1dz

where the line integral is computed along a circle included in the ROC.As shown below, this formula is impractical and the inverse z-Transform isbest computed using a different procedure. As for the Laplace transformthis formula is insightful in that it reveals that the z-Transform is in fact adecomposition of a function using a basis of functions of type zn.

4.11. z-Transform of sampled data signals

It is interesting to see how the z-Transform and Laplace transform arein fact close relatives. Consider a CT signal f(t), its Laplace transform isgiven by:

F (s) =

∫ ∞0

f(τ)e−sτdτ.

Suppose we transform the signal into a DT sequence by taking a sampleevery T time units. This can be done by multiplying the signal by a sequenceof Dirac δ, each one extracting a sample:

fD(t) = f(t)

∞∑k=0

δ(t− kT ).

4.11. Z-TRANSFORM OF SAMPLED DATA SIGNALS 73

If we compute the Laplace transform of this signal, we find:

L (fD(t)) =

∫ ∞0

fD(τ)e−sτdτ

=

∫ ∞0

f(t)∞∑k=0

δ(t− kT )e−sτdτ

=

∫ ∞0

∞∑k=0

f(kT )δ(t− kT )e−sτdτ

=

∫ ∞0

∞∑k=0

f(kT )δ(t− kT )e−skTdτ

=

∞∑k=0

∫ kT

(k−1)Tf(kT )δ(t− kT )e−skTdτ

=∞∑k=0

f(kT )e−skT∫ kT

(k−1)Tδ(t− kT )dτ

=∞∑k=0

f(kT )e−skT

=∞∑k=0

f(kT )(e−sT )k

If we set esT = z we find

F (z) =

∞∑k=0

f(kT )zk,

we find the definition of the z-Transform.

CHAPTER 5

State space Realisation

75

CHAPTER 6

Structural Properties of Systems

77

CHAPTER 7

Linear Systems Control

79

APPENDIX A

Some mathematical definition of interest

We now list some mathematical definition of interest. In the text belowwe will refer to a set denoted as T and to its elements, denoted by lowercapitals letters such as t, t1, t2, . . ..

Definition A.1. Un insieme T si dice totalmente ordinato se su diessa e definita una relazione ≤ di ordinamento totale definita su coppiedegli elementi dell’insieme. Una definizione di ordinamento totale gode delleseguenti proprieta:

(1) Riflessivita: ∀t ∈ T : t ≤ t;(2) Antisimmetria: t1 ≤ t2 ∧ t2 ≤ t1 → t1 = t2(3) Transitivita: t1 ≤ t2 ∧ t2 ≤ t3 → t1 ≤ t3(4) Confrontabilita (tricotomia): ∀t1, t2 ∈ T o t1 ≤ t2 oppure t2 ≤ t1.

Se valgono solo le prime tre proprieta si parla di ordinamento parziale.

Come esempio tutti i possibili spazi rappresentati il temp nel nostrocorso sono almeno parzialmente ordinati.

Diamo ora la definizione di spazio metrico.

Definition A.2. Consideriamo una funzione d : T × T → R, dettadistanza, definita sulle coppie di elementi di un insieme T che associa a cias-cuna coppia di punti un numero reale. Tale funzione definisce una metricase:

• Positivita: ∀t1, t2 ∈ T , d(t1, t2) ≥ 0 e d(t1, t2) = 0 se e solo set1 = t2.• Simmetria: ∀t1, t2 ∈ T : d(t1, t2) = d(t2, t1).• Disuguaglianza triangolare: ∀t1, t2, t2 ∈ T : d(t1, t3) ≤ d(t1, t2) +d(t2, t3).

Definition A.3. Uno spazio metrico e un insieme su cui sia definitauna metrica.

Definition A.4. A point t is a limit point of a set T if for any ε > 0there exist another point t1 different from t such that d(t, t1) ≤ ε, whered(., .) is a metric.

Definition A.5. The space T is continuum if it is a closed connectedset. A connected set is one where no matter how it is divided in two disjointset, at least one of them will contain limit points of the other. A closed setis one that contains the limit poitns of all its subsets.

81

82 A. SOME MATHEMATICAL DEFINITION OF INTEREST

The time space T for CT systems is a continuum because we cannot poseconstraints on when events and observations occurs, just as it happens inthe physical world.

Bibliography

[Hes09] Joao P Hespanha, Linear systems theory, Princeton university press, 2009.[LSV98] Edward A Lee and Alberto Sangiovanni-Vincentelli, A framework for compar-

ing models of computation, Computer-Aided Design of Integrated Circuits andSystems, IEEE Transactions on 17 (1998), no. 12, 1217–1229.

[PPR95] Charles L Philips, John M Parr, and E Riskin, Signals, systems, and transforms,Prentice Hall, 1995.

[RI79] Antonio Ruberti and Alberto Isidori, Teoria dei sistemi, Boringhieri, 1979.

83

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