EE351: Spectrum Analysis and Discrete Time Systems ...homepage.usask.ca/~hhn404/EE351/slides/EE351-Fall05-Part1...EE351: Spectrum Analysis and Discrete Time Systems (Signals, Systems

EE351: Spectrum Analysis and Discrete Time Systems

(Signals, Systems and Transforms)

Dr. Ha H. Nguyen

Associate Professor

Department of Electrical Engineering

University of Saskatchewan

August 2005

EE351–Spectrum Analysis and Discrete Time Systems University of Saskatchewan

Introduction

• The concepts of signals and systems arise in a wide variety of areas, such as

communications, circuit design, biomedical engineering, power systems, speech

processing, etc.

• The ideas and techniques associated with these concepts play an important role

in such diverse areas.

• Although the physical nature of the signals and systems that arise in these

various disciplines may be drastically different, two basic features in common are:

– The signals, which are functions of one or more independent variables,

contain information about the behavior or nature of some phenomenon.

– The systems respond to particular signals by producing other signals or some

desired behavior.

• Examples of signals and systems:

– Voltages and currents as functions of time in an electrical circuit are

examples of signals. A circuit is itself an example of a system, which

responds to applied voltages and currents.

– A camera is a system that receives light from different sources and produces

a photograph.

Dr. H. Nguyen Page 1


• Example problems of signal and system analysis:

– Analyzing existing systems: We are presented with a specific system and are

interested in characterizing it in detail to understand how it will respond to

various inputs (analysis of a circuit).

– Designing systems to process signals in particular ways. For example, to

design systems to enhance or restore signals that have been degraded in

some way (image restoration, image enhancement).

– Designing systems to extract specific pieces of information from signals.

Examples include the estimation of heart rate from an electrocardiogram,

weather forecast.

– Designing of signals with particular properties. For example, the design of

communication signals must take into account the need for reliable reception

in the presence of distortion (due to transmission media) and interference

(such as noise).



Chapter I: Signals and Systems

There is an analytical framework–that is, a language for describing signals and

systems and an extremely powerful set of tools for analyzing them–that applies

equally well to problems in many fields.

This chapter begins the development of such an analytical framework for signals and

systems by introducing their mathematical description and representations.

Signals are represented mathematically as functions of one or more

independent variables.

Examples:

• A speech signal can be represented mathematically by acoustic pressure as a

function of time.

• A picture can be represented by brightness as a function of two spatial variables.

This course focuses only on signals involving a single independent variable.

For convenience, the independent variable will generally referred to as time, although

it may not in fact represent time in specific applications.



Examples of SignalsArterial Blood Pressure

0 0.5 1 1.5 2 2.5

60

70

80

90

100

110

120

Time (sec)

AB

P (m

mH

g)

Portland State University ECE 222 Signal Fundamentals Ver. 1.06 7

Microelectrode Recording

2 2.01 2.02 2.03 2.04 2.05 2.06

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

Time (sec)


Speech

1.2 1.25 1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

Time (sec)

Linus: Philosophy of Wet Suckers


Electrocardiogram

0 0.5 1 1.5 2 2.5

6.5

7

7.5

8

8.5

Time (sec)


31

1.4

.6Sta

bility

Consid

erbounded–input

bounded–output

(BIB

O)

stability

Stable

systemif

foran

ybounded

input

signal

|x(t)|

≤B

x<

∞,

∀t

the

outp

ut

signal

isbounded

|y(t)|

≤B

y<

∞,

∀t

Exam

ple

:

–Stab

lesystem

Averag

er:y[n

]=

1

2N

+1

N∑k=−

N

x[n

−k]

Bounded

input|x

[n]|

<B

x⇒

bounded

outp

ut|y

[n]|

<B

y=

Bx

–In

stable

system

Integ

rator:y(t)

=

t∫−∞

x(τ

)dτ

E.g

.bounded

inputx(t)

=u(t)

⇒unbounded

outp

uty(t)

=t

System

stability

isim

portan

tin

engin

eering

applicatio

ns,

unstab

le

systems

need

tobe

stabilized

.

Lam

pe,Schober:

Sig

nals

and

Com

munic

atio

ns

32

Exam

ple

:T

he

first

Taco

ma

Narrow

ssu

spen

sion

bridge

collap

seddue

tow

ind-in

duced

vibrations,

Novem

ber

1940.

(Photos from http :/ / w w w .e n m .b ris.a c .u k / rese a rch/ n on lin e a r/ ta com a / ta com a .htm l)

Lam

pe,Schober:

Sig

nals

and

Com

munic

atio

ns



Discrete-time & Continuous-time Signals

• The course will cover concepts and techniques associated both with

continuous-time and discrete-time signals (and systems).

• Continuous-time signals are defined for a continuum of values of the

independent variable (the independent variable is continuous).

– Will always be treated as a function of t.

– Parentheses are used to denote continuous-time functions, for example x(t).

– The independent variable t is a real-valued and continuous.

• Discrete-time signals are only defined at discrete times (the independent variable

takes on only a discrete set of values).

– Will always be treated as a function of n.

– Square brackets are used to denote discrete-time functions, for example x[n].

– The independent variable n is an integer.

• Examples:

– The speech signal as a function of time and atmospheric pressure as a

function of altitude are examples of continuous time signals.

– The daily closing stock market index is an example of discrete-time signal.



• It is often useful to represent the signals graphically:

Arterial Blood Pressure

0 0.5 1 1.5 2 2.5

60

70

80

90

100

110

120

Time (sec)

AB

P (m

mH

g)


Microelectrode Recording

2 2.01 2.02 2.03 2.04 2.05 2.06

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

Time (sec)


Speech

1.2 1.25 1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

Time (sec)

Linus: Philosophy of Wet Suckers


Electrocardiogram

0 0.5 1 1.5 2 2.5

6.5

7

7.5

8

8.5

Time (sec)

Portland State University ECE 222 Signal Fundamentals Ver. 1.06 6Figure 1: A continuous-time signal (electrocardiogram).

3

Discrete– time sig nals

– S ymbol n for independent variable

– U se brackets [·]

Discrete– time sig nal: x[n]

– Graph ical representation

x[0]

x[1]x[−1]

0 321

x[2]x[−2]

n−3 −2 −1

54

x[n]

Lampe , S c h o b e r: S ig n als an d C ommu n icatio n s

4

1.1.2 S ign a l E n e rgy a n d Powe r

O ften classifi cation of sig nals according to en erg y and power

– Terminolog y en erg y and power u sed for any sig nal x(t), x[n]

– N eed not necessarily h av e a ph y sical meaning

S ig nal energ y

– E nerg y of a possibly complex continu ou s– time sig nal x(t) in

interv al t1 ≤ t ≤ t2

E(t1, t2) =

t2∫

t1

|x(t)|2 d t

– E nerg y of a possibly complex discrete– time sig nalx[n] in interv al

n1 ≤ n ≤ n2

E(n1, n2) =

n2∑n=n1

|x[n]|2

– Total energ y

E∞ = E( − ∞,∞) =

∞∫

− ∞

|x(t)|2 d t

E∞ = E( − ∞,∞) =

∞∑n= − ∞

|x[n]|2


Figure 2: An example of discrete-time signal.



• Continuous-time signals (and systems) have very strong roots in problems

associated with physics, and, more recently, with electrical circuits and

communications.

• The techniques of discrete-time signals (and systems) have strong roots in

numerical analysis, statistics and time series analysis (associated with such

applications as the analysis of economic and demographic data).

• In the past decades, there has been a growing interrelationship between

continuous-time signals and systems and discrete-time signals and systems. This

has come from the dramatic advances in technology for the implementation of

systems and for the generation of signals. For example, it is increasingly

advantageous to consider processing continuous-time signals by representing

them with time samples.

• This course develops the concepts of continuous-time and discrete-time signals

and systems in parallel. Since many of the concepts are similar, by treating them

in parallel, insight and intuition can be shared and both the similarities and

differences between them become better focused.



Signal Energy and Power

• In many (but not all) applications, the signals are directly related to physical

quantities that capturing power and energy in a physical system.

• Example: Let v(t) and i(t) be the voltage and current across a resistor with

resistance R. Then the instantaneous power is

p(t) = v(t)i(t) =1

Rv2(t)

The total energy expended over the time interval t1 ≤ t ≤ t2 is∫ t2

t1

p(t)dt =

∫ t2

t1

1

Rv2(t)dt

The average power over this time interval is

1

t2 − t1

∫ t2

t1

p(t)dt =1

t2 − t1

∫ t2

t1

1

Rv2(t)dt

• For most of this course we will use a broad definition of power and energy that

applies to any signal x(t) or x[n]. Such definitions need not necessarily have a

physical meaning.



• Signal energy

– The total energy over the time interval t1 ≤ t ≤ t2 of a possibly complex

continuous-time signal x(t) is:

E(t1, t2) =

∫ t2

t1

|x(t)|2dt

– Similarly, the total energy of a possibly complex discrete-time signal x[n] over

the time interval n1 ≤ n ≤ n2 is:

E(n1, n2) =

n2∑

n=n1

|x[n]|2

– The total energy is the energy in a signal over an infinite time interval:

E∞ = E(−∞,∞) =

∫ ∞

−∞

|x(t)|2dt

E∞ = E(−∞,∞) =

∞∑

n=−∞

|x[n]|2



Example: Find the total energy of the following discrete-time signal:

x[n] =

an, n ≥ 0

0, n < 0

where |a| < 1. The answer is:

E∞ =∞∑

n=−∞

|x[n]|2 =∞∑

n=0

(|a|2)n =1

1− |a|2

• Signal power

– Consider the time-averaged signal power.

– The average powers of x(t) and x[n] over the intervals t1 ≤ t ≤ t2 and

n1 ≤ n ≤ n2 are:

P (t1, t2) =1

t2 − t1

∫ t2

t1

|x(t)|2dt and P (n1, n2) =1

n2 − n1 + 1

n2∑

n=n1

|x[n]|2

respectively.



– Analogously, when the limits are taken over an infinite time interval, then:

P∞ = limT→∞

1

2T

∫ T

−T

|x(t)|2dt, P∞ = limN→∞

1

2N + 1

N∑

n=−N

|x[n]|2

• Classification of signals based on their energy and power

– Signals with finite total energy, E∞ <∞, are known as energy signals.

∗ The energy signals have zero average power: P∞ = 0.

∗ Examples of energy signals: All signals seen previously, any signal with

finite amplitude and finite duration (x(t) = 0 for |t| > τ and

max(|x(t)|) <∞).

– Signals with finite average power, P∞ > 0, are known as power signals.

∗ The power signals have infinite total energy: E∞ =∞ if P∞ > 0.

∗ Examples of power signals: periodic signals such as x(t) = cos(t) and

x[n] = sin(5n).

– Signals with infinite power (P∞ =∞) and infinite energy (E∞ =∞).

∗ These signals are not desirable in engineering applications.

∗ Examples: x(t) = et and x[n] = n5.



Signal Transformations

• Transformation of a signal is a central concept in signal and system analysis. For

example, an audio system takes an input signal representing music recorded on a

compact disc, and modifies it to enhance desirable characteristics.

• This section focuses on a very limited but important class of elementary signal

transformations that involve simple modification of the independent variable

(i.e., the time axis).

• Elementary signal transformations:

– Time shift: x(t)→ x(t− t0) and x[n]→ x[n− n0]

∗ If t0 > 0 or n0 > 0, signal is shifted to the right (i.e., delayed)

∗ If t0 < 0 or n0 < 0, signal is shifted to the left (i.e., advanced)

– Time reversal : x(t)→ x(−t) and x[n]→ x[−n]

– Time scaling : x(t)→ x(αt) and x[n]→ x[αn]

∗ If α > 1, signal appears compressed

∗ If α < 1, signal appears stretched



Example: The following figure plots x(t) and its transformations x(−t),

x(t− 1), x(t + 2) and x(t/2).

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(−t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(t−1)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(t+2)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(t/2)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(2t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(2(t−1))



More Example:7

1.1.3 Tran sform ation s of th e In d e p e n d e n t Variab le

Time sh ift

– Replace t → t − t0 x(t) → x(t − t0)

n → n − n0⇒

x[n] → x[n − n0]

– D elay : t0, n0 > 0, A dvan ce: t0, n0 < 0

Time rev ersal

– Replace t → −t x(t) → x(−t)

n → −n⇒

x[n] → x[−n]

Time scalin g

– Replace t → αt , α ∈ IR x(t) → x(αt)

n → αn , α ∈ ZZ⇒

x[n] → x[αn]

– C o n tin u o u s– time case: |α| < 1 : sig n al is lin early stretch ed

|α| > 1 : sig n al is lin early compressed

Time sh ift, time rev ersal, an d time scalin g o peratio n s arise n atu rally

in th e pro cessin g o f sig n als


8

E x am p le :

n

nt

n

n

t

t

Time-scaled sig n als

Time-rev ersed sig n als

Time-sh ifted sig n als

S ig n als

t

x[2 n]

x[−n]

x[n − 4 ]

x(2 /3t)

x(−t)

x(t − t0)

x(t) x[n]

t0




• A more general signal transformation is x(t)→ y(t) = x(αt + β) . For this

transformation, a systematic approach to obtain the plot of y(t) is as follows:

– First shift x(t) in accordance with the value of β: The signal x(t) is shifted to

the right if β < 0, shifted to the left if β > 0.

– Then perform time scaling and/or time reversal on the resulting signal in

accordance with the value of α: The resulting signal is linearly stretched if

|α| < 1, linearly compressed if |α| > 1 and reversed in time if α < 0.

Example: The following figures draw x(t), x(2t) and x(2(t− 1)).

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(2t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(2(t−1))



Even & Odd Symmetry

There is a set of useful properties of signals that relate to their symmetry under time

reversal.

• Even signal: x(−t) = x(t) or x[−n] = x[n].

• Examples: The signal plotted below, cos[kω0n].

9

1.1.4 Pe riod ic S ign a ls

Periodic continu ou s– time sig na l

x(t) = x(t + T ) , ∀t

– T > 0: Perio d

– x(t) periodic w ith T ⇒ x(t) a lso periodic w ith mT , m ∈ IN

– S ma llest period of x(t): Fu ndamenta l perio d T0.

– E x ample (T0 = T ):

0

x(t)

t4T3T2T−3T −2T −T T

Periodic discrete– time sig na l

x[n] = x[n + N ] , ∀n

– Integ er N > 0: Perio d

– x[n] periodic w ith N ⇒ x[n] a lso periodic w ith mN , m ∈ IN

– S ma llest period of x[n]: Fu ndamenta l perio d N0.

– E x ample (N0 = 4 ):

n

3 6

x[n]

0 1

2

54

A sig na l th a t is not periodic is referred to a s a perio dic.


1 0

1.1.5 E v e n a n d O d d S ign a ls

E v en sig na l

x(−t) = x(t) o r x[−n] = x[n]

– E x ample:x(t)

t

O dd sig na l

x(−t) = −x(t) o r x[−n] = −x[n]

– E x ample:

n

x[n]

– Necessa rily : x(0) = 0 or x[0] = 0

D ecomposition of a ny sig na l into a n ev en a nd odd pa rt:

x(t) = E vx(t) + O dx(t) or x[n] = E vx[n] + O dx[n]

w ith

E vx(t) =1

2(x(t) + x(−t)) or E vx[n] =

1

2(x[n] + x[−n])

a nd

O dx(t) =1

2(x(t) − x(−t)) or O dx[n] =

1

2(x[n] − x[−n])


• Odd signal: x(−t) = −x(t) or x[−n] = −x[n]. Note that an odd signal must

be zero at t = 0 or n = 0.

• Examples: The signal plotted below, sin(kω0t).

9



x(t) = x(t + T ) , ∀t

– T > 0: Perio d




0

x(t)



x[n] = x[n + N ] , ∀n




– E x ample (N0 = 4 ):

n

3 6

x[n]

0 1

2

54



1 0


E v en sig na l

x(−t) = x(t) o r x[−n] = x[n]

– E x ample:x(t)

t

O dd sig na l

x(−t) = −x(t) o r x[−n] = −x[n]

– E x ample:

n

x[n]




w ith

E vx(t) =1

2(x(t) + x(−t)) or E vx[n] =

1

2(x[n] + x[−n])

a nd

O dx(t) =1

2(x(t) − x(−t)) or O dx[n] =

1

2(x[n] − x[−n])




• Any signal can be written as a sum of an odd signal and an even signal:

x(t) = xe(t) + xo(t), where

xe(t) =1

2[x(t) + x(−t)]

xo(t) =1

2[x(t)− x(−t)]

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(−t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

xe(t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

x(t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

−x(−t)

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

0

1

t

xo(t)



Periodic Signals

An important class of signals that are encountered frequently in this course is the

class of periodic signals.

A signal is periodic if there is a positive value of T or N such that

x(t) = x(t + T ) or x[n] = x[n + N ]

T and N are called the periods of x(t) and x[n], respectively.

• For any integer m, mT and mN are also the periods.

• For continuous-time signals, the fundamental period T0 is the smallest positive

value of T such that x(t) = x(t + T ). For a special case where x(t) is a

constant, the fundamental period is undefined.

• For discrete-time signals, the fundamental period N0 is the smallest positive

integer of N such that x[n] = x[n + N ].

• Signals that are not periodic are said to be aperiodic.



