E1.10 Fourier Series and Transforms
Mike Brookes
E1.10 Fourier Series and Transforms (2014-5509)
Sums and Averages: 1 – 1 / 14
Syllabus
⊲ Syllabus
Optical Fourier
Transform
Organization
1: Sums and
Averages
Main fact: Complicated time waveforms can be
expressed as a sum of sine and cosine waves.
Why bother? Sine/cosine are the only bounded
waves that stay the same when differentiated.
Any electronic circuit:
sine wave in ⇒ sine wave out (same frequency).
Joseph Fourier
1768-1830
Hard problem: Complicated waveform → electronic circuit→ output = ?
Easier problem: Complicated waveform → sum of sine waves
→ linear electronic circuit (⇒ obeys superposition)
→ add sine wave outputs → output = ?
Syllabus:
Preliminary maths (1 lecture)
Fourier series for periodic waveforms (4 lectures)
Fourier transform for aperiodic waveforms (3 lectures)
E1.10 Fourier Series and Transforms (2014-5509)
Sums and Averages: 1 – 2 / 14
Optical Fourier Transform
Syllabus
Optical Fourier
Transform
Organization
⊲
1: Sums and
Averages
A pair of prisms can split light up into its component frequencies (colours).
This is called Fourier Analysis.
A second pair can re-combine the frequencies.
This is called Fourier Synthesis.
Fourier Analysis
Fourier Synthesis
We want to do the same thing with mathematical signals instead of light.
E1.10 Fourier Series and Transforms (2014-5509)
Sums and Averages: 1 – 3 / 14
Organization
Syllabus
Optical Fourier
Transform
Organization
⊲
1: Sums and
Averages
• 8 lectures: feel free to ask questions
• Textbook: Riley, Hobson & Bence “Mathematical Methods for Physics
and Engineering”, ISBN:978052167971-8, Chapters [4], 12 & 13
• Lecture slides (including animations) and problem sheets + answers
available via Blackboard or from my website:
http://www.ee.ic.ac.uk/hp/staff/dmb/courses/E1Fourier/E1Fourier.htm
• Email me with any errors in slides or problems and if answers are
wrong or unclear
E1.10 Fourier Series and Transforms (2014-5509)
Sums and Averages: 1 – 4 / 14
Syllabus
Optical Fourier
Transform
Organization
1: Sums and
⊲ Averages
Geometric Series
Infinite Geometric
Series
Dummy Variables
Dummy Variable
Substitution
Averages
Average Properties
Periodic Waveforms
Averaging Sin and
Cos
Summary
1: Sums and Averages
E1.10 Fourier Series and Transforms (2014-5509)
Sums and Averages: 1 – 5 / 14
Geometric Series
Syllabus
Optical Fourier
Transform
Organization
1: Sums and
Averages
Geometric Series
Infinite Geometric
Series
Dummy Variables
⊲
Dummy Variable
Substitution
Averages
Average Properties
Periodic Waveforms
Averaging Sin and
Cos
Summary
A geometric series is a sum of terms that increase or decrease by a constant
factor, x:
S = a + ax + ax2 + . . . + axn
The sequence of terms themselves is called a geometric progression.
We use a trick to get rid of most of the terms:
S = a + ax + ax2 + . . . + axn−1 + axn
xS =
ax + ax2 + ax3 + . . . + axn + axn+1
Now subtract the lines to get: S − xS = (1 − x) S = a − axn+1
a = first term
Divide by 1 − x to get:
S =a×
1−xn+1
1−x
Example:
S = 3 + 6 + 12 + 24
=3×
1−24
1−2
=3×
E1.10 Fourier Series and Transforms (2014-5509)
n + 1 = number of terms
[a = 3, x = 2, n + 1 = 4]
−15
−1
= 45
Sums and Averages: 1 – 6 / 14
Infinite Geometric Series
Syllabus
Optical Fourier
Transform
Organization
1: Sums and
Averages
Geometric Series
Infinite Geometric
Series
Dummy Variables
⊲
Dummy Variable
Substitution
Averages
Average Properties
Periodic Waveforms
Averaging Sin and
Cos
Summary
2
n
A finite geometric series: Sn = a + ax + ax + · · · + ax =
1−xn+1
a 1−x
What is the limit as n → ∞?
If |x| < 1 then xn+1 −→ 0 which gives
a = first term
n→∞
1
=
S∞ = a + ax + ax2 + · · · = a 1−x
Example 1:
0.4 + 0.04 + 0.004 + . . . =
0.4
1−0.1
Example 2: (alternating signs)
2 − 1.2 + 0.72 − 0.432 + . . . =
Example 3:
1 + 2 + 4 + . . . 6=
1
1−2
=
1
−1
2
1−(−0.6)
x = factor
[a = 0.4, x = 0.1]
= 0.4̇
= 1.25
[a = 2, x = −0.6]
[a = 1, x = 2]
= −1
The formula S = a + ax + ax2 + . . . =
E1.10 Fourier Series and Transforms (2014-5509)
a
1−x
a
1−x
is only valid for |x| < 1
Sums and Averages: 1 – 7 / 14
Dummy Variables
Syllabus
Optical Fourier
Transform
Organization
1: Sums and
Averages
Geometric Series
Infinite Geometric
Series
Dummy Variables
⊲
Dummy Variable
Substitution
Averages
Average Properties
Periodic Waveforms
Averaging Sin and
Cos
Summary
Using a
P
sign, we can write the geometric series more compactly:
P
Sn = a + ax + ax2 + . . . + axn = nr=0 axr
[Note: x0 , 1 in this context even when x = 0]
Here r is a dummy variable: you can replace it with anything else
Pn
Pn
Pn
r
k
α
=
=
r=0 ax
k=0 ax
α=0 ax
Dummy variables are undefined outside the summation so they sometimes
get re-used elsewhere in an expression:
P3
P2
1−24
1−32
r
r
+ 3 × 1−3 = 15 + 12 = 27
r=0 2 +
r=1 3 = 1 × 1−2
The two dummy variables are both called r but they have no connection
with each other at all (or with any other variable called r anywhere else).
E1.10 Fourier Series and Transforms (2014-5509)
Sums and Averages: 1 – 8 / 14
Dummy Variable Substitution
Syllabus
Optical Fourier
Transform
Organization
1: Sums and
Averages
Geometric Series
Infinite Geometric
Series
Dummy Variables
Dummy Variable
⊲ Substitution
Averages
Average Properties
Periodic Waveforms
Averaging Sin and
Cos
Summary
We can derive the formula for the geometric series using
P
P
Sn = nr=0 axr and xSn = nr=0 axr+1
P
notation:
We need to manipulate the second sum to involve xr .
Use the substitution s = r + 1 ⇔ r = s − 1.
Substitute for r everywhere it occurs (including both limits)
Pn+1 s Pn+1 r
xSn = s=1 ax = r=1 ax
It is essential to sum over exactly the same set of values when substituting
for dummy variables.
Pn
Pn+1 r
r
Subtracting gives (1 − x)Sn = Sn − xSn = r=0 ax − r=1 ax
r ∈ [1, n] is common to both sums, so extract the remaining terms:
Pn
Pn
0
n+1
r
(1 − x)Sn = ax − ax
+ r=1 ax − r=1 axr
0
n+1
n+1
= ax − ax
=a 1−x
Hence:
Sn =
1−xn+1
a 1−x
E1.10 Fourier Series and Transforms (2014-5509)
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Averages
Syllabus
Optical Fourier
Transform
Organization
1: Sums and
Averages
Geometric Series
Infinite Geometric
Series
Dummy Variables
Dummy Variable
Substitution
Averages
Average Properties
Periodic Waveforms
Averaging Sin and
Cos
Summary
⊲
If a signal varies with time, we can plot its waveform, x(t).
The average value of x(t) in the range T1 ≤ t ≤ T2 is
R T2
1
hxi[T1 ,T2 ] = T2 −T1 t=T1 x(t)dt
<x>[T1,T2]
x(t)
T1
<x>[T1,T2]
T2
T1
x(t)
T2
The area under the curve x(t) is equal to the area of the rectangle
defined by 0 and hxi[T1 ,T2 ] .
Angle brackets alone, hxi, denotes the average value over all time
hx(t)i = limA,B→∞ hx(t)i[−A,+B]
E1.10 Fourier Series and Transforms (2014-5509)
Sums and Averages: 1 – 10 / 14
Average Properties
Syllabus
Optical Fourier
Transform
Organization
1: Sums and
Averages
Geometric Series
Infinite Geometric
Series
Dummy Variables
Dummy Variable
Substitution
Averages
Average Properties
Periodic Waveforms
Averaging Sin and
Cos
Summary
The properties of averages follow from the properties of integrals:
Addition:
Add a constant:
Constant multiple:
hx(t) + y(t)i = hx(t)i + hy(t)i
hx(t) + ci = hx(t)i + c
ha × x(t)i = a × hx(t)i
where the constants a and c do not depend on time.
⊲
For example:
hx(t) + y(t)i[T1 ,T2 ] =
1
T2 −T1
=
R T2
t=T1
1
T2 −T1
(x(t) + y(t)) dt
R T2
t=T1
x(t)dt +
1
T2 −T1
= hx(t)i[T1 ,T2 ] + hy(t)i[T1 ,T2 ]
R T2
t=T1
y(t)dt
But beware: hx(t) × y(t)i =
6 hx(t)i × hy(t)i.
E1.10 Fourier Series and Transforms (2014-5509)
Sums and Averages: 1 – 11 / 14
Periodic Waveforms
Syllabus
Optical Fourier
Transform
Organization
1: Sums and
Averages
Geometric Series
Infinite Geometric
Series
Dummy Variables
Dummy Variable
Substitution
Averages
Average Properties
A periodic waveform with period T repeats itself at intervals of T :
x(t + T ) = x(t)
⇒
x(t ± kT ) = x(t) for any integer k.
The smallest T > 0 for which x(t + T ) = x(t) ∀t is the fundamental
period. The fundamental frequency is F = T1 .
<x>
T
Periodic
⊲ Waveforms
Averaging Sin and
Cos
Summary
For a periodic waveform, hx(t)i equals the average over one period.
It doesn’t make any difference where in a period you start or how many
whole periods you take the average over.
Example:
x(t) = |sin t|
R
1 π
hxi = π t=0 |sin t| dt =
=
1
π
π
[− cos t]0 =
E1.10 Fourier Series and Transforms (2014-5509)
1
π
1
π
Rπ
t=0
sin t dt
(1 + 1) =
2
π
≈ 0.637
Sums and Averages: 1 – 12 / 14
[proof that x(t ± kT ) = x(t)]
Proof that x(t + T ) = x(t) ∀t ⇒ x(t ± kT ) = x(t) ∀t, ∀k ∈ Z
We use induction. Let Hk be the hypothesis that x(t + kT ) = x(t) ∀t. Under the assumption that
x(t + T ) = x(t) ∀t, we will show that if Hk is true, then so are Hk+1 and Hk−1 . Since we know that
H0 is definitely true, this implies that Hk is true for all integers k, i.e. for all k ∈ Z.
Suppose Hk is true, i.e. x(τ + kT ) = x(τ ) ∀τ . Now set τ = t + T . This gives x(t + T + kT ) =
x(t + T ) ∀t. But, we assume that x(t + T ) = x(t), so x (t + (k + 1)T ) = x(t + T + kT ) =
x(t + T ) = x(T ) ∀t. Hence Hk+1 is true.
Now suppose Hk is true as before but this time set τ = t − T . Substituting this into u(τ + kT ) =
u(τ ) gives u(t−T +kT ) = u(t−T ). Substituting it also into u(τ +T ) = u(τ ) gives u(t) = u(t−T ).
Finally, combining these two identities gives u (t + (k − 1)T ) = u(t) which is Hk−1 .
E1.10 Fourier Series and Transforms (2014-5509)
Sums and Averages: 1 – note 1 of slide 12
Averaging Sin and Cos
Optical Fourier
Transform
Organization
1: Sums and
Averages
Geometric Series
Infinite Geometric
Series
Dummy Variables
Dummy Variable
Substitution
Averages
Average Properties
Periodic Waveforms
Averaging Sin and
Cos
Summary
⊲
A sine wave, x(t) = sin 2πF t, has a frequency F and a period T = F1
1
so that, sin 2πF t + F = sin (2πF t + 2π) = sin 2πF t.
1
T
1
RT
hsin 2πF ti =
sin (2πF t) dt
t=0
=0
x(t)
Syllabus
F=1kHz
0
-1
0
0.5
1
Time (ms)
1.5
2
Also, hcos 2πF ti = 0 except for the case F = 0 since cos 2π0t ≡ 1.
Hence:
Also:
hsin 2πF ti = 0
and hcos 2πF ti =
(
0 F 6= 0
1 F =0
ei2πF t = hcos 2πF t + i sin 2πF ti
= hcos 2πF ti + i hsin 2πF ti
(
0 F 6= 0
=
1 F =0
E1.10 Fourier Series and Transforms (2014-5509)
Sums and Averages: 1 – 13 / 14
Summary
Syllabus
Optical Fourier
Transform
Organization
1: Sums and
Averages
Geometric Series
Infinite Geometric
Series
Dummy Variables
Dummy Variable
Substitution
Averages
Average Properties
Periodic Waveforms
Averaging Sin and
Cos
Summary
⊲
• Sum of geometric series (see RHB Chapter 4)
n+1
◦ Finite series: S = a × 1−x
1−x
a
◦ Infinite series: S = 1−x but only if |x| < 1
• Dummy variables
◦ Commonly re-used elsewhere in expressions
◦ Substitutions must cover exactly the same set of values
R T2
1
• Averages: hxi[T1 ,T2 ] = T2 −T1 t=T1 x(t)dt
• Periodic waveforms: x(t ± kT ) = x(t) for any integer k
◦ Fundamental period is the smallest T
◦ Fundamental frequency is F = T1
◦ For periodic waveforms, hxi is the average over any integer
number of periods
◦ hsin 2πF ti = 0
◦ hcos 2πF ti = ei2πF t =
E1.10 Fourier Series and Transforms (2014-5509)
(
0 F 6= 0
1 F =0
Sums and Averages: 1 – 14 / 14
⊲ 2: Fourier Series
Periodic Functions
Fourier Series
Why Sin and Cos
Waves?
Dirichlet Conditions
Fourier Analysis
Trigonometric
Products
Fourier Analysis
Fourier Analysis
Example
Linearity
Summary
E1.10 Fourier Series and Transforms (2014-5379)
2: Fourier Series
Fourier Series: 2 – 1 / 11
Periodic Functions
2: Fourier Series
Periodic Functions
Fourier Series
Why Sin and Cos
Waves?
Dirichlet Conditions
Fourier Analysis
Trigonometric
Products
Fourier Analysis
⊲
Fourier Analysis
Example
Linearity
Summary
A function, u(t), is periodic with period T if u(t + T ) = u(t) ∀t
• Periodic with period T ⇒ Periodic with period kT ∀k ∈ Z+
The fundamental period is the smallest T > 0 for which u(t + T ) = u(t)
T
If you add together functions with different periods the fundamental period
is the lowest common multiple (LCM) of the individual fundamental
periods.
Example:
2π
• u(t) = cos 4πt ⇒ Tu = 4π
= 0.5
• v(t) = cos 5πt ⇒ Tv = 2π
5π = 0.4
• w(t) = u(t) + 0.1v(t) ⇒ Tw = lcm(0.5, 0.4) = 2.0
Tu = 0.5
Tv = 0.4
E1.10 Fourier Series and Transforms (2014-5379)
Tw = 2.0
Fourier Series: 2 – 2 / 11
Fourier Series
2: Fourier Series
Periodic Functions
Fourier Series
Why Sin and Cos
Waves?
Dirichlet Conditions
Fourier Analysis
Trigonometric
Products
Fourier Analysis
⊲
If u(t) has fundamental period T and fundamental frequency F = T1 then,
in most cases, we can express it as a (possibly infinite) sum of sine and
cosine waves with frequencies 0, F , 2F , 3F , · · · .
T
u(t) =
Fourier Analysis
Example
Linearity
Summary
T
sin 2πF t
[b1 = 1]
−0.4 sin 2π2F t
[b2 = −0.4]
+0.4 sin 2π3F t
[b3 = 0.4]
−0.2 cos 2π4F t
[a4 = −0.2]
T/2
T/3
T/4
The Fourier series for u(t) is
P∞
a0
u(t) = 2 + n=1 (an cos 2πnF t + bn sin 2πnF t)
The Fourier coefficients of u(t) are a0 , a1 , · · · and b1 , b2 , · · · .
