fourier analysis techniques
DESCRIPTION
FOURIER ANALYSIS TECHNIQUES. LEARNING GOALS. FOURIER SERIES. Fourier series permit the extension of steady state analysis to general periodic signal. FOURIER TRANSFORM. Fourier transform allows us to extend the concepts of frequency domain to arbitrary non-periodic inputs. FOURIER SERIES. - PowerPoint PPT PresentationTRANSCRIPT
FOURIER ANALYSIS TECHNIQUES
Fourier series permit the extension of steady state analysis to general periodicsignal.
FOURIER SERIES
LEARNING GOALS
FOURIER TRANSFORM
Fourier transform allows us to extend the concepts of frequency domain toarbitrary non-periodic inputs
FOURIER SERIES
The Fourier series permits the representation of an arbitrary periodic signal as a sum of sinusoids or complex exponentials
Periodic signal
tTtftf
Ttf
),()(
0)( that such exists there iff periodic is signal The
The smallest T that satisfies the previous condition is called the (fundamental)period of the signal
tnccjtnccecec nnnntjn
ntjn
n 00 sin)(cos)(09
000
n
ac
For
FOURIER SERIES RESULTS
110
10
10
0
sincos)(
)(
Re)cos()(
)(,)(
0
non
non
n
tjnn
n
tjnnn
nnon
tnbtnaatf
ectf
eDatnDatf
tfTtf
o
forms equivalent
following the of one in expressed be can then period with periodic, is If
Cosine expansion
Complex exponential expansion
Trigonometric series
Phasor for n-th harmonic0
02
T
nnnnn jbacD 2
na nb
Relationship between exponential and trigonometric expansions
nnn
nnn
jbac
jbac
2
2
*)( nn cctf then valued-real is If
GENERAL STRATEGY: . Approximate a periodic signal using a Fourier series. Analyze the network for each harmonic using phasors or complex exponentials. Use the superposition principle to determine the response to the periodic signal
sin2
cos2
sincos
jee
ee
je
jj
jj
j
Original Periodic Signal
Approximation with 2 terms
Approximation with 4 terms
Approximation with 100 terms
2
1
2
102 sincos)(
non
non tnbtnaatf
100
1
100
10 sincos
non
non tnbtnaa
N
Nn
tjnnN ectf 0)(
EXAMPLE OFQUALITY OFAPPROXIMATION
EXPONENTIAL FOURIER SERIES
Any “physically realizable” periodic signal, with period To, can be represented overthe interval by the expression 011 Tttt
00
2;)( 0
Tectf
n
n
tjnn
The sum of exponential functions is always a continuous function. Hence, the righthand side is a continuous function.Technically, one requires the signal, f(t), to be at least piecewise continuous. In thatcase, the equality does not hold at the points where the signal is discontinuous
Computation of the exponential Fourier series coefficients
n
n
tjnnectf 0)(
01
1
Tt
t
tjke 0
dteecdtetf tjkTt
t
n
n
tjnn
Tt
t
tjk 0
01
1
0
01
1
0)(
knT
kndte
Tt
t
tknj
for
for
0
)( 001
1
0
01
1
0)(1
0
Tt
t
tjkk dtetf
Tc
simpler nscomputatio make to chosen be can andarbitray is 1t
LEARNING EXAMPLEDetermine the exponential Fourier series
00
2;)( 0
Tectf
n
n
tjnn
01
1
0)(1
0
Tt
t
tjnn dtetf
Tc
nintegratio the Do 3.
convenient aSelect 2.
and Determine 1.
:strategyA
0
1
0
t
T
TTT
200
2/1 Tt
2
4
4
2
4
4
2
2
000011
)(1
)(1
T
T
tjn
T
T
T
T
tjntjn
T
T
tjnn dtVe
TdtVe
TdteV
Tdtetv
Tc
2/
4/
4/
4/
4/
2/0
000
)(
T
T
tjnT
T
tjnT
T
tjnn eee
jnT
Vc
424424
0
000000T
jnT
jnT
jnT
jnT
jnT
jn
n eeeeeejTn
Vc
20 T
)sin(22
sin42
222
22
nn
n
Veeee
nj
Vc jnjn
nj
nj
n
sin2
j
ee jj
oddn
n
n
Vevenn
cn2
sin2
0
case this in
for is This
0
!0
0
c
n
LEARNING EXTENSIONDetermine the exponential Fourier series
nintegratio the Do 3.
convenient aSelect 2.
and Determine 1.
