fourier transform
DESCRIPTION
This presentation describes the Fourier Transform used in different mathematical and physical applications. The presentation is at an Undergraduate in Science (math, physics, engineering) level. Please send comments and suggestions to improvements to [email protected]. More presentations can be found at my website http://www.solohermelin.com.TRANSCRIPT
Fourier Transform
SOLO HERMELIN
Updated: 22.07.07Run This
Fourier Transform
dttjtftfF exp:F
SOLO
Jean Baptiste JosephFourier
1768 - 1830
F (ω) is known as Fourier Integral or Fourier Transformand is in general complex
jAFjFF expImRe
Using the identities
tdtj
2exp
we can find the Inverse Fourier Transform Ftf -1F
002
1
2exp
2expexp
2exp
tftfdtfdd
tjf
dtjdjf
dtjF
2exp:
dtjFFtf -1F
002
1
tftfdtf
If f (t) is continuous at t, i.e. f (t-0) = f (t+0)
This is true if (sufficient not necessary)f (t) and f ’ (t) are piecewise continue in every finite interval1
2 and converge, i.e. f (t) is absolute integrable in (-∞,∞)
dttf
Fourier TransformSOLO
tf-1F
F FProperties of Fourier Transform
Linearity 1
221122112211 exp: FFdttjtftftftf
F
Symmetry 2 tF
-1FF f2
tFdttjtFfdt
tjtFfd
tjFtft
F
exp22
exp2
exp
Proof:
Conjugate Functions3 tf *
-1FF *F
Proof:
tfd
tjFd
tjFtf ****
2exp
2exp 1-F
Fourier Transform
a
Faa
d
ajfdttjtaftaf
ta 1
expexp:F
FjdttjjtfFd
ddttjtftfF nn
n
n
FF expexp:
SOLO
tf-1F
F FProperties of Fourier Transform
Scaling4
Derivatives5
Proof:
taf-1F
F
a
Fa
1
Proof:
Corollary: for a = -1 tf
-1FF F
tftj n-1F
F
Fd
dn
n
tftd
dn
n
-1FF Fj n
Fj
dtjjFtf
td
ddtjFFtf nn
n
n1-1- FF
2
exp2
exp
Fourier TransformSOLO
tf-1F
F FProperties of Fourier Transform
Convolution6
Proof:
212121
212121
expexpexp
expexpexp:
FFFdjfdduujufjf
ddttjtfjfdtdtfftjdtff
ut
F
tftf 21-1F
F 21 * FF
dtfftftf 2121 :*-1F
F 21 FF
The animations above graphically illustrate the convolution of two rectangle functions (left) and two Gaussians (right). In the plots, the green curve shows the convolution of the blue and red curves as a function of t, the position indicated by the vertical green line. The gray region indicates the product as a function of g (τ) f (t-τ) , so its area as a function of t is precisely the convolution.
http://mathworld.wolfram.com/Convolution.html
Fourier TransformSOLO
tf-1F
F FProperties of Fourier Transform
dFFdttftf 2*
12*
1 2
1Parseval’s Formula7
Proof:
dttjtfF exp11
22exp
2exp 2
*
112*
2*
12*
1
dFF
ddttjtfFdt
dtjFtfdttftf
22exp
2exp 21122121
dFF
ddttjtfFdt
dtjFtfdttftf
2exp*
2
*
2
dtjFtf
dttjtfFdttjtfF expexp 1111
dFFdFFdttftf 212121 2
1
2
1
Signal Duration and BandwidthSOLO
tf-1F
F FRelationships from Parseval’s Formula
dFFdttftf 2*
12*
1 2
1Parseval’s Formula7
Choose tstjtftf m 21
,2,1,0
2
12
22
ndd
Sddttst
m
mm
tftj n-1F
F
Fd
dn
n
and use 5a
Choose n
n
td
tsdtftf 21 and use 5b tf
td
dn
n
-1FF Fj n
,2,1,02
1 22
2
ndSdttd
tsd mn
n
Choosec
,2,1,0,,2,1,0
2*
mndd
SdS
jdt
td
tsdtstj
m
mn
n
n
nmm
n
n
td
tsdtf 1
tstjtf m2
Fourier TransformSOLO
tf-1F
F FProperties of Fourier Transform
Modulation9
Shifting: for any a real 8
Proof:
ttf 0cos -1F
F 002
1 FF
Proof:
tjtjt 000 expexp2
1cos
atf -1F
F ajF exp tajtf exp-1F
F aF
Fajdajfdttjatfatfat
expexpexp:F
aFdttajtfdttjtajtftajtf
expexpexp:expF
use shifting property with a=±ω0
atf -1F
F ajF exp
Fourier TransformSOLO tf
-1FF FProperties of Fourier Transform (Summary)
Linearity 1 221122112211 exp: FFdttjtftftftf
F
Symmetry 2
tF-1F
F f2
Conjugate Functions3 tf *
-1FF *F
Scaling4 taf-1F
F
a
Fa
1
Derivatives5 tftj n-1F
F
Fd
dn
n
tftd
dn
n
-1FF Fj n
Convolution6
tftf 21-1F
F 21 * FF
dtfftftf 2121 :*-1F
F 21 FF
dFFdttftf 2*
12*
1 2
1
Parseval’s Formula7
Shifting: for any a real 8 tajtf exp
-1FF aF
Modulation9 ttf 0cos -1F
F 002
1 FF
dFFdFFdttftf 212121 2
1
2
1
Fourier TransformSOLO
dttjtftfF exp:F
2exp
dtjFFtf 1-F
-1FF
dttjtfF exp
* - complex conjugate
dttjtfF exp**
imaginarytf
realtf
0Re
0Im
tf
tf
tftf
tftf
*
*
*
*
FF
FF
realtf *FF
imaginarytf *FF
Therefore
Fourier Transform of Real or Imaginary Functions
Fourier TransformSOLO
realtf *FF
imaginarytf *FF
realtf
FF
FF
ImIm
ReRe
imaginarytf
FF
FF
ImIm
ReRe
dttjtftfFjFF exp:ImRe F
Fdttjtfdttjtftf expexpF
tftftftf eveneven 5.0: tftftftf oddodd 5.0:
realtf
FjFFtftftftf
FFFtftftftf
evenodd
eveneven
Im5.05.0:
Re5.05.0:
F
F
FFFtftftftf
FjFFtftftftf
evenodd
eveneven
Re5.05.0:
Im5.05.0:
F
F imaginarytf
Fourier Transform of Real or Imaginary Functions (continue – 1)
Fourier Transform
FRe
FIm
Real & Even
t
tfIm
tfRe
Real & Even
SOLO
FRe
FIm
Imaginary & Odd
t
tfIm
tfRe
Real & Odd
FRe
FIm
Imag. & Even
t
tfIm
tfRe
Imag. &Even
FRe
FIm
Real & Odd
t
tfIm
tfRe
Imag. & Odd
realtf
tftftftf even 5.0:
tftftftf even 5.0:
imaginarytf
FF
FFtftf even
ReRe
5.0FF
*FF
tftftftf odd 5.0:
FjFj
FFtftf even
ImIm
5.0FF
realtf *FF
*FF
tftftftf odd 5.0:
FjFj
FFtftf even
ImIm
5.0FF
imaginarytf *FF
FF
FFtftf even
ReRe
5.0FF
Fourier Transform of Real or Imaginary Functions (continue – 2)
Fourier TransformSOLO
dtttftfjdtttftf
dttjttftfdttjtftfF
oddevenoddeven
oddeven
sincos
sincosexp:F
tftftftf eveneven 5.0: tftftftf oddodd 5.0: tftftf oddeven
0000
0
cos2coscoscoscoscos dtttfdtttfdfdtttfdtttfdtttf eveneven
f
eveneven
t
eveneven
even
0coscoscoscoscos000
0
dtttfdfdtttfdtttfdtttf odd
f
oddodd
t
oddodd
odd
0sinsinsinsinsin000
0
dtttfdfdtttfdtttfdtttf even
f
eveneven
t
eveneven
even
0000
0
sin2sinsinsinsinsin dtttfdtttfdfdtttfdtttfdtttf oddodd
f
oddodd
t
oddodd
odd
Therefore
00
sin2cos2exp: dtttfjdtttfdttjtftfF oddeven F
0
cos25.0 dtttfFFF eveneven
0
sin25.0 dtttfjFFF oddodd
Odd and Even Parts
Fourier Transform
2cosImsinRe
2sinImcosRe
2sincosImRe