The nth harmonic of the fundamental is the component at a frequency nF .
E1.10 Fourier Series and Transforms (2014-5379)
Fourier Series: 2 – 3 / 11
Why Sin and Cos Waves?
2: Fourier Series
Periodic Functions
Fourier Series
Why Sin and Cos
Waves?
Dirichlet Conditions
Fourier Analysis
Trigonometric
Products
Fourier Analysis
⊲
Fourier Analysis
Example
Linearity
Summary
Why are engineers obsessed with sine waves?
Answer: Because ...
1. A sine wave remains a sine wave of the same frequency when you
(a) multiply by a constant,
(b) add onto to another sine wave of the same frequency,
(c) differentiate or integrate or shift in time
2. Almost any function can be expressed as a sum of sine waves
◦ Periodic functions → Fourier Series
◦ Aperiodic functions → Fourier Transform
3. Many physical and electronic systems are
(a) composed entirely of constant-multiply/add/differentiate
(b) linear: u(t) → x(t) and v(t) → y(t)
means that u(t) + v(t) → x(t) + y(t)
⇒ sum of sine waves → sum of sine waves
In these lectures we will use T for the fundamental period and F =
the fundamental frequency.
E1.10 Fourier Series and Transforms (2014-5379)
1
T
for
Fourier Series: 2 – 4 / 11
Dirichlet Conditions
2: Fourier Series
Periodic Functions
Fourier Series
Why Sin and Cos
Waves?
Dirichlet
Conditions
Fourier Analysis
Trigonometric
Products
Fourier Analysis
⊲
Fourier Analysis
Example
Linearity
Summary
Not all u(t) can be expressed as a Fourier Series.
Peter Dirichlet derived a set of sufficient conditions.
The function u(t) must satisfy:
• periodic and single-valued
RT
|u(t)| dt < ∞
•
0
• finite number of maxima/minima per period
• finite number of finite discontinuities per
period
Peter Dirichlet
1805-1859
Good:
t2
sin(t)
quantized
Bad:
tan (t)
E1.10 Fourier Series and Transforms (2014-5379)
sin
1
t
∞ halving steps
Fourier Series: 2 – 5 / 11
Fourier Analysis
2: Fourier Series
Periodic Functions
Fourier Series
Why Sin and Cos
Waves?
Dirichlet Conditions
Fourier Analysis
Trigonometric
Products
Fourier Analysis
⊲
Fourier Analysis
Example
Linearity
Summary
Suppose that u(t) satisfies the Dirichlet conditions so that
P
u(t) = a20 + ∞
n=1 (an cos 2πnF t + bn sin 2πnF t)
Question: How do we find an and bn ?
Answer: We use a clever trick that relies on taking averages.
hx(t)i equals the average of x(t) over any integer number of periods:
R
1 T
hx(t)i = T t=0 x(t)dt
Remember, for any integer n, hsin 2πnF ti = 0 (
0 n 6= 0
hcos 2πnF ti =
1 n=0
Finding an and bn is called Fourier analysis.
E1.10 Fourier Series and Transforms (2014-5379)
Fourier Series: 2 – 6 / 11
Trigonometric Products
2: Fourier Series
Periodic Functions
Fourier Series
Why Sin and Cos
Waves?
Dirichlet Conditions
Fourier Analysis
Trigonometric
Products
Fourier Analysis
⊲
Fourier Analysis
Example
Linearity
Summary
sin(x ± y) = sin x cos y ± cos x sin y
⇒ sin x cos y =
1
2
sin(x + y) +
1
2
sin(x − y)
cos(x ± y) = cos x cos y ∓ sin x sin y
⇒ cos x cos y = 12 cos(x + y) + 21 cos(x − y)
sin x sin y = 12 cos(x − y) − 12 cos(x + y)
Set x = 2πmF t, y = 2πnF t (with m + n 6= 0) and take time-averages:
• hsin (2πmF t) cos (2πnF t)i
= 21 sin(2π (m + n) F t) + 12 sin(2π (m − n) F t) = 0
• hcos (2πmF t) cos (2πnF t)i
=
1
2
cos(2π (m + n) F t) +
1
2
cos(2π (m − n) F t) =
(
0
1
2
m 6= n
m=n
cos(2π (m + n) F t) =
(
0
m 6= n
m=n
• hsin (2πmF t) sin (2πnF t)i
=
1
2
cos(2π (m − n) F t) −
1
2
Summary: hsin cosi = 0
hsin sini = hcos cosi =
E1.10 Fourier Series and Transforms (2014-5379)
1
2
1
2
[provided that m + n 6= 0]
if m = n or otherwise = 0.
Fourier Series: 2 – 7 / 11
[Trigonometric Products Proofs]
Proof that cos x cos y =
1
2
cos(x + y) +
1
2
cos(x − y)
We know that
cos(x + y) = cos x cos y − sin x sin y
cos(x − y) = cos x cos y + sin x sin y
Adding these two gives
cos(x + y) + cos(x − y) = 2 cos x cos y
From which: cos x cos y = 12 cos(x + y) + 21 cos(x − y)
Subtracting instead of adding gives: sin x sin y = 12 cos(x − y) −
Proof that
1
2
cos(2π (m + n) F t) +
1
2
1
2
cos(x + y)
(
0 m 6= n
cos(2π (m − n) F t) = 1
m=n
2
We are assuming that m and n are integers with m + n 6= 0 and we use the result that hcos 2πf ti is
zero unless f = 0 in which case hcos 2π0ti = 1 . The frequency of the first term, cos(2π (m + n) F t),
is (m + n) F which is definitely non-zero (because of our assumption that m + n 6= 0) and so the
average of this cosine wave is zero. The frequency of the second term is (m − n) F and this is zero
only if m = n. So it follows that the entire expression is zero unless m = n in which case the second
term gives a value of 12 . Since m and n are integers, we can take the averages over a time interval T
and be sure that this includes an integer number of periods for both terms.
E1.10 Fourier Series and Transforms (2014-5379)
Fourier Series: 2 – note 1 of slide 7
Fourier Analysis
a0
2
P∞
2: Fourier Series
Periodic Functions
Fourier Series
Why Sin and Cos
Waves?
Dirichlet Conditions
Fourier Analysis
Trigonometric
Products
Fourier Analysis
Find an and bn in u(t) =
Fourier Analysis
Example
Linearity
Summary
Proof [an , n > 0]:
2 hu(t) cos (2πnF t)i
P∞
a0
= 2 2 cos (2πnF t) + r=1 2 har cos (2πrF t) cos (2πnF t)i
P
+ ∞
r=1 2 hbr sin (2πrF t) cos (2πnF t)i
⊲
+
(an cos 2πnF t + bn sin 2πnF t)
R
2 T
Answer: an = 2 hu(t) cos (2πnF t)i , T 0 u(t) cos (2πnF t) dt
R
2 T
bn = 2 hu(t) sin (2πnF t)i , T 0 u(t) sin (2πnF t) dt
n=1
Proof [a0 ]: 2 hu(t) cos (2π0F t)i = 2 hu(t)i = 2 ×
Term 1: 2
a0
2
a0
2
= a0
cos (2πnF t) = 0
(
an r = n
Term 2: 2 har cos (2πrF t) cos (2πnF t)i =
0
r 6= n
P∞
⇒ r=1 2 har cos (2πrF t) cos (2πnF t)i = an
Term 3: 2 hbr sin 2πrF t cos (2πnF t)i = 0
Proof [bn , n > 0]: same method as for an
E1.10 Fourier Series and Transforms (2014-5379)
Fourier Series: 2 – 8 / 11
Fourier Analysis Example
2: Fourier Series
Periodic Functions
Fourier Series
Why Sin and Cos
Waves?
Dirichlet Conditions
Fourier Analysis
Trigonometric
Products
Fourier Analysis
Fourier Analysis
Example
Linearity
Summary
⊲
Truncated Series:
P
uN (t) = a20 + N
n=1 (an cos 2πnF t + bn sin 2πnF t)
Pulse: T = 20, width W = T4 , height A = 8
R
2 T
an = T 0 u(t) cos 2πnt
T dt
R
W
= T2 0 A cos 2πnt
T dt
2AT
2πnt W
= 2πnT sin T 0
A
A
nπ
= nπ
sin 2πnW
=
sin
T
nπ
2
R
2 T
bn = T 0 u(t) sin 2πnt
T dt
2AT
2πnt W
= 2πnT − cos T 0
A
nπ
= nπ 1 − cos 2
n
0
1
2
3
4
5
6
10
5
0
0
5
10
10
15
20
25
15
20
25
15
20
25
15
20
25
15
20
25
15
20
25
15
20
25
N=0
5
0
0
5
10
10
N=1
5
0
0
5
10
10
N=2
5
0
0
5
10
10
N=5
5
0
0
5
10
10
N=10
5
an
bn
4
8
π
8
π
0
16
2π
−8
3π
8
3π
0
0
8
5π
8
5π
0
16
6π
0
0
10
10
N=20
5
0
0
E1.10 Fourier Series and Transforms (2014-5379)
5
5
10
Fourier Series: 2 – 9 / 11
[Small Angle Approximation]
In the previous example, we can obtain a0 by noting that a20 = hu(t)i, the average value of the
waveform which must be AW
= 2. From this, a0 = 4. We can, however, also derive this value from
T
the general expression.
A
The expression for am is am = nπ
sin nπ
. For the case, n = 0, this is difficult to evaluate because both
2
the numerator and denominator are zero. The general way of dealing with this situation is L’Hôpital’s
rule (see section 4.7 of RHB) but here we can use a simpler and very useful technique that is often
referred to as the “small angle approximation”. For small values of θ we can approximate the standard
trigonometrical functions as: sin θ ≈ θ, cos θ ≈ 1 − 0.5θ 2 and tan θ ≈ θ. These approximations are
obtained by taking the first three terms of the Taylor series; they are accurate to O(θ 3 ) and are exactly
correct when θ = 0. When m = 0 we can therefore make an exact approximation to an by writing
A
A
sin nπ
≈ nπ
× nπ
= A
= 4. What we have implicitly done here is to assume that n is a
an = nπ
2
2
2
real number (instead of an integer) and then taken the limit of an as n → 0.
E1.10 Fourier Series and Transforms (2014-5379)
Fourier Series: 2 – note 1 of slide 9
Linearity
2: Fourier Series
Periodic Functions
Fourier Series
Why Sin and Cos
Waves?
Dirichlet Conditions
Fourier Analysis
Trigonometric
Products
Fourier Analysis
Fourier Analysis
Example
Linearity
Summary
⊲
Fourier analysis maps a function of time onto a set of Fourier coefficients:
u(t) → {an , bn }
This mapping is linear which means:
(1) For any γ, if u(t) → {an , bn } then γu(t) → {γan , γbn }
(2) If u(t) → {an , bn } and u′ (t) → {a′n , b′n } then
(u(t) + u′ (t)) → {an + a′n , bn + b′n }
Proof for an : (proof for bn is similar)
R
2 T
(1) If T 0 u(t) cos (2πnF t) dt = an , then
R
2 T
cos (2πnF t) dt
T 0 (γu(t))
R
T
= γ T2 0 u(t) cos (2πnF t) dt = γan
R
2 T
(2) If T 0 u(t) cos (2πnF t) dt = an and
R
2 T ′
′
u
(t)
cos
(2πnF
t)
dt
=
a
n then
T 0
R
2 T
+ u′ (t)) cos (2πnF t) dt
T 0 (u(t)
R
R
2 T ′
2 T
= T 0 u(t) cos (2πnF t) dt + T 0 u (t) cos (2πnF t) dt
= an + a′n
E1.10 Fourier Series and Transforms (2014-5379)
Fourier Series: 2 – 10 / 11
Summary
2: Fourier Series
Periodic Functions
Fourier Series
Why Sin and Cos
Waves?
Dirichlet Conditions
Fourier Analysis
Trigonometric
Products
Fourier Analysis
• Fourier Series: P
u(t) = a20 + ∞
n=1 (an cos 2πnF t + bn sin 2πnF t)
Fourier Analysis
Example
Linearity
Summary
• Fourier Analysis = “finding an and bn ”
⊲
• Dirichlet Conditions: sufficient conditions for FS to exist
◦ Periodic, Single-valued, Bounded absolute integral
◦ Finite number of (a) max/min and (b) finite discontinuities
◦ an = 2 hu(t) cos (2πnF t)i
R
2 T
, T 0 u(t) cos (2πnF t) dt
◦ bn = 2 hu(t) sin (2πnF t)i
R
2 T
, T 0 u(t) sin (2πnF t) dt
• The mapping u(t) → {an , bn } is linear
For further details see RHB 12.1 and 12.2.
E1.10 Fourier Series and Transforms (2014-5379)
Fourier Series: 2 – 11 / 11
3: Complex
⊲ Fourier Series
Euler’s Equation
Complex Fourier
Series
Averaging Complex
Exponentials
Complex Fourier
Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd Symmetry
3: Complex Fourier Series
Antiperiodic ⇒ Odd
Harmonics Only
Symmetry Examples
Summary
E1.10 Fourier Series and Transforms (2014-5543)
Complex Fourier Series: 3 – 1 / 12
Euler’s Equation
3: Complex Fourier
Series
Euler’s Equation
Euler’s Equation: eiθ = cos θ + i sin θ
Complex Fourier
Series
Averaging Complex
Exponentials
Hence: cos θ =
⊲
Complex Fourier
Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd Symmetry
Antiperiodic ⇒ Odd
Harmonics Only
Symmetry Examples
Summary
sin θ =
eiθ +e−iθ
2
iθ
e −e−iθ
2i
[see RHB 3.3]
= 12 eiθ + 21 e−iθ
= − 12 ieiθ + 21 ie−iθ
Most maths becomes simpler if you use eiθ instead of cos θ and sin θ
The Complex Fourier Series is the Fourier Series but written using eiθ
Examples where using eiθ makes things simpler:
Using eiθ
Using cos θ and sin θ
ei(θ+φ) = eiθ eiφ
cos (θ + φ) = cos θ cos φ − sin θ sin φ
eiθ eiφ = ei(θ+φ)
d iθ
dθ e
cos θ cos φ =
= ieiθ
E1.10 Fourier Series and Transforms (2014-5543)
1
2
d
dθ
cos (θ + φ) +
1
2
cos (θ − φ)
cos θ = − sin θ
Complex Fourier Series: 3 – 2 / 12
Complex Fourier Series
3: Complex Fourier
Series
Euler’s Equation
Complex Fourier
Series
Averaging Complex
Exponentials
⊲
Complex Fourier
Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd Symmetry
Antiperiodic ⇒ Odd
Harmonics Only
Symmetry Examples
Summary
Fourier Series: u(t) =
a0
2
+
P∞
n=1
(an cos 2πnF t + bn sin 2πnF t)
Substitute: cos θ = 21 eiθ + 21 e−iθ
u(t) =
=
=
a0
2
a0
2
+
P∞
+
n=1
an
P∞
1 iθ
2e
+
1 −iθ
2e
+
sin θ = − 12 ieiθ + 12 ie−iθ
bn − 21 ieiθ
1
1
i2πnF t
a
−
n
2
2 ibn e
−i2πnF t 1
1
2 an + 2 ibn e
n=1
P∞
+ n=1
P∞
i2πnF t
U
e
n
n=−∞
where

1
1

 2 an − 2 ibn
Un = 12 a0

1
1
a
+
|n|
2
2 ib|n|
and
+
1 −iθ
2 ie
[θ = 2πnF t]
[b0 , 0]
n≥1
n=0
n ≤ −1
⇔
U±n =
1
2
a|n| ∓ ib|n|
∗
The Un are normally complex except for U0 and satisfy Un = U−n
Complex Fourier Series: u(t) =
E1.10 Fourier Series and Transforms (2014-5543)
P∞
i2πnF t
U
e
n
n=−∞
[simpler ,]
Complex Fourier Series: 3 – 3 / 12
Averaging Complex Exponentials
3: Complex Fourier
Series
Euler’s Equation
Complex Fourier
Series
Averaging Complex
Exponentials
⊲
Complex Fourier
Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd Symmetry
Antiperiodic ⇒ Odd
Harmonics Only
Symmetry Examples
Summary
If x(t) has period
T
n
for some integer n (i.e. frequency
R
1 T
hx(t)i , T t=0 x(t)dt
n
T
= nF ):
This is the average over an integer number of cycles.
