Fourier Integrals Nonperiodic waves Fourier Integrals
Nonperiodic waves • In optics and quantum mechanics all real waves are pulses. • In order to generate a pulse out of harmonic functions that have a certain width and shape, we need to know – what frequency elements to add – How much of each frequency element to add
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
2
Addition of waves: different frequencies To generate beats we added two frequencies ω1 and ω2 E = 2 E01 cos [ km x − ωmt ] × cos ⎡⎣ kx − ωt ⎤⎦ 1 (ω1 + ω2 ) 2 1) If we add more frequency elements symmetrically around ω, then the carrier frequency is average of the added frequencies ω =
ω will not change. 2)) If we reduce the spacing p g between the frequency q y elements,, then 1 frequency of modulation ωm = (ω1 − ω2 ) will decrease. This is the 2 frequency of the envelope meaning the beats will have more separation separation. 3) When number of frequency elements go to infinity, we will have a single pulse. The λm goes to infinity.
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
3
Beats with different frequency intervals Superposition of two waves
Superposition of two waves
2
2
1
1
05 0.5
05 0.5
Δλ=30
Amplitude
1.5
Amplitude
1.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
0
1
2
3 Distance
4
5
-2
6 -5
Δλ=60
0
1
3 Distance
4
5
6 -5
x 10
Superposition of two waves
2
2
1.5
1.5
1
1
0.5
0.5 Amplitude
Amplitude
Superposition of two waves
Δλ=10
2
x 10
0 -0.5
Δλ=1 Δλ 1
0 -0.5
-1
-1
-1.5
-1.5 5 -2
0
1
2
3 Distance
4
5
6 -5
x 10
-2
0
1
2
3 Distance
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
4
5
6 -5
x 10
4
Square Wave For the following square wave even
2 4 ⎛ m2π ⎞ + ∑ sin c ⎜ ⎟ cos((mkx) a m =1 a a ⎝ ⎠ Frequency spectrum is amplitude
-0.5 -0.5
vs. k. It expresses weighing factors
Fourier ccoefficients amplitud de
∞
of each frequency component in f ( x) of each harmonic component at
1.5
2
0.4 0.2
4π
8π
0 -0.2 0
10
20
30
50 60 mk Square wave; time domain
−λ/8 λ/8
40
70
80
90
λ=2
0.5
-0.5 -1 Fourier coefficients amplitude
number of harmonics gets larger.
1 x Square wave; frequency domain
0
while increasing the wavelength. at constant mk. As k gets smaller because of large wavelength, m, the
0.5
1
Let's keep width of the peaks constant Zeros of the sinc function happen
0
0.6
1.5
f(x)
any spatial frequency present in the synthesis.
0.5 0
The Fourier components are f ( x) =
λ=1
λ/4 Square wave; time domain
1 f f(x)
-λ / a < x ≤ λ / a λ / a < x ≤ 3λ / a
f ( x) = {0+1
−λ/4
1.5
-0.5
0
0.5
1
1.5 2 2.5 x Square wave; frequency domain
3
3.5
4
0.4
0.2
0 0
4π 10
8π 20
30
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
40
50 mk
60
70
80
90
5
λ=4
Square wave; time domain 1.5
f(x)
Frequency spectrum off a pulse l
1 0.5 0
−λ/16 λ/16
At the limit we have a pulse in time domain and a continuous Fourier series in frequency domain known as Fourier integral
-1
0
1
2 3 x Square wave; frequency domain
4
5
6
0.2
0.1
4π
8π
0 0
10
20
30
40
50 60 mk Square wave; time domain
70
80
90
λ=8
1.5
−λ/32 λ/32
1 f(x)
As the wavelength goes to infinity, the wave gets closer to expression i off a pulse, l th the F Fourier i components get closer and leading to a continuum.
0.5 0 -0.5 -4 4
Fourier coefficients amplitude
As we increase the separation p between the pulses, the Fourier components get closer in the frequency space.