2exp:
dtFtFj
dtFtF
dtjtFjF
dtjFFtf -1F
SOLO
00: ttfCausal
Causal Functions
A causal functions is a equal zero for negative t
tftftftf eveneven 5.0: tftftftf oddodd 5.0:
Since
and 0 ttf we have 022 ttftftf oddeven
realtf
2sinImcosRe
dtFtFFtf -1F
causalrealtf &
02
sinIm42
cosRe4
2sinIm22
2cosRe22
00
td
tFd
tF
dtFtf
dtFtftf oddeven
FF
FF
ImIm
ReRe
0sinIm2
cosRe2
00
tdtFdtFtf
Fourier TransformSOLO
00: ttfCausalReal & Causal Functions 022 ttftftf oddeven
causalrealtf & 0sinIm2
cosRe2
00
tdtFdtFtf
dttjttfFjFF sincosImRe
dtduttuuFdttdutuuFdtttfF
cossinIm
2cossinIm
2cosRe
00
dvdtttvvFdttdvtvvFdtttfF
sincosRe
2sincosRe
2sinIm
00
Therefore
dtduttuuFF
cossinIm2
Re0
dtdvttvvFF
sincosRe2
Im0
But also
dtduttuuFF
coscosRe2
Re0
dtdvttvvFF
sinsinRe2
Im0
Real & Causal Functions
Real & Causal Functions
FF
FF
ImIm
ReRe
Fourier TransformSOLO
2sin
2cos
dtFj
dtFtftftf oddeven
tftftf
tf eveneven
2
: tftftf
tf oddodd
2
:
tfd
tFtftf
tf eveneven
2cos
2
tfd
tFjtftf
tf oddodd
2sin
2 FjFF ImRe
2sinRe
2sinIm
2cosIm
2cosReImRe
dtFj
dtF
dtFj
dtFtfjtftf
2sinIm
2cosReRe
dtF
dtFtf
2sinRe
2cosImIm
dtF
dtFtf
2sin
2cos
dtFj
dtFtftftftftf oddevenoddeven
Fourier TransformSOLO
Examples of Fourier Transform
Fourier TransformSOLO
Examples of Fourier Transform
Fourier TransformSOLO
Examples of Fourier Transform
Fourier TransformSOLO
Examples of Fourier Transform
Fourier TransformSOLO
Examples of Fourier Transform
Fourier TransformSOLO
Examples of Fourier Transform
Fourier Transform
2
10
2
11
t
tt
2
12
1
t
t
Rectangle1
t
tt
t
t
2/1
2/
2/1
lim
Limiter
tt sgnlimlim20
0
tlim
t
2/1
2/1
SOLO
Special Symbols
10
11
t
ttt
11
t
t
Triangle1
00
01
t
ttH
0
tH
t
Heavisideunit step
1
01
01sgn
t
tt
0
tsgn
t
Signum1
1
0
ttd
dlim
t
2/1
Area = 1td
d
Fourier Transform
00
0
0
2/1lim
2/1
2/
2/1
limlimlim:000 t
t
t
t
t
tt
t
td
dt
td
dt
SOLO
Special Symbols
0
ttd
dlim
t
2/1
Area = 1td
d
δ (t) function
Since tt sgn2
1limlim
0
we have also
δ (t) function is defined as:
ttd
dt sgn
2
1
0
ttd
dlim
t
2/1
Area = 1
0 t
tArea = 1
0
t
tt
t
t
2/1
2/
2/1
lim
Limiter
tt sgnlimlim20
0
tlim
t
2/1
2/1
Fourier TransformSOLO
Special Symbols
Properties of δ (t) function
0
ttd
dlim
t
2/1
Area = 1
0 t
tArea = 1
0
tt δ (t) is a even function: 2
00
0
0
2/1lim
0 t
t
t
tt
1
3
002
1
ffdtttfdtttfuu
Proof:
002
1
002
1lim
2
1lim
sgn2
1limsgn
2
1limsgnlim
2
1lim
0
0
sgn2
1
ff
fTfTffTfTftfdtfdTfTf
tfdtttftdtfdtttf
T
T
TT
T
TT
T
TT
T
TT
tdt
dtT
TT
4 Fourier Transform 10exp2
10exp
2
1exp
jjdttjtt F
Fourier Transform
xLn
x
xJ
xAi
x
x
x
x
xx
2exp
1lim
11lim
1lim
sin1
lim
4exp
2
1lim
lim
lim1
2
0
/10
0
0
2
0
1
0
220
SOLO
Special Symbols
δ (t) function The δ (t) function can be defined as the following limit as ε→0
Ai is the Airry function,
0
3
3cos
1dttx
txAi
dxnjxJ n sinexp2
1
Friedrich WilhelmBessel
1784 - 1846
Edmond NicolasLaguerre
1834 - 1886
Jn (x) is the Bessel function of the first kind,
and Ln (x) is the Laguerre polynomial of arbitrary positive order.
Fourier TransformSOLO
Special Symbols
δ (t) function The δ (t) function can be defined also by the limit n→∞
x
xn
xn
2
1sin
2
1sin
2
1lim
tnsincn
tnn
xnnx
n
n
n
lim
lim
explim 22
2/10
2/11
x
xx
x
xxsinc
sin
Fourier TransformSOLOδ (t) function
0
ttd
dlim
t
/1Area = 1
0 t
tArea = 1
0
2/2/
20
2
1
:
02
t
tet
tfj
Use
It’s Fourier Transform is
ff
ff
ffj
edtedtetf
tffjtffjtfj
0
0
2/
2/0
22/
2/
22 sin
2
11 0
0
For any function φ (t), defined at t=0- and t=0+, we have
002
11lim
1lim
1lim
1lim
1limlim
0
2/0
2/
00
0
2/
2
0
2/
0
2
0
2/
2/
2
00
000
tttt
dttedttedttedttt tfjtfjtfj
tt
0lim
Fourier Transform
20
2
1
:
02
t
tet
tfj
SOLOδ (t) function
ff
fff
0
0sin
tt
0lim
1sin
limlim0
0
00
ff
fff
0
0.25
1.00
89.0
3 dBf
1
4 dBf
2
nnf
1
0 f1
0 f0f
0.50
0.75
ff
ff
0
0sin
fdet tfj 2
Fourier Transform
2/0
2/1
:
0
0
fff
fffffS f
SOLOδ (f) function
Define:0f
f
f/1 Area =1
f
0ff Area = 1
0 f
2/0 ff 2/0 ff 0f
In the time domain we obtain:
tfj
ff
ff
tfjff
ff
tfjtfjff e
tf
tf
tj
e
ffde
ffdefSts 0
0
0
0
0
2
2/
2/
22/
2/
22 sin
2
11
For any function Φ (f), defined at f=f0- and f=f 0+ , we have
00
2/0
2/
0
2/0
2/
0
2/
2/00
2
11lim
1lim
1lim
1lim
1limlim
0
0
0
0
0
0
0
0
0
0
fffff
fff
dfff
dfff
dfff
dfffS
f
fff
ff
ff
f
fff
ff
ff
ff
fff
ff
00
lim fffS ff
00
:lim fffS ff
tfj
ff
ets 02
0lim
tdeff tffj 020
Fourier Transform
N
NnN Tntftf :
SOLOfN (f) N-Periodic Extension of a function f (t)
DefineN- extension of f (t)
0
N
NnN Tntftf
t
0 T1
T1 + T2 > T
-T2
tf
t
T
Fourier Transform
N
NnN Tntt :
SOLOδN (f) function
Define
Let find the Fourier transform of δN (f)
Tf
TNf
ee
j
j
ee
eee
eee
e
ee
etdenTttdetf
TfjTfj
TNfjTNfj
TfjTfjTfj
TNfjTNfjTfj
Tfj
NTfjTNfj
N
Nn
TnfjN
Nn
tfjtfjNN
sin
21
2sin2
2
1
1
2
12
2
12
2
12
2
12
2
1222
222
We can see that
,2,1,0
sin
12sin
sin
1212sin
kf
Tf
TNf
kTf
NkTNf
T
kf NN
N-extension of δ (t)
Fourier TransformSOLOδN (f) function (continue – 1)
N
NnN Tntt :
N
Nn
TnfjN e
Tf
TNff
2
sin
12sin
NT 0
fN12 N
TN 12
1
Tf
10 TN 12
1
T2
1
T2
1
T
1
f
0-NT T 2T-T
tN
t
δN (t) is a periodic function with a time period of T .
ΔN (f) is a periodic function with a frequency period of f0 = 1/T .
Tn
n
TTnj
edfedff
N
Nn
nn
N
Nn
T
T
TnfjN
Nn
T
T
TnfjT
T
N
1sin1
2
0001
2/1
2/1
22/1
2/1
22/1
2/1
Fourier TransformSOLOδN (f) function (continue – 2)
NT 0
fN12 N
TN 12
1
Tf
10 TN 12
1
T2
1
T2
1
T
1
f
0-NT T 2T-T
tN
t
When N → ∞ the peak goes to infinity and the null-to-null bandwidth goes to zero.This resembles to a delta function. To prove that this is the case let compute:
ΔN (f) is a periodic function with a frequency period of f0 = 1/T , with peak amplitude of(2 N+1) and null-to-null bandwidth of 2/ [(2N+1) T].