For a complex exponential:
ei2πnF t = hcos (2πnF t) + i sin (2πnF t)i
= hcos (2πnF t)i + i hsin (2πnF t)i
(
1 + 0i n = 0
=
0 + 0i n 6= 0
Hence:
ei2πnF t =
(
1 n=0
0 n=
6 0
E1.10 Fourier Series and Transforms (2014-5543)
[,]
Complex Fourier Series: 3 – 4 / 12
Complex Fourier Analysis
P∞
Un ei2πnF t
3: Complex Fourier
Series
Euler’s Equation
Complex Fourier Series: u(t) =
Complex Fourier
Series
Averaging Complex
Exponentials
To find the coefficient, Un , we multiply by something that makes all the
terms involving the other coefficients average to zero.
P∞
−i2πnF t
i2πrF t −i2πnF t
u(t)e
=
e
r=−∞ Ur e
P∞
i2π(r−n)F t
=
r=−∞ Ur e
P∞
= r=−∞ Ur ei2π(r−n)F t
Complex Fourier
⊲ Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd Symmetry
Antiperiodic ⇒ Odd
Harmonics Only
Symmetry Examples
Summary
n=−∞
All terms in the sum are zero, except for the one when n = r which equals
Un :
Un = u(t)e−i2πnF t
[,]
This shows that the Fourier series coefficients are unique: you cannot have
two different sets of coefficients that result in the same function u(t).
Note the sign of the exponent: “+” in the Fourier Series but “−” for
Fourier Analysis (in order to cancel out the “+”).
E1.10 Fourier Series and Transforms (2014-5543)
Complex Fourier Series: 3 – 5 / 12
Fourier Series ↔ Complex Fourier Series
3: Complex Fourier
Series
Euler’s Equation
a0
2
P∞
u(t) =
+ n=1 (an cos 2πnF t + bn sin 2πnF t)
P
i2πnF t
= ∞
U
e
n
n=−∞
Complex Fourier
Series
Averaging Complex
Exponentials
We can easily convert between the two forms.
Complex Fourier
Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd Symmetry
Fourier Coefficients → Complex Fourier Coefficients:
1
U±n = 2 a|n| ∓ ib|n|
⊲
Antiperiodic ⇒ Odd
Harmonics Only
Symmetry Examples
Summary
∗
]
[Un = U−n
Complex Fourier Coefficients → Fourier Coefficients:
an = Un + U−n = 2ℜ (Un )
bn = i (Un − U−n ) = −2ℑ (Un )
[ℜ = “real part”]
[ℑ = “imaginary part”]
The formula for an works even for n = 0.
E1.10 Fourier Series and Transforms (2014-5543)
Complex Fourier Series: 3 – 6 / 12
[Complex functions of time]
In these lectures, we are assuming that u(t) is a periodic real-valued function of time. In this case we
can represent u(t) using either the Fourier Series or the Complex Fourier Series:
u(t) =
a0
2
+
P∞
n=1 (an cos 2πnF t + bn sin 2πnF t) =
P∞
n=−∞
Un ei2πnF t
We have seen that the Un coefficients are complex-valued and that Un and U−n are complex conjugates
so that we can write U−n = Un∗ .
In fact, the complex Fourier series can also be used when u(t) is a complex-valued function of time
(this is sometimes
in the fields of communications and signal processing). In this case, it is still
Puseful
∞
true that u(t) = n=−∞ Un ei2πnF t , but now Un and U−n are completely independent and normally
U−n 6= Un∗ .
E1.10 Fourier Series and Transforms (2014-5543)
Complex Fourier Series: 3 – note 1 of slide 6
Complex Fourier Analysis Example
3: Complex Fourier
Series
Euler’s Equation
T = 20, width W =
Complex Fourier
Series
Averaging Complex
Exponentials
Method 1:
U±n = 21 an ∓ i 12 bn
Complex Fourier
Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd Symmetry
⊲
Antiperiodic ⇒ Odd
Harmonics Only
Symmetry Examples
Summary
T
4,
height A = 8
10
W
5
A
0
0
Method 2:
Un = u(t)e−i2πnF t
R
1 T
= T 0 u(t)e−i2πnF t dt
R
1 W
= T 0 Ae−i2πnF t dt
−i2πnF t W
A
= −i2πnF T e
0
A
−i2πnF W
= i2πn 1 − e
=
=
=
Ae−iπnF W
eiπnF W
i2πn
Ae−iπnF W
sin (nπF W )
nπ
−i nπ
8
nπ
4
nπ sin 4 e
E1.10 Fourier Series and Transforms (2014-5543)
−iπnF W
−e
n
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
5
10
an
15
20
bn
4
8
π
−8
3π
8
π
16
2π
8
3π
0
0
8
5π
8
5π
16
6π
0
0
25
Un
8
i 6π
4
4
5π + i 5π
0
−4
4
+
i
3π
3π
8
i 2π
4
4
+
i
π
π
2
−4
4
π +i π
i −8
2π
−4
−4
+
i
3π
3π
0
4
−4
+
i
5π
5π
−8
i 6π
Complex Fourier Series: 3 – 7 / 12
Time Shifting
P∞
Un ei2πnF t
3: Complex Fourier
Series
Euler’s Equation
Complex Fourier Series: u(t) =
Complex Fourier
Series
Averaging Complex
Exponentials
If v(t) is the same as u(t) but delayed by a time τ : v(t) = u(t − τ )
i2πnF t
P∞
P∞
i2πnF (t−τ )
−i2πnF τ
v(t) = n=−∞ Un e
= n=−∞ Un e
e
P∞
= n=−∞ Vn ei2πnF t
Complex Fourier
Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd Symmetry
⊲
Antiperiodic ⇒ Odd
Harmonics Only
Symmetry Examples
Summary
n=−∞
where Vn = Un e−i2πnF τ
Example:
u(t) = 6 cos (2πF t)
Fourier: a1 = 6, b1 = 0
Complex: U±1 = 21 a1 ∓ 12 ib1 = 3
v(t) = 6 sin (2πF t) = u(t − τ )
Time delay: τ = T4 ⇒ F τ = 14
−i π
Complex: V1 = U1 e 2 = −3i
π
V−1 = U−1 ei 2 = +3i
5
u(t)
0
-5
0
0.5
1
1.5
2
1
1.5
2
5
v(t)
0
-5
0
0.5
Note: If u(t) is a sine wave, U1 equals half the corresponding phasor.
E1.10 Fourier Series and Transforms (2014-5543)
Complex Fourier Series: 3 – 8 / 12
Even/Odd Symmetry
3: Complex Fourier
Series
Euler’s Equation
Complex Fourier
Series
Averaging Complex
Exponentials
Complex Fourier
Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd
⊲ Symmetry
Antiperiodic ⇒ Odd
Harmonics Only
Symmetry Examples
Summary
∗
]
(1) u(t) real-valued ⇔ Un conjugate symmetric [Un = U−n
(2) u(t) even [u(t) = u(−t)]
⇔ Un even [Un = U−n ]
(3) u(t) odd [u(t) = −u(−t)] ⇔ Un odd [Un = −U−n ]
∗
= U−n ]
(1)+(2) u(t) real & even ⇔ Un real & even [Un = U−n
∗
= −U−n ]
(1)+(3) u(t) real & odd ⇔ Un imaginary & odd [Un = U−n
Proof of (2): u(t) even ⇒ Un even
R
1 T
U−n = T 0 u(t)e−i2π(−n)F t dt
R
1 −T
= T x=0 u(−x)e−i2πnF x (−dx)
R
1 0
= T x=−T u(−x)e−i2πnF x dx
R
1 0
= T x=−T u(x)e−i2πnF x dx = Un
Proof of (3): u(t) odd ⇒ Un odd
Same as before, except for the last line:
R
1 0
= T x=−T −u(x)e−i2πnF x dx = −Un
E1.10 Fourier Series and Transforms (2014-5543)
[substitute x = −t]
[reverse the limits]
[even: u(−x) = u(x)]
[odd: u(−x) = −u(x)]
Complex Fourier Series: 3 – 9 / 12
Antiperiodic ⇒ Odd Harmonics Only
3: Complex Fourier
Series
Euler’s Equation
Complex Fourier
Series
Averaging Complex
Exponentials
Complex Fourier
Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd Symmetry
Antiperiodic ⇒
Odd Harmonics
Only
Symmetry Examples
Summary
⊲
A waveform, u(t), is anti-periodic if u(t + 21 T ) = −u(t).
If u(t) is anti-periodic then Un = 0 for n even.
Proof:
Define v(t) = u(t +
T
2 ),
then
(1) v(t) = −u(t) ⇒ Vn = −Un
(2) v(t) equals u(t) but delayed by − T2
(
⇒ Vn = Un ei2πnF
T
2
= Un einπ =
Un
−Un
n even
n odd
Hence for n even: Vn = −Un = Un ⇒ Un = 0
Example:
U0:5 = [0, 3 + 2i, 0, i, 0, 1]
Odd harmonics only ⇔
Second half of each period is the
negative of the first half.
E1.10 Fourier Series and Transforms (2014-5543)
U[0:5]=[0, 3+2j, 0, j, 0, 1]
5
0
-5
-1
-0.5
0
0.5
1
Complex Fourier Series: 3 – 10 / 12
Symmetry Examples
3: Complex Fourier
Series
Euler’s Equation
∗
.
All these examples assume that u(t) is real-valued ⇔ U−n = U+n
Complex Fourier
Series
Averaging Complex
Exponentials
(1) Even u(t) ⇔ real Un
Complex Fourier
Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd Symmetry
Antiperiodic ⇒ Odd
Harmonics Only
Symmetry
Examples
Summary
⊲
U0:2 = [0, 2, −1]
U[0:2]=[0, 2, -1]
2
0
-2
-4
-6
-1
(2) Odd u(t) ⇔ imaginary Un
U0:3 = [0, −2i, i, i]
-0.5
0
0.5
1
U[0:3]=[0, -2j, j, j]
5
0
-5
(3) Anti-periodic u(t)
⇔ odd harmonics only
U0:1 = [0, −i]
(4) Even harmonics only
⇔ period is really 12 T
U0:4 = [2, 0, 2, 0, 1]
E1.10 Fourier Series and Transforms (2014-5543)
-1
-0.5
0
U[0:1]=[0, -j]
0.5
1
-1
-0.5
0
0.5
1
2
0
-2
U[0:4]=[2, 0, 2, 0, 1]
8
6
4
2
0
-1
-0.5
0
0.5
1
Complex Fourier Series: 3 – 11 / 12
Summary
3: Complex Fourier
Series
Euler’s Equation
Complex Fourier
Series
Averaging Complex
Exponentials
Complex Fourier
Analysis
Fourier Series ↔
Complex Fourier
Series
Complex Fourier
Analysis Example
Time Shifting
Even/Odd Symmetry
Antiperiodic ⇒ Odd
Harmonics Only
Symmetry Examples
Summary
⊲
• Fourier Series: P
u(t) = a20 + ∞
n=1 (an cos 2πnF t + bn sin 2πnF t)
P∞
• Complex Fourier Series: u(t) = n=−∞ Un ei2πnF t
R
1 T
−i2πnF t
◦ Un = u(t)e
, T 0 u(t)e−i2πnF t dt
∗
◦ Since u(t) is real-valued, Un = U−n
◦ FS→CFS: U±n = 21 a|n| ∓ i 21 b|n|
◦ CFS→FS: an = Un + U−n
bn = i (Un − U−n )
• u(t) real and even ⇔ u(−t) = u(t)
⇔ Un is real-valued and even ⇔ bn = 0 ∀n
• u(t) real and odd ⇔ u(−t) = −u(t)
⇔ Un is purely imaginary and odd ⇔ an = 0 ∀n
• u(t) anti-periodic ⇔ u(t + T2 ) = −u(t)
⇔ odd harmonics only ⇔ a2n = b2n = U2n = 0 ∀n
For further details see RHB 12.3 and 12.7.
E1.10 Fourier Series and Transforms (2014-5543)
Complex Fourier Series: 3 – 12 / 12
4: Parseval’s
Theorem and
Convolution
Parseval’s Theorem
(a.k.a. Plancherel’s
Theorem)
Power Conservation
Magnitude Spectrum
and Power Spectrum
Product of Signals
⊲
Convolution
Properties
Convolution Example
Convolution and
Polynomial
Multiplication
Summary
4: Parseval’s Theorem and Convolution
E1.10 Fourier Series and Transforms (2014-5543)
Parseval and Convolution: 4 – 1 / 9
Parseval’s Theorem (a.k.a. Plancherel’s Theorem)
4: Parseval’s
Theorem and
Convolution
Parseval’s
Theorem (a.k.a.
Plancherel’s
Theorem)
Power Conservation
Magnitude Spectrum
and Power Spectrum
Product of Signals
⊲
Convolution
Properties
Convolution Example
Convolution and
Polynomial
Multiplication
Summary
Suppose we have two signals with the same period, T = F1 ,
P
i2πnF t
u(t) = ∞
n=−∞ Un e
P∞
∗
[u(t) = u∗ (t) if real]
⇒ u (t) = n=−∞ Un∗ e−i2πnF t
P∞
v(t) = n=−∞ Vn ei2πnF t
Now multiply u∗ (t) and v(t) together and take the average over [0, T ].
[Use different “dummy variables”, n and m, so they don’t get mixed up]
P∞
P∞
∗
∗ −i2πnF t
i2πmF t
hu (t)v(t)i =
n=−∞ Un e
m=−∞ Vm e
P∞
P∞
∗
= n=−∞ Un m=−∞ Vm e−i2πnF t ei2πmF t
P∞
P∞
∗
= n=−∞ Un m=−∞ Vm ei2π(m−n)F t
The quantity h· · · i equals 1 if m = n and 0 otherwise, so the only non-zero
element in the second sum is when m = n, so the second sum equals Vn .
P∞
∗
Hence Parseval’s theorem: hu (t)v(t)i = n=−∞ Un∗ Vn
D
E P
P∞
2
∞
∗
|u(t)| = n=−∞ Un Un = n=−∞ |Un |2
If v(t) = u(t) we get:
E1.10 Fourier Series and Transforms (2014-5543)
Parseval and Convolution: 4 – 2 / 9
[Manipulating sums]
If you have a multiplicative expression involving two or more sums, then you must use different dummy
variables for each of the sums:
P
P
af
(n)
m bg(m)
n
(1) You can always move any quantities to the right
P
P
P
P
a
bg(m)
=
af
(n)
m bf (n)g(m)
n
m
n
P P
= n m abf (n)g(m)
(2) You can move quantities to the left past a summation provided that they do not involve the dummy
variable of the summation:
P
P
P P
af
(n)
abf
(n)g(m)
=
m bg(m)
n
n
m
P
P
6= n af (n)g(m) m b
P
The last expression doesn’t make sense in any case since m is undefined outside m
(3) You can swap the summation order if the sum converges absolutely
P P
P P
P P
provided
that
h(n,
m)
h(n,
m)
=
n
m |h(n, m)| < ∞
m
n
n
m
The equality on the left is not necessarily true if the sum does not converge absolutely. Of course,
if the sum has only a finite number of terms, it is bound to converge absolutely.
E1.10 Fourier Series and Transforms (2014-5543)
Parseval and Convolution: 4 – note 1 of slide 2
Power Conservation
⊲
Convolution
Properties
Convolution Example
Convolution and
Polynomial
Multiplication
Summary
D
E
The average power of a periodic signal is given by Pu , |u(t)| .
This is the average electrical power that would be dissipated if the
signal represents the voltage across a 1 Ω resistor.
D
E P
2
∞
2
Parseval’s Theorem: Pu = |u(t)| = n=−∞ |Un |
P∞
2
= |U0 | + 2 n=1 |Un |2
[assume u(t) real]
P∞
n
]
= 14 a20 + 12 n=1 a2n + b2n
[U+n = an −ib
2
Parseval’s theorem ⇒ the average power in u(t) is equal to the sum of the
average powers in each of its Fourier components.