Examples of periodic signals are shown below:

9



x(t) = x(t + T ) , ∀t

– T > 0: Perio d




0

x(t)



x[n] = x[n + N ] , ∀n




– E x ample (N0 = 4 ):

n

3 6

x[n]

0 1

2

54



1 0


E v en sig na l

x(−t) = x(t) o r x[−n] = x[n]

– E x ample:x(t)

t

O dd sig na l

x(−t) = −x(t) o r x[−n] = −x[n]

– E x ample:

n

x[n]




w ith

E vx(t) =1

2(x(t) + x(−t)) or E vx[n] =

1

2(x[n] + x[−n])

a nd

O dx(t) =1

2(x(t) − x(−t)) or O dx[n] =

1

2(x[n] − x[−n])


9



x(t) = x(t + T ) , ∀t

– T > 0: Perio d




0

x(t)



x[n] = x[n + N ] , ∀n




– E x ample (N0 = 4 ):

n

3 6

x[n]

0 1

2

54



1 0


E v en sig na l

x(−t) = x(t) o r x[−n] = x[n]

– E x ample:x(t)

t

O dd sig na l

x(−t) = −x(t) o r x[−n] = −x[n]

– E x ample:

n

x[n]




w ith

E vx(t) =1

2(x(t) + x(−t)) or E vx[n] =

1

2(x[n] + x[−n])

a nd

O dx(t) =1

2(x(t) − x(−t)) or O dx[n] =

1

2(x[n] − x[−n])




Elementary Signals: Complex Exponential and Sinusoidal

Several classes of signals play prominent role:

• They model many physical signals.

• They serve as building blocks for many other signals.

• They serve for system analysis.

1. Continuous-Time Complex Exponential Signal: x(t) = Ceat where, in

general, both C and a are complex numbers.

• If both C and a are real ⇒ Real exponential signal.

Example: The following figures plot Ceat with C = 1 and a = ± 110 .

−10 −5 0 5 10 15 20 25 300

5

10

15

20

25

t−10 −5 0 5 10 15 20 25 300

0.5

1

1.5

2

2.5

3

t



• If a is imaginary (i.e., a = jω0) ⇒ Periodic complex exponential.

To see that the signal is indeed periodic, let C = Aejφ. Then

x(t) = Aej(ω0t+φ) ?= Aej(ω0(t+T )+φ) = Aej(ω0t+φ)ejω0T

where T is chosen such that ejω0T = 1. Excluding the trivial solution of

ω0 = 0, the fundamental period is T0 =2π

|ω0|.

Example: The following figures plot Ceat with C = 1 and a = j.

−100

1020

30 −1

0

1−1

−0.5

0

0.5

1

Imaginary PartTime (s)

Rea

l Par

t



• A harmonically related set of complex periodic exponentials is a set of

exponentials with fundamental frequencies that are all multiples of a single

positive frequency ω0:

φk(t) = ejkω0t where k = 0,±1,±2, . . .

– For k = 0, φ0(t) is a constant

– For all other values of k, φk(t) is periodic with fundamental frequency

|k|ω0 and fundamental period

2π

|k|ω0=

T0

|k|

– This is consistent with how the term harmonic is used in music

– Sets of harmonically related complex exponentials are used to

represent many other periodic signals (Fourier series)



Example: The figure below plots the real parts of several harmonically

related complex exponentials.

−20 −15 −10 −5 0 5 10 15 20−1

0

1φ 0(t

)

−20 −15 −10 −5 0 5 10 15 20−1

0

1

φ 1(t)

−20 −15 −10 −5 0 5 10 15 20−1

0

1

φ 2(t)

−20 −15 −10 −5 0 5 10 15 20−1

0

1

φ 3(t)

−20 −15 −10 −5 0 5 10 15 20−1

0

1

φ 4(t)

Time (s)



• General complex exponential signal: For the most general case, C = Aejφ

and a = r + jω0, then

x(t) = Ceat = Aertej(ω0t+φ) = Aert cos(ω0t + φ) + jAert sin(ω0t + φ)

– If r > 0, then x(t) is exponentially growing signal

– If r < 0, then x(t) is exponentially decaying signal

Example: The following figures plot the real parts of Ceat with C = 1 and

a = ±0.1 + j0.5.

−20 −15 −10 −5 0 5 10 15 20−8

−6

−4

−2

0

2

4

6

8

t

Rea

l Par

t of

x(t)

−20 −15 −10 −5 0 5 10 15 20−8

−6

−4

−2

0

2

4

6

8

t

Rea

l Par

t of

x(t)



2. Continous-Time Sinusoidal Signals

xc(t) = A cos(ω0t + φ) = ReAej(w0t+φ)

xs(t) = A sin(ω0t + φ) = ImAej(w0t+φ)Of course, both xc(t) and xs(t) also have fundamental period T0 = 2π

|ω0| .Periodic signals have infinite total energy, but finite average power. This

can be seen for the exponential x(t) = Aejω0t (assuming A is real) as follows:

• The energy over one period T0 is

E(0, T0) =

∫ T0

0

A2|ejω0t|2dt = A2T0

Thus, the total energy is E∞ =∞.

• The average power over one period is

P (0, T0) =E(0, T0)

T0= A2

• The average power is

P∞ = limT→∞

1

2T

∫ T

−T

A2|ejω0t|2dt = A2 2T

2T= A2.



3. Discrete-Time Complex Exponential and Sinusoidal Signals

• Complex exponential signal:

x[n] = Cαn = Ceβn (where α = eβ)

– Real complex signal if both C and α are real.

– General complex exponential signal: With C = Aejφ and α = |α|ejω0 , then

x[n] = A|α|nej(ω0n+φ) = A|α|n cos(ω0n + φ) + jA|α|n sin(ω0n + φ)

∗ If |α| > 1, then x[n] is exponentially growing signal

∗ If |α| < 1, then x[n] is exponentially decaying signal

∗ If |α| = 1, then

x[n] = Aej(ω0n+φ) = A cos(ω0n + φ) + jA sin(ω0n + φ)

• Sinusoidal signal

xc[n] = A cos(ω0n + φ) = ReAej(ω0n+φ)

xs[n] = A sin(ω0n + φ) = ImAej(ω0n+φ)The functions Aej(ω0n+φ), A cos(ω0n + φ) and A sin(ω0n + φ) are

discrete-time signals with finite average power but infinite total energy.



Example: The figures below plot the real parts of Ceβn for C = 1 and

β = ±0.1 + j0.5.

−20 −15 −10 −5 0 5 10 15 20−10

−5

0

5

10

Time Index (n)

Rea

l Par

t

−20 −15 −10 −5 0 5 10 15 20−10

−5

0

5

10

Time Index (n)

Rea

l Par

t



• DT complex exponential vs. CT complex exponential: There are

important differences between the properties of continuous-time and

discrete-time exponential signals ejω0t and ejω0n.

– The DT exponential signals are not distinct for distinct values of ω0:

x[n] = ejω0n = ej(ω0+k2π)n, k = 0,±1,±2, . . .

∗ Need only consider a frequency interval of 2π for ω0, typically

0 ≤ ω0 < 2π.

∗ As ω0 increases from 0, the signals oscillate more and more rapidly until

ω0 = π. As we continue to increase ω0, we decrease the rate of

oscillation until ω0 = 2π.

∗ Low-frequency exponentials have ω0 near 0, 2π and other even multiples

of π.

∗ High-frequencies are near ±π and other odd multiples of π.

– The exponential ejω0n is periodic if ω0/2π is a rational number:

x[n] = ejω0n = ejω0(n+N) ⇒ ω0

2π=

m

N, for some integer m

– If ω0 6= 0 and if N and m have no factors in common, then N is the

fundamental period and the fundamental frequency is 2πN = ω0

m .



Examples:

(a) Let x(t) = cos(8πt/35).

Then ω0 =

The fundamental period T0 = 2π/ω0 =

(b) Let x[n] = cos(8πn/35).

Then ω0 =

If the signal is periodic?

The fundamental period N0 = m(2π/ω0) =

for m =

(c) x[n] = cos(n/6).

Then ω0 =

If the signal is periodic?

The fundamental period N0 = m(2π/ω0) =

for m =



• Discerete-Time Complex Exponential Harmonics: A harmonically related set

of discrete-time complex exponentials is a set of exponentials with a common

period N :

φk[n] = ejk(2π/N)n where k = 0,±1,±2, . . .

– All the harmonics are not all distinct for all the values of k:

φk+N [n] = ej(k+N)(2π/N)n = ejk(2π/N)nej2πn = ejk(2π/N)n = φk[n]

– There are only N distinct periodic exponentials:

φ0[n] = 1

φ1[n] = ej2πn/N

φ2[n] = ej4πn/N

· · ·φN−1[n] = ej2π(N−1)n/N



Discrete-Time Unit Impulse and Unit Step

• The discrete-time unit impulse is defined as δ[n] =

0, n 6= 0

1, n = 0

-

6r

r r r r r r r r

1

δ[n]

n

– It is also known as the unit sample or Kronecker delta

– It is an even function: δ[n] = δ[−n]

• The discrete-time unit step is defined as u[n] =

0, n < 0

1, n ≥ 0

-

6r r r r r

r r r r

1

u[n]

n



• There is a close relationship between δ[n] and u[n]

– First order difference:

δ[n] = u[n]− u[n− 1]

– Running sum:

u[n] =

n∑

k=−∞

δ[k]

• The unit impulse can be used to sample the discrete time signal x[n] (sampling

property):

x[n]δ[n− n0] = x[n0]δ[n− n0]

This ability to use the unit impulse to extract a single value of x[n] through

multiplication will play an important role later.



Continuous-Time Unit Step

-

61

u(t)

t

The continuous-time unit step is defined as

u(t) =

0, t < 0

1, t > 0

• Discontinuous at t = 0.

• u(0) is not defined.

• Not of consequence because it is undefined for an infinitesimal period of time.



Continuous-Time Unit ImpulseUnit Step for Switches

vs

LinearCircuit

t=0

vsu(t)LinearCircuit

LinearCircuit

t=0

is

isu(t)LinearCircuit

• u(t) useful for representing the opening or closing of switches

• We will often solve for or be given initial conditions at t = 0

• We can then represent independent sources as though they wereimmediately applied at t = 0. More later.


Discrete-Time Basis Functions

• There is a close relationship between δ[n] and u[n]

δ[n] = u[n] − u[n − 1]

u[n] =n

∑

k=−∞

δ[k]

u[n] =∞∑

k=0

δ[n − k]

• The unit impulse can be used to sample a discrete-time signalx[n]:

x[0] =∞∑

k=−∞

x[k]δ[k] x[n] =∞∑

k=−∞

x[k]δ[n − k]

• This ability to use the unit impulse to extract a single value of x[n]through multiplication will play an important role later in the term


Continuous-Time Unit Impulse

t

ue(t)

t-e e -e e

t

u(t)

t

1 1

δe(t)

δ(t)

• δe(t) ≡due(t)

dt

• As e → 0 ,

– ue(t) → u(t)

– δe(t) for t = 0 becomes very large

– δe(t) for t = 0 becomes zero

• δ(t) ≡ lime→0 δe(t)


Continuous-Time Unit Step

t

u(t)

1

u(t) ≡

0 t < 0

1 t > 0

• Sometimes known as the Heaviside function

• Discontinuous at t = 0

• u(0) is not defined

• Not of consequence because it is undefined for an infinitesimalperiod of time


• Define δe(t) ≡ due(t)dt

• As e→ 0:

ue(t)→ u(t)

δe(t) for t = 0 becomes very large

δe(t) for t 6= 0 becomes zero

• The continuous-time unit impulse function (also known as Dirac delta impulse)

is defined as δ(t) = lime→0 δe(t).



Continuous-Time Unit Impulse (Continued)

-

66

1

δ(t)

t

• An equivalent definition of the unit-impulse function:

δ(t) ≡

0, t 6= 0

∞, t = 0and

∫ e

−e

δ(t)dt = 1 for any e > 0

• The function is zero everywhere, except zero.

• The most important property of an impulse is its area. The impulse area serves

as a measure of the impulse amplitude.



• Graphical representations:

19

Remark :

We u se th e sh ort-h and notation:

dx(t)

dt= x(t)

R elation between δ(t) and u(t)

– F irst ord er d eriv ativ e

δ(t) = u(t)

– R u nning integ ral

u(t) =

t∫

−∞

δ(τ ) dτ

Forma l d iffi cu lty: u(t) is not d iff erentiable in th e conventional sense

becau se of its d iscontinu ity at t = 0.

S ome more th ou g h ts on δ(t)

– C onsid er fu nctions u∆(t) and δ∆(t) instead of u(t) and δ(t):

u∆(t)

∆ t

δ∆(t)

1

∆

t∆

1

wh ereδ∆(t) = u∆(t)

u∆(t) =

t∫

−∞

δ∆(τ ) dτ


2 0

– L imit ∆ → 0

∗ u(t) = lim∆→0

u∆(t)

∗ δ(t) :

t

δ∆1(t)

δ∆3(t)

δ∆2(t)

∆3 ∆2 ∆1

1

∆1

1

∆3

1

∆2

O bserv e: A rea u nd er δ∆(t) always 1

⇒ δ(t) is an infi nitesimally narrow impu lse with area 1.

δ(t) = lim∆→0

δ∆(t)∞∫

−∞

δ(τ ) dτ = 1

– R epresentation

a

t

aδ(t)

t0

1

t

δ(t − t0)

1

t

δ(t)

Lampe , S c h o b e r: S ig n als an d C ommu n icatio n s• Real systems do not respond instantaneously. Most systems will respond nearly

the same to sharp pulses regardless of their shape–if

– They have the same amplitude.

– Their duration is much briefer than the system’s response.

• The unit impulse is an idealization of such pulses, which is short enough for any

system.



Properties of Unit Impulse

• The relationship between δ(t) and u(t):

– First order derivative: δ(t) =du(t)

dt

– Running integral: u(t) =

∫ t

−∞

δ(τ)dτ

• Sampling properties:

x(t)δ(t− t0) = x(t0)δ(t− t0)∫ ∞

−∞

x(t)δ(t− t0)dt =

∫ ∞

−∞

x(t0)δ(t− t0)dt = x(t0)

∫ ∞

−∞

δ(t− t0)dt = x(t0)

• Time scaling:

δ(at) =1

|a|δ(t), (a 6= 0)

This is because

∫ ∞

−∞

δ(at)dt =

∫ ∞

−∞

1

|a|δ(ν)dν =1

|a| .



Continuous-Time and Discrete-Time Systems

x(t) - CTsystem

- y(t) x[n] - DTsystem

- y[n]

System: A process in which input signals are transformed by the system or cause the

system to respond in some way, resulting in other signals as outputs.

• All of the systems that we will consider have a single input and a single output

• Continuous-time system transforms continuous-time signals.

• Discrete-time system transforms discrete-time signals.

• We will use the notation x(t) −→ y(t) to mean the input signal x(t) causes the

output signal y(t).

• Similar meaning is used for the notation x[n] −→ y[n].



• Simple examples of systems are given below.

– Quadratic system:

x(t) −→ y(t) = (x(t))2

– Delay system:

x[n] −→ y[n] = x[n− 1]

– System represented by a first order differential equation:

dy(t)

dt+ ay(t) = bx(t)

where a and b are constants.

– System described by a first order difference equation:

y[n] = ay[n− 1] + bx[n]

where a and b are constants.



Interconnections of Systems

• Many real systems are built as interconnections of several subsystems.

• It is useful to use the understanding of the component systems and of how they

are interconnected to analyze the operation and behavior of the overall system.

• Basic system interconnections:

23

1.3 C on tin u ou s–T im e a n d D isc re te –T im e S y ste m s

Unified representation of ph ysical processes by systems

S y ste m : E ntity th at transforms inpu t sig nals into new ou tpu t sig nals

– O n e or more inpu t and ou tpu t sig nals

– C o n tin u o u s– time system transforms continu ou s– time sig nals

– D iscrete– time system transforms discrete– time sig nals

Formal representation of inpu t– ou tpu t relation

– C ontinu ou s– time system

x(t) −→ y(t)

– D iscrete– time system

x[n] −→ y[n]

R emark : A noth er popu lar notation th at you may find in book s is

y(t) = Sx(t), wh ere S· represents th e system operator.

P ictorial representation of systems

Continuous−timesystemx(t) y(t)

systemDiscrete−time

x[n] y[n]


24

1.3.1 S im p le E x a m p le s of S y ste m s

Q u adratic system

y(t) = (x(t))2

S ystem represented by a first order diff erential eq u ation

y(t) + ay(t) = bx(t)

with constants a and b

D elay system

y[n] = x[n − 1 ]

S ystem described by a first order diff erence eq u ation

y[n] = ay[n − 1 ] + bx[n]

with constants a and b

1.3.2 In te rcon n e c tion s of S y ste m s

O ften convenient: break down a complex system into smaller su bsystems

S eries (cascade) interconnection

System 1 System 2Input Output

E xamples: C ommu nication ch annel and receiv er, detector and de-

coder in commu nications


25

Parallel interconnection

System 1

System 2

OutputInput

E x ample: Diversity tran smissio n : transmission of th e same sig nal

ov er two antennas and receiv ing it with one antenna

Feedback interconnection

OutputInput

System 2

System 1

E x amples: Closed-loop freq u ency/ ph ase/ timing synch ronization in

commu nications, h u man motion control

1.4 B a sic S yste m P rope rtie s

S imple math ematical formu lation of basic (ph ysical) system proper-

ties

Classifi cation of systems

For conciseness: only defi nitions for continu ou s-time systems

R eplacing “(t)” by “[n]” ⇒ defi nitions for discrete-time systems


26

1.4.1 L in e a rity

L et x1(t) −→ y1(t) and x2(t) −→ y2(t)

L in ear system if

1 . A dditiv ity

x1(t) + x2(t) −→ y1(t) + y2(t)

2 . H omog eneity

ax1(t) −→ ay1(t) , ∀a ∈ C

L inear systems possess property of su perposition

L et xk(t) −→ yk(t), th en

K∑

k= 1

akxk(t) = x(t) −→ y(t) =K∑

k= 1

akyk(t)

“N ot linear” systems are referred to as n o n lin ear.