Tdff
T
T
N
12/1
2/1
01
limlim2/1
2/1
22/1
2/1
T
dffedfffN
Nn
T
T
Tnfj
N
T
T
NN
Fourier Transform
0
m Tmf
Tf
11
Tf
10
T
1
f
0 nTT 2T-T
n
Tnt
t
T
1
T
1
T
2
T
2T
m
SOLOδN (f) function (continue – 3)
Tdff
T
T
N
12/1
2/1
01
limlim2/1
2/1
22/1
2/1
T
dffedfffN
Nn
T
T
Tnfj
N
T
T
NN
Therefore
m
T
T
NN T
mf
Tdfff 1
lim:2/1
2/1
Fourier TransformSOLOδN (f) function (continue – 4)
Let compute the convolution between f (t) and δN (f)
tfTntfdTntfdTntfttf N
N
Nn
N
Nn
N
NnN
:
Therefore ttftf NN Using this relation the Fourier Transform of fN (t) is given by
Tf
TfNfFffFfF NN
sin
12sin Tntfttftf
N
NnNN
If N → ∞ then
mm
m
T
mf
T
mF
TT
mffF
T
T
mf
TfFffFfF
11
1
m
tT
mj
m
n
eT
mF
T
T
mf
T
mF
T
Tntftf
2
1
1
1F
Fourier TransformSOLOδN (f) function (continue – 4)
m T
mf
T
mF
TfF 1
m
tT
mj
n
eT
mF
TTntftf
21
f∞ (t) is a periodic function with a time period of T .
F∞ (f) is a periodic function with a frequency period of f0 = 1/T .
We obtained the Fourier Series description of a periodic function
tdetf
TT
mF
Taeatf
tT
mj
mm
tT
mj
m
22 11
If we define
2/0
2/0
Tt
Tttftf
2/
2/
20
T
T
tfj tdetffF
then
2/
2/
22 1 T
T
tT
mj
mm
tT
mj
m tdetfT
aeatf
Fourier Transform
xxxf
SOLO
Simple Fourier Series
0cos1
cos1
dxxnxdxxnxfan
nn
xn
n
xnx
dxxnxdxxnxfb
n
n
1
0
2
0
12
sincos2
sin1
sin1
http://en.wikipedia.org/wiki/Fourier_series
Square wave equation
xNN
xxxfSN 1sin1
13sin
3
1sin
Sawtooth wave equation
http://en.wikipedia.org/wiki/Sawtooth_wave
http://en.wikipedia.org/wiki/Square_wave
xnn
xxxfSn
N sin1
22sinsin21
http://mathworld.wolfram.com/FourierSeries.html
Fourier Transform
122
sin
2sin
8
kTriangle k
xkkxf
SOLO
Simple Fourier Series
Triangular wave equation
Fourier Transform
122
sin
2sin
8k
Triangle k
xkkxf
SOLO
Simple Fourier Series
Triangular wave equation
http://mathworld.wolfram.com/FourierSeries.html
SignalsSOLO
Signal Duration and Bandwidth
then
tdetsfS tfi 2
fdefSts tfi 2
t
t2
t
2ts
ff
f2
2fS
2/1
2
22
:
tdts
tdtstt
t
tdts
tdtst
t2
2
:
Signal Duration Signal Median
2/1
2
2224
:
fdfS
fdfSff
f
fdfS
fdfSf
f2
22
:
Signal Bandwidth Frequency Median
Fourier
Signals
fdefSts tfi 2
SOLO
Signal Duration and Bandwidth (continue – 1)
dffSfSdfdesfS
dfdesfSdfdefSsdss
tfi
tfitfi
2
22
fdefSts tfi 2
fdefSfitd
tsdts tfi 22'
dffSfSfdfdesfSfi
dfdesfSfidfdefSfsidss
tfi
tfitfi
222
22
2'2
'2'2''
dffSds 22
Parseval Theorem
From
From
dffSfdtts2222
4'
Signals
dffS
fdfd
fSdfS
i
dffS
fdtdetstfS
dffS
tdfdefStst
dffS
tdtstst
tdts
tdtst
t
fifi
22
2
2
2
22
2
2:
SOLO
Signal Duration and Bandwidth
tdetsfS tfi 2
fdefSts tfi 2Fourier
tdetstifd
fSd tfi 22
fdefSfitd
tsd tfi 22
tdts
tdtd
tsdtsi
tdts
tdfdefSfts
tdts
fdtdetsfSf
tdts
fdfSfSf
fdfS
fdfSf
f
fifi
22
2
2
2
22
2 2222
:
Signals
dffSfdttstdttsdttstdtts
222222
2
2 4'4
1
dffSdts22
SOLO
Signal Duration and Bandwidth (continue – 1)
0&0 ftChange time and frequency scale to get
From Schwarz Inequality:
dttgdttfdttgtf22
Choose tstd
tsdtgtsttf ':&
dttsdttstdttstst22
''we obtain
dttstst 'Integrate by parts
sv
dtstsdu
dtsdv
stu '
'
dttststdttsstdttstst '' 2
0
2
dttsdttstst 2
2
1'
dffSfdtts2222
4'
dffS
dffSf
dtts
dttst
dtts
dffSf
dtts
dttst
2
222
2
2
2
222
2
244
4
1
assume 0lim
tstt
SignalsSOLO
Signal Duration and Bandwidth (continue – 2)
22
2
222
2
24
4
1
ft
dffS
dffSf
dtts
dttst
Finally we obtain ft 2
1
0&0 ftChange time and frequency scale to get
Since Schwarz Inequality: becomes an equalityif and only if g (t) = k f (t), then for:
dttgdttfdttgtf22
tftsteAttd
sdtgeAts tt 222:
22
we have ft 2
1
Laplace’s Transform
sdej
jdej
fdet tsf
js
tjf
js
tfj
2
1
2
1 2:
:
2:
:
2
f
ts dtetftfsF0
L
SOLO
Laplace L-Transform
Laplace’s Transform
To find the Inverse Laplace’s Transform (L -1) we use:
00
dsdefdsedefdsesFj
j
tsj
j
tssj
j
ts
tfdtf
j
j
ts dsesFj
tf2
1
For a signal f (t) we define the Laplace’s Transform (L)
Pierre-Simon Laplace1749-1827
0
2 dtfj tfj2
Laplace’s Transform
C2
f
a
0t
00
t
js s - plane
SOLO
Laplace L-Transform (continue – 1)
The Inverse Laplace’s Transform (L -1) is given by:
j
j
ts dsesFj
tf2
1
Using Jordan’s Lemma (see “Complex Variables” presentation or the end of this one)
Jordan’s Lemma GeneralizationIf |F (z)| ≤ M/Rk for z = R e iθ where k > 0 and M are constants, then
for Γ a semicircle arc of radius R, and center at origin:
00lim
mzdzFe zm
R
where Γ is the semicircle, in the left part of z plane.
x
y
R
we can write
j
j
tstsf
f
dsesFj
dsesFj
sFtf
2
1
2
11-L
dsesFj
dsesFj
dsesFj
sFtf ts
C
tsj
j
ts
2
1
2
1
2
1
0
1-L
If the F (s) has no poles for σ > σf+, according to Cauchy’s Theoremwe can use a closed infinite region to the left of σf+, to obtain
Laplace’s TransformSOLO
Properties of Laplace L-Transform
s - Domaint - Domain
tf
f
st sdtetfsF Re0
1 if
M
iii zsFc maxRe
1
Linearity
M
iii tfc
1
3 000 1121 nnnn ffsfssFs Differentiation n
n
td
tfd
4
t
tdf
ss
sF 0
lim1Integration
t
df
5 s
sFReal DefiniteIntegration
t
df0
t
ddf0 0
2s
sF
2
a
sF
a
1Scaling taf
Laplace’s TransformSOLO
Properties of Laplace L-Transform (continue – 1)
s - Domaint - Domain
tf
f
st sdtetfsF Re0
6 n
n
sd
sFdMuliplicity by tn tft n
7
0
dssFDivision by t t
tf
8 sFe sTime shifting tutf
9 asF Complex Translations
tfe ta
10 sHsF Convolutiont - plane
0
dthfthtf
11
j
j
dsHFj
sHsFj
2
1
2
1Convolution s - plane
thtf
Laplace’s TransformSOLO
Properties of Laplace L-Transform (continue – 2)
s - Domaint - Domain
tf
f
st sdtetfsF Re0
12 Initial Value Theorem sFstfst
limlim0
13 Final Value Theorem sFstfst 0limlim
14 Parseval’s Theorem
j
j
j
j
ts
j
j
ts
dssGsFj
dsdtetgsFj
dttgdsesFj
dttgtf
2
1
2
1
2
1
0
00
tf
0n
T Tntt
0
*
n
T TntTnfttftf
tf *
tfT t
f
ts dtetftfsF0
L
SOLO
Sampling and z-Transform
0
1
1
00sT
n
sTn
n
T eeTnttsS LL
0
00**
1
1
2
1
f
j
j
tsT
n
sTn
n
de
Fj
ttf
eTnfTntTnf
tfsF
L
LL
tse
ofPoleststs
FofPoles
tsts
n
nsT
e
FResd
e
F
j
e
FResd
e
F
j
eTnf
sF
1
1
0
*
112
1
112
1
2
1
Poles of
Tse 1
1
Poles of
F
planes
Tnsn
2
j
j
0s
Laplace Transforms
The signal f (t) is sampled at a time period T.