Example:
2
u(t) = 2 + 2 cos 2πF t + 4 sin 2πF t − 2 sin 6πF t
E
2
1
2
2
2
|u(t)| = 4 + 2 2 + 4 + (−2) = 16
u(t)
D
U[0:3]=[2, 1-2j, 0, j]
U[0:3]=[2, 1-2j, 0, j]
8
6
4
2
0
-2
-4
u2(t)
4: Parseval’s
Theorem and
Convolution
Parseval’s Theorem
(a.k.a. Plancherel’s
Theorem)
Power
Conservation
Magnitude Spectrum
and Power Spectrum
Product of Signals
-1
-0.5
0
Time (s)
U0:3 = [2, 1 − 2i, 0, i]
E1.10 Fourier Series and Transforms (2014-5543)
0.5
1
⇒
60
40
20
0
Pu=<u2>=16
-1
|U0 |2 + 2
-0.5
0
Time (s)
P∞
0.5
1
2
|U
|
= 16
n
n=1
Parseval and Convolution: 4 – 3 / 9
Magnitude Spectrum and Power Spectrum
4: Parseval’s
Theorem and
Convolution
Parseval’s Theorem
(a.k.a. Plancherel’s
Theorem)
Power Conservation
Magnitude
Spectrum and
Power Spectrum
Product of Signals
⊲
Convolution
Properties
Convolution Example
Convolution and
Polynomial
Multiplication
Summary
The spectrum of a periodic signal is the values of {Un } versus nF .
n q
o
The magnitude spectrum is the values of {|Un |} = 21 a2|n| + b2|n| .
n
o n o
2
The power spectrum is the values of |Un | = 41 a2|n| + b2|n| .
Example:
u(t) = 2 + 2 cos 2πF t + 4 sin 2πF t − 2 sin 6πF t
Fourier Coefficients: a0:3 = [4, 2, 0, 0]
b1:3 = [4, 0, −2]
Spectrum: U−3:3 = [−i, 0, 1 + 2i, 2, 1 − 2i, 0, i]
√
√
Magnitude Spectrum: |U−3:3 | = 1, 0, 5, 2, 5, 0, 1
P
2
= [1, 0, 5, 4, 5, 0, 1]
Power Spectrum: U−3:3
[ = u2 (t) ]
5
1
0
Σ=16
|U2n|
|Un|
2
-3
-2
-1
0
1
Frequency (Hz)
2
3
0
-3
-2
-1
0
1
Frequency (Hz)
2
3
The magnitude and power spectra of a real periodic signal are symmetrical.
2
A one-sided power power spectrum shows U0 and then 2 |Un | for n ≥ 1.
E1.10 Fourier Series and Transforms (2014-5543)
Parseval and Convolution: 4 – 4 / 9
Product of Signals
4: Parseval’s
Theorem and
Convolution
Parseval’s Theorem
(a.k.a. Plancherel’s
Theorem)
Power Conservation
Magnitude Spectrum
and Power Spectrum
Product of Signals
⊲
Convolution
Properties
Convolution Example
Convolution and
Polynomial
Multiplication
Summary
Suppose we have two signals with the same period, T = F1 ,
P
i2πnF t
u(t) = ∞
n=−∞ Un e
P∞
v(t) = m=−∞ Vn ei2πmF t
P∞
If w(t) = u(t)v(t) then Wr = m=−∞ Ur−m Vm , Ur ∗ Vr
Proof:
P∞
P∞
i2πnF t
i2πmF t
w(t) = u(t)v(t) = n=−∞ Un e
m=−∞ Vm e
P∞
P∞
= n=−∞ m=−∞ Un Vm ei2π(m+n)F t
Now we change the summation variable to use r instead of n:
r =m+n⇒n=r−m
This is a one-to-one mapping: every pair (m, n) in the range ±∞
corresponds to exactly one pair (m, r) in the same range.
P∞
P∞
P∞
i2πrF t
w(t) = r=−∞ m=−∞ Ur−m Vm e
= r=−∞ Wr ei2πrF t
P∞
where Wr = m=−∞ Ur−m Vm , Ur ∗ Vr .
Wr is the sum of all products Un Vm for which m + n = r.
The spectrum Wr = Ur ∗ Vr is called the convolution of Ur and Vr .
E1.10 Fourier Series and Transforms (2014-5543)
Parseval and Convolution: 4 – 5 / 9
Convolution Properties
4: Parseval’s
Theorem and
Convolution
Parseval’s Theorem
(a.k.a. Plancherel’s
Theorem)
Power Conservation
Magnitude Spectrum
and Power Spectrum
Product of Signals
Convolution
⊲ Properties
Convolution Example
Convolution and
Polynomial
Multiplication
Summary
Convolution behaves algebraically like multiplication:
1) Commutative: Ur ∗ Vr = Vr ∗ Ur
2) Associative: Ur ∗ Vr ∗ Wr = (Ur ∗ Vr ) ∗ Wr = Ur ∗ (Vr ∗ Wr )
3) Distributive over addition: Wr ∗(
(Ur + Vr ) = Wr ∗ Ur + Wr ∗ Vr
1 r=0
4) Identity Element or “1”: If Ir =
, then Ir ∗ Ur = Ur
0 r 6= 0
Proofs: (all sums are over ±∞)
1) Substitute
for m:Pn = r − m ⇔ m = r − n
P
m Ur−m Vm =
n Un Vr−n
[1 ↔ 1 for any r]
2) Substitute for n: k = r + m − n ⇔ n = r + m − k
P P
P P
n ((
m Un−m Vm ) Wr−n ) =
k ((
m Ur−k Vm ) Wk−m )
P P
P
P
= k m Ur−k Vm Wk−m = k (Ur−k ( m Vm Wk−m ))
P
P
P
3) m Wr−m (Um + Vm ) = m Wr−m Um + m Wr−m Vm
P
4) Ir−m Um = 0 unless m = r. Hence m Ir−m Um = Ur .
E1.10 Fourier Series and Transforms (2014-5543)
[1 ↔ 1]
Parseval and Convolution: 4 – 6 / 9
Convolution Example
Convolution and
Polynomial
Multiplication
Summary
-1
0
Time (s)
1
2
0
w(t)
0
-50
-2
-1
0
Time (s)
1
2
-2
|Vn|
2
5
-1
0
1
Frequency (Hz)
-1
0
Time (s)
1
-2
-1
0
1
Frequency (Hz)
2
2
20
1
0
[0 = V0 ]
50
|Wn|
-2
4
2
0
-2
-4
10
|Un|
⊲
Convolution
Example
10
0
Convolution
Properties
v(t)
u(t) = 10 + 8 sin 2πt
v(t) = 4 cos 6πt
U−1:1 = [4i, 10, −4i] V−3:3 = [2, 0, 0, 0, 0, 0, 2]
u(t)
4: Parseval’s
Theorem and
Convolution
Parseval’s Theorem
(a.k.a. Plancherel’s
Theorem)
Power Conservation
Magnitude Spectrum
and Power Spectrum
Product of Signals
-3
-2
-1
0
1
Frequency (Hz)
2
3
10
0
-4
-3
3
4
w(t) = u(t)v(t) = (10 + 8 sin 2πt) 4 cos 6πt
= 40 cos 6πt + 32 sin 2πt cos 6πt
= 40 cos 6πt + 16 sin 8πt − 16 sin 4πt
W−4:4 = [8i, 20, −8i, 0, 0, 0, 8i, 20, −8i]
To convolve Un and Vn :
Replace each harmonic in Vn by a scaled copy of the entire {Un }
(or vice versa) and sum the complex-valued coefficients of any
overlapping harmonics.
E1.10 Fourier Series and Transforms (2014-5543)
Parseval and Convolution: 4 – 7 / 9
Convolution and Polynomial Multiplication
4: Parseval’s
Theorem and
Convolution
Parseval’s Theorem
(a.k.a. Plancherel’s
Theorem)
Power Conservation
Magnitude Spectrum
and Power Spectrum
Product of Signals
Convolution
Properties
Convolution Example
Convolution and
Polynomial
Multiplication
Summary
⊲
Two polynomials: u(x) = U3 x3 + U2 x2 + U1 x + U0
v(x) =
V2 x2 + V1 x + V0
Now multiply the two polynomials together:
w(x) = u(x)v(x)
= U3 V2 x5 + (U3 V1 + U2 V2 ) x4 + (U3 V0 + U2 V1 + U1 V2 ) x3
+ (U2 V0 + U1 V1 + U0 V2 ) x2 + (U1 V0 + U0 V1 ) x +U0 V0
The coefficient of xr consists of all the coefficient pair from U and V where
the subscripts add up to r. For example, for r = 3:
P2
W3 = U3 V0 + U2 V1 + U1 V2 = m=0 U3−m Vm
If all the missing coefficients are assumed to be zero, we can write
P∞
Wr = m=−∞ Ur−m Vm , Ur ∗ Vr
So, to multiply two polynomials, you convolve their coefficient sequences.
Actually, the complex Fourier Series is iust a polynomial:
P∞
P∞
i2πnF t
i2πF t n
u(t) = n=−∞ Un e
= n=−∞ Un e
E1.10 Fourier Series and Transforms (2014-5543)
Parseval and Convolution: 4 – 8 / 9
Summary
4: Parseval’s
Theorem and
Convolution
Parseval’s Theorem
(a.k.a. Plancherel’s
Theorem)
Power Conservation
Magnitude Spectrum
and Power Spectrum
Product of Signals
Convolution
Properties
Convolution Example
Convolution and
Polynomial
Multiplication
Summary
⊲
P∞
∗
• Parseval’s Theorem: hu (t)v(t)i = n=−∞ Un∗ Vn
D
E P
2
2
◦ Power Conservation: |u(t)| = ∞
|U
|
n
n=−∞
◦ or in terms of an and bn :
D
E
2
|u(t)| = 14 a20 +
1
2
P∞
2
n=1 an
+
b2n
• Linearity: w(t) = au(t) + bv(t) ⇔ Wn = aUn + bVn
• Product of signals ⇔ Convolution of complex
P∞ Fourier coefficients:
w(t) = u(t)v(t) ⇔ Wn = Un ∗ Vn , m=−∞ Un−m Vm
• Convolution acts like multiplication:
◦ Commutative: U ∗ V = V ∗ U
◦ Associative: U ∗ V ∗ W is unambiguous
◦ Distributes over addition: U ∗ (V + W ) = U ∗ V + U ∗ W
◦ Has an identity: Ir = 1 if r = 0 and = 0 otherwise
• Polynomial multiplication ⇔ convolution of coefficients
For further details see RHB Chapter 12.8.
E1.10 Fourier Series and Transforms (2014-5543)
Parseval and Convolution: 4 – 9 / 9
5: Gibbs
⊲ Phenomenon
Discontinuities
Discontinuous
Waveform
Gibbs Phenomenon
Integration
Rate at which
coefficients decrease
with m
Differentiation
Periodic Extension
t2 Periodic
5: Gibbs Phenomenon
Extension: Method
(a)
t2 Periodic
Extension: Method
(b)
Summary
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – 1 / 11
Discontinuities
Rate at which
coefficients decrease
with m
Differentiation
Periodic Extension
t2 Periodic
Extension: Method
(a)
t2 Periodic
Extension: Method
(b)
Summary
A function, v(t), has a discontinuity of amplitude b at t = a if
lime→0 (v(a + e) − v(a − e)) = b 6= 0
Conversely, v(t), is continuous at t = a if the limit, b, equals zero.
Examples:
b
b
v(t)
⊲
u(t)
5: Gibbs Phenomenon
Discontinuities
Discontinuous
Waveform
Gibbs Phenomenon
Integration
0
0
a–e
a
Time (t)
a+e
Continuous
a–e
a
Time (t)
a+e
Discontinuous
We will see that if a periodic function, v(t), is discontinuous, then its
Fourier series behaves in a strange way.
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – 2 / 11
Discontinuous Waveform
5: Gibbs Phenomenon
Discontinuities
Discontinuous
Waveform
Gibbs Phenomenon
Integration
⊲
Rate at which
coefficients decrease
with m
Differentiation
Periodic Extension
t2 Periodic
Extension: Method
(a)
t2 Periodic
Extension: Method
(b)
Summary
1
F
= 20, width= 12 T , height A = 1
R
1 0.5T
Ae−i2πmF t dt
Um = T 0
−i2πmF t 0.5T
i
= 2πmF T e
0
((−1)m −1)i
i
−iπm
= 2πm e
−1 =
2πm


m 6= 0, even
0
= 0.5 m = 0

 −i
m odd
mπ
Pulse: T =
So, u(t) =
1
2
+ π2 sin 2πF t + 31 sin 6πF t
+ 15 sin 10πF t + . . .
PN
E1.10 Fourier Series and Transforms (2014-5559)
0
0
5
10
1
15
20
max(u0)=0.500
0.5
N=0
0
0
5
10
1
15
20
max(u1)=1.137
0.5
N=1
0
0
5
10
1
15
20
max(u3)=1.100
0.5
N=3
0
0
5
10
1
15
20
max(u5)=1.094
0.5
N=5
0
0
5
10
1
15
20
max(u41)=1.089
0.5
N=41
0
0
i2πmF t
Define: uN (t) = m=−N Um e
uN (0) = 0.5 ∀N
R
1 π sin t
1
maxt uN (t) −→ 2 + π 0 t dt ≈ 1.0895
N →∞
1
0.5
1
5
10
15
20
max(u41)=1.089
0.5
0
-1
-0.5
0
0.5
1
[Enlarged View: u41 (t)]
Gibbs Phenomenon: 5 – 3 / 11
[Powers of −1 and i]
Expressions involving (−1)m or, less commonly, im arise quite frequently and it is worth becoming
familiar with them. They can arise in several guises:
m
e−iπm = eiπm = eiπ
= cos (πm) = (−1)m
1 m
1 πm
i2
e
= im
= ei 2 π
m
1 πm
−i 1
π
−i 2
= (−i)m
= e 2
e
As m increases these expressions repeat with periods of 2 or 4. Simple expressions involving these
quantities make useful sequences.
m
−4
−3
−2
−1
0
1
2
3
4
(−1)m = cos πm = eiπm
1
−1
1
−1
1
−1
1
−1
1
im = ei0.5πm
1
i
−1
−i
1
i
−1
−i
1
(−i)m = e−i0.5πm
1
−i
−1
i
1
−i
−1
i
1
(1 + (−1)m )
1
0
1
0
1
0
1
0
1
(1 − (−1)m )
0
1
0
1
0
1
0
1
0
1
0
−1
0
1
0
−1
0
1
1
0
0
0
1
0
0
0
1
1
2
1
2
1 m
(i + (−i)m ) = cos 0.5πm
2
1
(1 + (−1)m + im + (−i)m )
4
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – note 1 of slide 3
Gibbs Phenomenon
5: Gibbs Phenomenon
Discontinuities
Discontinuous
Waveform
Gibbs Phenomenon
Integration
⊲
Rate at which
coefficients decrease
with m
Differentiation
Periodic Extension
t2 Periodic
Extension: Method
(a)
t2 Periodic
Extension: Method
(b)
Summary
PN
Truncated Fourier Series: uN (t) = m=−N Um ei2πmF t
If u(t) has a discontinuity of height b at t = a then:
(1) uN (a) −→ lime→0
N →∞
u(a−e)+u(a+e)
2
(2) uN (t) has an overshoot of about 9% of b at the discontinuity. For
large N the overshoot moves closer to the discontinuity but does
phenomenon).
In the limit the overshoot
not get smaller (Gibbs
R
π
equals − 21 + π1 0 sint t dt b ≈ 0.0895b.
(3) For large m, the coefficients, Um decrease no faster than |m|
Example:
uN (0) −→ 0.5
N →∞
maxt uN (t) −→ 1.0895 . . .
N →∞


m 6= 0, even
0
Um = 0.5 m = 0

 −i
m odd
mπ
E1.10 Fourier Series and Transforms (2014-5559)
−1
.
1
0.5
0
0
5
10
1
15
20
max(u41)=1.089
0.5
N=41
0
0
1
5
10
15
20
max(u41)=1.089
0.5
0
-1
-0.5
0
0.5
1
Gibbs Phenomenon: 5 – 4 / 11
[Origin of Gibbs Phenomenon]
This topic is included for interest but is not examinable.