E x a m ple :

1 . S ystem y(t) = tx(t) is linear.

To see th is let

x1(t) −→ y1(t) = tx1(t)

x2(t) −→ y2(t) = tx2(t)

and

x3(t) = ax1(t) + bx2(t) ,


25

Parallel interconnection

System 1

System 2

OutputInput

E x ample: Diversity tran smissio n : transmission of th e same sig nal

ov er two antennas and receiv ing it with one antenna

Feedback interconnection

OutputInput

System 2

System 1

E x amples: Closed-loop freq u ency/ ph ase/ timing synch ronization in

commu nications, h u man motion control

1.4 B a sic S yste m P rope rtie s

S imple math ematical formu lation of basic (ph ysical) system proper-

ties

Classifi cation of systems

For conciseness: only defi nitions for continu ou s-time systems

R eplacing “(t)” by “[n]” ⇒ defi nitions for discrete-time systems


26

1.4.1 L in e a rity

L et x1(t) −→ y1(t) and x2(t) −→ y2(t)

L in ear system if

1 . A dditiv ity

x1(t) + x2(t) −→ y1(t) + y2(t)

2 . H omog eneity

ax1(t) −→ ay1(t) , ∀a ∈ C

L inear systems possess property of su perposition

L et xk(t) −→ yk(t), th en

K∑

k= 1

akxk(t) = x(t) −→ y(t) =K∑

k= 1

akyk(t)

“N ot linear” systems are referred to as n o n lin ear.

E x a m ple :

1 . S ystem y(t) = tx(t) is linear.

To see th is let

x1(t) −→ y1(t) = tx1(t)

x2(t) −→ y2(t) = tx2(t)

and

x3(t) = ax1(t) + bx2(t) ,




Basic System Properties

• The basic system properties of continuous-time and discrete-time systems have

important physical interpretations.

• They also have relatively simple mathematical descriptions using the signals and

systems language.

1. Systems with and without memory: A system is said to be memoryless if and

only if the output y(t) at any time t0 depends only on the input x(t) at the

same time, i.e., x(t0).

• Memory indicates the system has the capability to store (remember)

information about input values at times other than the current time.

• In many physical systems, memory is directly associated with the storage of

energy.

• As examples, capacitors and inductors store energy and therefore create

systems with memory. In contrast, resistors have no such mechanism and

therefore create memoryless systems.



• Other examples of memoryless systems:

(a) Limiter: y[n] =

x[n], −A ≤ x[n] ≤ A

−A, x[n] < −A

A, x[n] > A

(b) Amplifier: y(t) = Ax(t)

• Other examples of systems with memory:

(a) Accumulator: y[n] =∑n

k=−∞x[k] = x[n] + y[n− 1]

(b) Delay: y(t) = x(t− t0)

• While the concept of memory in a system typically suggest storing past input

and output values, our formal definition of systems with memory also

includes the ones whose current output is dependent on the future values of

the input and output.

• Such systems can be found in applications in which the independent variable

is not time, such as in image processing, in processing signals that have been

recorded previously (speech, geophysical, meteorological signals).



2. Invertibility: A system is invertible if distinct inputs lead to distinct outputs.

• If the system is invertible, then an inverse system exists.

• When the inverse system is cascaded with the original system, the output is the

same as the input:

x[n] - System -y[n] Inverse

system- x[n]

• Examples of invertible systems:

(a) Amplifier: y(t) = Ax(t), A 6= 0. The inverse system is w(t) = 1Ay(t).

(b) Accumulator: y[n] = y[n− 1] + x[n]. The inverse system is

w[n] = y[n]− y[n− 1] (which is the diffirentiator)

• Examples of non-invertible systems:

(a) Limiter: y[n] =

x[n], −A ≤ x[n] ≤ A

−A, x[n] < −A

A, x[n] > A

(b) Slicer: y[n] =

1, x[n] ≥ 0

−1, else



3. Causal: A system is causal if the output y(t) at any time t0 depends on values

of the input x(t) at only the present and past times, −∞ < t ≤ t0.

• If two inputs to a causal system are identical up to some point in time, the

corresponding outputs must also be equal up to this same time:

If x1(t) = x2(t) for t ≤ t0 then y1(t) = y2(t) for t ≤ t0, ∀t0

• All analog circuits are causal.

• All memoryless systems are causal, since the systems only respond to the

current value of the input.

• Not all causal systems are memoryless (very few are).

• Note that causality is not often an essential constraint in applications in which

the independent variable is not time, such as in image processing, in processing

data that have been recorded previously.

• Examples:

(a) Accumulator is a causal system: y[n] =∑n

k=−∞x[k] = x[n] + y[n− 1]

(b) A smoothing averager is a noncausal system: y[n] =1

2M + 1

M∑

k=−M

x[n− k]



4. Stability: A system is bounded-input bounded-output (BIBO) stable if all

bounded inputs (|x(t)| ≤ Bx <∞, ∀t) result in bounded outputs

(|y(t)| ≤ By <∞, ∀t).• Informally, stable systems are those in which small inputs do not lead to

outputs that diverge (grow without bound).

• All physical circuits are technically stable.

• Ideal op amp without negative feedback are usually unstable.

• Examples:

(a) The smoothing averager is a stable system: y[n] =1

2M + 1

M∑

k=−M

x[n− k].

For bounded input |x[n]| ≤ Bx, the output |y[n]| ≤ By = Bx, which is also

bounded.

(b) The integrator is unstable system: y(t) =

∫ t

−∞

x(τ)dτ .

Let the bounded input be x(t) = u(t), then the output y(t) = t is

unbounded.

• System stability is important in engineering applications. Unstable systems need

to be stabilized.


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31

1.4.6 S ta b ility

Consider bo u n d ed – in p u t bo u n d ed – o u tp u t (B IB O ) stability

S ta ble system if for any bou nded inpu t sig nal

|x(t)| ≤ Bx < ∞ , ∀t

th e ou tpu t sig nal is bou nded

|y(t)| ≤ By < ∞ , ∀t

E x a m p le :

– S table system

A v erag er: y[n] =1

2N + 1

N∑k=−N

x[n − k]

B ou nded inpu t |x[n]| < Bx ⇒ bou nded ou tpu t |y[n]| < By =

Bx

– Instable system

Integ rator: y(t) =

t∫

−∞

x(τ ) d τ

E .g . bou nded inpu t x(t) = u(t) ⇒ u nbou nded ou tpu t y(t) = t

S ystem stability is important in eng ineering applications, u nstable

systems need to be stabilized.


32

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46


5. Time Invariance: A system is time invariant if it produces identical response to

the same input signal no matter when input signal is applied. Mathematically:

x(t)→ y(t) implies x(t− t0)→ y(t− t0).

x[n]→ y[n] implies x[n− n0]→ y[n− n0].

• In other words, a system is time invariant if a time shift in the input signal

results in a corresponding time shift in the output signal.

• Circuits that have non-zero energy stored on capacitors or in inductors at time

t = 0 are generally not time-invariant (i.e., they are time-variant).

• Memoryless does not imply time-invariant. For example, y(t) = x(t)× f(t).

• Examples:

(a) The system y(t) = (x(t))2 is?

(b) The system y[n] = nx[n] is?



6. Linearity: Consider a system with x1(t)→ y1(t) and x2(t)→ y2(t). The

system is said to be linear if:

a1x1(t) + a2x2(t)→ a1y1(t) + a2y2(t)

for any constant complex coefficients a1 and a2.

x(t) - CTsystem

- y(t) x[n] - DTsystem

- y[n]

a1x1(t) + a2x2(t)→ a1y1(t) + a2y2(t)

a1x1[n] + a2x2[n]→ a1y1[n] + a2y2[n]

• There are two related properties:

– Additive: x1(t) + x2(t)→ y1(t) + y2(t)

– Scaling : ax1(t)→ ay1(t)

• Linear systems enable the application of superposition: If the input consists of a

linear combination of different inputs, the output is the same linear combination

of the corresponding outputs.

• “Not linear” systems are referred to as nonlinear.



Linear Time-Invariant (LTI) Systems

A system is said to be linear time invariant (LTI) if it is both linear and time

invariant.

• The linearity and time-invariance properties play a fundamental role in signal

and system analysis because of the following two main reasons:

– Many physical processes posses these properties ⇒ can be modeled as LTI

systems.

– LTI systems can be analyzed in considerable detail, providing both insight

into their properties and a set of powerful tools for signal and system analysis.

• Key idea: If one can represent the input to an LTI system in terms of a linear

combination of a set of basic signals, one can apply the superposition principle to

compute the output of the system in terms of its responses to these basic signals.

• As will be seen shortly, the basic signals can be chosen to be the delayed

impulses ⇒ an LTI system is completely characterized by its response to a unit

impulse, i.e., the impulse response.



Impulse Response of an LTI System

• Impulse response is the system’s response to a unit impulse.

• The impulse response is denoted by h(t) or h[n]. Thus:

δ(t) −→ h(t)

δ[n] −→ h[n]

• For any input x(t) (or x[n]), it is possible to use the impulse response h(t) (or

h[n]) to find the output y(t) (or y[n]):

x(t) - h(t) - y(t) x[n] - h[n] - y[n]

• This method is called convolution sum in the discrete-time case and convolution

integral in the continuous-time case.

• Impulse response is an important concept (for example, it is used to implement

digital filters).



Discrete-Time Convolution Sum

• Any discrete-time input signal x[n] can be expressed as a sum of scaled and

delayed unit impulses (sampling property of the unit-impulse):

x[n] =

∞∑

k=−∞

x[k]δ[n− k]

• By linearity and time-invariance properties, the output of an LTI system is the

corresponding scaled sum of the outputs due to the delayed impulses:

x[k]δ[n− k]→ x[k]h[n− k]

x[n] =

∞∑

k=−∞

x[k]δ[n− k]→ y[n] =

∞∑

k=−∞

x[k]h[n− k] = x[n] ∗ h[n]

• The above operation is called the discrete-time convolution sum.

• Observe that the impulse response h[n] completely characterizes a discrete-time

LTI system: If we know h[n] then we can calculate the output y[n] for any

input x[n].



Example: The impulse response and the input of a discrete-time LTI system are

given below. Find the output.

h[n] = anu[n]

an, n ≥ 0

0, n < 0

(where a = 0.5)x[n] =

1, n = 0

−1, n = 2

2, n = 5

0, otherwise

0 5 10−1

0

1

2

x[n]

0 5 10−1

0

1

2

h[n]

• In terms of unit-impulses, the input signal can be expressed as

x[n] = 1 · δ[n]− 1 · δ[n− 2] + 2 · δ[n− 5]



• Since x1[n] = δ[n]→ y1[n] = h[n], x2[n] = −δ[n− 2]→ y2[n] = −h[n− 2] and

x3[n] = 2 · δ[n− 5]→ y3[n] = 2 · h[n− 5]. Therefore,

y[n] = y1[n] + y2[n] + y3[n] = h[n]− h[n− 2] + 2 · h[n− 5]

0 5 10−1

0

1

2x 1[n

]Input

0 5 10−1

0

1

2

y 1[n]

Output

0 5 10−1

0

1

2

x 2[n]

0 5 10−1

0

1

2

y 2[n]

0 5 10−1

0

1

2

x 3[n]

0 5 10−1

0

1

2

y 3[n]

0 5 10−1

0

1

2

x[n]

Time (n)0 5 10

−1

0

1

2

y[n]

Time (n)



Technique to Find Discrete-Time Convolution Sum

x[n] - h[n] - y[n] y[n] =∞∑

k=−∞

x[k]h[n− k]

Consider the evaluation of the output value at some specific time n, say n = n0:

y[n0] =

∞∑

k=−∞

x[k]h[n0 − k]

• Plot the two signals (or two sequences) x[k] and h[n0 − k] as functions of k.

• Multiplying these two functions to obtain a sequence g[k] = x[k]h[n0 − k].

• Summing all the samples in the sequence g[k] yields the output value at the

selected time n0.

To plot h[n0−k] as a function of k, it is convenient to follow the following two steps:

• Plot the signal h[−k] first.

• Obtain h[n0 − k] simply by shifting h[−k] to the right (by n0) if n0 is positive,

or to the left (by |n0|) if n0 is negative.



Example: Compute y[6] for the previous example.

−5 0 5 10−2

0

2

x[k]

−5 0 5 100

0.5

1

h[k]

−5 0 5 100

0.5

1

h[−

k]

−5 0 5 100

0.5

1

h[6−

k]

Time (k)

y[6] = h[6]− h[4] + 2h[1] = 0.56 − 0.54 + 2 · 0.51 = 0.9531 ≈ 1.



Convolution Sum Derivation: Summary

- h[n] -x[n] =

∞∑

k=−∞

x[k]δ[n− k] y[n] =

∞∑

k=−∞

x[k]h[n− k]

Linearity

- h[n] -x[k]δ[n− k] x[k]h[n− k]

Linearity

- h[n] -δ[n− k] h[n− k]

Time Invariance

- h[n] -δ[n] h[n]

Definition of h[n]

- h[n] -x[n] y[n]

LTI SystemInput Output



Continuous-Time Convolution Integral

x(t) - h(t) - y(t)

• Recall that if the input x(t) = δ(t), the output of the system is called the

impulse response, denoted by h(t).

• The goal is to obtain a complete characterization of a continuous-time LTI

system in terms of its impulse response.

• This means that, for any input x(t), we must be able to use the impulse

response h(t) to find the output y(t). This method is called convolution integral.

• To derive the convolution integral, we shall decompose and approximate the

input signal by rectangular pulses (or rectangles).



Derivation of the Convolution Integral

x(t)

0

( )rx t

( )x kw

kw( )1

2k w+( )12k w−

w

( )( ) ( )( )1 12 2( )x kw u t k w u t k w − − − − +

t

Approximate the input signal x(t) by a sum of weighted delayed rectangular pulses:

x(t) ≈ xr(t) =

∞∑

k=−∞

w · x(kw)

[u(t−

(k − 1

2

)w)− u

(t−

(k + 1

2

)w)]

w

Since δ(t− kw) =du(t− kw)

dt= lim

w→0

[u(t−

(k − 1

2

)w)− u

(t−

(k + 1

2

)w)]

w



So for very small w, x(t) can also be approximated as a sum of impulses:

x(t) ≈ xw(t) =

∞∑

k=−∞

w · x(kw) · δ(t− kw)

By linearity and time-invariance properties, we have

y(t) ≈ yw(t) =

∞∑

k=−∞

w · x(kw) · h(t− kw)

Finally, in the limit w → 0 the above approximations become exact representations

and the summations become the integrals:

x(t) = limw→0

xw(t) =

∫ ∞

−∞

x(τ)δ(t− τ)dτ

y(t) = limw→0

yw(t) =

∫ ∞

−∞

x(τ)h(t− τ)dτ

The last equation above is the continuous-time convolution integral.



Example: Approximations of a triangle with rectangular pulses and impulses:

0 1 20

0.5

1

w=0.50

Approximation with Rectangles

0 1 20

0.5Approximation with Impulses

0 1 20

0.5

1

w=0.25

0 1 20

0.1

0.2

0 1 20

0.5

1

w=0.10

0 1 20

0.05

0.1

0 1 20

0.5

1

w=0.05

Time (s)0 1 2

0

0.05

Time (s)



Example:R

x(t) m

r

r+

−+− y(t)C

The impulse response of the above RC circuit can be shown to be:

h(t) = RC · e− t

RC u(t)

= e−tu(t) =

e−t, t ≥ 0

0, t < 0(for RC = 1 s)

Let the input signal to the circuit be the triangle considered in the previous page :

x(t) =

t, 0 ≤ t ≤ 1

−(t− 2), 1 ≤ t ≤ 2

0, otherwise

Approximations of this signal by impulses were also shown in the previous page.



The following figures show the responses of the circuit to different approximations of

the input signal by impulses.

−5 0 5 100

0.2

0.4

0.6

0.8Input

−5 0 5 100

0.2

0.4

0.6

0.8Output

True Approximation

−5 0 5 100

0.1

0.2

0.3

0.4

−5 0 5 100

0.2

0.4

0.6

0.8True Approximation

−5 0 5 100

0.05

0.1

Time (s)−5 0 5 100

0.2

0.4

0.6

0.8

Time (s)

True Approximation



Convolution Integral: An Alternative Form

- h(t) -x(t) y(t) =

∫ ∞

−∞

x(τ)h(t− τ)dτ

Let u ≡ t− τ , then τ = t− u and du = −dτ .

y(t) =

∫ −∞

∞

x(t− u)h(u)(−du)

=

∫ ∞

−∞

x(t− u)h(u)du

=

∫ ∞

−∞

x(t− τ)h(τ)dτ

Both forms are called the convolution integral. It is often written as:

y(t) = x(t) ∗ h(t) =

∫ ∞

−∞

x(τ)h(t− τ)dτ =

∫ ∞

−∞

x(t− τ)h(τ)dτ



Convolution Integral: Solving Graphically

- h(t) -x(t) y(t) = x(t) ∗ h(t) =

∫ ∞

−∞

x(τ)h(t− τ)dτ

• To calculate this integral graphically, perform the following steps for any specific

value of t:

1. First obtain the signal h(t− τ) (regarded as a function of τ with t fixed)

from h(τ) by reflection about the origin and a shift to the right by t if t > 0

or a shift to the left by |t| for t < 0.

