6.003 Lecture 14: Fourier Representations

6.003: Signals and Systems

Fourier Representations

March 30, 2010

Mid-term Examination #2

Wednesday, April 7,

No recitations on the day of the exam.

Coverage: Lectures 1–15

Recitations 1–15

Homeworks 1–8

Homework 8 will not collected or graded. Solutions will be posted.

18Closed book: 2 pages of notes ( 2× 11 inches; front and back).

Designed as 1-hour exam; two hours to complete.

Review sessions during open office hours.

7:30-9:30pm.

Fourier Representations

Fourier series represent signals in terms of sinusoids.

→ leads to a new representation for systems as filters.

Fourier Series

Representing signals by their harmonic components.

DC

→

fundam

enta

l →

second

harm

onic

→

third

harm

onic

→

ω ω0 2ω0 3ω0 4ω0 5ω0 6ω0

0 1 2 3 4 5 6 ← harmonic #

fourt

h h

arm

onic

→

fift

h h

arm

onic

→

sixth

harm

onic

→

bassoon

t

Musical Instruments

Harmonic content is natural way to describe some kinds of signals.

Ex: musical instruments (http://theremin.music.uiowa.edu/MIS)

piano cello bassoon

t t t

oboe horn altosax

t t t

violin

t 1 252

seconds

(http://theremin.music.uiowa.edu/MIS)

Musical Instruments



piano cello bassoon

k k k

oboe horn altosax

k k k

violin

k


Musical Instruments



piano piano

t

k violin

violin

t

k

bassoon

t

k

bassoon


Harmonics

Harmonic structure determines consonance and dissonance.

octave (D+D’) fifth (D+A) D+E�

–1

0

1

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

t ime(per iods of "D")

D' A E 0 1 2 3 4 5 6 7 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

D D harmonics D

Harmonic Representations

What signals can be represented by sums of harmonic components?

ω ω0 2ω0 3ω0 4ω0 5ω0 6ω0

T= 2π ω0

T= 2π ω0

t

t

Only periodic signals: all harmonics of ω0 are periodic in T = 2π/ω0.

Harmonic Representations

Is it possible to represent ALL periodic signals with harmonics?

What about discontinuous signals?

2π ω0

t

2π ω0

t

Fourier claimed YES — even though all harmonics are continuous!

Lagrange ridiculed the idea that a discontinuous signal could be

written as a sum of continuous signals.

We will assume the answer is YES and see if the answer makes sense.

��

Separating harmonic components

Underlying properties.

1. Multiplying two harmonics produces a new harmonic with the

same fundamental frequency:

jkω0t jlω0t = e j(k+l)ω0t e × e .

2. The integral of a harmonic over any time interval with length

equal to a period T is zero unless the harmonic is at DC:

= 0t0+T jkω0tdt ≡ jkω0tdt =

0, k �t0

eT e

T, k = 0

= Tδ[k]

� �

Separating harmonic components

Assume that x t is periodic in T and is composed of a weighted sum( )

of harmonics of ω0 = 2π/T . ∞ �

x(t) = x(t + T ) = ake jω0kt

k=−∞

Then � � ∞ � x(t)e −jlω0tdt = ake

jω0kt e −jω0ltdt T T k=−∞

∞ � = � ak e jω0(k−l)tdt

k=−∞ T

∞ � = akTδ[k − l] = Tal k=−∞

Therefore1 1

x(t)e−jπT2 ktx(t)e −jω0ktdt= = dtak T T T T

�

Fourier Series

Determining harmonic components of a periodic signal.

ak = 1

x(t)e− j2Tπktdt (“analysis” equation)

T T

∞ � 2π x(t)= x(t + T ) = ake

j T kt (“synthesis” equation) k=−∞

Check Yourself

Let ak represent the Fourier series coefficients of the following

square wave.

t

1 2

−1 2

0 1

How many of the following statements are true?

1. ak = 0 if k is even

2. ak is real-valued

3. |ak| decreases with k2

4. there are an infinite number of non-zero ak 5. all of the above

� �

�

Check Yourself

Let ak represent the Fourier series coefficients of the following square

wave.