12
R
R
Poles of
Tse 1
1
Poles of
F
plane
Tnsn
2
j
j
0s
Z Transform
Fourier Transform
tf
0n
T Tntt
0
*
n
T TntTnfttftf
tf *
tfT t
SOLO
Sampling and z-Transform (continue – 1)
nnTse
nts
T
njs
T
njs
e
ofPolests
T
njsF
TeT
Tn
jsF
T
njsF
eT
njs
e
FRessF
ts
n
ts
212
lim
2
1
2
lim1
1
2
21
1
*
Poles of
F
j
0s
T
2
T
2
T
2
Poles of
*F plane
js
The signal f (t) is sampled at a time period T.
The poles of are given by tse 1
1
T
njsnjTsee n
njTs 221 2
n T
njsF
TsF
21*
Fourier TransformSOLO
F F-1
frequency-B/2 B/2B
F F-1
-B/2 B/2
B
1/Ts-1/Ts frequency
Sample
Sampling a function at an interval Ts (in time domain)
Anti-aliasing filters is used to enforce band-limited assumption.
causes it to be replicated at 1/ Ts intervals in the other (frequency) domain.
Sampling and z-Transform (continue – 2)
Bandlimited Continuous Time Signal
1/B sec
ampl
itud
e
time (sec)
-0.4
-0.2
0.2
0
0.4
0.6
0.8
1
0 5 10 15-15 -10 -5
Discrete-Time (Sampled) Signal
ampl
itud
e
sample
-0.4
-0.2
0.2
0
0.4
0.6
0.8
1
0 10 20-20 -10
Fourier Transform
tf
0n
T Tntt
0
*
n
T TntTnfttftf
tf *
tfT t
SOLO
Sampling and z-Transform (continue – 3)
0z
planez
Poles of
zF
C
The signal f (t) is sampled at a time period T.
The z-Transform is defined as:
iF
iF
iiF
Ts
FofPoles
T
F
n
n
ze
ze
F
zTnf
zFsFtf
1
0*
1
lim:Z
00
02
1 1
n
RzndzzzFjTnf
fCC
n
Fourier TransformSOLO
Sampling and z-Transform (continue – 4)
0
* 21
n
nsT
n
eTnfT
njsF
TsF
We found
The δ (t) function we have:
1
dtt fdtttf
The following series is a periodic function: n
Tnttd :
therefore it can be developed in a Fourier series:
n
n
n T
tnjCTnttd 2exp:
where: T
dtT
tnjt
TC
T
T
n
12exp
12/
2/
Therefore we obtain the following identity:
nn
TntTT
tnj 2exp
Second Way
Fourier Transform
dttjtftfF 2exp:2 F
0
* 21
n
nsT
n
eTnfT
njsF
TsF
dtjFFtf 2exp2:2-1F
SOLOSampling and z-Transform (continue – 5)
We found
Using the definition of the Fourier Transform and it’s inverse:
we obtain
dTnjFTnf 2exp2
0
111
0
* exp2exp2expnn
n sTndTnjFsTTnfsF
111
* 2exp22 dTnjFjsFn
nn T
nF
Td
T
n
TFjsF 2
1122 111
*
We recovered (with –n instead of n)
n T
njsF
TsF
21*
Second Way (continue)
Making use of the identity: with 1/T instead of T
and ν - ν 1 instead of t we obtain:
nn T
n
TTnj 11
12exp
nn
TntTT
tnj 2exp
Claude Elwood Shannon 1916 – 2001
http://en.wikipedia.org/wiki/Claude_E._Shannon
Fourier TransformSOLO
Henry Nyquist1889 - 1976
http://en.wikipedia.org/wiki/Harry_Nyquist
Nyquist-Shannon Sampling Theorem
The sampling theorem was implied by the work of Harry Nyquist in 1928 ("Certain topics in telegraph transmission theory"), in which he showed that up to 2B independent pulse samples could be sent through a system of bandwidth B; but he did not explicitly consider the problem of sampling and reconstruction of continuous signals. About the same time, Karl Küpfmüller showed a similar result, and discussed the sinc-function impulse response of a band-limiting filter, via its integral, the step response Integralsinus; this band-limiting and reconstruction filter that is so central to the sampling theorem is sometimes referred to as a Küpfmüller filter (but seldom so in English).
The sampling theorem, essentially a dual of Nyquist's result, was proved by Claude E. Shannon in 1949 ("Communication in the presence of noise"). V. A. Kotelnikov published similar results in 1933 ("On the transmission capacity of the 'ether' and of cables in electrical communications", translation from the Russian), as did the mathematician E. T. Whittaker in 1915 ("Expansions of the Interpolation-Theory", "Theorie der Kardinalfunktionen"), J. M. Whittaker in 1935 ("Interpolatory function theory"), and Gabor in 1946 ("Theory of communication").
http://en.wikipedia.org/wiki/Nyquist-Shannon_sampling_theorem
Fourier Transform
Poles of
S
j
0s
plane
js
fB2
SOLO
Nyquist-Shannon Sampling Theorem (continue – 1)
• Signal can be recovered if Fourier spectrum of the sampling signal do not overlap.
sB
2/sB2/sBf
A
fB
2/fB2/fB
f
A
sTA /
sT/1
Poles of
S
j
0s
sT
2Poles of
*Splane
js
fB2sT
2
sT
2
• Start with a band limited signal s (t) 2
0 fBfforfS
• Sample s (t) at a time period Ts, replicates spectrum every 1/Ts Hz.
k sTkfjSfS
12*
fjs 2
nsTnttsts *
k sTjksSsS
2*
L-1
L
FF-1
Fourier Transform
2
1
2
B
T
B
s
SOLO
Nyquist-Shannon Sampling Theorem (continue – 2)
• Signal can be recovered if Fourier spectrum of the sampling signal do not overlap.
BB
Ts
22
1
(Nyquist Sampling Rate)
• Complex signal band-limited to B/2 Hz requires B complex samples/second, or 2 B real samples/seconds (twice the highest frequency)
sB
2/sB2/sBf
A
fB
2/fB2/fB
f
A
sTA /
sT/1
• Start with a band-limited signal f (t) 2
0 fBfforfF • Sample f (t) at a time period Ts,
replicates spectrum every 1/Ts Hz.
Nyquist-Shannon Sampling Theorem:
fB
2/fB2/fB
f
sTA /
sT/1
BandlimitedFilter
Fourier TransformSOLO
The Discrete Time Fourier Transform (DTFT)
sB
2/sB2/sBf
A
fB
2/fB2/fB
f
A
sTA /
sT/1
• Start with a band limited signal s (t) 2
0 fBfforfS
• Sample s (t) at a time period Ts, replicates spectrum every 1/Ts Hz.
k sTkfSfS
1*
nss
ns
TntTns
Tnttsts
*
tdetsfS tfj 2
fdefSts tfj 2F
F-1
Continuous Fourier Transform
F
F-1
sT
ts*
tsq
t ts
Discretization of a Continuous Signal
fdefSTnts sTnfjs
2
n
nf
fj
s
Tf
n
TnfjsDTFT
ss
s
s eTnseTnsfS
2
1
2:
DTFT provides an approximation of the continuous-time Fourier transform.
Discrete Time Fourier Transform (DTFT)Define
Fourier TransformSOLO
The Discrete Time Fourier Transform (DTFT) (continue-1)
• Signal can be recovered if Fourier spectrum of the sampling signal do not overlap.