Fourier coefficients amplitude
-0.5 -2
-2 2
0
2
4
6 8 10 x Square wave; frequency domain
12
14
16
0.1
0.05
4π 8π
0 0
10
20
30
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
40
50 mk
60
70
80
90
6
Fourier Transforms As λ → ∞ or k → 0 the Fourier series transforms to Fourier integrals Finite λ →
∞
f ( x) = ∑ Am cos(mkx) + m =0
∞
∑B
m =0
m
sin(mkx)
Infinite λ →
⎛ ∞ f ( x) = lim ⎜ ∑ Am cos(mkx)Δk + k →0 ⎝ m =0
⎞ Bm sin(mkx)Δk ⎟ ∑ m =0 ⎠
Infinite λ →
∞ ∞ ⎤ 1⎡ f ( x) = ⎢ ∫ A ( k ) cos(kx) dk + ∫ B ( k ) sin(kx)dk ⎥ π ⎣0 0 ⎦
∞
Using the orthognality of sine and cosine functions we can find: ∞
A(k ) =
∫
−∞
∞
f ( x) cos(kx)dx ;
B(k ) =
∫
f ( x) sin( kx) dx
−∞
A(k ) and B(k ) are called Fourier cosine and sine transforms of the function f ( x). PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
7
Complex representation of the Fourier transforms 1 f ( x) = 2π
∞
∫
F ( k ) e − ikx dk
−∞
∞
F (k ) =
∫
f ( x)eikx dx
−∞
The function F (k ) is spoken of as the Fourier transform of f ( x) F (k ) = A(k ) + iB(k ) = F { f ( x )} The sine and cosine transforms are: F { f ( x )} = FC { f ( x )} + iFS { f ( x )} And inverse Fourier transform of F (k ) is: f ( x) = F −1 { F ( k )} PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
8
Two dimensional Fourier Transform
f ( x, y ) =
1 (2π )
F (k x , k y ) = ∫ ∫
2
∞
−∞
∫∫
∞
−∞
F ( kx , k y ) e
f ( x, y ) e
−i ( kx x + k y y )
i ( kx x + k y y )
dk x dk y
dxdyy
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
9
Gibbs Phenomena • Overshoot of a synthesized f(x) by 9% of the amplitude at discontinuities, known as Gibbs Phenomena, is due to the limited number of Fourier components used to create f(x). • When N actually goes to infinity, f(x) would be 100% accurate. • For limited N there is oscillations in f(x) with frequency of Nf0. • When the N is large enough the width of the oscillations 1/2Nf0 goes to zero and the overshoot contains zero power. power • This allows usage of Fourier series even if there is a discrepancy with the actual function.
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
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Exercise 4 1) 4.1) a) Calculate Fourier transform of the square pulse given in this figure figure. b) Plot the F { f ( x )} c) Use FFT function in MATLAB to plot the F { f ( x )} What happens to the zeros of F as L increases? kL f(x) Ans: F { f ( x )} = FC { f ( x )} = E0 L sin c 2 E0
-L/2 PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
0
+L/2
x 11
Exercise 4 2) 4.2) a) Calculate complex Fourier transform of the function given in this figure. b) Plot the F { f ( x )} c) Use FFT function in MATLAB to plot the F { f ( x )} sin 2 (kd / 2) Ans: F { f ( x )} = iFS { f ( x )} = 2iE0 d kd / 2 Use F { f ( x )} = FC { f ( x )} + iFS { f ( x )} or the exponential representation Note FC and FS are real.
f(x)
E0
-d
0
d
x
-E0 PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
12
Exercise 4.3)) a)) Determine Fourier transform of the wave train g given by y E1 ( x) = P( x) cos k p x and E2 ( x) = P ( x) cos 2 k p x where P( x) is the unit square pulse. k p is the spatial frequency of the oscillatory region of the pulse. b) Plot the Fourier transform for both E1 And E2 . c) Sketch the transforms in the limit as width of the P ( x) extend to infinity. d) Use FFT function in MATLAB to plot the F { E ( x )} for both functions. Ans: F { E1 ( x )} = FC { E1 ( x )} = L[sin c(k p + k ) L + sin c(k p − k ) L]; F { E2 ( x )} = FC { E2 ( x )} = L sin ckL +
E1(x)
L L sin c(k + 2k p ) L + sin c(k − 2k p ) L; 2 2 P(x) E2(x)
1
-L
0
1
1
+L
x
-L
0
+L
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
-L
0
+L
x
x 13
Time domain Fourier transform The same wavepacket of exercise 4 4.3 3 in time domain: ⎧ E0 cos ω p t E (t ) = ⎨ ⎩0 1 f (t ) = 2π F (ω ) =
∞
∫
∞
∫
-T ≤ t ≤ T | t |> T
F (ω ) e −iωt dω
−∞
f (t )eiωt dt
−∞
F { E ( t )} = A(ω ) = T [sin c(ω p + ω )T + sin c(ω p − ω )T ]; A(k)
A(ω)
ωp ω +π/T p
kp k +π/L p kp-π/L
k
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
ωp-π/T
ω
14
Frequency bandwidth In time domain temporal width of the pulse is Δt = 2T while 0 < ω < ∞ 2π and width of the transform is said to be Δω ≈ . Thus Δt Δω = 4π T Product of the width of the pakage in t -space and ω -space is constant. In frequency q y domain spatial p width of the p pulse is Δx = 2 L while 0 < k < ∞ 2π . Thus ΔxΔk = 4π and width of the transform is said to be Δk ≈ L A(k)