1 2

0 1 t

−1 2

� � � 12 kt 1 0 1 2πT −j2πktdt + e−j2πktdt−jx(t)= dt =−ak e e2 1− 2

2 0T

= 1 2− e jπk − e −jπk j4πk

1 ; if k is odd = jπk0 ; otherwise

�

Check Yourself

Let ak represent the Fourier series coefficients of the following square

wave. 1 ; if k is odd

ak = jπk 0 ; otherwise

How many of the following statements are true?√ 1. ak = 0 if k is even

2. ak is real-valued X

3. |ak| decreases with k2 X √ 4. there are an infinite number of non-zero ak 5. all of the above X

Check Yourself

Let ak represent the Fourier series coefficients of the following

square wave.

t

1 2

−1 2

0 1

How many of the following statements are true? 2

1. ak = 0 if k is even √

2. ak is real-valued X

3. |ak| decreases with k2 X

4. there are an infinite number of non-zero ak √

5. all of the above X

Fourier Series Properties

If a signal is differentiated in time, its Fourier coefficients are multi

plied by j 2 Tπ k.

Proof: Let ∞ � 2π

x(t) = x(t+ T ) = akej T kt

k=−∞

then ∞ � �

x(t) = x(t+ T ) = �

j 2πk ak e

j2Tπkt

T k=−∞

Check Yourself

Let bk represent the Fourier series coefficients of the following

triangle wave.

t

1 8

−1 8

0 1

How many of the following statements are true?

1. bk = 0 if k is even

2. bk is real-valued

3. |bk| decreases with k2

4. there are an infinite number of non-zero bk 5. all of the above

Check Yourself

The triangle waveform is the integral of the square wave.

t

12

−12

0 1

t

18

−18

0 1

Therefore the Fourier coefficients of the triangle waveform are j2πk

times those of the square wave. 1 1 −1

1

bk = jkπ × j2πk = 2k2π2 ; k odd

Check Yourself

Let bk represent the Fourier series coefficients of the following tri

angle wave. −1

bk = 2k2π2 ; k odd

How many of the following statements are true?√ 1. bk = 0 if k is even √ 2. bk is real-valued √ 3. |bk| decreases with k2 √ 4. there are an infinite number of non-zero bk√ 5. all of the above

Check Yourself

Let bk represent the Fourier series coefficients of the following

triangle wave.

t

1 8

−1 8

0 1

How many of the following statements are true? 5

1. bk = 0 if k is even √

2. bk is real-valued √

3. |bk| decreases with k2 √

4. there are an infinite number of non-zero bk √

5. all of the above √

�

Fourier Series

One can visualize convergence of the Fourier Series by incrementally

adding terms.

Example: triangle waveform

0 −1 j2πkt

2k2π2 e

k = −0

t

k odd 18

−18

0 1

�

Fourier Series


adding terms.


1 −1 j2πkt

2k2π2 e

k = −1

t

k odd 18

−18

0 1

�

Fourier Series


adding terms.


3 −1 j2πkt

2k2π2 e

k = −3

t

k odd 18

−18

0 1

�

Fourier Series


adding terms.


5 −1 j2πkt

2k2π2 e

k = −5

t

k odd 18

−18

0 1

�

Fourier Series


adding terms.


7 −1 j2πkt

2k2π2 e

k = −7

t

k odd 18

−18

0 1

�

Fourier Series


adding terms.


9 −1 j2πkt

2k2π2 e

k = −9

t

k odd 18

−18

0 1

�

Fourier Series


adding terms.


19 −1 j2πkt

2k2π2 e

k = −19

t

k odd 18

−18

0 1

�

Fourier Series


adding terms.


29 −1 j2πkt

2k2π2 e

k = −29

t

k odd 18

−18

0 1

�

Fourier Series


adding terms.


39 −1 j2πkt

2k2π2 e

k = −39 k odd

t

18

−18

0 1

Fourier series representations of functions with discontinuous slopes

converge toward functions with discontinuous slopes.

�

Fourier Series


adding terms.

Example: square wave

0 1 j2πkt e

k = −0 jkπ

k odd

0 1 t

−12

12

�

Fourier Series


adding terms.


1 1 j2πkt e

k = −1 jkπ

t

k odd 12

−12

0 1

�

Fourier Series


adding terms.


3 1 j2πkt e

k = −3 jkπ

t

k odd 12

−12

0 1

Fourier Series


adding terms.


5 1 j2πkt e

t

�

k = −5 k odd

jkπ

12

−12

0 1

�

Fourier Series


adding terms.


7 1 j2πkt e

k = −7 jkπ

t

k odd 12

−12

0 1

�

Fourier Series


adding terms.


9 1 j2πkt e

k = −9 jkπ

t

k odd 12

−12

0 1

�

Fourier Series


adding terms.


19 1 j2πkt e

k = −19 jkπ

t

k odd 12

−12

0 1

�

Fourier Series


adding terms.