Discretization of a Continuous Signal
fdefSTnts sTnfjs
2
DTFT-
1
DTFT
Discrete Time Fourier Transform (DTFT)
n
nf
fj
s
Tf
n
TnfjsDTFT
ss
s
s eTnseTnsfS
2
1
2:
We can see that
n
DTFTnkj
nf
fj
sn
nf
fkfj
ssDTFT fSeeTnseTnsfkfS ss
s
1
222
The Discrete Time Fourier Transform SDTFT (fs) is periodic with period fs.Let compute
ns
sn
nmnm
ss
f
fs
nmf
fj
s
n
f
f
nmf
fj
s
f
f n
nmf
fj
s
f
f
mf
fj
DTFT
TmsTnm
nmfTns
fnm
j
eTns
fdeTnsdfeTnsdfefS
s
s
s
s
s
s
s
s
s
s
s
s
1sin
2
10
2/
2/
2
2/
2/
22/
2/
22/
2/
2
n
TnfjsDTFT
seTnsfS 2:
s
s
s
T
T
nTfjDTFTss dfefSTTns
2/1
2/1
2
Fourier TransformSOLO
The Discrete Time Fourier Transform (DTFT) (continue-2)
Normalization of the frequency
DTFT-1
DTFT
n
TnfjsDTFT
seTnsfS 2:
s
s
s
T
T
nTfjDTFTss dfefSTTns
2/1
2/1
2
2/1,2/1
2/1,2/1
:
*
*
f
TTf
Tff
ss
s
n
nfjDTFT ensfS *2* :
DTFT-1
DTFT
2/1
2/1
*2 ** dfefSns nfjDTFT
Example 1,,1,002 NneAns nfj
1*
0
0
*
*
**
**
*2
*21
0
*2*
0
0
0
00
00
0
0
0
*sin
*sin
1
1
Nffj
ffj
Nffj
ffjffj
NffjNffj
ffj
NffjN
n
nffjDTFT
eff
NffA
e
e
ee
eeA
e
eAeAfS
0.50.40.30.20.1-0.1 0-0.2-0.3-0.4-0.5
N = 20
20
15
10
5
0
13.2 dB
f0 = 0.25|SDTFT(f*)|
Normalized Frequency
Fourier TransformSOLO
The Discrete Time Fourier Transform (DTFT) (continue-3)
n
nfjDTFT ensfS *2* :
DTFT-1
DTFT
2/1
2/1
*2 ** dfefSns nfjDTFT
Example
22&8,,00
21,,10,902
nn
nens
nfj
0
5
10
15
0 50 100 150 200 250 300
Δf
SPECTRUM OF 12-SAMPLE PULSE
DTFT
Sig
nal a
mpl
itud
e
27&4,,00
26,,10,302
nn
nens
nfj
0
10
20
30
0 50 100 150 200 250 300
Δf /2
SPECTRUM OF 24-SAMPLE PULSE
Sign
al a
mpl
itude
DTFT
0
0.5
1
0 5 10 15 20 25 30
12-SAMPLE PULSE
Signal sample
Sign
al am
plitu
de
0
0.5
1
0 5 10 15 20 25 30
24-SAMPLE PULSE
Signal sample
Sign
al am
plitu
de
Frequency Resolution Increases with Observation Time N Ts
DTFT
DTFT
Fourier Transform
1
0
2
:N
n
nkN
j
sDFT eTnskS
SOLO
The Discrete Fourier Transform (DFT)
Assume a periodic sequence, sampled at a time period Ts, such that s (n Ts) = s [(n+kN) Ts]
The Discrete Fourier Transform (DFT) requires an input function that is discrete and whose non-zero values have a limited (finite) duration.
Unlike the Discrete-time Fourier transform (DTFT), it only evaluates enough frequency components to reconstruct the finite segment that was analyzed. Its inverse transform cannot reproduce the entire time domain, unless the input happens to be periodic (forever). Therefore it is often said that the DFT is a transform for Fourier analysis of finite-domain discrete-time functions
For the sequence s (0), s (Ts),…,s [(N-1) Ts] we define the Discrete Fourier Transform:
ts*
sT sT tsT
N Ts N Ts N Ts
Fourier Transform
1
0
1
0
2
:N
n
nks
N
n
nkN
j
sDFT WTnseTnskS
SOLO
The Discrete Fourier Transform (DFT) (continue – 1)
For the sequence s (0), s (Ts),…,s [(N-1) Ts] we define the Discrete Fourier Transform:
where is a primitive N'th root of unityand is periodic
Nj
eW2
:
7
8
2
j
e
0
8
2
j
e
1
8
2
j
e
2
8
2
j
e3
8
2
j
e
4
8
2
j
e
5
8
2
j
e 6
8
2
j
e
1
n
Nm
Nj
n
Nj
Nmn
Nj
Nmn WeeeW
1
222
N
N
N s
s
s
s
s
s
W
NNNNNNN
NNNNNNN
NN
NN
NN
S
DFT
DFT
DFT
DFT
DFT
TNs
TNs
Ts
Ts
Ts
WWWWW
WWWWW
WWWWW
WWWWW
WWWWW
NS
NS
S
S
S
1
2
2
1
0
1
2
2
1
0
1121211101
1222221202
1222221202
1121211101
1020201000
NNN sWS NW is a Vandermonde type of Matrix
Fourier TransformSOLO
The Discrete Fourier Transform (DFT) (continue – 2)
nNmn WW
NH
NN IN
WW1
Nj
eW2
1
2* WeW N
j
1121211101
1222221202
1222221202
1121211101
1020201000
NNNNNNN
NNNNNNN
NN
NN
NN
N
WWWWW
WWWWW
WWWWW
WWWWW
WWWWW
W
1112121110
2122222120
2122222120
1112121110
0102020100
*
NNNNNNN
NNNNNNN
NN
NN
NN
TN
HN
WWWWW
WWWWW
WWWWW
WWWWW
WWWWW
WW
Let multiply those two matrices
mkN
mkW
W
W
WW
WWWWWWWWWW
mk
mk
N
mk
NmkN
j
jmk
mNNkmjjkmkmkmk
HNN
01
1
1
1
1
1
0
111100
,
Where IN is the NxN identity matrix
Fourier Transform
1
0
1
0
2
:N
n
nks
N
n
nkN
j
sDFT WTnseTnskS
SOLO
The Discrete Fourier Transform (DFT) (continue – 3)
For the sequence s (0), s (Ts),…,s [(N-1) Ts] we defined the Discrete Fourier Transform:
7
8
2
j
e
0
8
2
j
e
1
8
2
j
e
2
8
2
j
e3
8
2
j
e
4
8
2
j
e
5
8
2
j
e 6
8
2
j
e
1
NNN sWS NW is a Vandermonde type of Matrix
We found that
NH
NN IN
WW1
Where IN is the NxN identity matrix
Therefore the Inverse Discrete Fourier Transform (IDFT) is
NH
NN SWN
s1
1
0
21
0
11 N
n
nkN
j
DFT
N
k
nkDFTs ekS
NWkS
NTns
D.F.T.
I.D.F.T.
Fourier TransformSOLO
The Discrete Fourier Transform (DFT) (continue – 4)
Second way to find the Inverse Discrete Fourier Transform (IDFT). Let compute:
1
0
1
0
21
0
1
0
21
0
2 N
n
N
k
rnkN
j
s
N
k
N
n
rnkN
j
s
N
k
rkN
j
DFT eTnseTnsekS
Nmrn
NmrnN
rnN
jrnN
rnjrn
rnN
rnN
rn
rnN
rnN
jrnN
rnjrn
rnN
rn
rnN
jrnN
rnjrn
e
e
e
e
ern
Nj
rnj
rnN
j
Nrn
Nj
N
k
rnkN
j
0cossin
cossin
sin
sin
cossin
cossin
sin
sin
2sin
2cos1
2sin2cos1
1
1
1
1
2
2
2
2
1
0
2
,2,1,01
0
2
mTmNrsNekS s
N
k
rkN
j
DFT
Fourier Transform
1
0
1
0
2
:N
n
nks
N
n
nkN
j
sDFT WTnseTnskS
SOLO
The Discrete Fourier Transform (DFT) (continue – 1)
For the sequence s (0), s (Ts),…,s [(N-1) Ts] we define the Discrete Fourier Transform:
where is a primitive N'th root of unityand is periodic
Nj
eW2
:
7
8
2
j
e
0
8
2
j
e
1
8
2
j
e
2
8
2
j
e3
8
2
j
e
4
8
2
j
e
5
8
2
j
e 6
8
2
j
e
1
n
Nm
Nj
n
Nj
Nmn
Nj
Nmn WeeeW
1
222
s
s
s
s
s
NN
NN
NN
NN
DFT
DFT
DFT
DFT
DFT
TNs
TNs
Ts
Ts
Ts
WWWWW
WWWWW
WWWWW
WWWWW
WWWWW
NS
NS
S
S
S
1
2
2
1
0
1
2
2
1
0
12210
23320
23420
12210
00000
Fourier TransformSOLO
The Discrete Fourier Transform (DFT) (continue – 5)
The DFT ant Inverse DFT (IDFT) are given by
1
0
21 N
k
nkN
j
DFTs ekSN
Tns
1
0
2
:N
n
nkN
j
sDFT eTnskS
IDFT
DFT
with the periodic properties
,2,1,0
m
TnsTmNns ss
,2,1,0
m
kSNmkS DFTDFT
The sequence s (0), s (Ts),…,s [(N-1) Ts] can be interpreted to be a sequence of finitelength, given for r = 0, 1,…,N-1, and zero otherwise or a periodic sequence, defined for all r.
ts*
sT sT tsT
N Ts N Ts N Ts
ts*
sT tN Ts
Fourier Transform
1
0
2
:N
n
nkN
j
sDFT eTnskS
SOLO
The Discrete Fourier Transform (DFT) (continue – 6)
The DFT ant Inverse DFT (IDFT) are given by
1
0
21 N
k
nkN
j
DFTs ekSN
Tns
IDFT
DFT
n
nfjDTFT ensfS *2* :
2/1
2/1
*2 ** dfefSns nfjDTFT
IDTFT
DTFT
The DTFT ant Inverse DTFT (IDTFT) where given by
We can see that DFT is a sampled version of DTFT by tacking:
2/1,2/1
2/1,2/1
1,,1,0
*
*
f
TTf
NkTN
kf
N
kfTf
ss
ss
1,,1,0:1
0
2
NkfSeTnskSsTN
kfDTFT
N
n
nkN
j
sDFT
Fourier TransformSOLO
The Discrete Fourier Transform (DFT) (continue –7)
We can see that DFT is a sampled version of DTFT :
1,,1,0:1
0
2
NkfSeTnskSsTN
kfDTFT
N
n
nkN
j
sDFT
10 12 14 16 18 0 2 4 6 8 k
f0 0.1 0.2 0.3 0.4 0.5-0.1-0.2-0.3-0.4-0.50
5
10
15
20
25
s [n]= exp (j 2 π f0 n), 0 ≤ n ≤ 19N = 20
f0 = 0.25
|SD
TF
T (
f)| a
nd
|SD
TF (
k)|
10 12 14 16 18 0 2 4 6 8 k
f0 0.1 0.2 0.3 0.4 0.5-0.1-0.2-0.3-0.4-0.50
5
10
15
20
25s [n]= exp (j 2 π f0 n), 0 ≤ n ≤ 19
N = 20
f0 = 0.275
|SD
TF
T (
f)| a
nd
|SD
FT (
k)|
By changing f0 from 0.25 to 0.275 we move |SDTFT (f)| to the right, and since the samplingpoints didn’t change, we obtain different |SDFT (k)| values.