2. Next, multiply together the signals x(τ) and h(t− τ) to obtain the function

g(τ) = x(τ)h(t− τ).

3. Finally, y(t) is obtained by integrating the function g(τ) from τ = −∞ to

τ =∞.

• Generally, it is sufficient to plot both x(τ) and h(t− τ) on the same axis.



Example: Let x(t) =

1, 0 ≤ t ≤ 2T

0, otherwiseand h(t) =

2, 0 ≤ t ≤ T

3, T ≤ t ≤ 2T

0, otherwise

.

Find y(t) = x(t) ∗ h(t) =

∫ ∞

−∞

x(τ)h(t− τ)dτ .

41

Gra ph ica l in terpreta tio n

– T ime-rev erse h(τ ) ⇒ h(−τ )

– S h ift h(−τ ) by t ⇒ h(t − τ )

– In teg ra te pro d u cts o f o v erla ppin g compo n en ts

Example:

∗ In pu t: x(t) =

1, 0 ≤ t ≤ 2T

0, o th e rw ise

Impu lse respo n se: h(t) =

2, 0 ≤ t < T

3, T ≤ t ≤ 2T

0 o th e rw ise

∗ t < 0

t − 2T 2TTt

h(t − τ)

x(τ)

τ

N o o v erla p: y(t) = 0

∗ 0 ≤ t < T

t T 2T τ

y(t) =

t∫

0

1 · 2 d t = 2t


42

∗ T ≤ t < 2T

tT 2T τ

y(t) =

t−T∫

0

3 d t +

t∫

t−T

2 d t = 3(t − T ) + 2T

∗ 2T ≤ t < 3T

tT 2T τ

y(t) =

2T∫

t−T

2 d t +

t−T∫

t−2T

3 d t = 2(3T − t) + 3T

∗ 3T ≤ t < 4T

tT 2T τ

y(t) =

2T∫

t−2T

3 d t = 3(4T − t)


41



– S h ift h(−τ ) by t ⇒ h(t − τ )


Example:

∗ In pu t: x(t) =

1, 0 ≤ t ≤ 2T

0, o th e rw ise


2, 0 ≤ t < T

3, T ≤ t ≤ 2T

0 o th e rw ise

∗ t < 0

t − 2T 2TTt

h(t − τ)

x(τ)

τ


∗ 0 ≤ t < T

t T 2T τ

y(t) =

t∫

0

1 · 2 d t = 2t


42

∗ T ≤ t < 2T

tT 2T τ

y(t) =

t−T∫

0

3 d t +

t∫

t−T

2 d t = 3(t − T ) + 2T

∗ 2T ≤ t < 3T

tT 2T τ

y(t) =

2T∫

t−T

2 d t +

t−T∫

t−2T

3 d t = 2(3T − t) + 3T

∗ 3T ≤ t < 4T

tT 2T τ

y(t) =

2T∫

t−2T

3 d t = 3(4T − t)


41



– S h ift h(−τ ) by t ⇒ h(t − τ )


Example:

∗ In pu t: x(t) =

1, 0 ≤ t ≤ 2T

0, o th e rw ise


2, 0 ≤ t < T

3, T ≤ t ≤ 2T

0 o th e rw ise

∗ t < 0

t − 2T 2TTt

h(t − τ)

x(τ)

τ


∗ 0 ≤ t < T

t T 2T τ

y(t) =

t∫

0

1 · 2 d t = 2t


42

∗ T ≤ t < 2T

tT 2T τ

y(t) =

t−T∫

0

3 d t +

t∫

t−T

2 d t = 3(t − T ) + 2T

∗ 2T ≤ t < 3T

tT 2T τ

y(t) =

2T∫

t−T

2 d t +

t−T∫

t−2T

3 d t = 2(3T − t) + 3T

∗ 3T ≤ t < 4T

tT 2T τ

y(t) =

2T∫

t−2T

3 d t = 3(4T − t)


41



– S h ift h(−τ ) by t ⇒ h(t − τ )


Example:

∗ In pu t: x(t) =

1, 0 ≤ t ≤ 2T

0, o th e rw ise


2, 0 ≤ t < T

3, T ≤ t ≤ 2T

0 o th e rw ise

∗ t < 0

t − 2T 2TTt

h(t − τ)

x(τ)

τ


∗ 0 ≤ t < T

t T 2T τ

y(t) =

t∫

0

1 · 2 d t = 2t


42

∗ T ≤ t < 2T

tT 2T τ

y(t) =

t−T∫

0

3 d t +

t∫

t−T

2 d t = 3(t − T ) + 2T

∗ 2T ≤ t < 3T

tT 2T τ

y(t) =

2T∫

t−T

2 d t +

t−T∫

t−2T

3 d t = 2(3T − t) + 3T

∗ 3T ≤ t < 4T

tT 2T τ

y(t) =

2T∫

t−2T

3 d t = 3(4T − t)




41



– S h ift h(−τ ) by t ⇒ h(t − τ )


Example:

∗ In pu t: x(t) =

1, 0 ≤ t ≤ 2T

0, o th e rw ise


2, 0 ≤ t < T

3, T ≤ t ≤ 2T

0 o th e rw ise

∗ t < 0

t − 2T 2TTt

h(t − τ)

x(τ)

τ


∗ 0 ≤ t < T

t T 2T τ

y(t) =

t∫

0

1 · 2 d t = 2t


42

∗ T ≤ t < 2T

tT 2T τ

y(t) =

t−T∫

0

3 d t +

t∫

t−T

2 d t = 3(t − T ) + 2T

∗ 2T ≤ t < 3T

tT 2T τ

y(t) =

2T∫

t−T

2 d t +

t−T∫

t−2T

3 d t = 2(3T − t) + 3T

∗ 3T ≤ t < 4T

tT 2T τ

y(t) =

2T∫

t−2T

3 d t = 3(4T − t)


43

∗ t ≥ 4T

tT 2T τ

No overlap: y(t) = 0

∗ In su mmary

y(t) =

0, t < 0

2t, 0 ≤ t < T

2T + 3(t − T ), T ≤ t < 2T

2(3T − t) + 3T, 2T ≤ t < 3T

3(4T − t), 3T ≤ t < 4T

0, t ≥ 4T

5

4

3

2

1

y(t)/ T

3 421 t/ T


44

2.3 P rope rtie s of Line a r T im e –Inv a ria nt S yste m s

L in earity an d time in v arian ce ⇒ complete ch aracterization by im-

pu lse respon se

y[k] =∞

∑

k= − ∞

x[k]h[n − k] = x[k] ∗ h[k]

y(t) =

∞∫

− ∞

x(τ )h(t − τ ) d τ = x(t) ∗ h(t)

Fu rth er properties based on an d in terms of impu lse respon se repre-

sen tation

2.3.1 C om m u ta tiv e , D istrib u tiv e , a nd A ssoc ia tiv e P rope rty

of C onvolu tion — Inte rconne ctions of LT I S yste m s

Commu ta tiv e property: order of th e sig n als to be con volv ed can be

ch an g ed

– C on tin u ou s– time case

x(t) ∗ h(t) = h(t) ∗ x(t) =

∞∫

− ∞

h(τ )x(t − τ ) d τ

– D iscrete– time case

x[n] ∗ h[n] = h[n] ∗ x[n] =∞

∑

k= − ∞

h[k]x[n − k]


Combine all the cases, the final expression for y(t) is:

43

∗ t ≥ 4T

tT 2T τ


∗ In su mmary

y(t) =

0, t < 0

2t, 0 ≤ t < T

2T + 3(t − T ), T ≤ t < 2T

2(3T − t) + 3T, 2T ≤ t < 3T

3(4T − t), 3T ≤ t < 4T

0, t ≥ 4T

5

4

3

2

1

y(t)/ T

3 421 t/ T


44



pu lse respon se

y[k] =∞

∑

k= − ∞

x[k]h[n − k] = x[k] ∗ h[k]

y(t) =

∞∫

− ∞

x(τ )h(t − τ ) d τ = x(t) ∗ h(t)


sen tation




ch an g ed


x(t) ∗ h(t) = h(t) ∗ x(t) =

∞∫

− ∞

h(τ )x(t − τ ) d τ


x[n] ∗ h[n] = h[n] ∗ x[n] =∞

∑

k= − ∞

h[k]x[n − k]


43

∗ t ≥ 4T

tT 2T τ


∗ In su mmary

y(t) =

0, t < 0

2t, 0 ≤ t < T

2T + 3(t − T ), T ≤ t < 2T

2(3T − t) + 3T, 2T ≤ t < 3T

3(4T − t), 3T ≤ t < 4T

0, t ≥ 4T

5

4

3

2

1

y(t)/ T

3 421 t/ T


44



pu lse respon se

y[k] =∞

∑

k= − ∞

x[k]h[n − k] = x[k] ∗ h[k]

y(t) =

∞∫

− ∞

x(τ )h(t − τ ) d τ = x(t) ∗ h(t)


sen tation




ch an g ed


x(t) ∗ h(t) = h(t) ∗ x(t) =

∞∫

− ∞

h(τ )x(t − τ ) d τ


x[n] ∗ h[n] = h[n] ∗ x[n] =∞

∑

k= − ∞

h[k]x[n − k]




Convolution Integral Derivation: Summary

- h(t) -x(t) =

∫ ∞

−∞

x(τ)δ(t− τ)dτ y(t) =

∫ ∞

−∞

x(τ)h(t− τ)dτ

Linearity

- h(t) -x(τ)δ(t− τ) x(τ)h(t− τ)

Linearity

- h(t) -δ(t− τ) h(t− τ)

Time Invariance

- h(t) -δ(τ) h(t− τ)

Definition of h(t)

- h(t) -x(t) y(t)

LTI SystemInput Output



Summary of Convolution Integral

- h(t) -x(t) y(t) = x(t) ∗ h(t) =

∫ ∞

−∞

x(τ)h(t− τ)dτ

• The convolution integral describes how the output y(t) is related to the input

signal x(t) and the impulse response h(t).

• Only two assumptions were made about the system:

– Linear

– Time Invariant

• Key points:

– The impulse response h(t) completely defines the behavior of a

continuous-time LTI system.

– If h(t) is known, then the output of a continuous-time LTI system can be

found for any input x(t) using the convolution integral.



Properties of LTI Systems

1 Commutative property: The order of the signals to be convolved can be

interchanged:

y[n] = x[n] ∗ h[n] = h[n] ∗ x[n]

y(t) = x(t) ∗ h(t) = h(t) ∗ x(t)

Implications:

– The output of an LTI system with input x(t) and impulse response h(t) is

identical to the output of an LTI system with input h(t) and impulse

response x(t).

– Irrelevant whether h(t) or x(t) is reflected and shifted to compute the

convolution integral.



2 Distributive property:

x[n] ∗ (h1[n] + h2[n]) = x[n] ∗ h1[n] + x[n] ∗ h2[n]

x(t) ∗ (h1(t) + h2(t)) = x(t) ∗ h1(t) + x(t) ∗ h2(t)

Implication: A parallel combination of LTI systems can be replaced by a single

LTI system whose unit impulse response is the sum of the individual unit impulse

responses in the parallel combination.

45

– Implication s:

∗ ou tpu t of system with impu lse respon se h(t) to in pu t x(t)

iden tical to ou tpu t of system with impu lse respon se x(t) to

in pu t h(t)

∗ Irrelev an t wh eth er h(t) or x(t) is refl ected an d sh ifted for

con volu tion

Distributive property


x(t) ∗ (h1(t) + h2(t)) = x(t) ∗ h1(t) + x(t) ∗ h2(t)


x[n] ∗ (h1[n] + h2[n]) = x[n] ∗ h1[n] + x[n] ∗ h2[n]

– Implication

Parallel con n ection of two LT I systems represen ted by sin g le

eq u iv alen t LT I system

h2(t)

h1(t)

h1(t) + h2(t)

≡

y(t)x(t)

x(t) y(t)


46

A sso cia tive Property


x(t) ∗ (h1(t) ∗ h2(t)) = (x(t) ∗ h1(t)) ∗ h2(t)


x[n] ∗ (h1[n] ∗ h2[n]) = (x[n] ∗ h1[n]) ∗ h2[n]

– C on seq u en tly

y[n] = x[n] ∗ h1[n] ∗ h2[n]

– Implication s

∗ C h ose con v en ien t order of con volu tion

∗ C ascade con n ection of two LT I system represen ted by sin g le

LT I system

h1(t) h2(t)

≡

h1(t) ∗ h2(t) y(t)x(t)

x(t) y(t)

A ssociativ e + commu tativ e property ⇒ order in a cascade of LT I

systems irrelev an t

y(t) = (x(t) ∗ h1(t)) ∗ h2(t) = (x(t) ∗ h2(t)) ∗ h1(t)

≡

h2(t)

h2(t)

h1(t)

h1(t)

y(t)x(t)

x(t) y(t)




3 Associative property:

x[n] ∗ (h1[n] ∗ h2[n]) = (x[n] ∗ h1[n]) ∗ h2[n]

x(t) ∗ (h1(t) ∗ h2(t)) = (x(t) ∗ h1(t)) ∗ h2(t)

Implications:

– The order of convolution is not important.

– Cascade connection of two LTI system is represented by a single LTI system.

– Using both the associative and commutative properties ⇒ The order in a

cascade of LTI systems is irrelevant.

45

– Implication s:



in pu t h(t)


con volu tion



x(t) ∗ (h1(t) + h2(t)) = x(t) ∗ h1(t) + x(t) ∗ h2(t)


x[n] ∗ (h1[n] + h2[n]) = x[n] ∗ h1[n] + x[n] ∗ h2[n]

– Implication



h2(t)

h1(t)

h1(t) + h2(t)

≡

y(t)x(t)

x(t) y(t)


46



x(t) ∗ (h1(t) ∗ h2(t)) = (x(t) ∗ h1(t)) ∗ h2(t)


x[n] ∗ (h1[n] ∗ h2[n]) = (x[n] ∗ h1[n]) ∗ h2[n]


y[n] = x[n] ∗ h1[n] ∗ h2[n]

– Implication s



LT I system

h1(t) h2(t)

≡

h1(t) ∗ h2(t) y(t)x(t)

x(t) y(t)



y(t) = (x(t) ∗ h1(t)) ∗ h2(t) = (x(t) ∗ h2(t)) ∗ h1(t)

≡

h2(t)

h2(t)

h1(t)

h1(t)

y(t)x(t)

x(t) y(t)


45

– Implication s:



in pu t h(t)


con volu tion



x(t) ∗ (h1(t) + h2(t)) = x(t) ∗ h1(t) + x(t) ∗ h2(t)


x[n] ∗ (h1[n] + h2[n]) = x[n] ∗ h1[n] + x[n] ∗ h2[n]

– Implication



h2(t)

h1(t)

h1(t) + h2(t)

≡

y(t)x(t)

x(t) y(t)


46



x(t) ∗ (h1(t) ∗ h2(t)) = (x(t) ∗ h1(t)) ∗ h2(t)


x[n] ∗ (h1[n] ∗ h2[n]) = (x[n] ∗ h1[n]) ∗ h2[n]


y[n] = x[n] ∗ h1[n] ∗ h2[n]

– Implication s



LT I system

h1(t) h2(t)

≡

h1(t) ∗ h2(t) y(t)x(t)

x(t) y(t)



y(t) = (x(t) ∗ h1(t)) ∗ h2(t) = (x(t) ∗ h2(t)) ∗ h1(t)

≡

h2(t)

h2(t)

h1(t)

h1(t)

y(t)x(t)

x(t) y(t)




Proofs (for continuous-time case only):

• Commutative property:

x(t) ∗ h(t) =

∫ ∞

−∞

x(τ)h(t− τ)dτλ=t−τ

= −∫ −∞

∞

x(t− λ)h(λ)dλ

=

∫ ∞

−∞

x(t− λ)h(λ)dλ = h(t) ∗ x(t)

• Distributive property:

x(t) ∗ (h1(t) + h2(t)) =

∫ ∞

−∞

x(τ)(h1(t− τ) + h2(t− τ))dτ

=

∫ ∞

−∞

x(τ)h1(t− τ)dτ +

∫ ∞

−∞

x(τ)h2(t− τ)dτ

= x(t) ∗ h1(t) + x(t) ∗ h2(t)



• Associative property:

x(t) ∗ (h1(t) ∗ h2(t)) = x(t) ∗ (h2(t) ∗ h1(t))

= x(t) ∗∫ ∞

−∞

h2(τ)h1(t− τ)dτ

=

∫ ∞

−∞

x(λ)

∫ ∞

−∞

h2(τ)h1(t− λ− τ)dτdλ

=

∫ ∞

−∞

∫ ∞

−∞

x(λ)h1(t− λ− τ)dλh2(τ)dτ

=

[∫ ∞

−∞

x(λ)h1(t− λ)dλ

]

∗ h2(t)

= (x(t) ∗ h1(t)) ∗ h2(t)



LTI System’s Properties by the Impulse Response

• LTI systems with and without memory

– Recall that for memoryless systems, the output signal depends only on the

present value of the input signal.