Our first goal is to express the partial Fourier series, uN (t), in terms of the original signal, u(t). We
begin by substituting the integral expression for Un in the partial Fourier series
R
P+N
P+N
T
1
−i2πnF
τ
i2πnF
t
dτ ei2πnF t
= n=−N T 0 u(τ )e
uN (t) = n=−N Un e
Now we swap the order of the integration and the finite summation (OK if the integral converges ∀n)
P
RT
+N
1
i2πnF
(t−τ
)
dτ
uN (t) = T 0 u(τ )
n=−N e
Now apply the formula for the sum of a geometric progression with z = ei2πF (t−τ ) :
P+N
−N
z −(N +0.5) −z N +0.5
−z N +1
n = z
=
z
n=−N
1−z
z −0.5 −z 0.5
R
i2π(N +0.5)F (τ −t)
−e−i2π(N +0.5)F (τ −t)
dτ
uN (t) = T1 0T u(τ ) e
ei2π0.5F (τ −t) −e−i2π0.5F (τ −t)
R
sin π(2N +1)F (τ −t)
= T1 0T u(τ )
dτ
sin πF (τ −t)
So if we define the Dirichlet Kernel to be DN (x) =
R
uN (t) = T1 0T u(τ )DN (2πF (τ − t)) dτ
sin((N +0.5)x)
,
sin 0.5x
and set x = 2πF (τ − t), we obtain
So what we have shown is that uN (t) can be obtained by multiplying u(τ ) by a time-shifted Dirichlet
Kernel and then integrating over one period. Next we will look at the properties of the Dirichlet Kernel.
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – note 1 of slide 4
[Dirichlet Kernel]
This topic is included for interest but is not examinable.
PN
P
inx = 1 + 2
e
Dirichlet Kernel definition: DN (x) = +N
n=1 cos nx =
n=−N
sin((N +0.5)x)
sin 0.5x
DN (x) is plotted below for N = {2, 5, 10, 21}. The vertical red lines at ±π mark one period.
0
5
0
-6
-4
-2
0
x
2
4
6
-6
-4
-2
0
x
2
4
40
D10(x)
D21(x)
2
20
D5(x)
D10(x)
D5(x)
D2(x)
10
D2(x)
4
10
0
6
-6
-4
-2
0
x
2
4
D21(x)
20
6
0
-6
-4
-2
0
x
•
Periodic: with period 2π
•
Average value: hDN (x)i =
•
π
π
First Zeros: DN (x) = 0 at x = ± N +0.5
define the main lobe as − N +0.5
<x<
•
Peak value: 2N + 1 at x = 0. The main lobe gets narrower but higher as N increases.
•
Main Lobe semi-integral:
π
π
R N +0.5
R N +0.5
DN (x)dx = x=0
x=0
1
2π
R +π
−π
2
4
6
DN (x)dx = 1
sin((N +0.5)x)
dx
sin 0.5x
=
1
N +0.5
sin y
y
y=0 sin
2N +1
Rπ
π
N +0.5
dy[y = (N + 0.5)x]
where we substituted y = (N + 0.5)x. Now, for large N , we can approximate sin 2Ny+1 ≈ 2Ny+1 :
π
R N +0.5
R π sin y
Rπ
sin y
1
dy
=
2
D
(x)dx
≈
y
N
x=0
y=0 y dy ≈ 3.7038741 ≈ 2π × 0.58949
N +0.5 y=0
2N +1
We see that, for large enough N , the main lobe semi-integral is independent of N .
[In MATLAB DN (x) = (2N + 1) × diric(x, 2N + 1)]
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – note 2 of slide 4
[Gibbs Phenomenon Overshoot]
This topic is included for interest but is not examinable.
1
u41(t)
0.5 T=20
The partial Fourier Series, uN (t), can be obtained by multiplying u(t)
by a shifted Dirichlet Kernel and integrating over one period:
RT
1
uN (t) = T 0 u(τ )DN (2πF (τ − t)) dτ
For the special case when u(t) is a pulse of height 1 and width 0.5T :
R
uN (t) = T1 00.5T DN (2πF (τ − t)) dτ
Substitute x = 2πF (τ − t)
R πF T −2πF t
1
DN (x) dx =
uN (t) = 2πF
T −2πF t
•
•
R π−2πF t
−2πF t
DN (x) dx
For t = 0 (theRblue integral and the blue circle on the upper graph):
π
1
uN (0) = 2π
0 DN (x) dx = 0.5
-0.5
0
0.5
1
1.5
2
80 t=0
60
40
20
0
-20
-1
-0.5
0
0.5
1
1.5
2
80 t=0.24
60
40
20
0
-20
-1
-0.5
0
0.5
1
1.5
2
80 t=0.96
60
40
20
0
-20
-1
-0.5
0
0.5
1
1.5
2
T
2N +1
For t =
(the red integral and the red circle on the upper graph):
π
π
R π− N +0.5
R0
R π− N +0.5
T
1
1
1
DN (x) dx
uN 2N +1 = 2π − π
DN (x) dx= 2π − π DN (x) dx + 2π 0
N +0.5
•
1
2π
0
-1
N +0.5
For large N , we replace the first term by a constant (since it is the semi-integral of the main lobe)
and replace the upper limit of theR second term by π:
π
1
≈ 0.58949 + 2π
0 DN (x) dx = 1.08949
For 0 ≪ t ≪ 0.5T , uN (t) ≈ 1 (the green integral and the green circle on the upper graph).
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – note 3 of slide 4
Integration
5: Gibbs Phenomenon
Discontinuities
Discontinuous
Waveform
Gibbs Phenomenon
Integration
⊲
Rate at which
coefficients decrease
with m
Differentiation
Periodic Extension
t2 Periodic
Extension: Method
(a)
t2 Periodic
Extension: Method
(b)
Summary
P∞
Suppose u(t) = m=−∞ Um ei2πmF t
[boundedness requires U0 = 0]
Define v(t) to be the integral of u(t)
Rt
R t P∞
i2πmF τ
v(t) = u(τ )dτ =
dτ
m=−∞ Um e
R
R t i2πmF τ
P
P∞
[assume OK to swap and ]
= m=−∞ Um e
dτ
P∞
1
= c + m=−∞ Um i2πmF
ei2πmF t
P∞
= c + m=−∞ Vm ei2πmF t where c is an integration constant
Hence Vm =
−i
2πmF
Um except for V0 = c (arbitrary constant)
Example:
Square wave: Um = −2i
mπ for odd m (0 for even m)
−1
−i
× −2i
=
Triangle wave: Vm = 2πmF
2
mπ
π m2 F for odd m (0 for even m)
1
5
u7(t)
0
v7(t)
0
-1
-5
0
5
10
15
20
0
5
10
15
20
1
Um
Convergence: v(t) always converges if u(t) does since Vm ∝ m
vN (t) is a good approximation even for small N
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – 5 / 11
Rate at which coefficients decrease with m
5: Gibbs Phenomenon
Discontinuities
Discontinuous
Waveform
Gibbs Phenomenon
Integration
Rate at which
coefficients
decrease with m
Differentiation
Periodic Extension
t2 Periodic
⊲
Extension: Method
(a)
t2 Periodic
Extension: Method
(b)
Summary
−1
m
for odd m (0 for even m)
Square wave: Um = −2i
π
−2
for odd m (0 for even m)
Triangle wave: Vm = π−1
2F m
1
5
u7(t)
0
v7(t)
0
-1
-5
0
5
10
15
u(t) multiplies the Um by
20
−i
2πF
0
5
10
15
20
Integrating
× m−1 ⇒ they decrease faster.
The rate at which the coefficients, Um , decrease with m depends on the
lowest derivative that has a discontinuity:
• Discontinuity in u(t) itself (e.g. square wave)
−1
For large |m|, Um decreases as |m|
• Discontinuity in u′ (t) (e.g. triangle wave)
For large |m|, Um decreases as |m|−2
• Discontinuity in u(n) (t)
−(n+1)
For large |m|, Um decreases as |m|
• No discontinuous derivatives
For large |m|, Um decreases faster than any power (e.g. e−|m| )
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – 6 / 11
Differentiation
5: Gibbs Phenomenon
Discontinuities
Discontinuous
Waveform
Gibbs Phenomenon
Integration
Rate at which
coefficients decrease
with m
Differentiation
Periodic Extension
t2 Periodic
⊲
Extension: Method
(a)
t2 Periodic
Extension: Method
(b)
Summary
Integration multiplies Um by
−i
2πmF
.
Hence differentiation multiplies Um by
2πmF
−i
= i2πmF
If u(t) is a continuous differentiable function and w(t) = du(t)
dt then,
provided that w(t) satisfies the Dirichlet conditions, its Fourier coefficients
are:
(
0
m=0
Wm =
.
i2πmF Um m 6= 0
Since we are multiplying Um by m the coefficients Wm decrease more
slowly with m and so the Fourier series for w(t) may not converge (i.e.
w(t) may not satisfy the Dirichlet conditions).
d
dt
d
dt
−→
Um ∝ |m|
−2
−→
Um ∝ |m|
−1
Um ∝ |m|
−0
Differentiation makes waveforms spikier and less smooth.
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – 7 / 11
Periodic Extension
5: Gibbs Phenomenon
Discontinuities
Discontinuous
Waveform
Gibbs Phenomenon
Integration
Suppose y(t) is only defined over a finite interval (a, b).
You have two reasonable choices to make a periodic version:
(a) T = b − a,
u(t) = y(t) for a ≤ t < b
(
y(t)
a≤t≤b
(b) T = 2(b − a), u(t) =
y(2b − t) b ≤ t ≤ 2b − a
Rate at which
coefficients decrease
with m
Differentiation
Periodic Extension
2
t Periodic
⊲
Extension: Method
(a)
t2 Periodic
Extension: Method
(b)
Summary
Example:
y(t) = t2 for 0 ≤ t < 2
4
4
4
2
2
2
0
0
-2
0
2
4
y(t)
0
-2
0
2
4
(a) T = 2
-2
0
2
4
(b) T = 4
Option (b) has twice the period, no discontinuities, no Gibbs phenomenon
overshoots and if y(t) is continuous the coefficients decrease at least as fast
−2
as |m| .
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – 8 / 11
t2 Periodic Extension: Method (a)
5: Gibbs Phenomenon
Discontinuities
Discontinuous
Waveform
Gibbs Phenomenon
Integration
Rate at which
coefficients decrease
with m
Differentiation
Periodic Extension
t2 Periodic
Extension: Method
⊲ (a)
t2 Periodic
Extension: Method
(b)
Summary
y(t) = t2 for 0 ≤ t < 2
Method (a): T = F1 = 2
R
1 T 2 −i2πmF t
Um = T 0 t e
dt
h
2te−i2πmF t
1 t2 e−i2πmF t
= T −i2πmF − (−i2πmF
+
)2
Substitute e−i2πmF 0 = e−i2πmF T = 1
i
h
2
2T
T
= T1 −i2πmF
− (−i2πmF
)2
=
4
2i
πm
+
4
4
3
[for integer m]
4
K=3
2
0
4
K=6
2
0
2
0
t2 dt =
2
2
0
iT
2e−i2πmF t
(−i2πmF )3 0
RT
π 2 m2
K=1
-2
U0 =
1
T
0
-2
0
2
4
-2
0
2
4
U0:3 = [1.333, 0.203 + 0.637i, 0.051 + 0.318i, 0.023 + 0.212i]
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – 9 / 11
t2 Periodic Extension: Method (b)
5: Gibbs Phenomenon
Discontinuities
Discontinuous
Waveform
Gibbs Phenomenon
Integration
Rate at which
coefficients decrease
with m
Differentiation
Periodic Extension
t2 Periodic
Extension: Method
(a)
t2 Periodic
Extension: Method
⊲ (b)
Summary
y(t) = t2 for 0 ≤ t < 2
Method (b): T = F1 = 4
R
1 0.5T
Um = T −0.5T t2 e−i2πmF t dt
h 2 −i2πmF t
2te−i2πmF t
1 t e
= T −i2πmF − (−i2πmF )2 +
m
Substitute e±iπmF T = e±iπm = (−1)
h
i
(−1)m
−2T
= T
(−i2πmF )2
=
4
(−1)m T 2
2π 2 m2
=
2
4
K=3
0
2
4
4
3
[for integer m]
K=6
2
0
-2
−0.5T
t2 dt =
[all even powers of t cancel out]
2
0
2e
(−i2πmF )3 −0.5T
R 0.5T
(−1)m 8
π 2 m2
4
K=1
U0 =
i0.5T
−i2πmF t
1
T
0
-2
0
2
4
-2
0
2
4
U0:3 = [1.333, −0.811, 0.203, −0.090]
[u(t) real+even ⇒ Um real]
Convergence is noticeably faster than for method (a)
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – 10 / 11
Summary
5: Gibbs Phenomenon
Discontinuities
Discontinuous
Waveform
Gibbs Phenomenon
Integration
Rate at which
coefficients decrease
with m
Differentiation
Periodic Extension
t2 Periodic
Extension: Method
(a)
t2 Periodic
Extension: Method
(b)
Summary
⊲
• Discontinuity at t = a
◦ Gibbs phenomenon: uN (t) overshoots by 9% of iump
◦ uN (a) → mid point of iump
Rt
−i
Um
• Integration: If v(t) = u(τ )dτ , then Vm = 2πmF
and V0 = c, an arbitrary constant. U0 must be zero.
• Differentiation: If w(t) = du(t)
dt , then Wm = i2πmF Um provided
w(t) satisfies Dirichlet conditions (it might not)
• Rate of decay:
−(k+1)
◦ For large n, Un decreases at a rate |n|
where
the first discontinuous
derivative
D
E P
2
◦ Error power: (u(t) − uN (t)) = |n|>N |Un |2
dk u(t)
dtk
is
• Periodic Extension of finite domain signal of length L
◦ (a) Repeat indefinitely with period T = L
◦ (b) Reflect alternate repetitions for period T = 2L
no discontinuities or Gibbs phenomenon
For further details see RHB Chapter 12.4, 12.5, 12.6
E1.10 Fourier Series and Transforms (2014-5559)
Gibbs Phenomenon: 5 – 11 / 11
6: Fourier
⊲ Transform
Fourier Series as
T →∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta Function
Dirac Delta Function:
Scaling and
Translation
Dirac Delta Function:
Products and
Integrals
Periodic Signals
Duality
6: Fourier Transform
Time Shifting and
Scaling
Gaussian Pulse
Summary
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform: 6 – 1 / 12
Fourier Series as T → ∞
6: Fourier Transform
Fourier Series as
T → ∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta Function
Dirac Delta Function:
Scaling and
Translation
Dirac Delta Function:
Products and
Integrals
Periodic Signals
Duality
⊲
Time Shifting and
Scaling
Gaussian Pulse
Summary
Fourier Series: u(t) =
P∞
n=−∞
Un ei2πnF t
The harmonic frequencies are nF ∀n and are spaced F =
1
T
apart.
As T gets larger, the harmonic spacing becomes smaller.
e.g. T = 1 s ⇒ F = 1 Hz
T = 1 day ⇒ F = 11.57 µHz
If T → ∞ then the harmonic spacing becomes zero, the sum becomes an
integral and we get the Fourier Transform:
R +∞
u(t) = f =−∞ U (f )ei2πf t df
Here, U (f ), is the spectral density of u(t).
•
•
•
•
U (f ) is a continuous function of f .
U (f ) is complex-valued.
u(t) real ⇒ U (f ) is conjugate symmetric ⇔ U (−f ) = U (f )∗ .
Units: if u(t) is in volts, then U (f )df must also be in volts
⇒ U (f ) is in volts/Hz (hence “spectral density”).
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform: 6 – 2 / 12
Fourier Transform
6: Fourier Transform
Fourier Series as
T →∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta Function
Dirac Delta Function:
Scaling and
Translation
Dirac Delta Function:
Products and
Integrals
Periodic Signals
Duality
⊲
Time Shifting and
Scaling
Gaussian Pulse
Summary
P∞
Fourier Series:
u(t) = n=−∞ Un ei2πnF t
The summation is over a set of equally spaced frequencies
fn = nF where the spacing between them is ∆f = F = T1 .