:strategyA
0
1
0
t
T
00
2;)( 0
Tectf
n
n
tjnn
01
1
0)(1
0
Tt
t
tjnn dtetf
Tc
20 T 0
01 t
;2
1
2
1
2
1)(
2
1 1
0
1
0
2
0 nj
ee
njdtedttvc
njntjntj
n
0nFor
2
1
0
0
c
nFor
LEARNING EXTENSIONDetermine the exponential Fourier series
nintegratio the Do 3.
convenient aSelect 2.
and Determine 1.
:strategyA
0
1
0
t
T
00
2;)( 0
Tectf
n
n
tjnn
01
1
0)(1
0
Tt
t
tjnn dtetf
Tc
60 T30
21 t
2)(6
1 2
20
dttvc
2
1
31
1
31
2
34
2
4246
1)(
6
1dtedtedtedttvct
njt
njt
nj
n
3
233333
2
4422442
nj
nj
nj
nj
nj
nj
n eeeeeej
nc
3sin2
3
2sin4
nnncn
sin2
j
ee jj
TRIGONOMETRIC FOURIER SERIES
1102 sincos)(
non
non tnbtnaatf
tnccjtnccecec nnnntjn
ntjn
n 00 sin)(cos)(00
000
n
ac
For
na nb
Relationship between exponential and trigonometric expansions
nnn
nnn
jbac
jbac
2
2
*)( nn cctf then valued-real is If
sin2
cos2
sincos
jee
ee
je
jj
jj
j
0
02
T
01
1
0)(1
0
Tt
t
tjnn dtetf
Tc
01
1
01
1
01
1
00
00
00
sin)(2
cos)(2
)(1
Tt
tn
Tt
tn
Tt
t
dttntfT
b
dttntfT
a
dttfT
a
The trigonometric form permits the use of symmetry properties of the function to simplify the computation of coefficients
Even function symmetry )()( tftf
x
x
dttfdttf0
0
)()(
x x
Odd function symmetry )()( tftf
x
x
dttfdttf0
0
)()(
,2,1,0sin)(2 2
2
00
0
0
ndttntfT
b
T
Tn
,2,1,cos)(4 2
00
0
0
ndttntfT
a
T
n
20
1T
t
TRIGONOMETRIC SERIES FOR FUNCTIONS WITH EVEN SYMMETRY
01
1
01
1
01
1
00
00
00
sin)(2
cos)(2
)(1
Tt
tn
Tt
tn
Tt
t
dttntfT
b
dttntfT
a
dttfT
a
TRIGONOMETRIC SERIES FOR FUNCTIONS WITH ODD SYMMETRY
nnnnnn
nnn
accjbac
jbac
2
2
,1,0,0cos)(2 2
2
00
0
0
ndttntfT
a
T
Tn
,2,1,sin)(4 2
00
0
0
ndttntfT
b
T
n ,2,1,
0*
0
njbcc
c
nnn
3sin2
3
2sin4
nnncn
,2,1, nan
20 a
2
000
0
)(2T
dttfT
a
FUNCTIONS WITH HALF-WAVE SYMMETRY
tT
tftf
,
2)( 0
Each half cycle is an inverted copy of the adjacent half cycle
Examples of signals with half-wave symmetry
odd nfor
odd nfor
even nfor
2
00
0
2
00
0
0
0
0
sin)(4
cos)(4
0
0
T
n
T
n
nn
dttntfT
b
dttntfT
a
ba
a
There is further simplification if the functionis also odd or even symmetric
LEARNING EXAMPLEFind the trigonometric Fourier series coefficients
This is an even function with half-wave symmetry
,2,1,0 nbneven
even nfor
odd nfor symmetry wave-half
0
cos)(4 2
00
0
0
n
T
n
a
dttntfT
a
dttkVT
dttkVdttkVT
a
TT
T
T
k 0
4
00
2
4
0
4
012 12cos
812cos12cos
4
2
)12(sin
)12(
4
4)12sin(
)12(
80
012
k
k
VTk
kTa k
22
00 TT
LEARNING EXAMPLEFind the trigonometric Fourier series coefficients
This is an odd function with half-wave symmetry
,2,1,0;0 nan odd
odd nfor
odd nfor symmetry wave-half
0
sin)(4 2
00
0
0
n
T
n
b
dttntfT
b
2
4
00
0
4
00
0012
0
0
0
])12sin[()2
(4
])12sin[(44
T
T
T
k dttkT
tT
Vdttkt
T
V
Tb
Use change of variableto show that the twointegrals have the same value
