6.003: Signal Processing

6.003: Signal Processing

Signal Processing

• Overview of Subject

• Signals: Definitions, Examples, and Operations

• Time and Frequency Representations

• Fourier Series

September 9, 2021


Signals are functions that contain and convey information.

Examples:

• the MP3 representation of a sound

• the JPEG representation of a picture

• an MRI image of a brain

Signal Processing develops the use of signals as abstractions:

• identifying signals in physical, mathematical, computation contexts,

• analyzing signals to understand the information they contain, and

• manipulating signals to modify and/or extract information.


Signal Processing is widely used in science and engineering to ...

• model some aspect of the world,

• analyze the model, and

• interpret results to gain a new or better understanding.

model result

world new understanding

make model

analyze

(math, computation)

interpret results

Signal Processing provides a common language across disciplines.


Signal Processing is widely used in science and engineering to ...

• model some aspect of the world,

• analyze the model, and

• interpret results to gain a new or better understanding.

model result

world new understanding

make model

analyze

(math, computation)

interpret results

Signal Processing provides a common language across disciplines.

Classical analyses use a variety of maths, especially calculus. We will also

use computation to solve real-world problems that are difficult or impos-

sible to solve analytically.

→ strengthens ties to the real world

Course Mechanics

Schedule

Lecture: Tue. and Thu. 2-3pm in 3-270

Recitation: Tue. and Thu. 3-4pm in 5-234 or 36-156

Office Hours: Tue. and Thu. 4-5pm in 5-234 or 36-156

and other times TBD

Homework – issued Tuesdays, due following Tuesday at noon

• Exercises: study aids; not counted in grade

− online with immediate feedback

• Problems: focus on developing problem solving skills

– pencil and paper problems taken from previous exams

– simple computational extensions to real-world data

– completely specified, unambiguous, self-contained

• Labs: focus on applications of 6.003 to authentic problems

– more open-ended, multiple approaches, multiple solutions

– deepen understanding and demonstrate wide applicability

– issued Tuesday, required check-in Friday, due following Tuesday

Two Midterms and a Final Exam

Signals


– may have 1 or 2 or 3 or even more independent variables

t

sou

nd

pre

ssu

re(t

)

x

y brightness (x, y)

Signals


– dependent variable can be a scalar or a vector

x

yscalar: brightness

at each point (x, y)

x

y

vector: (red,green,blue)at each point (x, y)

Signals


– dependent variable can be real, imaginary, or complex-valued

t

x(t) = ej2πt = cos 2πt +j sin 2πt

1

0

−1

1 2

Signals


– continuous domain versus discrete domain

t

x(t)

0 0.1 0.2 0.3 0.4 0.5n

x[n]

0 2 4 6 8 10

Signals from physical systems are often of continuous domain:

• continuous time – measured in seconds

• continuous spatial coordinates – measured in meters

Computations usually manipulate functions of discrete domain:

• discrete time – measured in samples

• discrete spatial coordinates – measured in samples

Signals

Relating continuous and discrete representations enables application of

computational methods to solve problems that are intrinsically continuous.

Sampling: converting CT signals to DT

t

x(t)

0T 2T 4T 6T 8T 10Tn

x[n] = x(nT )

0 2 4 6 8 10

T = sampling interval

Important for computational manipulation of physical data.

• digital representations of audio signals (as in MP3)

• digital representations of images (as in JPEG)

Signals



Reconstruction: converting DT signals to CT

zero-order hold

n

x[n]

0 2 4 6 8 10t

x(t)

0 2T 4T 6T 8T 10T


commonly used in audio output devices

Signals



Reconstruction: converting DT signals to CT

piecewise linear

n

x[n]

0 2 4 6 8 10t

x(t)

0 2T 4T 6T 8T 10T


commonly used in rendering images

Signals

Periodic signals consist of repeated cycles (periods).

t

x(t) = x(t+ T )

0 T

periodic

t

x(t)

0

aperiodic

n

x[n] = x[n+N ]

0 Nn

x[n]

0

Useful for modeling periodic or nearly-periodic systems

• vibrating strings

• planetary motions

Signals

Right-sided signals are zero before some starting time.

Left-sided signals are zero after some ending time.

right-sided left-sided

t

x(t)

0 Tst

x(t)

0Te

n

x[n]

0 Ns

n

x[n]

0Ne

Useful for modeling systems that have a well-defined starting point:

• piano note

• striking a cymbal

Signals

Signals can be symmetric or antisymmetric about time zero.

symmetric antisymmetric

t

x(t) = x(−t)

0

t

x(t) = −x(−t)

n

x[n] = x[−n]

0

n

x[n] = −x[−n]

0

Check Yourself

Computer generated speech (by Robert Donovan)

t

f(t)

Listen to the following four manipulated signals:

f1(t), f2(t), f3(t), f4(t).