R 0.5T
−i2πnF t
Un = u(t)e
= ∆f t=−0.5T u(t)e−i2πnF t dt
Spectral Density: If u(t) has finite energy, Un → 0 as ∆f → 0. So we
Un
, on the set of frequencies {fn }:
define a spectral density, U (fn ) = ∆f
R 0.5T
Un
U (fn ) = ∆f = t=−0.5T u(t)e−i2πfn t dt
we can writeP
[Substitute Un = U (fn )∆f ]
i2πfn t
u(t) = ∞
∆f
n=−∞ U (fn )e
Fourier Transform: Now if we take the limit as ∆f → 0, we get
R∞
u(t) = −∞ U (f )ei2πf t df
[Fourier Synthesis]
R∞
U (f ) = t=−∞ u(t)e−i2πf t dt
[Fourier Analysis]
Un
remains
For non-periodic signals Un → 0 as ∆f → 0 and U (fn ) = ∆f
finite. However, if u(t) contains an exactly periodic component, then the
corresponding U (fn ) will become infinite as ∆f → 0. We will deal with it.
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform: 6 – 3 / 12
Fourier Transform Examples
u(t)
a=2
0.5
|U(f)|
0
-5
0.5
0.4
0.3
0.2
0.1
-5
0
Time (s)
5
0
Frequency (Hz)
5
0
Frequence (Hz)
5
0.5
0
-0.5
-5
1
a+i2πf
1
a=2
0.5
Example 2:
0
-5
0
5
−a|t|
Time (s)
v(t) = e
1
R∞
0.5
V (f ) = −∞ v(t)e−i2πf t dt
0
-5
0
5
Frequency (Hz)
R ∞ −at −i2πf t
R0
at −i2πf t
dt
dt + 0 e e
= −∞ e e
(a−i2πf )t 0
(−a−i2πf )t ∞
1
−1
= a−i2πf e
+ a+i2πf e
−∞
0
1
2a
1
= a−i2πf + a+i2πf = a2 +4π2 f 2
[v(t) real+symmetric
⇒ V (f ) real+symmetric]
|V(f)|
Time Shifting and
Scaling
Gaussian Pulse
Summary
1
<U(f) (rad/pi)
⊲
Example(1:
e−at t ≥ 0
u(t) =
0
t<0
R∞
U (f ) = −∞ u(t)e−i2πf t dt
R ∞ −at −i2πf t
dt
= 0 e e
R ∞ (−a−i2πf )t
dt
= 0 e
(−a−i2πf )t ∞
−1
= a+i2πf e
=
0
v(t)
6: Fourier Transform
Fourier Series as
T →∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta Function
Dirac Delta Function:
Scaling and
Translation
Dirac Delta Function:
Products and
Integrals
Periodic Signals
Duality
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform: 6 – 4 / 12
Dirac Delta Function
6: Fourier Transform
Fourier Series as
T →∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta
⊲ Function
Dirac Delta Function:
Scaling and
Translation
Dirac Delta Function:
Products and
Integrals
Periodic Signals
Duality
Time Shifting and
Scaling
Gaussian Pulse
Summary
We define a unit area pulse of width w as
(
1
−0.5w ≤ x ≤ 0.5w
w
dw (x) =
0 otherwise
This pulse has the property that its integral equals
1 for all values of w:
R∞
d (x)dx = 1
x=−∞ w
δ 0.2(x)
4
δ 0.5(x)
2
δ 3(x)
0
-3
-2
-1
0
x
1
2
3
1
2
3
1
δ (x)
0.5
0
-3
-2
-1
0
x
If we make w smaller, the pulse height increases to preserve unit area.
We define the Dirac delta function as δ(x) = limw→0 dw (x)
• δ(x) equals zero everywhere except at x = 0 where it is infinite.
R∞
• However its area still equals 1 ⇒ −∞ δ(x)dx = 1
• We plot the height of δ(x) as its area rather than its true height of ∞.
δ(x) is not quite a proper function: it is called a generalized function.
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform: 6 – 5 / 12
Dirac Delta Function: Scaling and Translation
6: Fourier Transform
Fourier Series as
T →∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta Function
Dirac Delta
Function: Scaling
and Translation
Dirac Delta Function:
Products and
Integrals
Periodic Signals
Duality
⊲
Time Shifting and
Scaling
Gaussian Pulse
Summary
Translation: δ(x − a)
δ(x) is a pulse at x = 0
δ(x − a) is a pulse at x = a
bδ(x)R has an area of b since
R∞
∞
(bδ(x)) dx = b −∞ δ(x)dx = b
−∞
δ (x-2)
0
-0.5δ(x+2)
-1
-3
Amplitude Scaling: bδ(x)
R∞
δ(x) has an area of 1 ⇔ −∞ δ(x)dx = 1
δ (x)
1
-2
-1
0
x
1
2
3
1
δ (4x) = 0.25δ (x)
0
-1
-3
-3δ(-4x-8) = -0.75δ(x+2)
-2
-1
0
1
2
x
3
b can be a complex number (on a graph, we then plot only its magnitude)
Time Scaling: δ(cx)
R∞
R∞
c > 0: x=−∞ δ(cx)dx = y=−∞ δ(y) dy
c
R
1 ∞
1
= c y=−∞ δ(y)dy = 1c = |c|
R −∞
R∞
c < 0: x=−∞ δ(cx)dx = y=+∞ δ(y) dy
c
R
+∞
= −1
δ(y)dy = −1
c
c =
y=−∞
In general, δ(cx) =
E1.10 Fourier Series and Transforms (2014-5559)
1
|c| δ(x)
[sub y = cx]
[sub y = cx]
1
|c|
for c 6= 0
Fourier Transform: 6 – 6 / 12
Dirac Delta Function: Products and Integrals
6: Fourier Transform
Fourier Series as
T →∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta Function
Dirac Delta Function:
Scaling and
Translation
Dirac Delta
Function:
Products and
Integrals
Periodic Signals
Duality
⊲
Time Shifting and
Scaling
Gaussian Pulse
Summary
If we multiply δ(x − a) by a function of x:
y = x2 × δ(x − 2)
The product is 0 everywhere except at x = 2.
So δ(x − 2) is multiplied by the value taken by
x2 at x = 2:
2
2
x × δ(x − 2) = x x=2 × δ(x − 2)
= 4 × δ(x − 2)
6
y=x2
4
2
0
-1
0
1
2
1
2
x
6
y=δ(x-2)
4
2
0
-1
0
x
In general for any function, f (x), that is
continuous at x = a,
f (x)δ(x − a) = f (a)δ(x − a)
Integrals:
R∞
−∞
Example:
6
y=22×δ(x-2) = 4δ(x-2)
4
2
0
-1
0
1
2
x
f (a)δ(x − a)dx
R∞
= f (a) −∞ δ(x − a)dx
= f (a)
[if f (x) continuous at a]
2
2
3x − 2x δ(x − 2)dx = 3x − 2x x=2 = 8
f (x)δ(x − a)dx =
R∞
−∞
R∞
E1.10 Fourier Series and Transforms (2014-5559)
−∞
Fourier Transform: 6 – 7 / 12
Periodic Signals
6: Fourier Transform
Fourier Series as
T →∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta Function
Dirac Delta Function:
Scaling and
Translation
Dirac Delta Function:
Products and
Integrals
Periodic Signals
Duality
⊲
Time Shifting and
Scaling
Gaussian Pulse
Summary
R∞
Fourier Transform: u(t) = −∞ U (f )ei2πf t df
R∞
U (f ) = t=−∞ u(t)e−i2πf t dt
Example: U (f ) = 1.5δ(f + 2) + 1.5δ(f − 2)
R∞
u(t) = −∞ U (f )ei2πf t df
R∞
= −∞ 1.5δ(f + 2)ei2πf t df
R∞
+ −∞ 1.5δ(f − 2)ei2πf t df
i2πf t i2πf t + 1.5 e
= 1.5 e
f =+2
f =−2 = 1.5 ei4πt + e−i4πt = 3 cos 4πt
[Fourier Synthesis]
[Fourier Analysis]
1.5
1.5δ(f+2)
0.5
0
-3
-2
-1
0
1
Frequency (Hz)
2
3
4 3cos(4πt)
2
0
-2
0
1
2
3
Time (s)
If u(t) is periodic then U (f ) is a sum of Dirac delta functions:
P∞
P∞
i2πnF t
u(t) = n=−∞ Un e
⇒ U (f ) = n=−∞ Un δ (f − nF )
R∞
Proof: u(t) = −∞ U (f )ei2πf t df
R ∞ P∞
= −∞ n=−∞ Un δ (f − nF ) ei2πf t df
R∞
P∞
= n=−∞ Un −∞ δ (f − nF ) ei2πf t df
P∞
= n=−∞ Un ei2πnF t
E1.10 Fourier Series and Transforms (2014-5559)
1.5δ(f-2)
1
4
5
Fourier Transform: 6 – 8 / 12
Duality
6: Fourier Transform
Fourier Series as
T →∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta Function
Dirac Delta Function:
Scaling and
Translation
Dirac Delta Function:
Products and
Integrals
Periodic Signals
Duality
⊲
Time Shifting and
Scaling
Gaussian Pulse
Summary
R∞
Fourier Transform: u(t) = −∞ U (f )ei2πf t df
R∞
U (f ) = t=−∞ u(t)e−i2πf t dt
Dual transform:
Suppose v(t) = U (t), then
R∞
V (f ) = t=−∞ v(t)e−i2πf t dτ
R∞
V (g) = t=−∞ U (t)e−i2πgt dt
R∞
= f =−∞ U (f )e−i2πgf df
So:
[Fourier Synthesis]
[Fourier Analysis]
[substitute f = g, v(t) = U (t)]
[substitute t = f ]
= u(−g)
v(t) = U (t) ⇒ V (f ) = u(−f )
Example:
u(t) = e−|t|
v(t) =
2
1+4π 2 t2
E1.10 Fourier Series and Transforms (2014-5559)
⇒
⇒
U (f ) =
2
1+4π 2 f 2
[from earlier]
V (f ) = e−|−f | = e−|f |
Fourier Transform: 6 – 9 / 12
Time Shifting and Scaling
6: Fourier Transform
Fourier Series as
T →∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta Function
Dirac Delta Function:
Scaling and
Translation
Dirac Delta Function:
Products and
Integrals
Periodic Signals
Duality
Time Shifting and
⊲ Scaling
Gaussian Pulse
Summary
R∞
Fourier Transform: u(t) = −∞ U (f )ei2πf t df
R∞
U (f ) = t=−∞ u(t)e−i2πf t dt
[Fourier Synthesis]
[Fourier Analysis]
Time Shifting and Scaling:
Suppose v(t) = u(at + b), then
R∞
V (f ) = t=−∞ u(at + b)e−i2πf t dt
R∞
−i2πf ( τ −b
a ) 1 dτ
= sgn(a) τ =−∞ u(τ )e
a
[now sub τ = at + b]
note that ±∞ limits swap if a < 0 hence sgn(a) =
=
=
b
1 i 2πf
a
|a| e
b
i 2πf
a
1
|a| e
So: v(t) = u(at + b)
E1.10 Fourier Series and Transforms (2014-5559)
R∞
τ =−∞
U
−i2π fa τ
u(τ )e
(
1
a>0
−1 a < 0
dτ
f
a
⇒ V (f ) =
b
i 2πf
a
1
|a| e
U
f
a
Fourier Transform: 6 – 10 / 12
Gaussian Pulse
2
t
− 2σ
2
√ 1
e
2πσ 2
Gaussian Pulse: u(t) =
R∞
This is a Normal (or Gaussian) probability distribution, so −∞ u(t)dt = 1.
R ∞ − t2 −i2πf t
R∞
1
−i2πf t
dt = √2πσ2 −∞ e 2σ2 e
U (f ) = −∞ u(t)e
dt
R ∞ − 1 (t2 +i4πσ2 f t)
1
dt
=√
e 2σ2
2πσ 2
−∞
2
2
− 2σ1 2 t2 +i4πσ 2 f t+(i2πσ 2 f ) −(i2πσ 2 f )
R∞
√ 1
e
2πσ 2 −∞
2
R ∞ − 1 (t+i2πσ2 f )2
2
1
i2πσ
f
1
)
(
2
√
2σ 2
= e 2σ
dt
e
2
−∞
2πσ
2
1
(i)
(i2πσ2 f )
− 21 (2πσf )2
2σ 2
=
Time Shifting and
Scaling
Gaussian Pulse
Summary
⊲
u(t)
0.4
σ=1
-2
0
Time (s)
1/(2πσ)=0.159
1
0.2
0
-4
dt
=e
=e
[(i) uses a result from complex analysis theory that:
R ∞ − 1 (t+i2πσ2 f )2
R ∞ − 1 t2
1
1
2
√
√
2σ
2σ 2
dt
=
dt = 1]
e
e
2
2
−∞
−∞
2πσ
2πσ
|U(f)|
6: Fourier Transform
Fourier Series as
T →∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta Function
Dirac Delta Function:
Scaling and
Translation
Dirac Delta Function:
Products and
Integrals
Periodic Signals
Duality
2
4
0.5
0
-0.6
-0.4
-0.2
0
0.2
Frequency (Hz)
0.4
0.6
Uniquely, the Fourier Transform of a Gaussian pulse is a Gaussian pulse.
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform: 6 – 11 / 12
Summary
6: Fourier Transform
Fourier Series as
T →∞
Fourier Transform
Fourier Transform
Examples
Dirac Delta Function
Dirac Delta Function:
Scaling and
Translation
Dirac Delta Function:
Products and
Integrals
Periodic Signals
Duality
Time Shifting and
Scaling
Gaussian Pulse
Summary
⊲
• Fourier Transform:
R∞
◦ Inverse transform (synthesis): u(t) = −∞ U (f )ei2πf t df
R∞
◦ Forward transform (analysis): U (f ) = t=−∞ u(t)e−i2πf t dt
⊲ U (f ) is the spectral density function (e.g. Volts/Hz)
• Dirac Delta Function:
R∞
◦ δ(t) is a zero-width infinite-height pulse with −∞ δ(t)dt = 1
R∞
◦ Integral: −∞ f (t)δ(t − a) = f (a)
1
|c| δ(t)
P
u(t) = ∞
n=−∞
◦ Scaling: δ(ct) =
• Periodic Signals:
Un ei2πnF t
P∞
⇒ U (f ) = n=−∞ Un δ (f − nF )
• Fourier Transform Properties:
◦ v(t) = U (t)
⇒ V (f ) = u(−f )
b
1 i 2πf
◦ v(t) = u(at + b)
⇒ V (f ) = |a| e a U fa
◦ v(t) =
− 21 ( σt )
√ 1
e
2πσ 2
2
1
⇒ V (f ) = e− 2 (2πσf )
2
For further details see RHB Chapter 13.1 (uses ω instead of f )
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform: 6 – 12 / 12
7: Fourier
Transforms:
Convolution and
Parseval’s
Theorem
Multiplication of
Signals
⊲
Multiplication
Example
Convolution Theorem
Convolution Example
Convolution
Properties
Parseval’s Theorem
Energy Conservation
Energy Spectrum
Summary
7: Fourier Transforms: Convolution and
Parseval’s Theorem
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform - Parseval and Convolution: 7 – 1 / 10
Multiplication of Signals
7: Fourier
Transforms:
Convolution and
Parseval’s Theorem
Multiplication of
Signals
⊲
Multiplication
Example
Convolution Theorem
Convolution Example
Convolution
Properties
Parseval’s Theorem
Energy Conservation
Energy Spectrum
Summary
Question: What is the Fourier transform of w(t) = u(t)v(t) ?