Fourier TransformSOLO
The Discrete Fourier Transform (DFT) (continue – 8)
We can see that DFT is a sampled version of DTFT :
1,,1,0:1
0
2
NkfSeTnskSsTN
kfDTFT
N
n
nkN
j
sDFT
10 12 14 16 18 0 2 4 6 8 k
f0 0.1 0.2 0.3 0.4 0.5-0.1-0.2-0.3-0.4-0.50
5
10
15
20
25
s [n]= exp (j 2 π f0 n), 0 ≤ n ≤ 19N = 20
f0 = 0.25
|SD
TF
T (
f)| a
nd
|SD
TF (
k)|
10 12 14 16 18 0 2 4 6 8 k
f0 0.1 0.2 0.3 0.4 0.5-0.1-0.2-0.3-0.4-0.50
5
10
15
20
25s [n]= exp (j 2 π f0 n), 0 ≤ n ≤ 59
N = 60
f0 = 0.25
|SD
TF
T (
f)| a
nd
|SD
FT (
k)|
Increase sampling density from N=20 to N=60.
SOLOProperties of The Discrete Fourier Transform (DFT) (continue – 9)
mns mkN
j
DFT ekS2
Linearity 1 nsns 2211
Shift of a Sequence2
3
4
5
Periodic Convolution
6
7
Conjugate
8
9
IDFTDFT
1
0
2
:N
n
nkN
j
DFT enskS
1
0
21 N
k
nkN
j
DFT ekSN
ns
kSkS DFTDFT 2211
nsns 21 , Periodic Sequence(Period N)
kSkS DFTDFT 21 , DFT(Period N)
nlN
jens
2 lkSDFT
1
021
N
m
mnsms kSkS DFTDFT 21
nsns 21
1
021
1 N
lDFTDFT lkSlS
N
ns kSDFT
ns kSDFT
Real & Imaginary nsRe
nsImj
2/kSkSkS DFTDFTeven
2/kSkSkS DFTDFTodd
SOLOProperties of The Discrete Fourier Transform (DFT) (continue – 10)
2/: nsnsnseven kSDFTReEven Part10
11
12 Symmetric Proprties)only when s (n) is real(
Parseval’s Formula
IDFTDFT
1
0
2
:N
n
nkN
j
DFT enskS
1
0
21 N
k
nkN
j
DFT ekSN
ns
nsns 21 , Periodic Sequence(Period N)
kSkS DFTDFT 21 , DFT(Period N)
lkSDFT
kSkS
kSkS
kSmkSm
kSkS
kSkS
DFTDFT
DFTDFT
DFTDFT
DFTDFT
DFTDFT
II
ReRe
Odd Part 2/: nsnsnsodd
Fourier TransformSOLO
Fast Fourier Transform (FFT)
John Wilder Tukey 1915 – 2000
http://en.wikipedia.org/wiki/John_Tukey
James W. Cooley1926 -
http://www.ieee.org/portal/pages/about/awards/bios/2002kilby.html
The Cooley-Tukey algorithm, is the most common fast Fourier transform (FFT) algorithm. It re-expresses the discrete Fourier transform (DFT) of an arbitrary composite size N = N1N2 in terms of smaller DFTs of sizes N1 and N2, recursively, in order to reduce the computation time to O(N log N) for highly-composite N (smooth numbers).
FFTs became popular after J. W. Cooley of IBM and John W. Tukey of Princeton published a paper in 1965 reinventing the algorithm (first invented by Gauss) and describing how to perform it conveniently on a computer
Fourier TransformSOLO
Fast Fourier Transform (FFT)
The radix-2 DIT Algorithm
The radix-2 decimation-in-time (DIT) FFT is the simplest and most common form of the Cooley-Tukey algorithm, although highly optimized Cooley-Tukey implementations typically use other forms of the algorithm as described below. Radix-2 DIT divides a DFT of size N into two interleaved DFTs (hence the name "radix-2") of size N/2 with each recursive stage.
1
0
1
0
2
:N
n
nks
N
n
nkN
j
sDFT WTnseTnskS
For the sequence s (0), s (Ts),…,s [(N-1) Ts] we define the Discrete Fourier Transform:
1,1, 22/12
*2
jNjevenN
NNj
Nj
eWeWWeWeW
Suppose N is a power of 2; i.e. N=2L (L is integer). Since N is a even integer, let compute SDFT (k) by separate s (nTs) into two (N/2)-point sequences consisting of the even-numberedpoints (n=2r) and odd numbered points (n=2r+1).
12/
0
212/
0
2
12/
0
1212/
0
2
122
122
N
n
kr
Nk
N
N
n
kr
N
N
n
krN
N
n
krNDFT
WrsWWrs
WrsWrskS
Fourier TransformSOLO
Fast Fourier Transform (FFT)
The radix-2 DIT Algorithm (continue – 1)
2/2/
2222
NN
jN
j
N WeeW
We divided the N-point DFT into two N/2-points DFTs.
kH
N
n
krN
kN
kG
N
n
krN
N
n
krN
kN
N
n
krNDFT
WrsWWrs
WrsWWrskS
12/
02/
12/
02/
12/
0
212/
0
2
122
122
Since
N/2 pointsDFT
N/2 pointsDFT
0s
2s 4s
2/Ns
1s
3s
5s
12/ Ns
ΣWNN/2
WNN/2+1
WNN/2+2
WNN-1
WNN/2-1
WN0
WN1
WN2
Σ
Σ
Σ
Σ
Σ
Σ
Σ
2/NSDFT
12/ NSDFT
22/ NSDFT
1NSDFT
1DFTS 2DFTS
0DFTS
12/ NSDFT
1G
2G
0G
12/ NG
1H
2H
0H
12/ NH
Fourier TransformSOLO
Fast Fourier Transform (FFT)
The radix-2 DIT Algorithm (continue – 2)
We divided the N-point DFT into two N/2-points DFTs.
4 pointsDFT
4 pointsDFT
0s
2s
4s
6s
1s
3s
5s
7s
ΣW84
W85
W86
W87
W83
Σ
Σ
Σ
Σ
Σ
Σ
Σ
4DFTS
5DFTS
6DFTS
7DFTS
1DFTS
2DFTS
0DFTS
3DFTS
1G
2G
0G
3G
1H
2H
0H
3H
W80
W81
W82
Reduction of an 8-points FFT to two4-points FFTs
2 pointsDFT
0s
2s
4s
6s
Σ
Σ
Σ
Σ
W40
W41
W42
2 pointsDFT
1G
2G
0G
3GW43
1DFTS
0DFTSΣ
Σ 1s
0s
A 2-points FFT
Reduction of an 4-points FFT to two2-points FFTs
Fourier TransformSOLO
Fast Fourier Transform (FFT)
The radix-2 DIT Algorithm (continue – 3)
ΣW84
W85
W86
W87
W83
Σ
Σ
Σ
Σ
Σ
Σ
Σ
4DFTS
5DFTS
6DFTS
7DFTS
1DFTS
2DFTS
0DFTS
3DFTS
1G
2G
0G
3G
1H
2H
0H
3H
W80
W81
W82
2 pointsDFT
Σ
Σ
Σ
Σ
W40
W41
W42
2 pointsDFT
W43
2 pointsDFT
Σ
Σ
Σ
Σ
W40
W41
W42
2 pointsDFT
W43
0s
2s
4s
6s
1s
3s
5s
7s
1st Stage 2nd Stage 3th Stage
Flow Diagram for an 8-points FFT
Fourier TransformSOLO
Fast Fourier Transform (FFT)
The radix-2 DIT Algorithm (continue – 2) kkj
kN
N
jNk
N eeW 12
2/
We divided the N-point DFT into two N/2-points DFTs.