– For discrete-time LTI systems:

y[n] =

∞∑

k=−∞

x[k]h[n− k]?= Kx[n], (K constant)

⇒ h[n] = Kδ[n] = h[0]δ[n]

– For continuous-time LTI systems:

y(t) =

∫ ∞

−∞

x(τ)h(t− τ)dτ?= Kx(t)

⇒ h(t) = Kδ(t) = h(0)δ(t)

– Identity systems (K = 1): x[n] = x[n] ∗ δ[n] and x(t) = x(t) ∗ δ(t).



• Invertibility of LTI systems

– An Invertible LTI system has LTI inverse


h[n] ∗ hinv[n] = δ[n]

– For continuous-time LTI systems:

h(t) ∗ hinv(t) = δ(t)

Example: For an accumulator, h[n] = u[n], the output is:

y[n] =

∞∑

k=−∞

x[k]u[n− k] =

n∑

k=−∞

x[k]

Note that u[n]− u[n− 1] = h[n]− h[n− 1] = δ[n]

⇒ hinv[n] = δ[n]− δ[n− 1]



• Causality of LTI systems

– Recall that for causal systems, the output at anytime depends on only past and

present values of the input.


y[n] =

∞∑

k=−∞

x[k]h[n− k] =

∞∑

k=−∞

h[k]x[n− k]?=

∞∑

k=0

h[k]x[n− k]

⇒ h[n] = 0, for n < 0

– Similarly, for continuous-time LTI system:

⇒ h(t) = 0, for t < 0

• Stability of LTI systems

– Recall that a system is stable if it produces bounded output for any bounded

input.


∗ Bounded input |x[n]| ≤ Bx, for all n



∗ Output:

|y[n]| =

∣∣∣∣∣

∞∑

k=−∞

h[k]x[n− k]

∣∣∣∣∣≤

∞∑

k=−∞

|h[k]||x[n− k]| ≤ Bx

∞∑

k=−∞

|h[k]|?≤ By

⇒ Sufficient condition for BIBO stability is that h[n] is absolutely summable:

∞∑

n=−∞

|h[n]| <∞

It can be shown that the above condition is also necessary.

– For continuous-time LTI systems, the sufficient and necessary condition for

BIBO stability is that h(t) is absolutely integrable:

∫ ∞

−∞

|h(τ)|dτ ≤ ∞

Example: For an integrator, h(t) = u(t), one has∫ ∞

−∞

|u(τ)|dτ =

∫ ∞

0

1dτ =∞

This system is therefore not stable.



LTI System’s Properties by the Impulse Response: Summary

Property Discrete-time LTI systems Continuous-time LTI systems

Memoryless h[n] = δ[n] h(t) = δ(t)

Invertibility h[n] ∗ hinv[n] = δ[n] h(t) ∗ hinv(t) = δ(t)

Causal h[n] = 0, for n < 0 h(t) = 0, for t < 0

Stability

∞∑

n=−∞

|h[n]| <∞∫

∞

−∞

|h(τ)|dτ <∞



Fourier Representation of Signals and Systems

• The representation and analysis of LTI systems through the convolution

operation developed before are based on representing signals as linear

combinations of shifted impulses:

x[n] =

∞∑

k=−∞

x[k]δ[n− k]

x(t) =

∫ ∞

−∞

x(τ)δ(t− τ)dτ

• An alternative representation for signals and LTI systems is considered by using

complex exponentials as the basic signals. Such representations are known as

the continuous-time and discrete-time Fourier series and transform.

– Fourier series and transform can be used to construct broad and useful

classes of signals.

– They provide another convenient expression for the input-output relationship

of LTI systems.

– They allow for insightful characterization and analysis of signals and LTI

systems.



A Brief History

1748 Euler observed that if a vibrating string could be written as a sum of normal

modes, this expression is also true for future times.

1753 Bernoulli argued that all physical positions of a string could be written as this

type of sum.

1759 Lagrange criticized this representation based on the belief that it could not

represent signals with corners, so it was of limited use.

1807 Fourier claimed any periodic signal could be represented as a sum of sinusoids.

Many of his ideas were developed by others. Lacroix, Monge and Laplace were

in favor, but Lagrange fervently opposed.

1822 Fourier finally published a book.

1829 Dirichlet provided precise conditions under which periodic signals could be

represented.

1965 Fast Forier Transform (FFT) published independently by Cooley & Tukey.

1984 Grossman & Morlet introduced wavelets as a specialized field.



Jean Baptiste Joseph Fourier (1768-1830)

Engineers need historical perspective. For background material on J.B.J. Fourier see,

for example,

www-history.mcs.st-andrews.ac.uk/history/Mathematicians/Fourier.html



Fourier Series Representation of Periodic Signals

As mentioned before, it is advantageous in the study of LTI systems to represent

signals as linear combinations of basic signals that possess the following two

properties:

1. The set of basic signals can be used to construct a broad and useful class of

signals.

2. The response of an LTI system to each basic signal is simple enough in structure

so that the representation for the response of the system to any signal

constructed as a linear combination of the basic signals can be easily obtained.

Fourier representation and analysis are developed mainly from the fact that both of

the above properties are provided by the set of complex exponential signals (in

continuous and discrete time).

Recall that complex exponential signals are the signals of the form est in continuous

time and zn in discrete time, where s and z are complex numbers.



Response of an LTI System to a Complex Exponential

• The importance of complex exponential signals in the study of LTI systems

stems from the fact that the response of an LTI system to a complex exponential

input is the same complex exponential with only a change in amplitude:

– Continuous-time: est −→ H(s)est

– Discrete-time: zn −→ H(z)zn

where the complex amplitude factor H(s) or H(z) is, in general, a function of

the complex variable s or z. As will be seen shortly, the amplitude factor H(s)

and H(z) are directly related to the impulse responses h(t) and h[n],

respectively.

• A signal for which the system output is a (possibly complex) constant times the

input is referred to as an eigenfunction of the system, and the amplitude factor

is referred to as the system’s eigenvalue.

• Thus, est and H(s) are the eigenfunction and eigenvalue of a continuous-time

LTI system. Similarly, zn and H(z) are the eigenfunction and eigenvalue of a

discrete-time LTI system.



Derivation of the Continuous-Time Eigenvalue

For the input x(t) = est, the output can be found through the convolution integral:

y(t) = x(t) ∗ h(t) =

∫ +∞

−∞

x(t− τ)h(τ)dτ

=

∫ +∞

−∞

es(t−τ)h(τ)dτ = est

∫ +∞

−∞

h(τ)e−sτdτ

︸︷︷︸

H(s)

= H(s)est

where H(s)4=

∫ +∞

−∞

h(τ)e−sτdτ =

∫ +∞

−∞

h(t)e−stdt

is a complex constant whose value depends on the complex variable s and the

system’s impulse response h(t).

Fourier analysis only involves the variable s that is purely imaginary, i.e., s = jω.

This means that we consider only complex exponentials of the form ejωt. With this

restriction, one has:

ejωt −→[∫ +∞

−∞

h(t)e−jωtdt

]

ejωt = H(jω)ejωt



Derivation of the Discrete-Time Eigenvalue

Similarly, the response of a discrete-time LTI system to a complex exponential input

can be computed through the convolution sum:

y[n] = x[n] ∗ h[n] =

+∞∑

k=−∞

x[n− k]h[k]

=

+∞∑

k=−∞

zn−kh[k] = zn+∞∑

k=−∞

h[k]z−k

︸︷︷︸

H(z)

= H(z)zn

where H(z)4=

+∞∑

k=−∞

h[k]z−k =

+∞∑

n=−∞

h[n]z−n is a complex constant whose value

depends on the complex variable z and the system’s impulse response h[n].

For the discrete-time case, Fourier analysis only involves variable z that has unit

magnitude, i.e., z = ejω. Thus we focus on complex exponentials of the form ejωn:

ejωn −→[

+∞∑

n=−∞

h[n]e−jωn

]

ejωn = H(ejω)ejωn



Response to A Linear Combination of Complex Exponentials

By linearity and time-invariance properties of LTI systems, one easily obtains:

x(t) =∑

k

akeskt −→ y(t) =∑

k

ak

[H(sk)eskt

]

x[n] =∑

k

ak(zk)n −→ y[n] =∑

k

ak [H(zk)(zk)n]

• If the input to an LTI system can be expressed as a linear combination of

complex exponentials, then the output can also be represented as a linear

combination of the same complex exponential signals.

• Note that each coefficient in the representation of the output is obtained as the

product of the corresponding coefficient ak of the input and the system’s

eigenvalue H(sk) or H(zk) associated with the eigenfunction eskt or znk ,

respectively.

• But what types of signals can be represented in this form?

• Virtually all of the (periodic) signals that we are interested in!

• This is important and interesting idea.



Fourier Series Representation of CT Periodic Signals

• Recall that a continuous-time signal x(t) is periodic if there exists a positive

constant T (T > 0) such that:

x(t + T ) = x(t) for all t

The fundamental period of x(t) is the minimum value of T for which the

above is satisfied. The fundamental period is often denoted as T0, while the

value ω0 = 2πT0

(rad/s) is referred to as the fundamental frequency.

• Note that when the fundamental frequency and/or the fundamental period is

clear from the context, the subscript 0 in ω0 and/or T0 might be dropped to

simplify notation.

• Introduced earlier are two basic periodic signals, namely the sinusoidal signal

x(t) = cos(ω0t) and the periodic complex exponential x(t) = ejω0t. Both of

these signals are periodic with fundamental frequency ω0 and fundamental

period T = 2π/ω0.



• The set of harmonically related complex exponentials is:

φk(t) = ejkω0t = ejk(2π/T )t, k = 0,±1,±2, ...

where each of these signals has a fundamental frequency that is a multiple of ω0.

• The kth harmonic components φk(t) and φ−k(t) have a fundamental frequency

of |k|ω0 and a fundamental period of T/|k|.

• Observe that a linear combination of harmonically related exponentials is also

periodic with fundamental period T (i.e., fundamental frequency ω0 = 2π/T ):

x(t) =

∞∑

k=−∞

akejkω0t (1)

This is simply because each term in the above sum has one or more complete

cycles every T = 2πω0

seconds.

• The representation of a periodic signal as a linear combination of harmonically

related complex exponentials in the form of (1) is referred to as Fourier series

representation.

• The question is how to find the coefficients ak.



Finding the CT Fourier Series Coefficients

Assume that the periodic signal x(t) with fundamental period T0 can be written as

x(t) =

+∞∑

k=−∞

akejkωt, where ω = 2π/T0.

Then: x(t)e−jnωt =+∞∑

k=−∞

akejkωte−jnωt

∫

T0

x(t)e−jnωtdt =

∫

T0

+∞∑

k=−∞

akej(k−n)ωtdt =

+∞∑

k=−∞

ak

∫

T0

ej(k−n)ωtdt

=

+∞∑

k=−∞

ak

∫

T0

[cos((k − n)ωt) + j sin((k − n)ωt)]dt = T0an

Thus an =1

T0

∫

T0

x(t)e−jnωtdt, or ak =1

T0

∫

T0

x(t)e−jkωtdt .

Remark : The notation∫

T0

implies that the integration is performed over any interval

of length T0.



• To summarize, if a periodic signal x(t) has a Fourier series representation:

x(t) = x(t + T0) =

+∞∑

k=−∞

akejkωt (synthesis equation)

then the coefficients are given by

ak =1

T0

∫

T0

x(t)e−jkωtdt (analysis equation)

where ω = 2π/T0.

• The coefficients ak are called the spectral coefficients or Fourier series

coefficients of x(t).

• These complex coefficients measure the portions of the signal x(t) at each

harmonic of the fundamental component.

• The coefficient a0 is the dc (or constant) component of x(t), which is simply the

average value of x(t) over one period:

a0 =1

T0

∫

T0

x(t)dt



Example: Consider the continuous-time periodic signal:

x(t) = 2 sin(2πt− 3) + sin(6πt) (2)

It is clear that the fundamental frequency is ω0 = 2π. Using Euler’s relation one has:

2 sin(2πt− 3) = 2 sin(ω0t− 3) =2

2j

(

ej(ω0t−3) − e−j(ω0t−3))

=e−3j

jejω0t − e3j

je−jω0t

sin(6πt) = sin(3ω0t) =1

2j

(ej3ω0t − e−j3ω0t

)=

1

2jej3ω0t − 1

2je−j3ω0t

Hence,

x(t) = − 1

2je−j3ω0t − e3j

je−jω0t +

e−3j

jejω0t +

1

2jej3ω0t (3)



From Equation (3), the nonzero Fourier series coefficients ak are identified to be:

a−3 = − 1

2j=

j

2=

1

2ej( π

2)

a−1 = −e3j

j= je3j = ej( π

2+3) = ej(3− 3π

2) = e−1.7124j

a1 =e−3j

j= −je−3j = ej[−( π

2+3)] = ej( 3π

2−3) = e1.7124j

a3 =1

2j= − j

2=

1

2ej(−π

2)

With ak = |ak|ejθk , then |ak| and θk are the magnitude and phase of ak,

respectively. Also, as a convention, the phase angles are always converted to the

range [−π, π] by adding (or subtracting) with multiples of 2π.

The magnitude and phase spectra of x(t) are sketched in Figure 3. Note that the

magnitude spectrum is even and the phase spectrum is odd. These are the common

properties for real-valued periodic signals (which will be discussed shortly).



−5 0 50

0.5

1Magnitude Spectrum

−5 0 5−2

−1

0

1

2Phase Spectrum

Normalized Frequency (ω/ω0)

Figure 3: Magnitude spectrum of x(t).



Example: Consider the periodic square wave:

t

( )x t

0 1

1

t

( )x t

0

1

2

1 21−2−

t

( )x t

0

K

1T T 2T1T−T−2T−

a0 =1

T

∫ T/2

−T/2

x(t)dt = K2T1

T

ak =1

T

∫ T/2

−T/2

x(t)e−jkω0tdt =1

T

∫ T1

−T1

Ke−jkω0tdt

= − K

jkω0Te−jkω0t

∣∣∣∣

T1

−T1

=2K

kω0T

[ejkω0T1 − e−jkω0T1

2j

]

= Ksin(kω0T1)

kπ= K

2T1

Tsinc

(

k2T1

T

)

where ω0 = 2πT and sinc(x) = sin(πx)

πx .



−20 −15 −10 −5 0 5 10 15 20−0.5

0

0.5

T=4T1a k

−20 −15 −10 −5 0 5 10 15 20−0.5

0

0.5

T=8T1a k

−20 −15 −10 −5 0 5 10 15 20−0.2

0

0.2

T=16T1

a k

Normalized frequency (k=ω/ω0)

Figure 4: FS coefficients of the square wave for different ratios T/T1: Combined

magnitude and phase spectrum.



−20 −15 −10 −5 0 5 10 15 20−0.2

0

0.2

T=16T1

a k

−20 −15 −10 −5 0 5 10 15 200

0.1

0.2

|ak|

−20 −15 −10 −5 0 5 10 15 20−5

0

5

∠a k

Normalized frequency (k=ω/ω0)

Figure 5: FS coefficients of the square wave for T/T1 = 16: Combined magni-

tude/phase spectrum and separate magnitude and phase spectra.



Properties of FS Coefficients for Real-Valued Signals

If x(t) is a real-valued periodic signal and can be represented as a Fourier series:

x(t) =

+∞∑

k=−∞

akejkωt ⇒ x∗(t) =+∞∑

k=−∞

a∗

ke−jkωt

Let l = −k, then one has

x∗(t) =

+∞∑

l=−∞

a∗

−lejlωt =

+∞∑

k=−∞

a∗

−kejkωt = x(t) =

+∞∑

k=−∞

akejkωt

Thus, by comparison of coefficients, it follows that

a∗

−k = ak, a−k = a∗

k

The above is known as the complex-conjugate symmetry of the Fourier series

coefficients of real-valued periodic signals. The complex-conjugate symmetry implies

that:

• The magnitude spectrum of the FS coefficients is even: |ak| = |a−k|.

• The phase spectrum of the FS coefficients is odd : ∠ak = −∠a−k.



Alternative Forms for the FS of Real Periodic Signals

Since a−k = a∗k for real periodic signals, one can write:

x(t) =

∞∑

k=−∞

akejkωt = a0 +∞∑

k=1

akejkωt +∞∑

k=1

a−ke−jkωt

= a0 +

∞∑

k=1

[(akejkωt) + (akejkωt)∗

]

= a0 + 2

∞∑

k=1

Reakejkωt = a0 + 2

∞∑

k=1

Re|ak|ejθkejkωt

= a0 + 2

∞∑

k=1

|ak|Reej(kωt+θk)

= a0 + 2∞∑

k=1

Akcos(kωt + θk) (amplitude-phase form)

where Ak4= |ak| is the magnitude of ak and θk

4= ∠ak is the phase angle of ak.

The above representation is known as the amplitude-phase form of the Fourier series.



Alternatively, one can also write:

x(t) =

∞∑

k=−∞

akejkωt

= a0 + 2

∞∑

k=1

Reakejkωt

= a0 + 2

∞∑

k=1

Reak cos(kωt) + jak sin(kωt)

= a0 + 2

∞∑

k=1

(Reak cos(kωt)− Imak sin(kωt))

= a0 + 2

∞∑

k=1

[Bk cos(kωt)− Ck sin(kωt)] (trigonometric form)

where Bk4= Reak and Ck

4= Imak.

The above is referred to as the trigonometric form of Fourier series.