R +∞
R +∞
i2πht
dh
and
v(t) = g=−∞ V (g)ei2πgt dg
Let u(t) = h=−∞ U (h)e
[Note use of different dummy variables]
w(t) = u(t)v(t)
R +∞
R +∞
i2πht
dh g=−∞ V (g)ei2πgt dg
= h=−∞ U (h)e
R +∞
R +∞
= h=−∞ U (h) g=−∞ V (g)ei2π(h+g)t dg dh
[merge e(··· ) ]
Now we make a change of variable in the second integral: g = f − h
R +∞
R +∞
= h=−∞ U (h) f =−∞ V (f − h)ei2πf t df dh
R
R +∞
R∞
i2πf t
dh df
[swap ]
= f =−∞ h=−∞ U (h)V (f − h)e
R +∞
= f =−∞ W (f )ei2πf t df
R +∞
R +∞
where W (f ) = h=−∞ U (h)V (f − h)dh h=−∞ U (h)V (f − h)dh ,
U (f ) ∗ V (f )
This is the convolution of the two spectra U (f ) and V (f ).
w(t) = u(t)v(t)
E1.10 Fourier Series and Transforms (2014-5559)
⇔
W (f ) = U (f ) ∗ V (f )
Fourier Transform - Parseval and Convolution: 7 – 2 / 10
Multiplication Example
u(t)
v(t) = cos 2πF t
V (f ) = 0.5 (δ(f + F ) + δ(f − F ))
|U(f)|
[from before]
w(t) = u(t)v(t)
W (f ) = U (f ) ∗ V (f )
0.5
= a+i2π(f
+F ) +
0.5
0.4
0.3
0.2
0.1
-5
E1.10 Fourier Series and Transforms (2014-5559)
0
Time (s)
5
0
Frequency (Hz)
5
0
Frequence (Hz)
5
0.5
0
-0.5
-5
0
-1
0.5
a+i2π(f −F )
|W(f)|
<W(f) (rad/pi)
If V (f ) consists entirely of Dirac impulses
then U (f ) ∗ V (f ) iust replaces each impulse
with a complete copy of U (f ) scaled by the
area of the impulse and shifted so that 0 Hz
lies on the impulse. Then add the
overlapping complex spectra.
-5
1
w(t)
Convolution
Properties
Parseval’s Theorem
Energy Conservation
Energy Spectrum
Summary
1
a+i2πf
a=2
0.5
0
<U(f) (rad/pi)
⊲
U (f ) =
1
t≥0
t<0
v(t)
Multiplication
Example
Convolution Theorem
Convolution Example
(
e−at
u(t) =
0
|V(f)|
7: Fourier
Transforms:
Convolution and
Parseval’s Theorem
Multiplication of
Signals
-5
0.4 F=2
0.2
0
-5
1
a=2, F=2
0.5
0
-0.5
-5
0.2
0.1
0
-5
0
Time (s)
0.5
5
0.5
0
Frequency (Hz)
0
Time (s)
5
5
0
Frequency (Hz)
5
0
Frequence (Hz)
5
0.5
0
-0.5
-5
Fourier Transform - Parseval and Convolution: 7 – 3 / 10
Convolution Theorem
7: Fourier
Transforms:
Convolution and
Parseval’s Theorem
Multiplication of
Signals
Multiplication
Example
Convolution
Theorem
Convolution Example
⊲
Convolution
Properties
Parseval’s Theorem
Energy Conservation
Energy Spectrum
Summary
Convolution Theorem:
w(t) = u(t)v(t) ⇔ W (f ) = U (f ) ∗ V (f )
w(t) = u(t) ∗ v(t) ⇔ W (f ) = U (f )V (f )
Convolution in the time domain is equivalent to multiplication in the
frequency domain and vice versa.
Proof of second line:
Given u(t), v(t) and w(t) satisfying
w(t) = u(t)v(t) ⇔ W (f ) = U (f ) ∗ V (f )
define dual waveforms x(t), y(t) and z(t) as follows:
x(t) = U (t) ⇔ X(f ) = u(−f )
y(t) = V (t) ⇔ Y (f ) = v(−f )
z(t) = W (t) ⇔ Z(f ) = w(−f )
Now the convolution property becomes:
w(−f ) = u(−f )v(−f ) ⇔ W (t) = U (t) ∗ V (t)
Z(f ) = X(f )Y (f ) ⇔ z(t) = x(t) ∗ y(t)
E1.10 Fourier Series and Transforms (2014-5559)
[duality]
[sub t ↔ ±f ]
[duality]
Fourier Transform - Parseval and Convolution: 7 – 4 / 10
Convolution Example
⊲
Convolution
Properties
Parseval’s Theorem
Energy Conservation
Energy Spectrum
Summary
w(t) = u(t) ∗ v(t)
R∞
= −∞ u(τ )v(t − τ )dτ
R min(t,1)
(1 − τ )eτ −t dτ
= 0
min(t,1)
u(t)
1
0.5
0
-2
0
2
Time t (s)
4
6
-2
0
2
Time t (s)
4
6
-2
0
2
Time t (s)
4
6
1
v(t)
Multiplication
Example
Convolution Theorem
Convolution
Example
(
1−t 0≤t<1
u(t) =
0
otherwise
(
e−t t ≥ 0
v(t) =
0
t<0
0.5
0
w(t)
7: Fourier
Transforms:
Convolution and
Parseval’s Theorem
Multiplication of
Signals
0.3
0.2
0.1
0
1
∫ = 0.307
0.5
0
= [(2 − τ ) eτ −t ]τ =0


t<0
0
= 2 − t − 2e−t 0 ≤ t < 1


(e − 2) e−t
t≥1
1
-2
∫ = 0.16
0.5
0
1
-2
∫ = 0.059
0.5
0
-2
u(τ)
0
u(τ)
0
2
Time τ (s)
4
2
Time τ (s)
4
v(2.5-τ)
2
Time τ (s)
6
t=1.5
v(1.5-τ)
u(τ)
0
t=0.7
v(0.7-τ)
4
6
t=2.5
6
Note how v(t − τ ) is time-reversed (because of the −τ ) and time-shifted to
put the time origin at τ = t.
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform - Parseval and Convolution: 7 – 5 / 10
Convolution Properties
7: Fourier
Transforms:
Convolution and
Parseval’s Theorem
Multiplication of
Signals
Multiplication
Example
Convolution Theorem
Convolution Example
Convolution
⊲ Properties
Parseval’s Theorem
Energy Conservation
Energy Spectrum
Summary
Convloution: w(t) = u(t) ∗ v(t) ,
R∞
−∞
u(τ )v(t − τ )dτ
Convolution behaves algebraically like multiplication:
1) Commutative: u(t) ∗ v(t) = v(t) ∗ u(t)
2) Associative:
u(t) ∗ v(t) ∗ w(t) = (u(t) ∗ v(t)) ∗ w(t) = u(t) ∗ (v(t) ∗ w(t))
3) Distributive over addition:
w(t) ∗ (u(t) + v(t)) = w(t) ∗ u(t) + w(t) ∗ v(t)
4) Identity Element or “1”: u(t) ∗ δ(t) = δ(t) ∗ u(t) = u(t)
5) Bilinear: (au(t)) ∗ (bv(t)) = ab (u(t) ∗ v(t))
Proof: In the frequency domain, convolution is multiplication.
Also, if u(t) ∗ v(t) = w(t), then
6) Time Shifting: u(t + a) ∗ v(t + b) = w(t + a + b)
1
w(at)
7) Time Scaling: u(at) ∗ v(at) = |a|
How to recognise a convolution integral:
the arguments of u(· · · ) and v(· · · ) sum to a constant.
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform - Parseval and Convolution: 7 – 6 / 10
Parseval’s Theorem
7: Fourier
Transforms:
Convolution and
Parseval’s Theorem
Multiplication of
Signals
Multiplication
Example
Convolution Theorem
Convolution Example
Lemma:
R
X(f ) = δ(f − g) ⇒ x(t) = δ(f − g)ei2πf t df = ei2πgt
R i2π(g−f )t
R i2πgt −i2πf t
dt = δ(g − f )
⇒ X(f ) = e
e
dt = e
Parseval’s Theorem:
Convolution
Properties
Parseval’s
⊲ Theorem
Energy Conservation
Energy Spectrum
Summary
R∞
∗
u (t)v(t)dt =
t=−∞
Proof:
R +∞
Let u(t) = f =−∞ U (f )ei2πf t df
and
R +∞
∗
U
(f )V (f )df
f =−∞
v(t) =
R +∞
i2πgt
V
(g)e
dg
g=−∞
[Note use of different dummy variables]
Now multiply u∗ (t) = u(t) and v(t) together and integrate over time:
R∞
∗
u
(t)v(t)dt
t=−∞
R +∞
R +∞
R∞
∗
−i2πf t
df g=−∞ V (g)ei2πgt dgdt
= t=−∞ f =−∞ U (f )e
R∞
R +∞
R +∞
∗
= f =−∞ U (f ) g=−∞ V (g) t=−∞ ei2π(g−f )t dtdgdf
R +∞
R +∞
∗
= f =−∞ U (f ) g=−∞ V (g)δ(g − f )dgdf
[lemma]
R +∞
= f =−∞ U ∗ (f )V (f )df
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform - Parseval and Convolution: 7 – 7 / 10
Energy Conservation
Convolution
Properties
Parseval’s Theorem
Energy
Conservation
Energy Spectrum
Summary
⊲
∗
u (t)v(t)dt =
t=−∞
R +∞
∗
U
(f )V (f )df
f =−∞
For the special case v(t) = u(t), Parseval’s theorem becomes:
R +∞
R∞
∗
∗
U
(f )U (f )df
u
(t)u(t)dt
=
f =−∞
t=−∞
R +∞
R∞
2
⇒ Eu = t=−∞ |u(t)| dt = f =−∞ |U (f )|2 df
Energy Conservation: The energy in u(t) equals the energy in U (f ).
Example: (
e−at
u(t) =
0
t≥0
t<0
⇒ Eu =
R
2
|u(t)| dt =
1
U (f ) = a+i2πf
[from before]
R
R
2
df
⇒
|U (f )| df = a2 +4π
2f 2
−1 2πf ∞
tan ( a )
1
π
=
=
=
2πa
2πa
2a
h
−e−2at
2a
1
u(t)
Multiplication
Example
Convolution Theorem
Convolution Example
Parseval’s Theorem:
R∞
i∞
0
=
1
2a
a=2
0.5
0
|U(f)|
7: Fourier
Transforms:
Convolution and
Parseval’s Theorem
Multiplication of
Signals
0.5
0.4
0.3
0.2
0.1
-5
-5
0
Time (s)
0
Frequency (Hz)
5
5
−∞
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform - Parseval and Convolution: 7 – 8 / 10
Convolution
Properties
Parseval’s Theorem
Energy Conservation
Energy Spectrum
Summary
⊲
W (f ) =
0.5
a+i2π(f +F )
=
+
0.5
a+i2π(f −F )
a+i2πf
|W(f)|
<W(f) (rad/pi)
Multiplication
Example
Convolution Theorem
Convolution Example
Example from before:
(
e−at cos 2πF t t ≥ 0
w(t) =
0
t<0
1
a=2, F=2
0.5
0
-0.5
-5
0.25
0.2
0.15
0.1
0.05
-5
|W (f )|2 =
2
2
a +4π f
2
(a2 −4π 2 (f 2 −F 2 ))2 +16π 2 a2 f 2
0
Time (s)
5
0
Frequency (Hz)
5
0
Frequence (Hz)
5
0
Frequency (Hz)
5
0.5
0
-0.5
-5
a2 +i4πaf −4π 2 (f 2 −F 2 )
|W(f)| 2
7: Fourier
Transforms:
Convolution and
Parseval’s Theorem
Multiplication of
Signals
w(t)
Energy Spectrum
0.06
0.04
0.02
0
-5
Energy Spectrum
2
• The units
R ∞of |W (f )|2 are “energy per Hz” so that its integral,
Ew = −∞ |W (f )| df , has units of energy.
2
• The quantity |W (f )| is called the energy spectral density of w(t) at
frequency f and its graph is the energy spectrum of w(t). It shows
how the energy of w(t) is distributed over frequencies.
2
• If you divide |W (f )| by the total energy, Ew , the result is non-negative
and integrates to unity like a probability distribution.
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform - Parseval and Convolution: 7 – 9 / 10
Summary
7: Fourier
Transforms:
Convolution and
Parseval’s Theorem
Multiplication of
Signals
Multiplication
Example
Convolution Theorem
Convolution Example
Convolution
Properties
Parseval’s Theorem
Energy Conservation
Energy Spectrum
Summary
⊲
• Convolution:
R∞
◦ u(t) ∗ v(t) , −∞ u(τ )v(t − τ )dτ
⊲ Arguments of u(· · · ) and v(· · · ) sum to t
◦ Acts like multiplication + time scaling/shifting formulae
• Convolution Theorem: multiplication ↔ convolution
◦ w(t) = u(t)v(t) ⇔ W (f ) = U (f ) ∗ V (f )
◦ w(t) = u(t) ∗ v(t) ⇔ W (f ) = U (f )V (f )
R +∞
R∞
∗
• Parseval’s Theorem: t=−∞ u (t)v(t)dt = f =−∞ U ∗ (f )V (f )df
• Energy Spectrum:
◦ Energy spectral Rdensity: |U (f )|R2 (energy/Hz)
2
2
◦ Parseval: Eu = |u(t)| dt = |U (f )| df
For further details see RHB Chapter 13.1
E1.10 Fourier Series and Transforms (2014-5559)
Fourier Transform - Parseval and Convolution: 7 – 10 / 10
⊲ 8: Correlation
Cross-Correlation
Signal Matching
Cross-corr as
Convolution
Normalized Cross-corr
Autocorrelation
Autocorrelation
example
Fourier Transform
Variants
Scale Factors
Summary
Spectrogram
E1.10 Fourier Series and Transforms (2015-5585)
8: Correlation
Fourier Transform - Correlation: 8 – 1 / 11
Cross-Correlation
8: Correlation
Cross-Correlation
Signal Matching
Cross-corr as
Convolution
Normalized Cross-corr
Autocorrelation
Autocorrelation
example
⊲
Fourier Transform
Variants
Scale Factors
Summary
Spectrogram
The cross-correlation between two signals u(t) and v(t) is
R∞ ∗
w(t) = u(t) ⊗ v(t) , −∞ u (τ )v(τ + t)dτ
R∞ ∗
= −∞ u (τ − t)v(τ )dτ
[sub: τ → τ − t]
The complex conjugate, u∗ (τ ) makes no difference if u(t) is real-valued
but makes the definition work even if u(t) is complex-valued.
Correlation versus Convolution:
R∞ ∗
u(t) ⊗ v(t) = −∞ u (τ )v(τ + t)dτ
[correlation]
R∞
u(t) ∗ v(t) = −∞ u(τ )v(t − τ )dτ
[convolution]
Unlike convolution, the integration variable, τ , has the same sign in the
arguments of u(· · · ) and v(· · · ) so the arguments have a constant
difference instead of a constant sum (i.e. v(t) is not time-flipped).
Notes: (a) The argument of w(t) is called the “lag” (= delay of u versus v).
(b) Some people write u(t) ⋆ v(t) instead of u(t) ⊗ v(t).
(c) Some swap u and v and/or negate t in the integral.
It is all rather inconsistent /.
E1.10 Fourier Series and Transforms (2015-5585)
Fourier Transform - Correlation: 8 – 2 / 11
Signal Matching
u(t)
0.5
0
0
Example 2:
y(t) is the same as v(t) with more noise
z(t) = u(t) ⊗ y(t) can still detect the
correct time delay (hard for humans)
200
400
600
1
v(t)
Example 1:
v(t) contains u(t) with an unknown delay
and added noise.
w(t) = u(t)
R ∗⊗ v(t)
= u (τ − t)v(τ )dt gives a peak
at the time lag where u(τ − t) best
matches v(τ ); in this case at t = 450
0.5
0
0
200
400
600
800
0
200
400
600
800
0
200
400
600
800
0
200
400
600
800
0
200
400
600
800
0
200
400
600
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20
w(t)
Fourier Transform
Variants
Scale Factors
Summary
Spectrogram
1
10
0
y(t)
⊲
Cross correlation is used to find where two
signals match: u(t) is the test waveform.
