ENE 208 Electrical Engineering Mathematics (2/2008)
Class 2, January 12, 2009
werapon.chi@kmutt.ac.th
1
Meet Mr. Fourier
• Jean Baptiste Joseph, Baron de
Fourier (March 21, 1768 - May 16,
1830), a French mathematical
physicist
• “No matter how complicated it is,
a wave that is periodic - with a
pattern that repeats itself consists of the sum of many
simple waves.”
2
1
Periodic Function
• y = f(x) = f(x+T) for the period T
• Frequency f = 1/T
y
y = f(x)
x
0
T
3
Time Domain: t ≥0
1
Period = T1 sec
Frequency = 1/T1 Hz
Amplitude
f1(t)=sin(t)
0
T1
T1/2
3T1/2
2T1
Time
(seconds)
-1
Period = T2 = T1/2 sec
Frequency = 2/T1 Hz
1
0
Nyquist Rate: 9 points
fsampling ≥ 2(2/T1)=4/T1 Hz
-1
Amplitude
f2(t)=sin(t)
T1/4
T1/2
3T1/4
T1
5T1/4
3T1/2
2T1
7T1/4
Time
(seconds)
4
2
Angle: Radian = (π/180)×Degree
1
T1=2π rad
0
Amplitude
f1(θ)=sin(θ)
π
2π
3π
-1
T2=2π rad
=T1 !?
1
0
T2=T1/2
T1=4π rad !?
4π Angle
(radians)
f2(θ)=sin(θ)
Amplitude
π
2π
3π
4π
5π
-1
6π
7π 8π
Angle
(radians) 5
Angular Velocity: ω=θ/t (rad/s)
1
Sinusoidal
ω1=2π/T1
0
=2πf1
Amplitude
f1(t)=sin(ω1t)
T1/2
T1
3T1/2
-1
1
ω2=2π/T2
=4π/T1 0
=4πf1
=2ω1
-1
“Fundamental
2T
Time Frequency”
(seconds)
1
Amplitude
f2(t)=sin(2ω1t)
T1/2
T1
3T1/2
2T1
“Harmonics”
Time
(seconds)
6
3
Frequency Domain => Spectrum
1
Amplitude
Amplitude
f1(t)=sin(2πf1t)
1
0
T1/2
T1
3T1/2
-1
1
0
-1
T1
Time
0
(seconds)
f1 2f1 3f1 4f1
Frequency (Hz)
f2(t)=sin(4πf1t)
Amplitude
=sin(2π(2f1)t)
1
Amplitude
T1/2
2T1
3T1/2
2T1
Time
(seconds) 0 f1 2f1 3f1 4f1
Frequency (Hz) 7
Prism = Fourier Transform
• How does it transforms or reveal the true beauty
and nature of visible light?
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4
Basis Function
• An element of the basis for a function space
• The Euclidean plane ℜ2
y
y1
∧
j
O ∧i
“Orthogonal”
A(x1,y1)
∧
∧
OA = x1i + y1j
“Projection”
x1
x
Inner product,
∧ ∧ ∧ ∧
i ⋅ j = | i | | j | cos(90°)
=1⋅1⋅0
=0
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Fourier Series
With sinusoidal basis function, sinθ ⋅ cosθ = 0
We can represent a periodic function,
f(t) = b1sinω1t + b2sinω2t + … + bnsinωnt
or
f(t) = a1cosω1t + a2cosω2t + … + ancosωnt
or
f(t) = a1cosω1t + b1sinω1t
+ a2cosω2t + b2sinω2t
+ … + ancosωnt + bnsinωnt
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5
Fourier Series (Cont’d)
f(t) = a0 + Σn=1→∞ [ ancos(nωt) + bnsin(nωt) ]
f(t) = a0 + a1cos(ωt) + b1sin(ωt)
+ a2cos(2ωt) + b2sin(2ωt)
+ … + ancos(nωt) + bnsin(nωt) + …
Known: Signal in time domain
f(t)
Unknown: Fourier Coefficients
a0, a1, b1, a2, b2, …, an, bn, …
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Trigonometric Formulas
sin(–θ) = –sinθ
cos(–θ) = cosθ
sin(θ1±θ2) = sinθ1cosθ2 ± sinθ2cosθ1
cos(θ1±θ2)= cosθ1cosθ2 sinθ1sinθ2
sin2θ + cos2θ = 1
tan2θ + 1 = sec2θ
1 + cot2θ = csc2θ
sin2θ = 2sinθcosθ
cos2θ = cos2θ – sin2θ = 2cos2θ – 1 = 1 – 2sin2θ
ejθ = cosθ + j sinθ where j= √–1
sinθ = 1/2j (ejθ – e–jθ )
cosθ = 1/2 (ejθ + e–jθ )
For sinusoidal signal, ω = 2π/T = 2πf