4
002
012
0
])12sin[(48
T
k dttktT
Vb
2
sincossin
ttt
dttt
4
02
0
0
0
02
012
0
)12(
])12sin[(
)12(
)12cos[(32T
kk
tk
k
tkt
T
Vb
2400
T
2
)12(sin
])12[(
8212
k
k
Vb k
LEARNING EXAMPLEFind the trigonometric Fourier series coefficients
symmetry wave-half with odd is 2
3)( tf
odd nfor
odd nfor symmetry wave-half
0
sin)(4 2
00
0
0
n
T
n
b
dttntfT
b
2
0
200
000
012
0
0
])12cos[()12(
2])12sin[(
2
14T
T
k tkkT
dttkT
b
)12(
2]))12cos[(1(
)12(
112
kk
kb k
tkk
tfk
00
)12sin()12(
2
2
3)(
,2,1,0;0 nan odd
LEARNING EXTENSION Determine the type of symmetry of the signals
LEARNING EXTENSIONDetermine the trigonometric Fourier series expansion
0 nb function even
,2,1,cos)(4 2
00
0
0
ndttntfT
a
T
n
2
000
0
)(2T
dttfT
a
60 T
2)42(3
10 a
dttndttnan 0
1
0
2
10 cos4
3
2cos2
3
2
30
3sin
3
2sin2
4
nn
nan
LEARNING EXTENSIONDetermine the trigonometric Fourier series expansion
odd nfor
odd nfor
even nfor
2
00
0
2
00
0
0
0
0
sin)(4
cos)(4
0
0
T
n
T
n
nn
dttntfT
b
dttntfT
a
ba
a
Half-wave symmetry
40 T20
odd nfor ,2
cos2
cos22
1
1
0
dttn
dttn
an
odd nfor ,2
sin2
sin22
1
1
0
dttn
dttn
bn
2
)12(sin
)12(
212
k
ka k
))12cos(2()12(
212
k
kb k
Fourier Series Via PSPICE Simulation
1. Create a suitable PSPICE schematic2. Create the waveform of interest3. Set up simulation parameters4. View the results
LEARNING EXAMPLE
Determine the Fourier Series of this waveform
1. PSPICE schematic
VPWL_FILE in PSPICE librarypiecewise linear periodic voltage source
File specifying waveform
Text file defining corners of piecewiselinear waveform
comments
Use transient analysis
Fundamentalfrequency (Hz)
Schematic used for Fourier series example
To view result:From PROBE menu
View/Output Fileand search until you find theFourier analysis data
Accuracy of simulation is affected by setup parameters.
Decreases with number of cyclesincreases with number of points
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(V_Vs)
DC COMPONENT = -1.353267E-08
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 1.000E+00 4.222E+00 1.000E+00 2.969E-07 0.000E+00 2 2.000E+00 1.283E+00 3.039E-01 1.800E+02 1.800E+02 3 3.000E+00 4.378E-01 1.037E-01 -1.800E+02 -1.800E+02 4 4.000E+00 3.838E-01 9.090E-02 4.620E-07 -7.254E-07 5 5.000E+00 1.079E-04 2.556E-05 1.712E-03 1.711E-03 6 6.000E+00 1.703E-01 4.034E-02 1.800E+02 1.800E+02 7 7.000E+00 8.012E-02 1.898E-02 -9.548E-06 -1.163E-05 8 8.000E+00 8.016E-02 1.899E-02 5.191E-06 2.816E-06 9 9.000E+00 5.144E-02 1.218E-02 -1.800E+02 -1.800E+02 10 1.000E+01 1.397E-04 3.310E-05 1.800E+02 1.800E+02 11 1.100E+01 3.440E-02 8.149E-03 -1.112E-04 -1.145E-04 12 1.200E+01 3.531E-02 8.364E-03 1.800E+02 