How many of the following relations are true?

• f1(t) = f(2t)• f2(t) = −f(t)• f3(t) = f(2t)• f4(t) = 1

3f(t)

Check Yourself

Computer generated speech (by Robert Donovan)

t

f(t)

Listen to the following four manipulated signals:

f1(t), f2(t), f3(t), f4(t).

How many of the following relations are true? 2

• f1(t) = f(2t)√

• f2(t) = −f(t) X

• f3(t) = f(2t) X

• f4(t) = 13f(t)

√

Musical Sounds as Signals


Example: a musical sound can be represented as a function of time.

t [seconds]

pressure

Although this time function is a complete description of the sound, it does

not expose many of the important properties of the sound.


Even though these sounds have the same pitch, they sound different.

t

piano

t

cello

t

bassoon

t

oboe

t

horn

t

altosax

t

violin

1262 sec.

It’s not clear how the differences relate to properties of the signals.

(audio clips from from http://theremin.music.uiowa.edu)

Musical Signals as Sums of Sinusoids

One way to characterize differences between these signals is express each

as a sum of sinusoids.

f(t) =∞∑k=0

(ck cos kωot+ dk sin kωot)

2πωo

t

cos(

0t)

2πωo

t

sin(0t)

2πωo

t

cos(ωot)

2πωo

t

sin(ω

ot)

2πωo

t

cos(

2ωot)

...2πωo

t

sin(2ωot)

...

Since these sounds are (nearly) periodic, the frequencies of the dominant

sinusoids are (nearly) integer multiples of a fundamental frequency ωo.

Harmonic Structure

The sum of sinusoids describes the distribution of energy across frequencies.

f(t) =∞∑k=0

(ck cos kωot+ dk sin kωot) =∞∑k=0

mk cos (kωot+ φk)

where m2k = c2

k + d2k and tanφk = dk

ck.

ωωo 2ωo 3ωo 4ωo 5ωo 6ωo

mk

0 1 2 3 4 5 6 ← harmonic #

DC→

fun

da

me

nta

l→

sec

on

dh

arm

on

ic→

thir

dh

arm

on

ic→

fou

rth

harm

on

ic→

fift

hh

arm

on

ic→

six

thh

arm

on

ic→

This distribution represents the harmonic structure of the signal.

Harmonic Structure

The harmonic structures of notes from different instruments are different.

t

piano

k

piano

t

bassoon

k

bassoon

t

violin

k

violin

Some musical qualities are more easily seen in time, others in frequency.

Consonance and Dissonance

Which of the following pairs is least consonant?

A1

t

A2

t

B1

t

B2

t

C1

t

C2

t

Obvious from the sounds ... less obvious from the waveforms.

Express Each Signal as a Sum of Sinusoids

f(t) =∞∑k=0

mk cos(kωot+ φk)

= m1 cos(ωot+φ1) +m2 cos(2ωot+φ2) +m3 cos(3ωot+φ3) + · · ·

time

freq

Two views: as a function of time and as a function of frequency

Express Each Signal as a Sum of Sinusoids

f(t) =∞∑k=0

mk cos(kωot+ φk)

= m1 cos(ωot+φ1) +m2 cos(2ωot+φ2) +m3 cos(3ωot+φ3) + · · ·

freq

The signal f(t) can be expressed as a discrete set of frequency components:

ω0: m1, φ12ω0: m2, φ23ω0: m3, φ3· · ·


Time functions do a poor job of conveying consonance and dissonance.

octave (D+D’) fifth (D+A) D+E[

t ime(per iods of "D")

harmonics

0 1 2 3 4 5 6 7 8 9 101112 0 1 2 3 4 5 6 7 8 9 101112 0 1 2 3 4 5 6 7 8 9 101112

–1

0

1

D

D'

D

A

D

E

0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7

Harmonic structure conveys consonance and dissonance better.

Fourier Representations of Signals

Fourier series are sums of harmonically related sinusoids.

f(t) =∞∑k=0

(ck cos(kωot) + dk sin(kωot))

where ωo = 2π/T represents the fundamental frequency.

Basis functions:

2πωo

t

cos(

0t)

2πωo

t

sin(0t)

2πωo

t

cos(ωot)

2πωo

t

sin(ω

ot)

2πωo

t

cos(

2ωot)

...2πωo

tsin

(2ωot)

...

Q1: Under what conditions can we write f(t) as a Fourier series?

Q2: How do we find the coefficients ck and dk.