±
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6
Finding a0
f(t) = a0+Σn=1→∞[ancos(nωt)+bnsin(nωt)]
∫t=0→T f(t) dt = ∫t=0→T a0 dt
+ Σn=1→∞ ∫t=0→T ancos(nωt) dt
+ Σn=1→∞ ∫t=0→T bnsin(nωt) dt
= a0 ∫t=0→T dt + 0 + 0
= a0T
a0 = 1/T ∫t=0→T f(t) dt
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Finding an
f(t) = a0+Σn=1→∞[ancos(nωt)+bnsin(nωt)]
f(t)cos(ωt) = a0cos(ωt) + a1cos2(ωt) + b1sin(ωt)cos(ωt)
+ a2cos(2ωt)cos(ωt) + b2sin(2ωt)cos(ωt)
+…
∫t=0→T f(t)cos(ωt) dt = a0 ∫t=0→T cos(ωt) dt
+ a1 ∫t=0→T cos2(ωt) dt
+ b1 ∫t=0→T sin(ωt)cos(ωt) dt
+ a2 ∫t=0→T cos(2ωt)cos(ωt) dt
+ b2 ∫t=0→T sin(2ωt)cos(ωt) dt
+…
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7
Finding an (Cont’d)
∫t=0→Tcos2(ωt)dt = 1/2 ∫t=0→T 1 + cos(2ωt) dt
= 1/2 ∫t=0→T dt + 1/2 ∫t=0→T cos(2ωt) dt
= T/2
∫t=0→Tsin(ωt)cos(ωt)dt = 1/ω ∫t=0→T sin(ωt) d(sin(ωt))
= 1/2ω sin2(ωt)|t=0→T
=0
∫t=0→Tcos(2ωt)cos(ωt)dt = 1/ω ∫t=0→T 1 – 2sin2(ωt) d(sin(ωt))
= 1/ω ∫t=0→T d(sin(ωt))
– 2/ω ∫t=0→T sin2(ωt) d(sin(ωt))
= 1/ω sin(ωt)|t=0→T – 2/3ω sin3(ωt)|t=0→T
=0
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Finding an (Cont’d)
∫t=0→Tsin(2ωt)cos(ωt)dt = 2 ∫t=0→T sin(ωt)cos2(ωt) dt
= –2/ω ∫t=0→T cos2(ωt) d(cos(ωt))
= –2/3ω cos3(ωt)|t=0→T
=0
then ∫t=0→Tcos(3ωt)cos(ωt)dt = 0
∫t=0→Tsin(3ωt)cos(ωt)dt = 0
…
Therefore,
∫t=0→T f(t)cos(ωt) dt = a0(0)+a1(T/2)+b1(0)+a2(0)+b2(0)+…
a1 = 2/T ∫t=0→T f(t)cos(ωt) dt
and we get
an = 2/T ∫t=0→T f(t)cos(nωt) dt
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8
Finding bn
f(t) = a0+Σn=1→∞[ancos(nωt)+bnsin(nωt)]
f(t)sin(ωt) = a0sin(ωt) + a1cos(ωt)sin(ωt) + b1sin2(ωt)
+ a2cos(2ωt)sin(ωt) + b2sin(2ωt)sin(ωt)
+…
∫t=0→T f(t)sin(ωt) dt = 0+0+ b1∫t=0→Tsin2(ωt)dt +0+0+…
= b1/2 ∫t=0→T 1 – cos(2ωt) dt
= b1/2 ∫t=0→T dt – b1/2 ∫t=0→T cos(2ωt) dt
= b1T/2 – 0
b1 = 2/T ∫t=0→T f(t)sin(ωt) dt
bn = 2/T ∫t=0→T f(t)sin(nωt) dt
and we get
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Finding bn (Cont’d)
cos(ωt)
cos(2ωt)
sin(2ωt)
sin(ωt)
sin(ωt)
sin(ωt)
cos(ωt)sin(ωt)
cos(2ωt)sin(ωt)
sin(2ωt)sin(ωt)
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9
Note:
For m and n are integers,
• ∫t=0→T sin(mωt)sin(nωt) dt = 0
,m≠n
= T/2 , m = n
• ∫t=0→T cos(mωt)cos(nωt) dt = 0 , m ≠ n
= T/2 , m = n
• ∫t=0→T sin(mωt)cos(nωt) dt = 0
e.g. prove
∫t=0→Tsin(mωt)sin(nωt)dt = 1/2 ∫t=0→T cos((m–n)ωt)–cos((m+n)ωt) dt
= 1/2ω(m–n) sin((m–n)ωt)|t=0→T – 1/2ω(m+n) sin((m+n)ωt)|t=0→T
= 1/2ω [ sin((m–n)ωT)/(m–n) – sin((m+n)ωT)/(m+n) ]
= 1/2ω [ sin(2π(m–n))/(m–n) – sin(2π(m+n))/(m+n) ]
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Fourier Series Coefficients
Decomposition:
know f(t), to find an and bn
• an = 2/T ∫t=0→T f(t) cos(2πnt/T) dt
= 2/T Σt=0→T f(t) cos(2πnt/T) ∆t
• bn = 2/T ∫t=0→T f(t) sin(2πnt/T) dt
= 2/T Σt=0→T f(t) sin(2πnt/T) ∆t
Reconstruction:
know an and bn, to find f(t)
• f(t) = a0/2 + Σn=0→∞ [ an cos(2πnt/T) + bn sin(2πnt/T) ]
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