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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
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
k k k
oboe horn altosax
k k k
violin
k
Musical Instruments
Harmonic content is natural way to describe some kinds of signals.
Ex: musical instruments (http://theremin.music.uiowa.edu/MIS)
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
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
1 −1 j2πkt
2k2π2 e
k = −1
t
k odd 18
−18
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
3 −1 j2πkt
2k2π2 e
k = −3
t
k odd 18
−18
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
5 −1 j2πkt
2k2π2 e
k = −5
t
k odd 18
−18
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
7 −1 j2πkt
2k2π2 e
k = −7
t
k odd 18
−18
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
9 −1 j2πkt
2k2π2 e
k = −9
t
k odd 18
−18
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
19 −1 j2πkt
2k2π2 e
k = −19
t
k odd 18
−18
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
29 −1 j2πkt
2k2π2 e
k = −29
t
k odd 18
−18
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
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
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
0 1 j2πkt e
k = −0 jkπ
k odd
0 1 t
−12
12
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
1 1 j2πkt e
k = −1 jkπ
t
k odd 12
−12
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
3 1 j2πkt e
k = −3 jkπ
t
k odd 12
−12
0 1
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
5 1 j2πkt e
t
�
k = −5 k odd
jkπ
12
−12
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
7 1 j2πkt e
k = −7 jkπ
t
k odd 12
−12
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
9 1 j2πkt e
k = −9 jkπ
t
k odd 12
−12
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
19 1 j2πkt e
k = −19 jkπ
t
k odd 12
−12
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
29 1 j2πkt e
k = −29 jkπ
t
k odd 12
−12
0 1
�
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
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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