12/
01
2/12/
0
2/ 2/2/N
n
knN
NkN
N
n
NnkN
knNDFT WWNnsnsWNnsWnskS
k
Since N/2 is an even integer (N=2L)
N/2 pointsDFT
N/2 pointsDFT
Σ
Σ
Σ
Σ
2DFTS 4DFTS
0DFTS
2NSDFT
1G
2G
0G
12/ NG
1H
2H
0H
12/ NH
1DFTS 3DFTS 5DFTS
1NSDFT
WN0
Σ
Σ
Σ
Σ
WN1
WN2
WNN/2-1
1s 2s
0s
12/ Ns
2/Ns 12/ Ns 22/ Ns
1Ns
tgofFFTN
N
n
nlN
WW
N
N
n
nlN
ng
DFT WngWNnsnslkSNN
L
2/
12/
02/
2
12/
0
2 2/2
2/2
thofFFTN
N
n
nlN
WW
N
N
n
nlN
nh
nNDFT WnhWWNnsnslkS
NN
L
2/
12/
02/
2
12/
0
2 2/2
2/12
Fourier TransformSOLO
Fast Fourier Transform (FFT)
The radix-2 DIT Algorithm (continue – 3)
We divided the N-point DFT into two N/2-points DFTs.
4 pointsDFT
4 pointsDFT
Σ
Σ
Σ
Σ
2DFTS
4DFTS
0DFTS
6DFTS
1G
2G
0G
3G
1H
2H
0H
3H
1DFTS
3DFTS
5DFTS
7DFTS
W80
Σ
Σ
Σ
Σ
W81
W82
W83
1s
2s
0s
3s
4s
5s
6s
7s
2- pointsDFT 2DFTS
1DFTS
0DFTS
1'G
0H
0'G
1H
Σ
Σ
Σ
Σ
1G
2G
0G
3G
2- pointsDFT 3DFTS
W40
W41
Reduction of an 8-points FFT to two4-points FFTs
Reduction of an 4-points FFT to two2-points FFTs
1DFTS
0DFTSΣ
Σ 1s
0s
A 2-points FFT(Butterfly)
Fourier TransformSOLO
Fast Fourier Transform (FFT)
The radix-2 DIT Algorithm (continue – 4)
Σ
Σ
Σ
Σ
W80
Σ
Σ
Σ
Σ
W81
W82
W83
1s
2s
0s
3s
4s
5s
6s
7s
2DFTS
4DFTS
0DFTS
6DFTS
1DFTS
3DFTS
5DFTS
7DFTS
Σ
Σ
Σ
Σ
W40
W41
Σ
Σ
Σ
Σ
W40
W41
Σ
Σ
Σ
Σ
Σ
Σ
Σ
Σ
1st Stage 2nd Stage 3th Stage
Flow Diagram for an 8-points FFT
Fourier Transform
1,,1,0:1
0
2
NkeTnskS
N
n
nkN
j
sDFT
86424648
162566425624
321024160102464
6440963844096160
1281638489616384384
SOLO
Fast Fourier Transform (FFT)
Arithmetic Operations for a Radix FFT versus DFT
For N = 2L we have L stages of Radix FFT and:
For N-point DFT we have:
For each row we have N complex additions and N complex multiplications, therefore for the N rows we have
Number of complex additions DFT = Number of complex multiplications DFT = NxN=N2
Number of complex additions FFT =N L=N log2 N
Number of complex additions FFT =N/2 (multiplications per stage) x L -1 =N/2 log2 (N/2)
Operation
Complex additions Complex multiplications
DFT DFTFFT FFTN=2L
Approximate number of Complex Arithmetic Operations Required for 2L-point DFT and FFT computations
SOLO Complex Variables
Laurent’s Series (1843)
Power Series
If f (z) is analytic inside and on the boundary of the ringshaped region R bounded by two concentric circles C1 andC2 with center at z0 and respective radii r1 and r2 (r1 > r2),
then for all z in R:
Pierre Alphonse Laurent1813 - 1854
C1
x
y
RC2R2
R1
z0
z
z'
r
P1
P0
z'
1 00
0
nn
n
n
n
nzz
azzazf
,2,1,0''
'
2
1
2
1
0
nzdzz
zf
ia
C
nn
,2,1,0''
'
2
1
1
1
0
nzd
zz
zf
ia
C
nn
Proof:
Since z is inside R we have R1 <|z-z0|=r < R2 , and |z’-z0|= R1 on C1 and R2 on C2.
Start with the Cauchy’s Integral Formula:
212
0
1
1
01
''
''
'
''
'
''
'
''
'
''
'
'
0
CCC
P
P
P
PC
dzzz
zfdz
zz
zfzfdzdz
zz
zfdz
zz
zfdz
zz
zfdz
zz
zfzf
SOLO Complex Variables
Laurent’s Series (continue - 1)
Power SeriesPierre Alphonse Laurent
1813 - 1854
C1
x
y
RC2R2
R1
z0
z z'
r
Proof (continue – 1):
Since z and z’ are inside R we have R1 >|z-z0|=r >R2, |z’-z0|=R1.
From Cauchy’s Integral Formula:
21
''
''
'
'
CC
dzzz
zfdz
zz
zfzf
Use the identity:
11
1
1 12n
n
For I integral:
nn
zz
zz
zz
zzzz
zz
zz
zz
zz
zz
zzzzzz 0
0
0
0
1
0
0
0
0
0
0
00 '
'1
1
''1
'
1
'1
1
'
1
'
1
n
n
n
R
C
n
n
n
zs
C
n
za
C
za
C
Rzzzazzzazazzzz
zdzf
i
zz
zzzz
zdzf
izz
zz
zdzf
izz
zdzf
i
n
n
0000100
0
0
1
0
0
02
00
2
0
2
01
2
00
2
''
''
2
'
''
2
1
'
''
2
1
'
''
2
1
1
''
'
2
1
C
zdzz
zf
i
We have:
n
n
n
C
n
n
n R
r
rR
MRdR
rRR
Mr
zzzz
zdzfzzR
11
1
2
0
1
110
0
2''
''
20
where |f (z)|<M in R and r/R1< 1, therefore: 0
n
nR
SOLO Complex Variables
Laurent’s Series (continue - 2)
Power SeriesPierre Alphonse Laurent
1813 - 1854
C1
x
y
RC2R2
R1
z0
z z'r
Proof (continue – 1):
Since z and z’ are inside R we have R1 >|z-z0|=r > R2, |z’-z0|=R2.