Alternative Forms of FS for Real Periodic Signals: Summary

Exponential: x(t) =∞∑

k=−∞

akejkωt

Amplitude-Phase: x(t) = a0 + 2

∞∑

k=1

Akcos(kωt + θk)

Trigonometric: x(t) = a0 + 2

∞∑

k=1

[Bk cos(kωt)− Ck sin(kωt)]

ak = Akejθk = Bk + jCk

Ak = |ak| =√

B2k + C2

k θk = ∠ak = tan−1

(Ck

Bk

)

Bk = Reak = Ak cos(θk) Ck = Imak = Ak sin(θk)

ak =1

T0

∫

T0

x(t)e−jkωtdt

Bk =1

T0

∫

T0

x(t) cos(kωt)dt, Ck = − 1

T0

∫

T0

x(t) sin(kωt)dt



Convergence & Discontinuities

• It has been shown that if a periodic signal can be represented by a Fourier series

then one can compute the Fourier coefficients.

• It is not clear, however, if any periodic signal could be represented by a Fourier

series.

• In general, the Fourier series representation x(t) of a periodic signal x(t) might

not always equal to x(t) for all t:

x(t) =

∞∑

k=−∞

akejkωt, with ak =1

T0

∫

T0

x(t)e−jkωtdt

• In fact, if x(t) is discontinuous, x(t) and x(t) are not equal for all t. Intuitively,

this should make sense because how can a linear combination of continuous

signals (sinusoids) represent a discontinuous signal?

• The question is when a periodic signal x(t) does in fact have a Fourier series

representation, i.e., when the infinite FS converges to the original signal x(t)?



• Consider the difference between the true signal and its FS representation:

e(t)4= x(t)− x(t)

• The Fourier series is said to converge if this error signal has zero energy over

one period:

E =

∫

T0

|e(t)|2dt = 0

• Note that, just because the zero error energy (E = 0), it does not imply

x(t) = x(t) for all t. It, however, does imply that any differences occur only at a

finite number of discrete (zero duration) points in time.

• In general, if t0 is a point of discontinuity, then:

x(t0) =1

2lim∆→0

[x(t0 + ∆) + x(t0 −∆)]

• At all other points the two signals x(t) and x(t) are equal.

• A sufficient condition for convergence is that the signal has a finite energy over a

single period:∫

T0

|x(t)|2dt <∞



– The above is true of all signals you could generate in the lab. Thus all of the

periodic signals generated by a function generator have equivalent FS

representations.

– If x(t) is a continuous signal, then it is safe to assume that x(t) = x(t). Note

that this is a stronger statement than merely stating that the FS converges.

Dirichlet Conditions for Convergence: An alternative sets of conditions developed by

Dirichlet guarantees that x(t) equals its FS representation, except at isolated values

of t for which x(t) is discontinuous. The Dirichlet conditions are as follows:

1. The signal x(t) must be absolutely integrable over any period:

∫

T0

|x(t)|dt <∞.

2. There must be a finite number of distinct maxima and minima during any single

period T0 of the signal.

3. In any finite interval of time, there are only finite number of discontinuities.

Furthermore, each of these discontinuities is finite.

• Figure 3.8 in textbook gives examples of signals that do not satisfy Dirichlet

condition(s).

• Note that the above are sufficient, but not necessary, conditions.



Gibbs Phenomenon

To gain some understanding of how the Fourier series converges for a periodic signal

with discontinuities, consider approximating a given periodic signal x(t) by the finite

Fourier series (with 2N + 1 terms):

xN (t) =

N∑

k=−N

akejkωt

For discontinuous signals x(t), it is observed that:

• There are ripples at the vicinity of the discontinuity.

• Gibbs showed that the peak amplitude of these ripples does not decrease with

increasing N .

• Specifically, there is an overshoot of 9% of the height of the discontinuity, no

matter how large N becomes.

• In fact, as N increases, the ripples in the finite Fourier series approximation

become compressed toward the discontinuity, but for any finite value of N , the

peak amplitude of the ripples remain constant.

The above behavior is known as the Gibbs phenomenon.



Example: Fourier series approximations of the periodic square wave.

−2 −1 0 1 2−0.5

0

0.5

1

1.5Fourier Series Approximation (N=5)

t (sec)−2 −1 0 1 2

−0.5

0

0.5

1


t (sec)

−2 −1 0 1 2−0.5

0

0.5

1


t (sec)−2 −1 0 1 2

−0.5

0

0.5

1


t (sec)



Example: Consider the continuous-time periodic waveform shown below:

t

( )x t

0 1

1

t

( )x t

0

1

2

1 21−2−

(a) Obviously, the fundamental period of x(t) is T0 = 2 and the fundamental

frequency is ω0 = 2πT0

= π.

(b) Next, we compute the trigonometric Fourier series coefficients Bk and Ck. To

this end, consider x(t) in one period, from 0 ≤ t ≤ 2. Then

x(t) =t

2, 0 ≤ t ≤ 2

Hence,

Bk =1

T0

∫

T0

x(t) cos(kω0t)dt =1

T0

∫

T0

t cos(kω0t)dt = 0



since t cos(kω0t) is an odd function. Similarly

Ck = − 1

T0

∫

T0

x(t) sin(kω0t)dt = −1

4

∫ 2

0

t sin(ω0kt)dt

= −1

4

[sin(ω0kt)

(ω0k)2− t

ω0kcos(ω0kt)

]∣∣∣∣

2

0

= −1

4× −2

πk=

1

2πk

The DC component of x(t) is simply

a0 =1

T0

∫

T0

x(t)dt =1

2

∫ 2

0

t

2dt =

t2

8

∣∣∣∣

2

0

=1

2

(c) The magnitude and phase spectrum for the FS coefficients of x(t) are plotted in

Figure 6. Observe that, since x(t) is a real function, it has an even amplitude

spectrum and an odd phase spectrum.



−5 0 5

0

0.2

0.4

0.6Magnitude Spectrum

−5 0 5−2

−1

0

1

2Phase Spectrum

Normalized Frequency (ω/ω0)

Figure 6: Magnitude and phase spectra of x(t).



(d) The following partial Fourier series approximation

x(t) = a0 + 2N∑

k=1

[Bk cos(kω0t)− Ck sin(kω0t)] of x(t) is also plotted below for

N = 5, 10, 50. Note the Gibbs phenomenon at the points of discontinuity.

−3 −2 −1 0 1 2 3

0

0.5

1N=5

−3 −2 −1 0 1 2 3

0

0.5

1N=10

−3 −2 −1 0 1 2 3

0

0.5

1N=50

t (sec)



Properties of CT Fourier Series

The Fourier series representation possesses a number of useful properties. Many

properties are listed in Table 3.1 of textbook (p. 206). Though we only discuss a few

of these properties in class, you should be familiar with all of them.

1. Linearity : If x1(t) and x2(t) are periodic signals with period T and they have

Fourier series representations:

x1(t)FS←→ ak, x2(t)

FS←→ bk

then

y(t) = α1x1(t) + α2x2(t)FS←→ α1ak + α2bk

2. Time Shifting : If x(t) is a periodic signal with fundamental period T that has

Fourier series coefficients ak, then

y(t) = x(t− t0)FS←→ bk = e−jkωt0ak = e−jk(2π/T )t0ak

Observe that time-shifting only changes the phase spectrum, not the magnitude

spectrum: |bk| = |ak|.



3. Time Reversal : If x(t) is a periodic signal with fundamental period T that has

Fourier series coefficients ak, then

y(t) = x(−t)FS←→ bk = a−k

It follows from the above relationship that:

• If x(t) is even, then bk = a−k = ak, i.e., the FS coefficients are even.

• If x(t) is odd, then bk = a−k = −ak, i.e., the FS coefficients are odd.

Furthermore, if x(t) is a real-valued function, then combining the above

time-reversal property and the complex-conjugate symmetry (a−k = a∗k)

discussed before leads to the following conclusions:

• If x(t) is real and even, then a−k = ak = a∗k. This means the FS coefficients

are real and even. Furthermore, the phases of ak can only be 0, π or −π.

• If x(t) is real and odd, then bk = a−k = −ak = a∗k. This implies that the FS

coefficients are purely imaginary and odd. Furthermore, the phase of ak can

only be 0, π/2 or −π/2.



4. Multiplication: If x1(t) and x2(t) are periodic signals with period T and they

have Fourier series representations:

x1(t)FS←→ ak, x2(t)

FS←→ bk

then

y(t) = x1(t)x2(t)FS←→ ck =

∞∑

l=−∞

albk−l = ak ∗ bk

Observe that

∞∑

l=−∞

albk−l is precisely the discrete-time convolution sum of two

sequences ak and bk.

5. Conjugation and Conjugate Symmetry : If x(t) is a periodic signal with

fundamental period T that has Fourier series coefficients ak, then

y(t) = x∗(t)FS←→ bk = a∗

−k

It follows from the above relationship that:

• If x(t) is real, i.e., x(t) = x∗(t), then bk = a∗

−k = ak ⇒ the FS coefficients

are conjugate symmetric.



• If x(t) is purely imaginary, i.e., x(t) = −x∗(t) then bk = a∗−k = −ak.

6. Parseval’s Relation: If the signal x(t) is a periodic signal with fundamental

period T that has Fourier coefficients ak, then its average power can be

computed as follows:1

T

∫

T

|x(t)|2dt =∞∑

k=−∞

|ak|2

Observe that the average power of the kth harmonic component is

1

T

∫

T

∣∣akejkωt

∣∣2dt =

1

T

∫

T

|ak|2dt = |ak|2

Thus Parseval’s relation simply states that the average total power of the signal

is the sum of the average powers in all of its harmonic components.

7. FS coefficients of symmetric signals: Recall that any signal x(t) can be written

as a sum of even and odd signals: x(t) = xe(t) + xo(t), where

xe(t) = x(t)+x(−t)2 and xo(t) = x(t)−x(−t)

2 . Consider the trigonometric form of

the Fourier series of a real periodic signal x(t):

x(t) = a0 + 2

∞∑

k=1

[Bk cos(kωt)− Ck sin(kωt)]



The coefficients Bk and Ck are determined by:

Bk =1

T0

∫

T0

x(t) cos(kωt)dt

=1

T0

∫ +T0/2

−T0/2

xe(t) cos(kωt)dt +1

T0

∫ +T0/2

−T0/2

xo(t) cos(kωt)dt

=1

T0

∫ +T0/2

−T0/2

xe(t) cos(kωt)dt =2

T0

∫ +T0/2

0

xe(t) cos(kωt)dt

and

Ck = − 1

T0

∫

T0

x(t) sin(kωt)dt

= − 1

T0

∫ +T0/2

−T0/2

xe(t) sin(kωt)dt− 1

T0

∫ +T0/2

−T0/2

xo(t) sin(kωt)dt

= − 1

T0

∫ +T0/2

−T0/2

xo(t) sin(kωt)dt = − 2

T0

∫ +T0/2

0

xo(t) sin(kωt)dt

It follows that:

• If x(t) is even, then xo(t) = 0 and therefore Ck = 0 and ak = Bk (real).

• If x(t) is odd, then xe(t) = 0 and hence Bk = 0 and ak = jCk (imaginary).



Fourier Series Representation of Discrete-Time Periodic Signals

• Recall that the set of all discrete-time complex exponential signals that are

periodic with period N is given by

φk[n] = ejkωn = ejk 2π

Nn, k = 0,±1,±2, ...

where ejωn is called the fundamental component, ej2ωn is called the 2nd

harmonic component, and, in general, ejkωn is called the kth harmonic

component.

• Unlike continuous-time exponentials, there are only N distinct harmonics. This

is because:

φk+lN [n] = ej(k+lN) 2π

Nn = ejk 2π

Nn+jl2πn = ejk 2π

Nn + ejl2πn = ejk 2π

Nn = φk[n]

• Wish to represent general DT periodic signals in terms of linear combinations of

harmonically related exponentials φk[n]:

x[n] =∑

k

akφk[n] =∑

k

akejkωn =∑

k

akejk 2π

Nn (4)

• The above representation yields a signal x[n] that is periodic with fundamental

period N .



• Since the signals (sequences) φk[n] are distinct only over a range of N

successive values of k, the summation in (4) needs only include terms in this

range. We use the notation < N > to indicate any range of k that contains N

successive integers. Thus the representation in (4) is rewritten as:

x[n] =∑

k=<N>

akφk[n] =∑

k=<N>

akejkωn =∑

k=<N>

akejk 2π

Nn (5)

• Equation (5) is referred to as the discrete-time Fourier series and the

coefficients ak as the Fourier series coefficients.

• Any finite-valued discrete-time periodic signal can be written exactly in this

form.

• The task now is to find the coefficients ak.

• In our derivation of the CT Fourier series coefficients we used the following

relation:∫

T

ejkωtdt =

∫

T

ejk(2π/T )tdt =

T k = 0,

0 k 6= 0



To solve for the DTFS coefficients, we need a similar relation:

∑

n=<N>

ejk(2π/N)n =

N k = lN,

0 k 6= lNfor any integer l.

The above essentially states that the sum over one period of the values of a

periodic complex exponential is zero, unless that complex exponential is a

constant.

• Now, since x[n] =∑

k=<N>

akejkωn, one has

∑

n=<N>

x[n]e−jlωn =∑

n=<N>

(∑

k=<N>

akejkωn

)

e−jlωn

=∑

k=<N>

ak

∑

n=<N>

ej(k−l)ωn = Nal

Thus al =1

N

∑

n=<N>

x[n]e−jlωn, or ak =1

N

∑

n=<N>

x[n]e−jkωn .



• To summarize, the Fourier series representation of a discrete-time periodic signal

x[n] with fundamental period N is given by:

x[n] =∑

k=<N>

akejkωn =∑

k=<N>

akejk(2π/N)n

where ak =1

N

∑

n=<N>

x[n]e−jkωn =1

N

∑

n=<N>

x[n]e−jk(2π/N)n

– The above two equations are known as the discrete-time Fourier series pair.

The first equation is called the synthesis equation, whereas the second

equation is called the analysis equation.

– The coefficients ak are called the spectral coefficients or the Fourier series

coefficients of x[n].

• Since φk[n] = φk+Nl[n] (there are only N distinct discrete-time exponential

harmonics) and the DTFS sum is over any N consecutive terms, one has:

x[n] = a0φ0[n] + a1φ1[n] + . . . + aN−1φN−1[n]

= aNlφ0[n] + a1+Nlφ1[n] + . . . + a(N−1)+NlφN−1[n]

By comparing terms by terms, it can be concluded that ak = ak+Nl , i.e., the

values of ak repeat periodically with period N .



Example: Consider the following discrete-time periodic square wave with period N :

83

Short-han d n otation

x[n]FS←→ ak

D en omin ation :

– ak: Fou rier series coeffi cien ts or spectral coeffi cien ts

D iff eren ces to con tin u ou s-time case

– D iscrete-time Fou rier series is finite

(R ecall from Section 1 .2 .2 : T here are on ly N distin ct discrete–

time complex expon en tial sig n als φk[n] = ejk(2π/N)n that are

periodic with period N (harmon ically related sig n als).)

– N o mathematical issu es with con v erg en ce — discrete–time Fou rier

series represen tation alway s exists

– ak = ak+N sin ce φk[n] = φk+N [n]

Remark :

T he set of coeffi cien ts

ak =1

N

N−1∑

n=0

x[n]e−jk(2π/N)n

is common ly referred to as the N -poin t d iscrete Fo u rier tra nsfo rm

(D FT ) of a fi n ite du ration sig n al x[n] with x[n] = 0 ou tside the

in terv al 0 ≤ n ≤ N . D u e to the existen ce of an extremely fast

alg orithm for the calcu lation of the D FT , called the fa st Fo u rier

tra nsfo rm (FFT ), the D FT (FFT ) is of u tmost importan ce in dig ital

sig n al processin g .


84

E x amp le:

– D iscrete–time periodic sq u are wave with period N

N1

1

N−N1 n

x[n]

– Fou rier series coeffi cien ts

ak =1

N

∑

n=〈N〉

x[n]e−jk(2π/N)n =1

N

N1∑

n=−N1

e−jk(2π/N)n

=

1

N

sin (2πk(N1 + 1/2)/N)

sin (πk/N), k 6= 0, ±N, ±2N, . . .

2N1 + 1

N, k = 0, ±N, ±2N, . . .

ak−→

k −→

– R emark : x[n] correspon ds to a finite n u mber of Fou rier coeffi -

cien ts ⇒ n o Gibbs phen omen on , n o con v erg en ce issu es




Example: Partial Fourier series for a discrete-time periodic square wave:

−15 −10 −5 0 5 10 15−1

−0.5

0

0.5

1


t (sec)−15 −10 −5 0 5 10 15−1

−0.5

0

0.5

1


t (sec)

−15 −10 −5 0 5 10 15−1

−0.5

0

0.5

1


t (sec)−15 −10 −5 0 5 10 15−1

−0.5

0

0.5

1


t (sec)



Example: A discrete-time periodic signal x[n] is real-valued and has a fundamental

period N = 5 (i.e., the fundamental frequency is ω0 = 2πN ). The nonzero Fourier

series coefficients for x[n] are:

a0 = 2, a2 = a∗

−2 = 2ejπ/6, a4 = a∗

−4 = 2ejπ/3

Express x[n] in the form:

x[n] = α0 +

∞∑

k=1

αk cos(ωkn + φk)

Solution: We know that x[n] can be represented by a linear combination of the set

of harmonically-related complex exponentials:

x[n] =∑

<N>

akejkω0n



Consider one period of ak with 0 ≤ k ≤ 4, then

a0 = 2

a1 = a1−N = a−4 = 2e−jπ/3

a2 = 2ejπ/6

a3 = a3−N = a−2 = 2e−jπ/6

a4 = 2ejπ/3

Hence,

x[n] =∑

<N>

akejkω0n =

4∑

k=0

akejk(2π/5)n

= 2 +(

a1ej(2π/5)n + a4e

j(8π/5)n)

+(

a2ej(4π/5)n + a3e

j(6π/5)n)

= 2 +(

a1ej(2π/5)n + a4e

−j(2π/5)n)

+(

a2ej(4π/5)n + a3e

−j(4π/5)n)

= 2 + 2(

ej[(2π/5)n−π/3] + e−j[(2π/5)n−π/3])

+2(

ej[(4π/5)n+π/6] + e−j[(4π/5)n+π/6])

= 2 + 4 cos(

2π

5n− π/3

)

+ 4 cos(

4π

5n + π/6

)



Compare and Contrast DTFS with CTFS

x(t) =

+∞∑

k=−∞

akejkωt, ak =1

T

∫

T

x(t)e−jkωtdt, (where ω =2π

T)

x[n] =∑

k=<N>

akejkωn, ak =1

N

∑

n=<N>

x[n]e−jkωn, (where ω =2π

N)

Unlike the continuous-time Fourier series (CTFS):

• There are only N terms in the sum of the discrete-time Fourier series (DTFS)

(since there are only N distinct DT exponential harmonics)

• The finite DTFS sum can be obtained using any N consecutive Fourier series

coefficients ak.

