Introduction to Digital Logic
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Transcript Introduction to Digital Logic
Advanced Digital Signal Processing
Prof. Nizamettin AYDIN
[email protected]
http://www.yildiz.edu.tr/~naydin
1
Fourier Series Coefficients
2
• Work with the Fourier Series Integral
ak
1
T0
T0
x (t )e
j ( 2 k / T0 ) t
dt
0
• ANALYSIS via Fourier Series
– For PERIODIC signals: x(t+T0) = x(t)
– Spectrum from the Fourier Series
3
HISTORY
• Jean Baptiste Joseph Fourier
– 1807 thesis (memoir)
• On the Propagation of Heat in Solid Bodies
– Heat !
– Napoleonic era
•
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Fourier.html
4
5
SPECTRUM DIAGRAM
• Recall Complex Amplitude vs. Freq
1
2
X
4e
*
k
j / 2
–250
7e
j / 3
10
7e
Xk Ake
–100
N
x(t ) Xa0 { aXkk e
k 1
4e
j k
0
1
2
j / 3
100
j 2 f k t
1
2
1
2
X k ak
j / 2
250
* j 2 f k t
aXkk e
f (in Hz)
6
Harmonic Signal
x (t )
ak e
j 2 k f 0 t
k
PERIOD/FREQUENCY of COMPLEX EXPONENTIAL:
2
2 f 0 0
T0
1
or T0
f0
7
Fourier Series Synthesis
x (t )
ak e
j 2 k f 0 t
k
ak X k Ak e
1
2
1
2
j k
N
x(t ) A0 Ak cos(2 kf0t k )
k 1
X k Ak e
j k
COMPLEX
AMPLITUDE
8
Harmonic Signal (3 Freqs)
a1
a3
a5
T = 0.1
9
SYNTHESIS vs. ANALYSIS
• SYNTHESIS
– Easy
– Given (k,Ak,fk) create
x(t)
• Synthesis can be HARD
– Synthesize Speech so that it
sounds good
• ANALYSIS
– Hard
– Given x(t), extract
(k,Ak,fk)
– How many?
– Need algorithm for
computer
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STRATEGY: x(t) ak
• ANALYSIS
– Get representation from the signal
– Works for PERIODIC Signals
• Fourier Series
– Answer is: an INTEGRAL over one period
ak
1
T0
T0
x (t )e
j 0k t
dt
0
11
INTEGRAL Property of exp(j)
• INTEGRATE over ONE PERIOD
T0
e
j ( 2 / T0 ) mt
0
T0
e
0
j ( 2 / T0 ) mt
T0
T0
j ( 2 / T0 ) mt
dt
e
j 2 m
0
T0
(e j 2 m 1)
j 2 m
dt 0
m0
2
0
T0
12
ORTHOGONALITY of exp(j)
• PRODUCT of exp(+j ) and exp(-j )
0
k
1
j ( 2 / T0 ) t j ( 2 / T0 ) kt
e
e
dt
T0 0
1 k
T0
T0
1
j ( 2 / T0 )( k ) t
e
dt
T0 0
13
Isolate One FS Coefficient
x (t )
ak e
j ( 2 / T0 ) k t
k
T0
1
T0
x (t )e
j ( 2 / T0 ) t
0
1
T0
T0
j ( 2 / T0 ) t
j
(
2
/
T
)
k
t
j
(
2
/
T
)
t
1 e
0
0
a
x
(
t
)
e
dt
a
e
dt
k
T0
k
0
0
Integralis zero
T0
1
T0
dt
j ( 2 / T0 ) k t j ( 2 / T0 ) t
e
ak e
dt
k
0
T0
ak
T0
1
T0
j ( 2 / T0 ) k t
x
(
t
)
e
dt
exceptfor k
0
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SQUARE WAVE EXAMPLE
1T
1
0
t
2 0
x (t )
1T t T
0
0
2 0
for T0 0.04 sec.
x(t)
1
–.02
0
.01
.02
0.04
t
15
FS for a SQUARE WAVE {ak}
1
ak
T0
T0
x (t )e
j ( 2 / T0 ) kt
dt
(k 0)
0
.02
.02
1
j ( 2 / .04) kt
j
(
2
/
.
04
)
kt
1
ak
1
e
dt
e
.04( j 2 k / .04)
0
.04 0
1
1 ( 1)
j ( ) k
(e
1)
( j 2 k )
j 2 k
k
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DC Coefficient: a0
1
ak
T0
1
a0
T0
T0
x (t )e
j ( 2 / T0 ) kt
(k 0)
dt
0
T0
1
x
(
t
)
dt
(
Area
)
T
0
0
.02
1
1
a0
1
dt
(.
