DSP First, 2/e
Lecture 7
Fourier Series Analysis
READING ASSIGNMENTS
 This Lecture:
 Fourier Series in Ch 3, Sect. 3-5
 Also, periodic signals, Sect. 3-4
 Other Reading:
 Appendix C: More details on Fourier Series
Aug 2016
© 2003-2016, JH McClellan & RW Schafer
3
LECTURE OBJECTIVES
 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)
 Draw spectrum from the Fourier Series coeffs
Aug 2016
© 2003-2016, JH McClellan & RW Schafer
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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/Biographies/Fourier.html
Aug 2016
© 2003-2016, JH McClellan & RW Schafer
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Aug 2016
© 2003-2016, JH McClellan & RW Schafer
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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
Aug 2016
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
© 2003-2016, JH McClellan & RW Schafer
f (in Hz)

7
Harmonic Signal->Periodic
x (t ) 

a e
k  
j 2 k F0 t
Sums of Harmonic
complex exponentials
are Periodic signals
k
PERIOD/FREQUENCY of COMPLEX EXPONENTIAL:
2
2 F0   0 
T0
Aug 2016
1
or T0 
F0
© 2003-2016, JH McClellan & RW Schafer
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Notation for Fundamental
Frequency in Fourier Series
 The k-th frequency is fk = kF0
x (t ) 

j 2 k F0 t
a
e

 k
k  

j 2 f k t
a
e
 k
k  
 Thus, f0 = 0 is DC
 This is why we use upper case F0 for the
Fundamental Frequency
Aug 2016
© 2003-2016, JH McClellan & RW Schafer
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Harmonic Signal (3 Freqs)
a1
a3
a5
T = 0.1
x (t ) 
5
j 2 (10k ) t
a
e
 k
k  5
Aug 2016
© 2003-2016, JH McClellan & RW Schafer
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Periodic signals->Harmonic?
x (t ) 

a e
k  
j 2 k F0 t
k
Can all periodic
signals be written as
harmonic signals?
 Fourier’s contribution was to postulate the answer is yes
 Called Fourier Series
 For heat transfer it is easy to solve PDE for sinusoidal
sources, but difficult for general sources
 Made formal by Dirichlet and Riemann
Aug 2016
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STRATEGY: x(t)  ak
 ANALYSIS
 Get representation from the signal
 Works for PERIODIC Signals
 Measure similarity between signal & harmonic
 Fourier Series
 Answer is: an INTEGRAL over one period
ak 
Aug 2016
1
T0

T0
x (t )e
 j0k t
dt
0
© 2003-2016, JH McClellan & RW Schafer
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CALCULUS for complex exp
d t
t
e  e
dt
b
e
a
t
dt 
1

d j t
t
e  je
dt

b
e
t

a

e

1
b
e
a

b
1 j t
1 j b
j a
a e dt  j e  j e  e 
a
b
Aug 2016
j t
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INTEGRAL Property of exp(j)
 INTEGRATE over ONE PERIOD
T0
e
T0
 j ( 2 / T0 ) mt
0
T0
 j ( 2 / T0 ) mt
dt 
e
 j 2 m
0
T0
 j 2 m

(e
 1)
 j 2 m
T0
e
 j ( 2 / T0 ) mt
dt  0
m0
0
Aug 2016
© 2003-2016, JH McClellan & RW Schafer
2
0 
T0
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ORTHOGONALITY of exp(j)
0
k



1

j ( 2 / T0 ) t  j ( 2 / T0 ) kt
e
e
dt  

T0 0

1
k



j ( 2 / T )(   k ) t
T0
e
0
m  k
T0
1
j ( 2 / T0 )(   k ) t
e
dt

T0 0
Aug 2016
© 2003-2016, JH McClellan & RW Schafer
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Fourier Series Integral
 Use orthogonality to determine ak from x(t)
ak 
T0
1
T0
 x (t )e
 j ( 2 / T0 ) k t
Fundamenta l Freq.
dt
F0  1 / T0
0
*
ak  ak
a0 
when x(t ) is real
T0
1
T0
 x(t )dt
(DC component)
0
Aug 2016
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Isolate One FS Coefficient
x (t ) 