Under what conditions can we write f(t) as a Fourier series?

Fourier series can only represent periodic signals.

Definition: a signal f(t) is periodic in T if

f(t) = f(t+T )for all t.

Note: if a signal is periodic in T it is also periodic in 2T , 3T , ...

The smallest positive number To for which f(t) = f(t + To) for all t is

sometimes called the fundamental period.

If a signal does not satisfy f(t) = f(t+T ) for any value of T , then the signal

is aperiodic.


Fourier series can only represent periodic signals.

ωωo 2ωo 3ωo 4ωo 5ωo 6ωo

T= 2πωo

t

T= 2πωo

t

All harmonics of ωo (cos(kωot) or sin(kωot)) are periodic in T = 2π/ωo.→ all sums of such signals are periodic in T = 2π/ωo.→ Fourier series can only represent periodic signals.

Calculating Fourier Coefficients

How do we find the coefficients ck and dk for all k?

Key idea: simplify by integrating over the period T of the fundamental.

Start with the general form:

f(t) = f(t+T ) = c0 +∞∑k=1


Integrate both sides over T :∫ T

0f(t) dt =

∫ T

0c0 dt+

∫ T

0

( ∞∑k=1

(ck cos(kωot) + dk sin(kωot)))dt

= Tc0 +∞∑k=1

(ck

∫ T

0cos(kωot) dt+ dk

∫ T

0sin(kωot) dt

)= Tc0

All but the first term integrates to zero, leaving

c0 = 1T

∫ T

0f(t) dt.

This k=0 term represents the average (“DC”) value.


Isolate the cl term by multiplying both sides by cos(lωot) before integrating.

f(t) = f(t+T ) = c0 +∞∑k=1


∫ T

0f(t) cos(lωot) dt =

∫ T

0c0 cos(lωot) dt

+∞∑k=1

∫ T

0ck cos(kωot) cos(lωot) dt

+∞∑k=1

∫ T

0dk sin(kωot) cos(lωot) dt

A product of sinusoids can be expressed as sum and difference frequencies.

cos(kωot) cos(lωot) = 12 cos((k−l)ωot) + 1

2 cos((k+l)ωot)

sin(kωot) cos(lωot) = 12 sin((k−l)ωot) + 1

2 sin((k+l)ωot)



f(t) = f(t+T ) = c0 +∞∑k=1


∫ T


∫ T

0c0 cos(lωot) dt

+∞∑k=1

∫ T

0ck

(12 cos((k−l)ωot) + 1

2 cos((k+l)ωot))dt

+∞∑k=1

∫ T

0dk

(12 sin((k−l)ωot) + 1

2 sin((k+l)ωot))dt

A product of sinusoids can be expressed as sum and difference frequencies.

cos(kωot) cos(lωot) = 12 cos((k−l)ωot) + 1

2 cos((k+l)ωot)

sin(kωot) cos(lωot) = 12 sin((k−l)ωot) + 1

2 sin((k+l)ωot)



f(t) = f(t+T ) = c0 +∞∑k=1


∫ T


∫ T

0c0 cos(lωot) dt

+∞∑k=1

∫ T

0ck

(12 cos((k−l)ωot) + 1

2 cos((k+l)ωot))dt

+∞∑k=1

∫ T

0dk

(12 sin((k−l)ωot) + 1

2 sin((k+l)ωot))dt

0

The c0 term is zero because the integral of cos(lωot) over T is zero.



f(t) = f(t+T ) = c0 +∞∑k=1


∫ T


∫ T

0c0 cos(lωot) dt

+∞∑k=1

∫ T

0ck

(12 cos((k−l)ωot) + 1

2 cos((k+l)ωot))dt

+∞∑k=1

∫ T

0dk

(12 sin((k−l)ωot) + 1

2 sin((k+l)ωot))dt

0

T2 cl 0

If k = l, then cos((k−l)ωot = 1 and the integral is T2 cl.

All of the other cos((k − l)ωot) terms in the sum integrate to zero.

All of the cos((k + l)ωot) terms in the integrate to zero.



f(t) = f(t+T ) = c0 +∞∑k=1


∫ T


∫ T

0c0 cos(lωot) dt

+∞∑k=1

∫ T

0ck

(12 cos((k−l)ωot) + 1

2 cos((k+l)ωot))dt

+∞∑k=1

∫ T

0dk

(12 sin((k−l)ωot) + 1

2 sin((k+l)ωot))dt

0

T2 cl 0

0 0

If k = l, then sin((k−l)ωot = 0 and the integral is 0.

All of the other dk terms are harmonic sinusoids that integrate to 0.