• The DTFS coefficients ak form a discrete-time periodic signal with the same

fundamental period N as that of x[n].

• The DTFS always converges for all periodic signals such that |x[n]| <∞

• There is no Gibbs phenomenon with DTFS.



DTFS Coefficients and The Fast Fourier Transform (FFT)

ak =1

N

∑

n=<N>

x[n]e−jk 2π

Nn

︸︷︷︸

DTFS coefficients

, X(k) =N−1∑

n=0

x[n]e−jk 2π

Nn

︸︷︷︸

discrete Fourier transform (DFT)

• X(k) defined as above is commonly referred to as the N -point discrete Fourier

transform (DFT) of a finite duration signal x[n], where x[n] = 0 outside the

interval 0 ≤ n ≤ N − 1.

• Observe from the above two equations that the DFT of one period of x[n] is

proportional to the Fourier series coefficients: X(k) = Nak

• The Fast Fourier Transform (FFT) is just a fast algorithm for the calculation of

DFT:

– The direct approach to find each of the N coefficients ak would require N2

computations.

– The FFT enables this to be solved using N · log2 N computations.

• FFT was developed by Tukey and Cooley in 1965.



Properties of Discrete-Time Fourier Series (DTFS)

• Many properties of DTFS are summarized in Table 3.2 in the textbook (page

221). You should be familiar with all of them.

• There are strong similarities between the properties of the discrete-time and

continuous-time Fourier series .

• As examples, several properties are listed below.

– Conjugate Symmetry for Real Signals: If x[n] is real, then a−k = a∗

k .

– Multiplication: If x1[n] and x2[n] are periodic signals with period N and they

have Fourier series representations:

x1[n]FS←→ ak, x2[n]

FS←→ bk

then y[n] = x1[n]x2[n] is periodic with period N and

y[n] = x1[n]x2[n]FS←→ ck =

∑

l=<N>

albk−l

The sum∑

l=<N>

albk−l is known as the periodic convolution sum between

the two periodic sequences ak and bk.



– Parseval’s Relation: If the signal x[n] is a periodic signal with fundamental

period N and it has FS representation x[n]FS←→ ak, then its average power

can be computed as follows:

1

N

∑

n=<N>

|x[n]|2 =∑

k=<N>

|ak|2

Observe that the average power of the kth harmonic component is

1

N

∑

n=<N>

∣∣∣akejk(2π/N)n

∣∣∣

2

= |ak|2

Thus Parseval’s relation simply states that the average total power of the

signal is the sum of the average powers in all of its harmonic components.



Fourier Series and LTI Systems

x(t) - h(t) - y(t) x[n] - h[n] - y[n]

• Let x(t) and x[n] be continuous-time and discrete-time periodic signals at the

inputs of LTI systems with impulse responses h(t) and h[n], respectively. Then

the outputs of the LTI systems can be easily found as follows:

x(t) =

+∞∑

k=−∞

akejkωt −→ y(t) =+∞∑

k=−∞

akH(jkω)ejkωt

x[n] =∑

k=<N>

akejkωn −→ y[n] =∑

k=<N>

akH(ejkω)ejkωn

where H(jω) and H(ejω) are called the frequency responses of continuous-time

and discrete-time LTI systems, respectively. They are defined as:

H(jω) =

∫ ∞

−∞

h(t)e−jωtdt, and H(ejω) =

∞∑

n=−∞

h[n]e−jωn



• Observations:

x(t) =

+∞∑

k=−∞

akejkωt −→ y(t) =+∞∑

k=−∞

akH(jkω)ejkωt

x[n] =∑

k=<N>

akejkωn −→ y[n] =∑

k=<N>

akH(ejkω)ejkωn

– The outputs y(t) and y[n] are also periodic with the same fundamental

frequencies as x(t) and x[n], respectively.

– If the set ak is the set of Fourier series coefficients for the input x(t), then

akH(jkω) is the set of the Fourier series coefficients for the output y(t):

x(t)FS←→ ak −→ y(t)

FS←→ akH(jkω)

Thus, the effect of the LTI system is to modify individually each of the

Fourier series coefficients of the input through multiplication by the value of

the frequency response at the corresponding frequency.

– For discrete-time case, the relationship between the Fourier series coefficients

of the input and output of an LTI system is exactly the same:

x[n]FS←→ ak −→ y[n]

FS←→ akH(

ejk(2π/N))



Continuous-Time Fourier Transform: Overview

• Fourier series enables us to represent periodic signals as linear combinations of

harmonically related complex exponentials. Such representation can be used to

clearly describe the effect of LTI systems on periodic signals.

• This chapter extends such representation to continuous-time signals that are

not periodic (i.e., aperiodic signals).

• We will see that, a rather large class of signals, including all energy signals, can

also be represented through a linear combination of complex exponentials.

• The main difference compared to FS representations of periodic signals is that

the complex exponentials in the representation of aperiodic signals are

infinitesimally close in frequency ⇒ The representation in terms of linear

combination takes the form of an integral rather than the sum.

• In the FS representation of a periodic signal, as the period increases, the

fundamental frequency decreases and the harmonically related components

become closer in frequency.

• Fourier reasoned that any aperiodic signal can be viewed as a periodic signal with

an infinite period. As the period becomes infinite, the frequency components

form a continuum and the FS sum becomes an integral (i.e., Fourier transform).



Derivation of Fourier Transform

• Consider an aperiodic signal x(t) that is of finite duration: x(t) = 0 if |t| > T1.

• Form a periodic signal x(t), with a period T , for which x(t) is one period.

• As T →∞ (or ω0 = 2πT → 0), x(t) is equal to x(t) for any finite value of t.

x(t)

t

…

t

1T1T− 0

1T1T− TT−T2− T2

( ) tjjX ωω e

( ) tjkjkX 0e0ωω

( ) 000eArea ωω ω tjkjkX=

0ωk0)1( ω+k

ω

0

0

)(~ tx

…

• We know that x(t) can be represented with a Fourier series:

x(t) =

+∞∑

k=−∞

akejkω0t, ak =1

T

∫ T/2

−T/2

x(t)e−jkω0tdt



The FS coefficients of x(t) is:

ak =1

T

∫ T/2

−T/2

x(t)e−jkω0tdt =1

T

∫ +∞

−∞

x(t)e−jkω0tdt =1

TX(jkω0)

where X(jω) =

∫ ∞

−∞

x(t)e−jωtdt is the envelope of Tak. Alternatively, Tak can be

obtained as the sample X(jω)|ω=kω0of the envelope function X(jω). Now:

x(t) =

+∞∑

k=−∞

1

TX(jkω0)e

jkω0t =1

2π

+∞∑

k=−∞

X(jkω0)ejkω0tω0

⇒ x(t) = limT→∞

x(t) = limω0→0

[

1

2π

+∞∑

k=−∞

X(jkω0)ejkω0tω0

]

=1

2π

∫ +∞

−∞

X(jω)ejωtdω

x(t)

t

…

t

1T1T− 0

1T1T− TT−T2− T2

( ) tjjX ωω e

( ) tjkjkX 0e0ωω

( ) 000eArea ωω ω tjkjkX=

0ωk0)1( ω+k

ω

0

0

)(~ tx

…



Fourier Transform: Summary

• Fourier transform pair :

Fx(t) = X(jω)4=

∫ +∞

−∞

x(t)e−jωtdt : Fourier transform

(Fourier integral, spectrum)

F−1X(jω) = x(t)4=

1

2π

∫ +∞

−∞

X(jw)ejωtdω : inverse Fourier transform

• The synthesis equation represents an aperiodic signal as a linear combination of

complex exponentials.

• For a periodic signal, the complex exponentials in its FS respresentation have

amplitudes ak occurring at a discrete set of harmonically related frequencies

kω0, k = 0,±1,±2, . . .

• For aperiodic signals, the complex exponentials occur at a continuum of

frequencies and have “amplitude” X(jω)(dω/2π). Thus X(jω) should be

interpreted as the amplitude density

• X(jω) provides information for describing x(t) as a linear combination of

sinusoidal signals at different frequencies.



Conditions for Convergence of Fourier Transforms

The Fourier transform of a signal x(t) exists if∫ +∞

−∞

|x(t)|2dt <∞

and any discontinuities are finite

• The above is true for all signals of finite amplitude and duration.

• Convergence does not imply that the inverse Fourier transform x(t) will be equal

to the original signal x(t) for all values of t.

• However, convergence implies that, although x(t) and x(t) may differ

significantly at individual values of t, there is no energy in their difference (i.e.,

the energy of the difference signal e(t) = x(t)− x(t) is zero.

• An alternative set of sufficient conditions (Dirichlet conditions) for the

convergence of Fourier transforms is stated in the textbook (page 290).

• Does a periodic signal have a Fourier transform? The answer is No, but we can

find one if we allow X(jω) to be expressed in terms of impulse functions ⇒ The

Fourier series and Fourier transform can be incorporated into a common

framework (convenient).



Example: Transform of a Rectangular Pulse

x(t) =

1 |t| < T1

0 otherwise

FT⇐⇒ X(jω) =

∫ T1

−T1

e−jωtdt =2 sin(ωT1)

ω= 2T1sinc

(ωT1

π

)

−5 0 50

0.5

1x(

t)

t (sec)−20 −10 0 10 20

−0.5

0

0.5

1

T1=0.5

X(j

ω)

ω (rad/sec)

−5 0 50

0.5

1

x(t)

t (sec)−20 −10 0 10 20

−1

0

1

2

T1=1

X(j

ω)

ω (rad/sec)

−5 0 50

0.5

1

x(t)

t (sec)−20 −10 0 10 20

−5

0

5

10

T1=5

X(j

ω)

ω (rad/sec)

If the signal stretches in time, its spectrum compresses in frequency.



Sinc Function

• The sinc function arises frequently in Fourier analysis: sinc(θ)4=

sin(πθ)

πθ

• sinc(0) = 1, sinc(n) = 0 if n is a nonzero integer.

•∫ ∞

−∞

sinc(t)dt = 1

−8 −6 −4 −2 0 2 4 6 8−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

sinc

(θ)

θ



More Examples

• Fourier transform of the unit impulse δ(t):

X(jω) =

∫ ∞

−∞

δ(t)e−jωtdt =

∫ ∞

−∞

δ(t)e−jω0dt =

∫ ∞

−∞

δ(t)dt = 1

The unit impulse has a FT consisting of equal contributions at all

frequencies.

Remark: In class, the above result was obtained by considering the FT of a

rectangular pulse over [−T1, T1] and with amplitude 12T1

, and then let T1 → 0.

• Fourier transform of a constant: Let x(t) = 1. The Fourier transform of x(t)

can be found as a limit of the FT of the rectangular pulse when T1 →∞.

Observe that the function 2T1sinc(

ωT1

π

)approaches an impulse at ω = 0 as

T1 →∞. Specifically, since∫ ∞

−∞

2T1sinc

(ωT1

π

)

dω = 2π, then

x(t) = 1FT⇐⇒ X(jω) = 2πδ(ω)

The above result is intuitively satisfying since x(t) only contains a DC

component at ω = 0.



• Transform of a decaying exponential : Let x(t) = e−atu(t) where a > 0. Then

the Fourier transform of x(t) is

X(jω) =

∫ ∞

0

e−ate−jωtdt = − 1

a + jωe−(a+jω)t

∣∣∣∣

∞

0

=1

a + jω, a > 0

The magnitude and phase spectra of x(t) are:

|X(jω)| = 1√a2 + ω2

, ∠X(jω) = − tan−1(ω

a

)

−25 −20 −15 −10 −5 0 5 10 15 20 250

0.2

0.4

0.6

0.8

1

(a=1)

Mag

nitu

de o

f X

(jω

)

ω (rad/sec)

−25 −20 −15 −10 −5 0 5 10 15 20 25−2

−1

0

1

2

(a=1)

Phas

e of

X(j

ω)

ω (rad/sec)



Inverse Fourier Transform of a Rectangular Pulse

X(jω) =

1 |w| < W

0 otherwise

FT⇐⇒ x(t) =1

2π

∫ W

−W

ejωtdω =sin(Wt)

πt

−10 −5 0 5 10

0

0.2

0.4

0.6

0.8

1X

(jω

)

ω−10 −5 0 5 10

−0.1

0

0.1

0.2

0.3

0.4

x(t)

t (sec)

−10 −5 0 5 10

0

0.2

0.4

0.6

0.8

1

X(j

ω)

ω−10 −5 0 5 10

−0.2

0

0.2

0.4

0.6

0.8

x(t)

t (sec)

If the signal’s spectrum compresses in frequency, the signal stretches in time.



Fourier Transforms of Periodic Signals

• Consider a signal x(t) with Fourier transform X(jω) = 2πδ(ω − ω0). Then x(t)

can be found by performing the inverse Fourier transform:

x(t) =1

2π

∫ ∞

−∞

2πδ(ω − ω0)ejωtdω = ejω0t

• The above implies the following Fourier transform pair:

ejω0t FT⇐⇒ 2πδ(ω − ω0)

• Now, if x(t) is a periodic signal, it has a Fourier series representation:

x(t) =

∞∑

k=−∞

akejkω0t

By the linearity of integral, it is straightforward to verify that

x(t) =∞∑

k=−∞

akejkω0t FT⇐⇒ X(jω) =∞∑

k=−∞

2πakδ(ω − kω0)



Properies of Fourier Transform

Linearity: If x1(t)FT⇐⇒ X1(jω) and x2(t)

FT⇐⇒ X2(jω), then:

a1x1(t) + a2x2(t)FT⇐⇒ a1X1(jω) + a2X2(jω)

This property follows directly from the linearity of integrals.

Time-Shifting:

x(t− t0)FT⇐⇒ e−jωt0X(jω)

Note that:

|X(jω)e−jωt0 | = |X(jω)|∠X(jω)e−jωt0 = ∠X(jω)− ωt0

• Thus, a shift in time does not affect the magnitude of the Fourier transform.

• The effect of a time shift is to introduce a phase shift −ωt0, which is a linear

function of the frequency ω.



Conjugation and Conjugate Symmetry:

x∗(t)FT⇐⇒ X∗(−jω)

• If x(t) is real, then

x(t) = x∗(t)⇒ X(jω) = X∗(−jω)

– The magnitude spectrum is even: |X(jω)| = |X(−jω)|– The phase spectrum is odd: ∠X(jω) = −∠X(−jω)

• Because of this symmetry

– X(jω) for a real signal is often only plotted for positive frequencies.

– X(jω) for w < 0 can be inferred from these plots.

• If x(t) is both real and even, then X(jω) is a real and even function of ω. This

implies that the value of phase spectrum ∠X(jω) can only be 0, π or −π.

• If x(t) is both real and odd, then X(jω) is a purely imaginary and odd function

of ω. This implies that the value of phase spectrum ∠X(jω) can only be 0, π/2

or −π/2.



Differentiation:

dx(t)

dt

FT⇐⇒ jωX(jω)

dnx(t)

dtnFT⇐⇒ (jω)nX(jω)

• This property can be easily derived by taking the derivative of both sides of the

synthesis equation. The key advantage is that ordinary differential equations

become algebraic in the frequency domain.

Integration:

∫ t

−∞

x(τ)dτFT⇐⇒ 1

jωX(jω) + πX(0)δ(ω)

• Differentiation in time = Multiplication by jω in frequency ⇒ Integration in

time = Division by jω in frequency. Also there is an impulse at ω = 0 to reflect

the possible existence of a DC (average) value.