1.800E+02 13 1.300E+01 2.343E-02 5.549E-03 1.800E+02 1.800E+02 14 1.400E+01 3.068E-02 7.267E-03 -3.545E-05 -3.960E-05 15 1.500E+01 3.379E-04 8.003E-05 -3.208E-03 -3.212E-03 16 1.600E+01 2.355E-02 5.579E-03 -1.800E+02 -1.800E+02 17 1.700E+01 1.309E-02 3.101E-03 2.905E-04 2.854E-04 18 1.800E+01 1.596E-02 3.781E-03 -5.322E-05 -5.856E-05 19 1.900E+01 1.085E-02 2.569E-03 -1.800E+02 -1.800E+02 20 2.000E+01 2.994E-04 7.092E-05 1.800E+02 1.800E+02
TOTAL HARMONIC DISTORTION = 3.378352E+01 PERCENT
* file pwl1.txt* example 14.5 * BECA 7* ORCAD 9.1* By J.L. Aravena0,00.1,10.2,30.3,60.4,30.5,00.6,-30.7,-60.8,-30.9,-11.0,0
RELEVANT SEGMENT OFOUTPUT FILE
TIME-SHIFTINGIt is easier to study the effect of time-shift with the exponentialseries expansion
n
n
tjnnectf 0)(
n
n
tjntjnn
n
n
ttjnn eececttf 00000 )(
0 )(
Time shifting the function only changes the phase of the coefficients
nctn
tft
for shift phase
for shift time
00
0 )(
240
000 nT
ntn
200 T
)4
()( 0Ttftv
)(tf
LEARNING EXAMPLE
0TT
oddn
n
n
Vevenn
cn2
sin2
0
n
n
tjnvnectv 0)(
oddne
n
n
V
evennc n
jvn 2
2sin
2
0
12
sin
2sin
2cos
2
2
n
njnejn
n
Vjc kv 2
12,
Time shifting and half-wave periodic signals
00
0
1
011
0
20
20)(
)(
2)()(
)(
TtT
Tttf
tf
Ttftftf
T
tf
with
as expressed be can , period,
with , signal, periodic wave-halfEvery
n
n
tjnnectf 0)(1
Assume
n
n
tjnT
jn
n eecT
tf 0
00 20
1 )2
(
200 T
ne njn ,1
k
tkjk ectf 0)12(
122)(
Only the odd coefficients of f1 are used
LEARNING EXTENSION
)1( tv
It was shown beforethat for v(t)
0,2
1
2
10
nnj
ec
c
jn
n
. of those from angle aby differ they
that show and for tscoefficien the compute
v(t)πn
)v(t-1
2
1)1(
2
1 2
00 dttvc d
function delayed theFor
0
002
2T
T
njn
nd cec
jn
jn
enj
e
12
2
1
2
0 2
1)1(
2
1dtedtetvc tjntjn
nd
njjn ee
nj2
2
1
0nFor
WAVEFORM GENERATION
If the Fourier series for f(t) is known then one caneasily determine the expansion for any time-shifted and time-scaled version of f(t)
n
n
tnjnectf 0)(
n
n
tjnnectf 0)( Assume Time scaling does not change the
values of the series expansion
n
n
tjntjnn
n
n
ttjnn eececttf 00000 )(
0)( Time-shifting modifies the phase ofthe coefficients
n
n
tnjtjnn eecttf 000)( 0
The coefficients of a linear combination of signals are the linear combination ofthe coefficients
nccc nnn 213
One can tabulate the expansions for some basic waveforms and use them to determinethe expansions or other signals
Signals withFourier seriestabulated in BECA 8
LEARNING EXTENSION
2,6,2
)2/()(
)(
0
1
1
TA
tftv
tvFor
tjn
n
en
ntv 3
1 3sin
2)(
2
0
0
sin)(
tjn
n
eT
n
n
Atf
From the table of Fourier series
00 T
Ac
3
20 c
Strictly speaking the value for n=0 mustbe computed separately.