02
0
)
.04 0
.04
1
2
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Fourier Coefficients ak
• ak is a function of k
– Complex Amplitude for k-th Harmonic
– This one doesn’t depend on the period, T0
1
j k
k
1 ( 1)
ak
0
j 2 k
1
2
k 1,3,
k 2,4,
k 0
18
Spectrum from Fourier Series
0 2 /(0.04) 2 (25)
j
k
ak 0
12
k 1,3,
k 2,4,
k 0
19
Fourier Series Integral
• HOW do you determine ak from x(t) ?
ak
T0
1
T0
x (t )e
0
a0 T1
0
T0
j ( 2 / T0 ) k t
dt
Fundamental Frequency f0 1 / T0
x
(
t
)
dt
*
ak ak
when x(t ) is real
(DC component)
0
20
Fourier Series & Spectrum
21
Example
x(t ) sin (3 t )
3
j j 9 t 3 j j 3 t 3 j j 3 t j j 9 t
x ( t ) e
e
e
e
8
8
8
8
22
Example
x(t ) sin (3 t )
3
j j 9 t 3 j j 3 t 3 j j 3 t j j 9 t
x ( t ) e
e
e
e
8
8
8
8
In this case, analysis
just requires picking
off the coefficients.
k 3
ak
k 1
k 1
k 3
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STRATEGY: x(t) ak
• ANALYSIS
– Get representation from the signal
– Works for PERIODIC Signals
• Fourier Series
– Answer is: an INTEGRAL over one period
ak
1
T0
T0
x (t )e
j 0k t
dt
0
24
FS: Rectified Sine Wave {ak}
T0
1
ak
T0
ak
j ( 2 / T0 ) kt
x
(
t
)
e
dt
( k 1)
0
Half-Wave Rectified Sine
T0 / 2
j ( 2 / T0 ) kt
2
sin(
t
)
e
dt
T
1
T0
0
0
T0 / 2
1
T0
0
e j ( 2 / T0 ) t e j ( 2 / T0 ) t j ( 2 / T0 ) kt
e
dt
2j
T0 / 2
1
j 2T0
e j ( 2 / T0 )( k 1) t dt
T0 / 2
1
j 2T0
0
e
j ( 2 / T0 )( k 1) t
e j ( 2 / T0 )( k 1) t dt
0
T0 / 2
j 2T0 ( j ( 2 / T0 )( k 1))
0
e
j ( 2 / T0 )( k 1) t
T0 / 2
j 2T0 ( j ( 2 / T0 )( k 1))
0
25
FS: Rectified Sine Wave {ak}
ak
e
j 2T0 ( j ( 2 / T0 )( k 1))
1
4 ( k 1)
1
4 ( k 1)
T0 / 2
j ( 2 / T0 )( k 1) t
e
e
k 1 ( k 1)
4 ( k 2 1)
0
j ( 2 / T0 )( k 1)T0 / 2
j ( k 1)
1
e
j 2T0 ( j ( 2 / T0 )( k 1))
0
1 4 (1k 1) e j ( 2 / T0 )( k 1)T0 / 2 1
1
4 ( k 1)
e
j ( k 1)
0
1
k
(1) 1 ?j 4
1
( k 2 1)
T0 / 2
j ( 2 / T0 )( k 1) t
1
k odd
k 1
k even
26
Fourier Coefficients ak
• ak is a function of k
– Complex Amplitude for k-th Harmonic
– This one doesn’t depend on the period, T0
1
j k
k
1 ( 1)
ak
0
j 2 k
1
2
k 1,3,
k 2,4,
k 0
27
Fourier Series Synthesis
• HOW do you APPROXIMATE x(t) ?
ak
T0
1
T0
x (t )e
j ( 2 / T0 ) k t
dt
0
• Use FINITE number of coefficients
x (t )
N
ak e
j 2 k f 0 t
ak ak*
when x(t ) is real
k N
28
Fourier Series Synthesis
29
Synthesis: 1st & 3rd Harmonics
1 2
2
y (t ) cos( 2 ( 25)t 2 )
cos( 2 (75)t 2 )
2
3
30
Synthesis: up to 7th Harmonic
1 2
2
2
2
y (t ) cos( 50 t 2 )
sin(150 t )
sin( 250 t )
sin( 350 t )
2
3
5
7
31
Fourier Synthesis
1 2
2
x N (t ) sin(0t )
sin( 30t )
2
3
32
Gibbs’ Phenomenon
• Convergence at DISCONTINUITY of x(t)
– There is always an overshoot
– 9% for the Square Wave case
33
Fourier Series Demos
• Fourier Series Java Applet
– Greg Slabaugh
• Interactive
– http://users.ece.gatech.edu/mcclella/2025/Fsdemo_Slabaugh/fourier.html
• MATLAB GUI: fseriesdemo
– http://users.ece.gatech.edu/mcclella/matlabGUIs/index.html
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