29 1 j2πkt e

k = −29 jkπ

t

k odd 12

−12

0 1

�

Fourier Series


adding terms.


39 1 j2πkt e

k = −39 jkπ

t

k odd 12

−12

0 1

Fourier Series

Partial sums of Fourier series of discontinuous functions “ring” near

discontinuities: Gibb’s phenomenon.

9%

t

12

−12

0 1

This ringing results because the magnitude of the Fourier coefficients

is only decreasing as k1

(while they decreased as

k12 for the triangle).

You can decrease (and even eliminate the ringing) by decreasing the

magnitudes of the Fourier coefficients at higher frequencies.

Fourier Series: Summary

Fourier series represent periodic signals as sums of sinusoids.

• valid for an extremely large class of periodic signals

• valid even for discontinuous signals such as square wave

However, convergence as # harmonics increases can be complicated.

Filtering

The output of an LTI system is a “filtered” version of the input.

Input: Fourier series → sum of complex exponentials.∞ � 2π

x(t) = x(t + T ) = ake j T kt

k=−∞

Complex exponentials: eigenfunctions of LTI systems. 2π 2π 2π ejT kt → H(j k)ejT kt

T

Output: same eigenfunctions, amplitudes/phases set by system. ∞ ∞ � 2π � 2π 2π

x(t) = j kt → y(t) = akH(j k) j ktTake T Te

k=−∞ k=−∞

Filtering

Notion of a filter.

LTI systems

• cannot create new frequencies.

• can scale magnitudes and shift phases of existing components.

Example: Low-Pass Filtering with an RC circuit

+ −

vi

+

vo

−

R

C

Lowpass Filter

Calculate the frequency response of an RC circuit.

KVL: vi(t) = Ri(t) + vo(t)R

+ C: i(t) = Cvo(t)

+ −

vi C Solving: vi(t) = RCvo(t) + vo(t)

vo Vi(s) = (1 + sRC)Vo(s)− H(s) =

Vo(s) = 1

Vi(s) 1 + sRC 1

0.1

|H(jω

)|

0.01 ω

0.01 0.1 1 10 100 1/RC 0

−π 2 ω∠H(jω

)|

0.01 0.1 1 10 100 1/RC

Lowpass Filtering

Let the input be a square wave.

t

12

−12

0 T

� 1 2π x(t) = e jω0kt ; ω0 =

k odd jπk T

1

|X(jω

)|

0.1

0.01 ω

∠X(jω

)| 0.01 0.1 1 10 100 1/RC0

−π 2 ω

0.01 0.1 1 10 100 1/RC

Lowpass Filtering

Low frequency square wave: ω0 << 1/RC.

t

12

−12

0 T

� 1 2π x(t) = e jω0kt ; ω0 =

k odd jπk T

1

|H(jω

)|

0.1

0.01 ω

0.01 0.1 1 10 100 1/RC 0

∠H(jω

)|

−π 2 ω

0.01 0.1 1 10 100 1/RC

Lowpass Filtering

Higher frequency square wave: ω0 < 1/RC.

t

12

−12

0 T

� 1 2π x(t) = e jω0kt ; ω0 =

k odd jπk T

1

|H(jω

)|

0.1

0.01 ω

0.01 0.1 1 10 100 1/RC 0

∠H(jω

)|

−π 2 ω

0.01 0.1 1 10 100 1/RC

Lowpass Filtering

Still higher frequency square wave: ω0 = 1/RC.

t

12

−12

0 T

x(t) = � 1

e jω0kt ; ω0 =2π

k odd jπk T

0.01

0.1

1

ω

|H(jω

)|∠H

(jω

)|

−π 2

0.01 0.1 1 10 100 1/RC 0

ω

0.01 0.1 1 10 100 1/RC

Lowpass Filtering

High frequency square wave: ω0 > 1/RC.

t

12

−12

0 T

x(t) = � 1

e jω0kt ; ω0 =2π

k odd jπk T

0.01

0.1

1

ω

|H(jω

)|∠H

(jω

)|

−π 2

0.01 0.1 1 10 100 1/RC 0

ω

0.01 0.1 1 10 100 1/RC

Fourier Series: Summary

Fourier series represent signals by their frequency content.

Representing a signal by its frequency content is useful for many

signals, e.g., music.

Fourier series motivate a new representation of a system as a filter.

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6.003 Signals and Systems Spring 2010

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6.003 Lecture 14: Fourier Representations

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