3
2,1,3,4
2)(
)(
00
02
2
TA
Ttftv
tvFor
3
40 c
21
213
32
0
2 3sin
4)(
tjn
nn
en
ntv
tjnjn
nn
een
ntv 3
2
0
2 3sin
4)(
njne )1(
Use the table of Fourier series to determine the expansionsof these functions
FREQUENCY SPECTRUM
The spectrum is a graphical display of the coefficients of the Fourier series.The one-sided spectrum is based on the representation
10
10 Re)cos()(
n
tjnnn
nnon
oeDatnDatf
The amplitude spectrum displays Dn as the function of the frequency.The phase spectrum displays the angle as function of the frequency.The frequency axis is usually drawn in units of fundamental frequency
n
n
tjnnectf 0)(
The two-sided spectrum is based on the exponential representation
In the two-sided case, the amplitude spectrum plots |cn| while the phase spectrumplots versus frequency (in units of fundamental frequency)nc
nnnnn jbacD 2*nn cc
Both spectra display equivalent information
LEARNING EXAMPLE
)cos40
sin20
)( 02201
tnn
tnn
tv
oddnn
The Fourier series expansion, when A=5,is given by
Determine and plot the first four termsof the spectrum
nnnnn jbacD 2
1225.72040
211 jD 1022.2
3
20
9
4023
jD
9592.0,973.1 7755 DD
Amplitude spectrum
Phase spectrum
LEARNING EXTENSIONDetermine the trigonometric Fourier series and plot thefirst four terms of the amplitude and phase spectra
10sin
2)(
n
tnn
AAtf
From the table ofseries
nnnnn jbacD 2
tnn
tv
2sin1
2
1)(
1
0;1
;2
10 n
njDD n n |Dn| Arg(Dn)
0 0.5 01 0.318309 -902 0.159155 -903 0.106103 -904 0.079577 -90
|Dn|
0
0.1
0.2
0.3
0.4
0.5
0.6
1 2 3 4 5
|Dn|
STEADY STATE NETWOK RESPONSE TO PERIODIC INPUTS
1. Replace the periodic signal by its Fourier series2. Determine the steady state response to each harmonic3. Add the steady state harmonic responses
)( jH
)( 1|| tjec ))((1
11|)(||| tjejHc
)cos(|| 1 tD ))(cos(|)(||| 111 tjHD
)(|)(||)(|)( )( jHejHjH j
}|Re{| 1tjeD })(|)(||Re{| 111
tejHD
0,5:)( TAtv
LEARNING EXAMPLE
oddnn
tnn
tnn
tv1
22 2cos40
2sin20
)(
j1
)1
||2(j)( jI
)(44
2)(
212
2
212
)(1
jVj
jV
j
jjV
)(1 jVj21
2
10 2
1VV
1
2tan
1616
1
44
1)(
jjH
1
2
tan)44(
1616|44|
j
j
nj
njbaD nnn
204022
22
22
2040||
nnDn
)2
tan180(2
tan180 11 nnDn
nn 2
)1802tan2
tan2cos(6416
12040)( 11
2
22
22
n
nnt
nnntvon
LEARNING EXTENSION )(ticurrent thefor expression statesteady the Find
12 2cos
)14(
4020)(
nS nt
ntv
nnn
DD nn 2;1,180)14(
40;
2020
jZ
2||21
)()()(
)()(
jVjYjZ
jVjI S
S
j
j
j
j
jj
jjZ
1
3
22
26
22
41
22
4
1)(
harmonic th-nfor phasor nn Dnj
njDnjYnjI
23
21)2()2(
))3/2(tan2tan(49
41)2(
3
1)0(
112
2
nnn
nnjY
Y
j
jjY
3
1)(
10 )2cos(||||)0()(
nnnn YntDYDYti
AVERAGE POWER
In a network with periodic sources (of the same period) the steady state voltageacross any element and the current through are all of the form
10
10
)cos()(
)cos()(
ninndc
nvnndc
tnIIti
tnVVtv
Tt
t
o
o
dttitvT
P )()(1
Power Average
)cos()cos(
)cos()cos()()(
2
1 2
121 01 1
01
10
10
inn n
vnnn
nvnndc
ninndcdcdc
tntnIV
tnVItnIVIVtitv
))()cos((
))()cos((
2)cos()cos(
21
21
1 2
2
1 2
121021
021
1 1
210
1 101
invn
invn
n n
nnin
n nvnnn tnn
tnnIVtntnIV
0
00)cos(
2 0
0
00 kT
kdttk
T
Tt
t
Tt
t
0
0
21)cos( nnforT invn
)cos(21
inn
vnnn
dcdcIV
IVP
Power Average
The average power is thesum of the average powersfor each harmonic
LEARNING EXTENSION
])[100754cos(2.1)45377cos(8.1)(
])[102754cos(24)60377cos(3664)(
Attti
Vtttv
Determine the average power
)180cos(cos
))100180102cos(2.124)15cos(8.136(5.0 P
][91.162
78.2859.62WP
LEARNING EXAMPLE
])[20754cos(12)30377cos(1642)( Vtttv
network theby
absorbedpower average the and , current, the Determine )(ti
jCjLRjZ
1)(
)(
)()(
jZ
jVjI
88.7964.053.2654.716
3016)377(
37710
1377020.016)377(
,3016)377(
377
4
jjjI
jjjZ
jV
49.2675.026.1308.1516
2012)754(
75410
1754020.016)754(