j ( 2 / T0 ) k t
a
e
 k
k  
T0
1
T0
 j ( 2 / T0 )  t
x
(
t
)
e
dt 

0
 
j ( 2 / T0 ) k t   j ( 2 / T0 )  t
ak e
dt
e
0  k
 

T0
1
T0
T0


 j ( 2 / T0 )  t
j ( 2 / T0 ) k t  j ( 2 / T0 )  t
1
1
dt   ak  T0  e
e
dt   a
T0  x ( t )e


k   
0
0

Integral is zero
T0
except for k  
 a  T1 x (t )e  j ( 2 / T0 )  t dt

T0
0

0
Aug 2016
 is dummy vari able, could be k
© 2003-2016, JH McClellan & RW Schafer
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Fourier Series: x(t)  ak
 ANALYSIS
 Given a PERIODIC Signal
 Fourier Series coefficients are obtained via
an INTEGRAL over one period
ak 
1
T0

T0
x (t )e
 j0k t
dt
0
 Next, consider a specific signal, the FWRS
 Full Wave Rectified Sine
Aug 2016
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Full-Wave Rectified Sine
x(t )  sin( 2 t / T1 )
Period is T0  12 T1
Absolute value flips the
negative lobes of a sine wave
Aug 2016
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Full-Wave Rectified Sine
{ak}
T0
1
ak   x(t )e j ( 2 / T0 ) kt dt
T0 0
ak 
Full-Wave Rectified Sine
x (t )  sin( 2 t / T1 )
T0
1
T0
 j ( 2 / T0 ) kt

sin(
t
)
e
dt
 T0
Period : T0  12 T1
0


e j ( / T0 ) t  e  j ( / T0 ) t  j ( 2 / T0 ) kt
e
dt
0
2j
T0
1
T0
T0
1
j 2 T0
 j ( / T0 )( 2 k 1) t
e
dt 

0

Aug 2016
e
 j (  / T0 )( 2 k 1) t
j 2 T0 (  j (  / T0 )( 2 k 1))
 x (t )  sin(  t / T0 )
T0
1
j 2 T0
 j ( / T0 )( 2 k 1) t
e
dt

0
T0

e
 j (  / T0 )( 2 k 1) t
T0
j 2 T0 (  j (  / T0 )( 2 k 1))
0© 2003-2016, JH McClellan & RW Schafer 0
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Full-Wave Rectified Sine
ak 
e
T0
 j ( / T0 )( 2 k 1) t

j 2T0 (  j ( / T0 )( 2 k 1))
e
T0
 j ( / T0 )( 2 k 1) t
j 2T0 (  j ( / T0 )( 2 k 1))
0

1
2 ( 2 k 1)
e
0
 j ( / T0 )( 2 k 1)T0

{ak}



 1  2 ( 21k 1) e  j ( / T0 )( 2 k 1)T0  1



  ( 21k 1) e  j ( 2 k 1)  1   ( 21k 1) e  j ( 2 k 1)  1

Aug 2016

2 k 1 ( 2 k 1)
 ( 4 k 2 1)

2
 (1)  1 
 (4k 2  1)
2k

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Fourier Coefficients: ak
 ak is a function of k
 Complex Amplitude for k-th Harmonic
2
ak 
2
 (4k  1)
 Does not depend on the period, T0
 DC value is a0  2 /   0.6336
Aug 2016
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Spectrum from Fourier Series
Plot a k for Full-Wave Rectified Sinusoid
2
ak 
 (4k 2  1)
Aug 2016
F0  1 / T0
© 2003-2016, JH McClellan & RW Schafer
and 0  2 F0
26
Reconstruct From Finite Number
of Harmonic Components
Full-Wave Rectified Sinusoid x(t )  sin(  t / T0 )
2
 (4k 2  1)
 F0  100 Hz
a0  2 /   0.6336
T0  10 ms
ak 

N
xN (t )  a0   ak e j 2 kF0 t  ak e  j 2 kF0t

k 1
How close is xN (t ) to x(t )  sin(  t / T0 ) ?
Aug 2016
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Reconstruct From Finite Number
of Spectrum Components
Full-Wave Rectified Sinusoid x(t )  sin(  t / T0 )
T0  10 ms
 F0  100 Hz
a0  2 /   0.6336
2
ak 
2

(
4
k
 1)
Aug 2016
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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
Aug 2016
© 2003-2016, JH McClellan & RW Schafer
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