The only non-zero term on the right side is T2 cl.

We can solve to get an expression for cl as

cl = 2T

∫ T

0f(t) cos(lωot) dt


Analogous reasoning allows us to calculate the dk coefficients, but this time

multiplying by sin(lωot) before integrating.

f(t) = f(t+T ) = c0 +∞∑k=1


∫ T

0f(t) sin(lωot) dt =

∫ T

0c0 sin(lωot) dt

+∞∑k=1

∫ T

0ck cos(kωot) sin(lωot) dt

+∞∑k=1

∫ T

0dk sin(kωot) sin(lωot) dt

A single term remains after integrating, allowing us to solve for dl as

dl = 2T

∫ T

0f(t) sin(lωot) dt


Summarizing . . .

If f(t) is expressed as a Fourier series

f(t) = f(t+T ) = c0 +∞∑k=1


the Fourier coefficients are given by

c0 = 1T

∫Tf(t) dt

ck = 2T

∫Tf(t) cos(kωot) dt; k = 1, 2, 3, . . .

dk = 2T

∫Tf(t) sin(kωot) dt; k = 1, 2, 3, . . .

Example of Analysis

Find the Fourier series coefficients for the following triangle wave:

t

f(t) = f(t+2)

0 1 2−1−2

1

T = 2

ωo = 2πT

= π

c0 = 1T

∫ T

0f(t) dt = 1

2

∫ 2

0f(t)dt = 1

2

ck = 2T

∫ T/2

−T/2f(t) cos 2πkt

Tdt = 2

∫ 1

0t cos(πkt) dt =

{− 4π2k2 k odd

0 k = 2, 4, 6, . . .

dk = 0 (by symmetry)

Example of Synthesis

Generate f(t) from the Fourier coefficients in the previous slide.

Start with the Fourier coefficients

f(t) = c0 −∞∑k=1

(ck cos(kωot) + dk sin(kωot)) = 12 −

∞∑k = 1k odd

4π2k2 cos(kπt)

f(t) = 12 −

0∑k = 1k odd

4π2k2 cos(kπt)

t

f(t)

0 1 2−1−2




f(t) = c0 −∞∑k=1


∞∑k = 1k odd

4π2k2 cos(kπt)

f(t) = 12 −

1∑k = 1k odd

4π2k2 cos(kπt)

t

f(t)

0 1 2−1−2




f(t) = c0 −∞∑k=1


∞∑k = 1k odd

4π2k2 cos(kπt)

f(t) = 12 −

3∑k = 1k odd

4π2k2 cos(kπt)

t

f(t)

0 1 2−1−2




f(t) = c0 −∞∑k=1


∞∑k = 1k odd

4π2k2 cos(kπt)

f(t) = 12 −

5∑k = 1k odd

4π2k2 cos(kπt)

t

f(t)

0 1 2−1−2




f(t) = c0 −∞∑k=1


∞∑k = 1k odd

4π2k2 cos(kπt)

f(t) = 12 −

7∑k = 1k odd

4π2k2 cos(kπt)

t

f(t)

0 1 2−1−2




f(t) = c0 −∞∑k=1


∞∑k = 1k odd

4π2k2 cos(kπt)

f(t) = 12 −

9∑k = 1k odd

4π2k2 cos(kπt)

t

f(t)

0 1 2−1−2




f(t) = c0 −∞∑k=1


∞∑k = 1k odd

4π2k2 cos(kπt)

f(t) = 12 −

19∑k = 1k odd

4π2k2 cos(kπt)

t

f(t)

0 1 2−1−2




f(t) = c0 −∞∑k=1


∞∑k = 1k odd

4π2k2 cos(kπt)

f(t) = 12 −

99∑k = 1k odd

4π2k2 cos(kπt)

t

f(t)

0 1 2−1−2

The synthesized function approaches original as number of terms increases.

Two Views of the Same Signal

The harmonic expansion provides an alternative view of the signal.

f(t) =∞∑k=0

(ck cos(kωot) + dk sin(kωot)) =∞∑k=0

mk cos(kωot+φk)

We can view the musical signal as

• a function of time f(t), or

• as a sum of harmonics with amplitudes mk and phase angles φk.

Both views are useful. For example,

• the peak sound pressure is more easily seen in f(t), while

• consonance is more easily analyzed by comparing harmonics.

This type of harmonic analysis is an example of Fourier Analysis,

which is a major theme of this subject.

Recitations

Reconvene in 10 minutes for recitation.

If the second character of your kerberos username is in ’abcdefghij’:

go to room 5-234

else:

go to room 36-156

6.003: Signal Processing

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Transcript of 6.003: Signal Processing