2
0
-2
20
10
z(t)
8: Correlation
Cross-Correlation
Signal Matching
Cross-corr as
Convolution
Normalized Cross-corr
Autocorrelation
Autocorrelation
example
0
-10
E1.10 Fourier Series and Transforms (2015-5585)
0
-1
0
q(t)
Example 3:
p(t) contains −u(t) so that
q(t) = u(t) ⊗ p(t) has a negative peak
p(t)
1
-10
-20
Fourier Transform - Correlation: 8 – 3 / 11
Cross-correlation as Convolution
8: Correlation
Cross-Correlation
Signal Matching
Cross-corr as
Convolution
Normalized Cross-corr
Autocorrelation
Autocorrelation
example
⊲
Fourier Transform
Variants
Scale Factors
Summary
Spectrogram
Correlation: w(t) = u(t) ⊗ v(t) =
R∞
∗
u
(τ − t)v(τ )dτ
−∞
If we define x(t) = u∗ (−t) then
R∞ ∗
R∞
x(t) ∗ v(t) , −∞ x(t − τ )v(τ )dτ = −∞ u (τ − t)v(τ )dτ
= u(t) ⊗ v(t)
Fourier Transform of x(t):
R∞ ∗
R∞
−i2πf t
dt = −∞ u (−t)e−i2πf t dt
X(f ) = −∞ x(t)e
∗
R
R∞ ∗
∞
= −∞ u (t)ei2πf t dt = −∞ u(t)e−i2πf t dt
= U ∗ (f )
So w(t) = x(t) ∗ v(t)
⇒ W (f ) = X(f )V (f ) = U ∗ (f )V (f )
Hence the Cross-correlation theorem:
w(t) = u(t) ⊗ v(t)
⇔
= u∗ (−t) ∗ v(t)
W (f ) = U ∗ (f )V (f )
Note that, unlike convolution, correlation is not associative or commutative:
v(t) ⊗ u(t) = v ∗ (−t) ∗ u(t) = u(t) ∗ v ∗ (−t) = w∗ (−t)
E1.10 Fourier Series and Transforms (2015-5585)
Fourier Transform - Correlation: 8 – 4 / 11
Normalized Cross-correlation
8: Correlation
Cross-Correlation
Signal Matching
Cross-corr as
Convolution
Normalized
Cross-corr
Autocorrelation
Autocorrelation
example
⊲
Fourier Transform
Variants
Scale Factors
Summary
Spectrogram
Correlation: w(t) = u(t) ⊗ v(t) =
R∞
∗
u
(τ − t)v(τ )dτ
−∞
If we define y(t) = u(t − t0 ) for some fixed t0 , then Ey = Eu :
R∞
R∞
2
2
Ey = −∞ |y(t)| dt = −∞ |u(t − t0 )| dt
R∞
= −∞ |u(τ )|2 dτ = Eu
[t → τ + t0 ]
Cauchy-Schwarz inequality:
2
⇒ |w(t0 )| =
R∞
R∞
−∞
∗
y (τ )v(τ )dτ
∗
≤ Ey Ev
2
≤ Ey Ev = Eu Ev
√
but t0 was arbitrary, so we must have |w(t)| ≤ Eu Ev for all t
−∞
u (τ − t0 )v(τ )dτ
2
We can define the normalized cross-correlation
z(t) =
u(t)⊗v(t)
√
Eu Ev
with properties: (1) |z(t)| ≤ 1 for all t
(2) |z(t0 )| = 1 ⇔ v(τ ) = αu(τ − t0 ) with α constant
E1.10 Fourier Series and Transforms (2015-5585)
Fourier Transform - Correlation: 8 – 5 / 11
[Cauchy-Schwarz Inequality Proof]
You do not need to memorize this proof
We want to prove the Cauchy-Schwarz Inequality:
R∞
−∞
u∗ (t)v(t)dt
2
≤ Eu Ev
where Eu ,
R∞
−∞
|u(t)|2 dt.
R∞ ∗
Suppose we define w , −∞
u (t)v(t)dt. Then,
R
[|· · · |2 always ≥ 0]
0 ≤ |Ev u(t) − w∗ v(t)|2 dt
R
= (Ev u∗ (t) − wv ∗ (t)) (Ev u(t) − w∗ v(t)) dt
[|z|2 = z ∗ z]
R
R
R
R
= Ev2 u∗ (t)u(t)dt + |w|2 v ∗ (t)v(t)dt − w∗ Ev u∗ (t)v(t)dt − wEv u(t)v ∗ (t)dt
R
2
2R
2
|v(t)|2 dt − Ev w∗ w − Ev ww∗
= Ev |u(t)| dt + |w|
[definition of w]
= Ev2 Eu + |w|2 Ev − 2 |w|2 Ev = Ev Eu Ev − |w|2
[|z|2 = z ∗ z]
Unless Ev = 0 (in which case, v(t) ≡ 0 and the C-S inequality is true), we must have |w|2 ≤ Eu Ev
which proves the C-S inequality.
Also, Eu Ev = |w|2 only if we have equality in the first line,
R
that is, |Ev u(t) − w∗ v(t)|2 dt = 0 which implies that the integrand is zero for all t.
This implies that u(t) =
w∗
v(t).
Ev
So we have shown that Eu Ev = |w|2 if and only if u(t) and v(t) are proportional to each other.
E1.10 Fourier Series and Transforms (2015-5585)
Fourier Transform - Correlation: 8 – note 1 of slide 5
Autocorrelation
8: Correlation
Cross-Correlation
Signal Matching
Cross-corr as
Convolution
Normalized Cross-corr
Autocorrelation
Autocorrelation
example
⊲
Fourier Transform
Variants
Scale Factors
Summary
Spectrogram
The correlation of a signal with itself is its autocorrelation:
R∞ ∗
w(t) = u(t) ⊗ u(t) = −∞ u (τ − t)u(τ )dτ
The autocorrelation at zero lag:
R∞ ∗
w(0) = −∞ u (τ − 0)u(τ )dτ
R∞ ∗
= −∞ u (τ )u(τ )dτ
R∞
= −∞ |u(τ )|2 dτ = Eu
The autocorrelation at zero lag, w(0), is the energy of the signal.
The normalized autocorrelation:
z(t) = u(t)⊗u(t)
Eu
satisfies z(0) = 1 and |z(t)| ≤ 1 for any t.
Wiener-Khinchin Theorem:
[Cross-correlation theorem when v(t) = u(t)]
w(t) = u(t) ⊗ u(t) ⇔ W (f ) = U ∗ (f )U (f ) = |U (f )|2
The Fourier transform of the autocorrelation is the energy spectrum.
E1.10 Fourier Series and Transforms (2015-5585)
Fourier Transform - Correlation: 8 – 6 / 11
Autocorrelation example
8: Correlation
Cross-Correlation
Signal Matching
Cross-corr as
Convolution
Normalized Cross-corr
Autocorrelation
Autocorrelation
example
⊲
Fourier Transform
Variants
Scale Factors
Summary
Spectrogram
Cross-correlation is used to find when two different signals are similar.
Autocorrelation is used to find when a signal is similar to itself delayed.
First graph shows s(t) a segment of the microphone signal from the initial
vowel of “early” spoken by me. The waveform is “quasi-periodic” =
“almost periodic but not quite”.
.
Second graph shows normalized autocorrelation, z(t) = s(t)⊗s(t)
Es
z(0) = 1 for t = 0 since a signal always matches itself exactly.
z(t) = 0.82 for t = 6.2 ms = one period lag (not an exact match).
z(t) = 0.53 for t = 12.4 ms = two periods lag (even worse match).
Norm Autocorr, z(t)
Speech s(t)
1
0.5
0
-0.5
-1
10
20
30
Time (ms)
E1.10 Fourier Series and Transforms (2015-5585)
40
50
1
0.82
Lag = 6.2 ms (161 Hz)
0.5
0
0
5
10
Lag: t (ms)
15
20
Fourier Transform - Correlation: 8 – 7 / 11
Fourier Transform Variants
8: Correlation
Cross-Correlation
Signal Matching
Cross-corr as
Convolution
Normalized Cross-corr
Autocorrelation
Autocorrelation
example
Fourier Transform
⊲ Variants
Scale Factors
Summary
Spectrogram
There are three different versions of the Fourier Transform in current use.
(1) Frequency version (we have used this in lectures)
R∞
R∞
−i2πf t
dt
u(t) = −∞ U (f )ei2πf t df
U (f ) = −∞ u(t)e
• Used in the communications/broadcasting industry and textbooks.
• The formulae do not need scale factors of 2π anywhere.
,,,
(2) Angular frequency version
R∞
e
u(t)e−iωt dt
U (ω) =
R∞
e (ω)eiωt dω
u(t) =
U
−∞
e (ω) = U (f ) = U ( ω )
Continuous spectra are unchanged: U
2π
However δ-function spectral components are multiplied by 2π so that
e (ω) = 2π × δ(ω − 2πf0 )
U (f ) = δ(f − f0 ) ⇒ U
• Used in most signal processing and control theory textbooks.
−∞
1
2π
(3) Angular frequency + symmetrical scale factor
R∞
R∞
1
1
−iωt
b
b (ω)eiωt dω
U (ω) = √2π −∞ u(t)e
dt
u(t) = √2π −∞ U
e (ω)
b (ω) = √1 U
In all cases U
2π
• Used in many Maths textbooks (mathematicians like symmetry)
E1.10 Fourier Series and Transforms (2015-5585)
Fourier Transform - Correlation: 8 – 8 / 11
Scale Factors
8: Correlation
Cross-Correlation
Signal Matching
Cross-corr as
Convolution
Normalized Cross-corr
Autocorrelation
Autocorrelation
example
Fourier Transform
Variants
Scale Factors
Summary
Spectrogram
⊲
Fourier Transform using Angular Frequency:
R∞
R∞
1
−iωt
e
e (ω)eiωt dω
dt
u(t) = 2π −∞ U
U (ω) = −∞ u(t)e
R
R
1
Any formula involving df will change to 2π dω
[since dω = 2πdf ]
Parseval’s Theorem:
R ∗
R ∗
1
e (ω)Ve (ω)dω
u (t)v(t)dt = 2π U
2
R
R
2
1
e
Eu = |u(t)| dt = 2π U (ω) dω
Waveform Multiplication: (convolution implicitly involves integration)
e
e
f (ω) = 1 U
w(t) = u(t)v(t) ⇒ W
2π (ω) ∗ V (ω)
Spectrum Multiplication: (multiplication ; integration)
f (ω) = U
e (ω)Ve (ω)
w(t) = u(t) ∗ v(t) ⇒ W
b
To obtain formulae for version (3) of the Fourier
√ Transform, U (ω),
e (ω) = 2π U
b (ω).
substitute into the above formulae: U
E1.10 Fourier Series and Transforms (2015-5585)
Fourier Transform - Correlation: 8 – 9 / 11
Summary
8: Correlation
Cross-Correlation
Signal Matching
Cross-corr as
Convolution
Normalized Cross-corr
Autocorrelation
Autocorrelation
example
Fourier Transform
Variants
Scale Factors
Summary
Spectrogram
⊲
R∞
• Cross-Correlation: w(t) = u(t) ⊗ v(t) = −∞ u∗ (τ − t)v(τ )dτ
◦ Used to find similarities between v(t) and a delayed u(t)
◦ Cross-correlation theorem: W (f ) = U ∗ (f )V (f )
√
◦ Cauchy-Schwarz Inequality: |u(t) ⊗ v(t)| ≤ Eu Ev
⊲ Normalized cross-correlation:
• Autocorrelation: x(t) = u(t) ⊗ u(t) =
u(t)⊗v(t)
√
Eu Ev
R∞
−∞
≤1
u∗ (τ − t)u(τ )dτ ≤ Eu
◦ Wiener-Khinchin: X(f ) = energy spectral density, |U (f )|2
◦ Used to find periodicity in u(t)
• Fourier Transform using ω:
◦ Continuous spectra unchanged; spectral impulses multiplied by 2π
R
R
1
◦ In formulae: df → 2π dω; ω-convolution involves an integral
For further details see RHB Chapter 13.1
E1.10 Fourier Series and Transforms (2015-5585)
Fourier Transform - Correlation: 8 – 10 / 11
Spectrogram
8: Correlation
Cross-Correlation
Signal Matching
Cross-corr as
Convolution
Normalized Cross-corr
Autocorrelation
Autocorrelation
example
Spectrogram of “Merry Christmas” spoken by Mike Brookes
Fourier Transform
Variants
Scale Factors
Summary
Spectrogram
⊲
E1.10 Fourier Series and Transforms (2015-5585)
Fourier Transform - Correlation: 8 – 11 / 11
[Complex Fourier Series]
All waveforms have period T = 1. δcondition is 1 whenever “condition” is true and otherwise 0.
Waveform
x(t) for |t| < 0.5
Xn
Square wave
2δ|t|<0.25 − 1
Pulse of width d
δ|t|<0.5d
Sawtooth wave
2t
Triangle wave
1 − 4 |t|
2 sin 0.5πn
× δn6=0
πn
sin πdn
πn
i(−1)n
× δn6=0
πn
2(1−(−1)n )
π 2 n2
E1.10 Fourier Series and Transforms (2015-5585)
Fourier Transform - Correlation: 8 – note 1 of slide 11
[Fourier Transform Properties A]
You need not memorize these properties. All integrals are
Property
x(t)
Forward
x(t)
Inverse
Spectral Zero
R
R∞
−∞
Xf )
X(f )ei2πf t df
R
x(t)dt
R
x(t)e−i2πf t dt
X(f )
Duality
X(t)
= X(0)
R
= X(f )df
Reversal
x(−t)
X(−f )
conjugate
x∗ (t)
X ∗ (−f )
dn
x(t)
dtn
(−i2πt)n x(t)
(i2πf )n X(f )
Temporal Zero
Temporal Derivative
x(0)
x(−f )
Time Shift
x(t − T )
dn
X(f )
df n
1
X(f ) + 21 X(0)δ(f )
i2πf
2iπf β
f
1
e α X( α
)
|α|
X(f )e−i2πf T
Frequency Shift
x(t)ei2πF t
X(f − F )
Spectral Derivative
Integral
Scaling
E1.10 Fourier Series and Transforms (2015-5585)
Rt
−∞
x(τ )dτ
x(αt + β)
Fourier Transform - Correlation: 8 – note 2 of slide 11
[Fourier Transform Properties B]
You need not memorize these properties. All integrals are
R∞
−∞
Property
x(t)
Xf )
Linearity
αx(t) + βy(t)
αX(f ) + βY (f )
Multiplication
x(t)y(t)
X(f ) ∗ Y (f )
Convolution
x(t) ∗ y(t)
X(f )Y (f )
Correlation
x(t) ⊗ y(t)
X ∗ (f )Y (f )
Autocorrelation
x(t) ⊗ x(t)
|X(f )|2
Parseval or
Plancherel
x∗ (t)y(t)dt
R
Ex = |x(t)|2 dt
P
n x(t − nT )
P
n x(nT )δ(t − nT )
Repetition
Sampling
Modulation
Convolution: x(t) ∗ y(t) =
R
x(t) cos(2πF t)
1
T
R
= X ∗ (f )Y (f )df
R
= |X(f )|2 df
P
k
k
kX T δ f − T
k
1 P
kX f − T
T
1
X(f
2
− F ) + 12 X(f + F )
R
x(τ )y(t − τ )dτ
R ∗
R ∗
Cross-correlation: x(t) ⊗ y(t) = x (τ )y(τ + t)dτ = x (τ − t)y(τ )dτ
E1.10 Fourier Series and Transforms (2015-5585)
Fourier Transform - Correlation: 8 – note 3 of slide 11
[Fourier Transform Pairs]
You need not memorize these pairs.
x(t)
X(f )
X(f )
δ(t)
1
1
δ(f )
rect(t)
sin(πf )
πf
sin2 (πf )
π2 f 2
1
1
δ
(f
+
α)
+
δ (f − α)
2
2
1
α+2πif
2α
α2 +4π 2 f 2
1
iπf 1 P∞
k
k=−∞ δ f − T
T
sin(t)
t
sin2 (t)
t2
πrect(πf )
tri(t)
cos(2παt)
e−αt u(t)
e−α|t|
sgn(t)
P∞
x(t)
n=−∞
δ(t − nT )
Elementary Functions:
(
1, |t| < 0.5
rect(t) =
0, elsewhere


−1, t < 0
sgn(t) = 0,
t=0


1,
t>0
sin(2παt)
πtri(πf )
i
δ (f
2
+ α) − 2i δ (f − α)
1
(α+2πif )2
2
e−πf
te−αt u(t)
e−πt
2
u(t)
1
δ(f )
2
+
1
2πif
(
1 − |t|,
tri(t) =
0,
E1.10 Fourier Series and Transforms (2015-5585)
u(t) =
1
2
|t| < 1
elsewhere


0,
(1 + sgn(t)) = 0.5,


1,
x<0
x=0
x>0
Fourier Transform - Correlation: 8 – note 4 of slide 11