,2012)754(
754
4
jjjI
jjjZ
jV
0)0(
0
jI
)(Zcircuit open as actscapacitor
)49.26754cos(75.0)88.79377sin(64.0)( ttti
2
)49.6cos(75.012)88.49cos(64.016)0(42
P
)cos(21
inn
vnnn
dcdcIV
IVP
Power Average
FOURIER TRANSFORM
dtetfF tj )()(
2)()(
deFtf tj
][ )()( tfF F )]([)( Ftf 1-F
A heuristic view of the Fourier transform
A non-periodic function can be viewed as the limit of a periodic function when theperiod approaches infinity
n
tT
jn
np ectf2
)(
2/
2/
2
)(1 T
T
tT
jn
n dtetfT
c
TTectf
n
tT
jn
np1
)(2
2/
2/
2
)(T
T
tT
jn
n dtetfTc
nnT
2
dtetf tj)(
2)(
deF tj
T
LEARNING EXAMPLE Determine the Fourier transform
2/
2/
)()(
dteVdtetvV tjtj
2/
2/
1)(
tjej
VV
j
eeV
jj 2/2/
2
2sin
)(
VV
2
2sin
0
T
Vcn
For comparison we show the spectrumof a related periodic function
50 T for Spectrum
LEARNING EXAMPLES
Determine the Fourier transform of theunit impulse function
2)()(
deFtf tj
dtetfF tj )()(
][ )()( tfF F
ajtj edteatt
)()]([F
impulse the
of property sampling or Sifting
)()()( afdttfat
time of function ingcorrespond the
find and Consider )(2)( 0 F
2)(2)( 0
detf tj
tjetf 0)(
)πδ(ω-ωeF tjω02)( 0 ][F
LEARNING EXTENSION
Determine t][ 0sin)( FF
j
eettf
tjtj
2sin)(
00
0
j
F2
)(2)(2)( 00
)()()( 00 jF
tj oetf )(
of transformFourier the Determine
Proof of the convolution property
dxxtfxftftftf )()()()()( 2121
dtedxxtfxf
dtetfF
tj
tj
)()(
)()(
21
dxdtextfxfF tj
)()()( 21
Exchanging orders of integration
Change integration variable
xutxtu
And limits of integration remain the same
dxdueufxfF xuj
)(
21 )()()(
dxdueeufxfF xjuj
)()()( 21
dueufdxexfF ujxj )()()( 21
)()()( 21 FFF
A Systems application of the convolution property
)(th)(tvi
dxxvxthtv io )()()(
LEARNING EXTENSION
)()( tuetv ti
)()( 2 tueth t
(And all initial conditions are zero)
)(tvo determine to transform Fourier Use
From the table of transforms
1
1
2
1
)()()(
jj
VHV io
12)1)(2(
1 21
s
A
s
A
ss
Use partial fractionexpansion!
1|)()1(
1|)()2(
12
21
so
so
sVsA
sVsA
)()(
2
1
1
1)(
2 tueetv
jjV
tto
o
The output (response) of a network canbe computed using the Fourier transform
PARSEVAL’S THEOREM
2|)(||)(| 22 d
Fdttf
Think of f(t) as a voltage applied to a one Ohm resistor2|)(|)()()( tftitvtp
By definition, the left hand side is the energy of the signal
time) in density energy (or power2|)(| tf
domain frequency the in density Energy2|)(| F
Intuitively, if the Fourier transform has a large magnitude over a frequency rangethen the signal has significant energy over that range
And if the magnitude of the Fourier transform is zero (or very small) then the signal has no significant energy in that range
Parseval’s theorem permits the determination of the energy of a signal in agiven frequency range
LEARNING BY APPLICATIONExamine the effect of this low-pass filter in thequality of the input signal
20
20)(
jVi
1
11
1
)(
RCjR
Cj
CjH
)()()( io VHV
One can use Bode plots to visualizethe effect of the filter
High frequencies in the input signal are attenuatedin the output
The effect is clearly visible in the time domain
The output signal is slower and with less energy than the input signal
EFFECT OF IDEAL FILTERS
Effect of low-pass filter
Effect of high-pass filter
Effect of band-pass filter
Effect of band-stop filter
EXAMPLEAMPLITUDE MODULATED (AM) BROADCASTING
Audio signals do not propagate well in atmosphere – they get attenuated very quickly
Original Solution: Move the audio signals to a different frequency range for broadcasting.The frequency range 540kHz – 1700kHz is reserved for AM modulated broadcasting
Audio signal
Carrier signals
Broadcastedsignal
AM receivers pick a faint copy of v(t)
1000 ; 900a cf kHz f kHz
( ) (1 ( ))cos2c
v t s t f t
Nothing in audio range!