Example: To see the useful application of the differentiation property of the Fourier

transform, consider finding the Fourier transform of the signal x(t) shown below.



t

1

1

0

x(t)

-1

-1

t10-1

-1

dx(t)/dt

)1( −tδ)1( +tδ

The first approach: Applying FT directly

X(jω) =

∫∞

−∞

x(t)e−jωtdt =

∫ 1

−1

(−t)[cos(jωt)− j sin(jωt)]dt

= −

∫ 1

−1

t cos(jωt)dt

︸︷︷︸

=0 since t cos(jωt) is an odd function

+j

∫ 1

−1

t sin(jωt)dt

= 2j

[sin(ωt)

ω2−

t cos(ωt)

ω

]∣∣∣∣

1

0

= 2j

[sin(ω)

ω2−

cos(ω)

ω

]

The second approach: Applying differentiation propertydx(t)

dt

FT←→

−2 sin(ω)

ω+ e−jω + ejω =

−2 sin(ω)

ω+ 2 cos(ω)

x(t)FT←→ X(jω) =

1

jω

[−2 sin(ω)

ω+ 2 cos(ω)

]

= 2j

[sin(ω)

ω2−

cos(ω)

ω

]



Remark: In general, impulses arise when a function to be differentiated is

discontinuous. The impulse occurs at the point of the discontinuity with a strength

equal to the ‘size’ of the jump.

t

( )x t

0 T

A

t

( )u t

0

1

T− t

( )x t

0 T

A

T−

A−

t0

1 2

t0

1 2

1 2−

)(sign2

1t

)(exp2

1at−

1exp( )

2at−

t

( )x t

01t 2t

Upward-going impulse in thederivative at t1 of strengthequal to the size of the jump

Downward-going (i.e., negativestrength) impulse in thederivative at t2 of strength equalto the negative size of the jump



Time & Frequency Scaling: x(at)FT⇐⇒ 1

|a|X(

jω

a

)

• This applies for both positive and negative values of a.

• If the signal is stretched out in time (|a| < 1), the Fourier transform is

compressed (high frequencies are moved to lower frequencies).

• If the signal is compressed in time (|a| > 1), the Fourier transform is expanded

(low frequencies are moved to higher frequencies).

• This time-frequency relationship can be seen in many examples of Fourier

transform pairs considered before.

−10 −5 0 5 10

0

0.2

0.4

0.6

0.8

1

X(j

ω)

ω−10 −5 0 5 10

−0.1

0

0.1

0.2

0.3

0.4

x(t)

t (sec)

−10 −5 0 5 10

0

0.2

0.4

0.6

0.8

1

X(j

ω)

ω−10 −5 0 5 10

−0.2

0

0.2

0.4

0.6

0.8

x(t)

t (sec)



Parseval’s Relation: The energy of a signal x(t) is defined as:

Wx4=

∫ +∞

−∞

|x(t)|2dt

If x(t) represents the voltage across a 1Ω resistor, then Wx is the energy dissipated

by the resistor.

Parseval’s relation states that:

∫ +∞

−∞

|x(t)|2dt =1

2π

∫ +∞

−∞

|X(jω)|2dω

The important of this theorem is that it tells us to think of |X(jω)|2 as the energy

spectral density (i.e., how the signal’s energy is distributed over frequency).

A more general version of Parseval’s relation is

∫ +∞

−∞

x(t)y∗(t)dt =1

2π

∫ +∞

−∞

X(jω)Y ∗(jω)dω

where x(t) and y(t) are two arbitrary continuous-time signals.



Example: To illustrate the application of Parseval’s relation, consider the following

problem: An engineer wishes to send a signal sinc(

Wtπ

)across a communications

channel, where W is a positive constant. The channel can be modeled as an ideal

low-pass filter with a transfer function

H(jω) =

1, ω ≤ 2π × 25× 103 rad/sec

0, otherwise

What range of values of W can the engineer guarantee that at least 95% of the

signal energy will reach the receiver?

Solution:

x(t) = sinc

(Wt

π

)

=sin(Wt)

Wt=

π

W

sin(Wt)

πt

Since

a(t) =sin(Wt)

πt

F⇐⇒ A(jω) =

1, ω ≤W

0, otherwise

Then

x(t) =π

W

sin(Wt)

πt

F⇐⇒ X(jω) =

πW , ω ≤W

0, otherwise



The energy of x(t) is

Ex =1

2π

∫ W

−W

|X(jω)|2dω =1

2π

∫ W

−W

( π

W

)2

dω =π

W

The Fourier transform of the output signal y(t) is Y (jω) = X(jω)H(jω). The

channel can be modeled as an ideal low-pass filter with a transfer function

H(jω) =

1, ω ≤ ωc

0, otherwise

where ωc = 2π × 25× 103 rad/sec is the cutoff frequency of the channel. Obviously

the effect of the channel on the input signal depends on the value of W :

• If W ≤ ωc then Y (jω) = X(jω) and therefore y(t) = x(t): The channel passes

the signal x(t) without any distortion and of course Ey = Ex.

• The more interesting situation is when W > ωc. In this case the transmitted

signal x(t) will be filtered by the channel and not all the energy of the



transmitted signal can be received at the receiver. Since

Y (jω) = X(jω)H(jω) =

πW , ω ≤ ωc

0, otherwise

the energy of the output signal y(t) is

Ey =1

2π

∫ ωc

−ωc

|Y (jω)|2dω =1

2π

∫ ωc

−ωc

( π

W

)2

dω =πωc

W 2

Thus if one requires Ey = 0.95Ex, then πωc

W 2 = 0.95 πW . It then follows that

W =ωc

0.95= 52.6π × 103 rad/sec

To conclude: 0 < W ≤ 52.6π × 103 rad/sec.



The Convolution Property

- h(t) -x(t) y(t) = x(t) ∗ h(t)

• Recall the response of an LTI system to a periodic input signal is:

+∞∑

k=−∞

akejkω0t −→+∞∑

k=−∞

akH(jkω0)ejkω0t

where H(jω) is the frequency response of the LTI system. It was defined as

H(jω) =

∫ ∞

−∞

h(t)e−jωtdt

• Thus the system’s frequency response is precisely the Fourier transform of the

system’s impulse response!

• The above says that the FS coefficients of the output are those of the input

multiplied by the frequency response of the system evaluated at the

corresponding harmonic frequencies.

• For aperiodic input signals, what is the effect of the LTI system in

frequency-domain (i.e., how does it change the spectrum of the input)?



• By the linearity property of LTI systems:

x(t) =1

2π

∫ +∞

−∞

X(jω)ejωtdω −→ y(t) =1

2π

∫ +∞

−∞

X(jω)[H(jω)ejωt]dω

since y(t) =1

2π

∫ +∞

−∞

Y (jω)ejωtdω

then Y (jω) = X(jω)H(jω)

• To summarize: x(t) ∗ h(t)FT⇐⇒ X(jω) ·H(jω)

• The Fourier transform maps the convolution of two signals into the product of

their Fourier transforms (a complicated convolution in the time-domain is

equivalent to a simple multiplication in the frequency-domain).

• For an LTI system, the frequency response captures the change in complex

amplitude of the Fourier transform of the input at each frequency ω.

• Since h(t) completely characterizes an LTI system, then so must H(jω).

• H(jω) is also known as the transfer function.

• |H(jω)| is the magnitude-frequency response: |Y (jω)| = |X(jω)||H(jω)|

• ∠H(jω) is the phase-frequency response: ∠Y (jω) = ∠X(jω) + ∠H(jω)



A formal proof of the convolution property :

Suppose y(t) = x(t) ∗ h(t). One can solve for Y (jω) in terms of X(jω) and H(jω)

as follows:

Y (jω) =

∫ +∞

−∞

(∫ +∞

−∞

x(τ)h(t− τ)dτ

)

e−jωtdt

=

∫ +∞

−∞

x(τ)

(∫ +∞

−∞

h(t− τ)e−jωtdt

)

dτ

=

∫ +∞

−∞

x(τ)

(∫ +∞

−∞

h(u)e−jω(u+τ)du

)

dτ

=

∫ +∞

−∞

x(τ)

(∫ +∞

−∞

h(u)e−jωudu

)

e−jωτdτ

= H(jω)

∫ +∞

−∞

x(τ)e−jωτdτ

= H(jω) ·X(jω)

= X(jω) ·H(jω)



Example: This example illustrates the use of convolution property. It also shows the

importance of the phase-frequency response of an LTI system.

Consider the continuous-time LTI system with frequency response

H(jω) =a− jω

a + jω, a > 0

(a) The magnitude and phase responses of the system are:

|H(jω)| =√

a2 + ω2

√a2 + ω2

= 1

∠H(jω) = (− tan−1 ω

a)− (tan−1 ω

a) = −2 tan−1 ω

a

−10 −5 0 5 10

0

0.5

1

|H(jω)|

ω (rad/sec)−10 −5 0 5 10−4

−2

0

2

4∠H(jω)

ω (rad/sec)



(b) Using the table of Fourier transform on page 329 of textbook, the impulse

response of the system is found to be

H(jω) = −1 +2a

a + jω

FT⇐⇒ h(t) = −δ(t) + 2ae−atu(t)

(c) Let a = 1. We wish to find the output of the system when the input is

x(t) = cos(t/√

3) + cos(t) + cos(√

3t)

Since a = 1, then |H(jω)| = 1 and ∠H(jω) = −2 tan−1 ω. First, it is

convenient to find the output of an LTI system to the sinusoidal input

cos(ω0t + θ). To this end, recall that ej(ω0t+θ) is an eigenfunction of the LTI

systems:

ej(ω0t+θ) ⇒ H(jω0)ej(ω0t+θ) = |H(jω0)|ej[(ω0t+θ)+∠H(jω0)]

e−j(ω0t+θ) ⇒ H(−jω0)e−j(ω0t+θ) = |H(jω0)|e−j[(ω0t+θ)−∠H(jω0)]

where we have used the fact that

H(−jω0) = |H(−jω0)|ej∠H(−jω0) = |H(jω0)|e−j∠H(jω0). Thus

cos(ω0t + θ)⇒ |H(jω0)| cos(ω0t + θ + ∠H(jw0))



Therefore,

y(t) = cos

(t√3− π

3

)

+ cos(

t− π

2

)

+ cos

(√3t− 2π

3

)

Plots of the input and output are shown below. Observe that since the system’s

phase-frequency response is not a linear function of ω, the output is not simply a

shifted version of the input. This example clearly shows that one cannot ignore

the phase-frequency response of an LTI system.

−20 −15 −10 −5 0 5 10 15 20−4

−2

0

2

4Input

t (sec)

−20 −15 −10 −5 0 5 10 15 20−4

−2

0

2

4Output

t (sec)



Filtering

• In may applications, it is of interest to change the relative amplitudes of the

frequency components in a signal, or to eliminate some frequency components

entirely. Such processes are referred to as filtering.

• LTI systems that change the shape of the input spectrum are often referred to as

frequency-shaping filters.

• LTI systems that are designed to pass some frequencies essentially undistorted

and significantly attenuate or eliminate other frequencies are referred to as

frequency-selective filters.

• From the convolution property of the Fourier transform, it is evident that

filtering can be conveniently accomplished through the use of LTI systems with

an appropriate chosen frequency response H(jω).

• Frequency-domain method thus provides ideal tools to examine this very

important class of applications.



Ideal Frequency-Selective Filters

0

1

ω

)( ωjH

cω

Lowpass

0

1

ω

)( ωjH

cω

Highpass

0

1

ω

)( ωjH

cω

Notch

0

1

ω

)( ωjH Bandpass

0

1

ω

)( ωjH Bandstop

1cω 2cω 1cω 2cω

The frequency components within the passband are passed without modification,

whereas the freq. components that fall into the stopband are completely eliminated.

• Lowpass filters pass low frequencies (ω < ωc).

• Highpass filters pass high frequencies (ω > ωc).

• Bandpass filters pass a range of frequencies (ωc1 < ω < ωc2).

• Bandstop filters pass two ranges of frequencies (ω < ωc1 and ω > ωc2).

• Notch filters pass all frequencies except ω ≈ ωc.



Example: Impulse Response of the Ideal Lowpass Filter

H(jω) =

1, |w| < W

0, otherwise

FT⇐⇒ h(t) =1

2π

∫ W

−W

ejωtdω =sin(Wt)

πt

−10 −8 −6 −4 −2 0 2 4 6 8 10

0

0.2

0.4

0.6

0.8

1

H(j

ω)

ω

−10 −8 −6 −4 −2 0 2 4 6 8 10−0.2

0

0.2

0.4

0.6

0.8

h(t)

t (sec)

The above ideal LPF cannot be implemented in practice since it is

non-causal. Moreover, the oscillatory behavior in the filter’s impulse response

is highly undesirable.



An Example of Frequency-Shaping Filter: A Practical LPF

( )iv t

R( )i t

C ( )ov t

+ +

−−

vi(t) = Ri(t) +1

C

∫ t

−∞

i(τ)dτ, vo(t) =1

C

∫ t

−∞

i(τ)dτ

Taking the Fourier transform of both sides of the above equations (and ignoring any DC

signals in the circuit) gives

Vi(jω) = RI(jω) +1

C

I(jω)

jω, Vo(jω) =

1

C

I(jω)

jω

⇒ H(jω) =Y (jω)

X(jω)=

1

1 + jωRC=

1

1 + jω/ωc

where ωc = 1RC

is called the cutoff frequency.

The impulse response of the filter is

h(t) = ωce−ωctu(t) =

1

RCe−

t

RC u(t) (Note that the filter is causal)



The magnitude-frequency and phase-frequency spectra are

|H(jω)| =1

√

1 + (ω/ωc)2, ∠H(jω) = − tan−1(ω/ωc)

−10 −5 0 5 100

0.5

1Magnitude Frequency−response

ω/ωc

−10 −5 0 5 10−2

−1

0

1

2Phase frequency−response

ω/ωc

The RC lowpass filter is a relatively crude approximation of the ideal lowpass filter. Better

approximations can be obtained with more-complicated circuits.



More Properties of Fourier Transform

Multiplication:

x(t) · w(t)FT⇐⇒ 1

2πX(jω) ∗W (jω)

Frequency-Shifting:

ejω0tx(t)FT⇐⇒ X(j(ω − ω0))

• Multiplication by a complex exponential shifts the Fourier transform to the

specified frequency.

• This is the basis of amplitude modulation (AM) technique used in

communications.

• This is also a convenient method for multiplexing multiple bandlimited signals

into a single channel.

• The signal can then be recovered by bandpass filtering and AM demodulation



Example: To illustrate the application of the frequency-shifting property, consider

finding the Fourier transform of a windowed cosine. Specifically, consider n cycles of

a cosine of frequency ω0 (rad/sec) that lasts only over the time interval −T/2 to

T/2 seconds, i.e., T = 2πω0

n. A typical plot of it is shown below.

−15 −10 −5 0 5 10 15−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t (sec)

Such a signal can be written as x(t) = A cos(ω0t)w(t), where the “window” function

w(t) = u(t + T/2)− u(t− T/2).

The Fourier transform of Aw(t) is AW (jω) = AT sin(ωT/2)(ωT/2) . Using the



frequency-shifting property of the Fourier transform one has

x(t) cos(ω0t) =1

2

[x(t)e−jω0t + x(t)ejω0t

] FT⇐⇒ 1

2X(j(ω + ω0)) +

1

2X(j(ω − ω0))

The above states that multiplying a function x(t) by a cos(ω0t) shifts its spectrum

‘up’ and ‘down’ by the frequency ω0 and scales the amplitude by 2. Thus the result is

G(jωn) = nAπ

ω0

[sin(nπ(ωn − 1))

nπ(ωn − 1)+

sin(nπ(ωn + 1))

nπ(ωn + 1)

]

where ωn = ω/ω0 is the normalized frequency. A further normalization by (Aπ/ω0)

yields:

Gnorm(jωn) =G(jω)

(Aπ/ω0)= n

[sin(nπ(ωn − 1))

nπ(ωn − 1)+

sin(nπ(ωn + 1))

nπ(ωn + 1)

]

Plots of Gnorm(jωn) for different values of n are shown below. Note that as more

and more cycles are taken the Fourier transform (i.e., the spectrum density) becomes

more and more concentrated around ±ω0 (rad/sec). In the limit as n→∞ or

equivalently T →∞ the density becomes 2 impulses located at ±ω0 (rad/sec).



−40 −20 0 20 40−0.5

0

0.5

1

1.5

ωn

Am

plitu

de D

ensi

ty S

pect

rum

n=1

−40 −20 0 20 40−2

0

2

4

6

ωn

Am

plitu

de D

ensi

ty S

pect

rum

n=5

−40 −20 0 20 40−5

0

5

10

ωn

Am

plitu

de D

ensi

ty S

pect

rum

n=10

−40 −20 0 20 40−20

−10

0

10

20

30

40

ωn

Am

plitu

de D

ensi

ty S

pect

rum

n=50



Example: Figure (a) below shows a system with input signal x(t) and output signal

y(t). The input signal has Fourier transform X(jω) shown in Figure (b). Determine

and sketch the spectrum Y (jω) of y(t).

-3W 3W 5W-5W

1x(t)

cos(5Wt) cos(3Wt)

3W-3W

1y(t)

(a)

-2W -2W

1

(b)

)( ωjX

ω

After modulated with carrier frequency 5W and passing through band-pass filter, the

output signal is shown in Figure (c) (shaded area). That signal is modulated one

more time with carrier frequency 3W . The modulated signal is plotted in Figure (d).

Therefore, Y (jω), the output of low-pass filter with cut-off frequency 3W can be

seen in Figure (e).



5W

1/2

ω

-3W 3W 5W-5W

1x(t)

cos(5Wt) cos(3Wt)

3W-3W

1y(t)

(a)

7W3W-5W -3W-7W

ω6W-6W 8W-8W 2W-2W

0

0

ω2W-2W 0

)( jwY

1/4

1/4

(c)

(d)

(e)

-2W 2W

1

(b)

)( ωjX

ω0

After the first modulation andpassing via band-pass filter

The second modulation

Output of the low-pass filter


Documents

EE351: Spectrum Analysis and Discrete Time Systems ...homepage.usask.ca/~hhn404/EE351/slides/EE351-Fall05-Part1...EE351: Spectrum Analysis and Discrete Time Systems (Signals, Systems