From Cauchy’s Integral Formula:
21
''
''
'
'
CC
dzzz
zfdz
zz
zfzf
Use the identity:
11
1
1 12n
n
For II integral:
nn
zz
zz
zz
zzzz
zz
zz
zz
zz
zz
zzzzzz 0
0
0
0
1
0
0
0
0
0
0
00
'
'1
1''1
1'
1
11
'
1
n
n
n
R
C
n
n
n
za
C
n
za
CC
Rzzzazzzazzzz
zdzfzz
i
zzzz
zdzf
izzzz
zdzf
izdzf
i
n
n
1
001
1
001
0
0
1
0
1
00
2
0
0
0
01
0
01
00
'
'''
2
1
1
'
''
2
11
'
''
2
1''
2
1
C
zdzz
zf
i'
'
'
2
1
We have:
n
n
n
C
n
n
n r
R
rR
RMdR
rRr
MR
zzzz
zdzfzzR
2
2
2
2
0
2
2
2
0
0
2'
'''
2
1
0
where |f (z)|<M in R and R2/r< 1, therefore: 0
n
nR Return to Table of Contents
Z2 TransformC1
x
y
R
C2r2 z0
z
r
z'
C
r1
SOLO
Z-Transform Two Sided
n
nzTnfzF
Example 1
TnaTnf
C
n dzzzFj
Tnf 1
2
1
1
0
0
0/1
/
01
1
n k
T
T
Tk
TT
nT
n
T
T
nT
n
nT
n
nTn
nazaz
az
a
z
a
z
z
a
nza
z
az
a
z
azazF
Z2 TransformSOLO
Z-Transform Two Sided
Example 2
gg
ff
rz
r
rr
gg rzr
gfgf rrzrr
C
dz
GFj
TngTnf
1
2
1Z
gfgf
fg
gf
rrzrr
nrrz
nrzr
0&/
0&/
C
rz
r
C n
n
n
n
n
rrC
nn
n
n
dz
GFj
dz
TngFj
zTngdzFj
zTngTnfTngTnf
gg
nf
nf
11
1
2
1
2
1
2
1
00
Z
Z2 TransformSOLO
Z-Transform Two Sided
Example 2 (continue – 1)
zban
ba
zba
z
b
zb
z
a
aResd
b
zb
z
a
aj
zban
z
ba
z
baResd
z
baj
ba
TT
TT
TT
CT
T
T
T
b
z
bb
z
T
T
T
T
C
TT
TTTTa
b
za
TT
TnTn
T
TT
T
TT
&01111
1
2
1
&01
1
1
11
1
1
1
11
2
1
Z
C
dz
GFj
TngTnf
1
2
1Z
gfgf
fg
gf
rrzrr
nrrz
nrzr
0/
0/
Z TransformSOLO
Properties of Z-Transform Functions
Z - Domaink - Domain
kf
ffk
k rzrzkfzF0
1
ii ff
M
iii rzrzFc minmax
1
Linearity
M
iii kfc
1
2 ,2,10 kkfmkf zFz mShifting
mkf
m
k
kmm zkfzFz1
mkf
m
k
kmm zkfzFz1
1kf 0fzFz
3 Scaling kfak
ffk
krazrazakfzaF
0
11
Z TransformSOLO
Properties of Z-Transform Functions (continue – 1)
4 Periodic Sequence kf
1111 ffN
N
rzrzFz
z
N = number of units in a period
Rf1- ,+ = radiuses of convergence in F(1) (z)
F(1) (z) = Z -Transform of the first period
5 Multiplication by k kfk
ff rzrzd
zFdz
6 Convolution
0
:m
mkhmfkhkf hfhf rrzrrzHzF ,min,max
7 Initial Value zFfz
lim0
8 Final Value existsfifzFzkfzk
1limlim1
Z - Domaink - Domain
kf
ffk
k rzrzkfzF0
Z TransformSOLO
Properties of Z-Transform Functions (continue – 2)
9 Complex Conjugate kf * ff rzrzF **
10 Product khkf hfhf
C
rrzrrz
zdzHzF
j,1,
2
1 1
12 Correlation
1,1,2
1 11
0
krrzrr
z
zdzzHzF
jkmhmfkhkf hfhf
C
k
m
11 Parceval’s Theorem
hfhf
Ck
rrzrrz
zdzHzF
jkhkf ,1,
2
1 1
0
Z - Domaink - Domain
kf
ffk
k rzrzkfzF0
Z TransformSOLO
Table of Z-Transform Functions
Z - Domaink - Domain
kf f
k
k RzzkfzF
0
1
mkf 110 11 mfzfzfzFz mm 2
mkf zFz m3
kfkfkf 1: 01 fzzFz 4
kfkfkfkf 122:2 1021 2 fzfzzzFz 5
kf3 2130331 23 fzfzzfzzzzFz 6
L2 Transform
0 aetf ta
SOLO
Laplace Transform Two Sided
j
j
j
j
ts
ts
j
j
tts
gg
tftf
dsGFj
ddtetgFj
dtetgdeFj
dtetgtftgtf
2
1
2
1
2
1
00
2L
Hence
ff
gf
j
j
ts
tfor
tfor
dsGFj
dtetgtftgtf
0
0
2
12L
Example 1
srealasreala
tastastastaststata
asasas
e
as
edtedtedteee
11
0
0
0
0
2L
0&1
0&1
2
tsrealaas
tasrealase
f
fta
L
aa sreal
simag
taetf
t
1
L2 Transform
0 aetf ta
SOLO
Laplace Transform Two Sided
Example 2
ab
dbsa
ee
fg
j
j
tbta
112L
0&1
0&1
2
tsrealaas
tasrealase
f
fta
L
0 betg tb
Find the two sided Laplace transform of f (t) g (t)
0&1
0&1
2
tsrealbbs
tbsrealbse
f
ftb
L
ba
dbsa
ee
gf
j
j
tbta
112L
basbsaRes
a
111
basbsaRes
a
111
C1
b a
0t
00
t
C2
ba
0t
00
t
SOLO
References
A. Papoulis, “The Fourier Integral and its Applications”, McGraw Hill, 1962
R.N. Bracewell, “The Fourier Transform and its Applications”, McGraw Hill, 1965, 1978
J.W. Goodman,“Introduction to Fourier Optics”, McGraw Hill, 1968
H. Stark, Ed. “Applications of Optical Fourier Transform”, Academic Press, 1982
A. Papoulis, “Systems and Transforms with Applications in Optics”, McGraw Hill, 1968
Fourier Transform
Athanasios Papoulis1921 - 2002 Ronald N. Bracewell
1921 -
Joseph W. GoodmanWilliam Ayer Professor, Emeritus
Packard 352Department of Electrical Engineering
Stanford UniversityStanford, CA 94305
Email :[email protected]
April 9, 2023 98
SOLO
TechnionIsraeli Institute of Technology
1964 – 1968 BSc EE1968 – 1971 MSc EE
Israeli Air Force1970 – 1974
RAFAELIsraeli Armament Development Authority
1974 – 2013
Stanford University1983 – 1986 PhD AA
Raymond Paley1907 - 1933
Norbert Wiener1894 - 1964
Paley – Wiener Condition
A necessary and Sufficient condition for a square-integrablefunction A (ω) ≥ 0 to be the Fourier spectrum of a causal functionis the convergence of the integral:
dA
21
ln
SOLO
The Mellin Transform
0
1: s
M exfsF
SOLO
Hjalmar Mellin1854 - 1933
Putting: tdexdex tt 11 sts ex
tdeefsF tst
M
We can see that the Mellin Transform of the function f (t) is identical to theBilateral Laplace Transform of f (e-t).
SOLO
Example
0
sindk
k
krLet compute:
x
y
R
A
B
C
D
E
F
G
H
Rx Rx
For this use the integral: 0ABCDEFGHA
zi
dzz
e
Since z = 0 is outside the region of integration
0
BCDEF
ziR xi
GHA
zi
R
xi
ABCDEFGHA
zi
dzz
edx
x
edz
z
edx
x
edz
z
e
00
0000
sin2
sin2
sinlim2limlimlim dk
k
rkidx
x
xidx
x
xidx
x
eedx
x
edx
x
eR
R
R xixi
R
R xi
RR
xi
R
idideideie
edz
z
e i
ii
eii
i
eiez
GHA
zi
00
1
0
0
00limlimlim
012
2
0
/2/2sin
0
sin
00
R
RRReRii
i
eRieRz
BCDEF
zi
eR
dedededeRieR
edz
z
e i
ii
Therefore: 0sin
20
idkk
rkidz
z
e
ABCDEFGHA
zi 2
sin
0
dkk
kr
Complex Variables
SOLO Complex Variables
Cauchy’s Theorem
C
x
y
R
Proof:
0C
dzzf
If f (z) is analytic with derivative f ‘ (z) which is continuous at all points insideand on a simple closed curve C, then:
yxviyxuzf ,, Since is analytic and has continuous first order derivative
y
ui
y
v
x
vi
x
u
zd
fdzf
iyzxz
'
y
u
x
v
y
v
x
u
& Cauchy - Riemann
0
00
RR
dydxy
v
x
uidydx
y
u
x
v
dyudxvidyvdxudyidxviudzzfCCCC
q.e.d.
Augustin Louis Cauchy ) 1789-1857(
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
Jordan’s LemmaIf |F (z)| ≤ M/Rk for z = R e iθ where k > 0 and M are constants, then
where Γ is the semicircle arc of radius R, center at origin, in theupper part of z plane, and m is a positive constant.
0lim
zdzFe zmi
R
x
y
R
Proof:
0lim
zdzFe zmi
R
using:
q.e.d.
0
deRieRFezdzFe iieRmieRz
zmi i
i
2/
0
sin
1
0
sin
1
0
sin
0
sincos
00
2
dReR
MdRe
R
MdReRFe
deRieRFedeRieRFedeRieRFe
Rm
k
Rm
k
iRm
iiRmRmiiieRmiiieRmi ii
2/0/2sin for2/
1sin
/2
Rm
k
Rm
k
Rm
k
iieRmi eR
Mde
R
Mde
R
MdeRieRFe
i
1
2222/
0
/2
1
2/
0
sin
1
0
012
limlim0
Rm
kR
iieRmi
Re
R
MdeRieRFe
i
Marie Ennemond Camille Jordan1838 - 1922
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
Jordan’s Lemma GeneralizationIf |F (z)| ≤ M/Rk for z = R e iθ where k > 0 and M are constants, then
for Γ a semicircle arc of radius R, and center at origin:
00lim
mzdzFe zmi
R
x
y
R
where Γ is the semicircle, in the upper part of z plane.
1
00lim
mzdzFe zmi
R
xy
R
where Γ is the semicircle, in the down part of z plane.
2
00lim
mzdzFe zm
R x
y
R
where Γ is the semicircle, in the right part of z plane.
3
00lim
mzdzFe zm
R
where Γ is the semicircle, in the left part of z plane.
4x
y
R
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
Integral of the Type (Bromwwich-Wagner)
jc
jc
ts sdsFei2
1
The contour from c - i ∞ to c + i ∞ is called Bromwich Contour
Thomas Bromwich1875 - 1929
x
y
0t
R
c
x
y0t
R c
0
0
2
1
lim2
1
2
1
tzFeRes
tzFeReszdzF
i
sdsFesdsFei
sdsFei
tf
tz
planezRight
tz
planezLeft
ts
ic
ic
ts
R
ic
ic
ts
where Γ is the semicircle, in the right part of z plane, for t < 0.
where Γ is the semicircle, in the left part of z plane, for t > 0.
This integral is also the Inverse Laplace Transform.