Audio signal has been AM modulated to the radio frequency range
1000 ; 900a cf kHz f kHz
( ) (1 ( ))cos2c
v t s t f t
EXAMPLE HARMONICS IN POWER SYSTEMS
0.2lineR
Harmonics account for 301.8W or 9.14% of the total power
LEARNING BY DESIGN “Tuning-out” an AM radio station
Fourier transform of signal broadcastby two AM stations
)()(
cos)](1[)(
cos)](1[)(
21
222
111
tsts
ttstv
ttstv
assume simplicity For
:2 station from signal
:1 station from signal
kHzfkHzf 960,900 21
)]()([)( 21 tvtvKtvr :antenna at signal
strength; equal with
receiver the reach stations both Assuming
Fourier transform of received signal
Ideal filter to tune out one AM station
Proposed tuning circuit
CsLsR
R
sv
svsG
r
ov 1)(
)()(
)/(
)/(
Ls
Ls
LCLRss
LRs
sGv 1)(
2 Next we show how to design the tuning circuit
by selecting suitable R,L,C
Designing the tuning circuit
LCLRss
LRs
sGv 1)(
2
LCfo 2
1 :frequency center
L
RBW
2
1 :bandwidth
Design equations
Design specifications
kHzorkHzf
kHzBW
o 960,900
60
Fourier transform of received signal
Ideal filter to tune out one AM station
More unknowns than equations. Makesome choices
10
10
R
kHzBWHL
L
RBW
2.159
2
1
kHzfpFC
kHzfpFC
LCf
o
o
9606.172
9004.196
2
10
Frequency response of circuit tuned to 960kHz
LEARNING EXAMPLE An example of band-pass filter
]102sin[1.0]102sin[01.0]102sin[1.0)( 543 ttttvS signalnoise noise
Proposedtwo-stageband-passwith twoidenticalstages
0V
01
:2
Cj
V
R
V Ao
-V KCL@
011
:1
Cj
V
Cj
VV
R
VV AAoAS
AV KCL@
212
2
2
11
12
1
)(
RRCj
CR
jCR
V
VjA
S
o
21 )()( jAV
VjA
S
out
S
o
V
VjA )(1
Design requirement: Make the signal 100 times stronger than the noise
202
0
o
o
jQ
jQ
AG
pass-band a of form General
212
2
2
11
12
1
)(
RRCj
CR
jCR
V
VjA
S
o
1
2
1
2
221
22
1
21
R
RA
R
RQ
CRQRRC
o
o
o
equations Design
Equating terms one gets a set of equations that can be used for design
filter the of factor quality
frequency center at gain
frequency center
Q
Ao
oIn a log scale, this filteris symmetric around the center frequency.Hence, focus on 1kHz
1kHz, 100kHz are noise frequencies10kHz is the signal frequency
Qj
QAj
oo
oo
10)
1001
1(
10: 2
2
2
10 at Gain o
1000
1
10)
1001
1(
102
2
2
Qj
Qj
oo
o
Design constraint
After two stagesthe noise gain is1000 timessmaller
Since center frequency is giventhis equation constrains the quality factor
1010001001 2 QQ
Q for Solving
We use the requirements to constraint Q
401
2 R
R
kRkR 401 21 Picking
nFCo 5.2102 4
Resulting Bode plot obtained with PSPICE AC sweep
Filter outputobtained withPSPICE
Fourier transform of output signal obtainedwith PSPICE
EXAMPLE MULTIPLE FREQUENCY “TRAP” CIRCUIT
BASIC NOTCH FILTER
1( )
( )1( )
O
in
jLV j C
HV R jL
j C
2
2
1
1
LC
LC j RC
1O
LC
Multiple frequency trap
1 2 310L L L H
Eliminates 10kHz, 20kHz and 30kHz
Tuned for 10kHz
Fourier