A(ω)
ωp ω +π/T p
kp k +π/L p kp-π/L π/L
k
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
ωp-π/T π/T
ω
15
Uncertainty relations Product of the width of the pakage in k -space and x-space is constant. Δω , Δν = Δω / 2π and Δk = 2πΔλ / λ 2 are known as frequency bandwidths. Choice of Δω and Δk are somewhat arbitrary arbitrary. Important fact is that Δt Δν ≈ 1 and ΔxΔk = constant These relations are known as uncertainty relations and have profound practical importance. They impose limits on accuracy of our measurements or precision achievements. Choosing a frequency bandwidth of Δν restricts us in receiving signals 1 that are shorter than or Δt = Δν
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
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Coherence length If a wavetrain has frequency bandwidth of Δν , it has been produced within Δt ≈ 1/ Δν time interval from a group of oscillators. These oscillators may have produced waves that have constant phase relationship with each other only during Δt. The next wavetrain has a completely different phase relationship relationship. Δtc is known as coherence time of the source. Δlc = cΔtc is known as the coherence length g of the light g p produced by the source. Δν is due to natural linewidth of the plus which is due to different b d i mechanisms broadening h i such h as th thermall (D (Doppler l effect), ff t) collision lli i etc.
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
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Meaning of negative spatial frequency • Complex representation of Fourier transforms gives rise to a symmetrical distributions of positive and negative spatial frequency terms. • Certain optical phenomena such as diffraction occur symmetrically in space space. • A relationship between these phenomena and spatial frequency spectrum can be constructed if we use both negative and positive spatial frequency terms. • Negative frequency becomes a useful mathematical device to describe physical systems that are symmetrical around a central point
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
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About the constants of the Fourier transforms f ( x) = α
∞
∫
F ( k ) e − ikx dk
−∞
F (k ) = β
∞
∫
f ( x)eikx dx
−∞
α and β can be anything as long as 1 αβ = for the 1D Fourier transforms 2π 1 αβ = 2 for the 2D Fourier transforms 4π When α =β we have symmetric Fourier transformation.
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
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Parseval’s identities for Fourier integrals Sum of the squares of the Fourier coefficients of a function is equal to the square integral of the function. A02 ∞ 1 L 2 + ∑ Am2 + Bm2 For Fourier series: ∫ { f ( x)} dx = L −L 2 m =1 If FS (k ) and GS (k ) are Fourier sine transforms of f ( x) and g ( x), then
(
∫
∞
0
)
∞
Fs (k )Gs (k )dk = ∫ f ( x) g ( x)dx 0
If FC (k ) and GC (k ) are Fourier cos transforms of f ( x) and g ( x), then
∫
∞
0
∞
FC (k )GC (k )dk = ∫ f ( x) g ( x)dx 0
If f ( x) = g ( x) then ∞
∫0 {Fs (k )}
2
∞
{ f ( x)} 0
dk = ∫
2
dx and
∞
∫0 {FC (k )}
2
∞
{ f ( x)} 0
dk = ∫
2
dx
For general Fourier transforms:
∫
∞
−∞
F (k )G (k ) dk = ∫ *
∞
−∞
f ( x) g ( x)* dx
G (k )* is the complex conjugate of the G (k ) PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
20
The convolution theorem
If F (k ) and G (k ) are the Fourier transforms of f ( x) and g ( x) then
∫
∞
−∞
F (k )G (k )e
− ikx
dk = ∫
∞
−∞
f (u ) g ( x − u )du
If we show the convolution of the functions f and g with f * g, 1 ∞ f *g = f (u ) g ( x − u )du ∫ −∞ ∞ 2π F { f * g} = F { f } F { g} Fourier Four ier transform of the convolution of two functions is equal to the product of their Fourier transforms.
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
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Discrete Fourier transform • Analytical Fourier transformation is possible for some functions. • If there is no functional representation of data such as image of a person or collection of data points, then how we perform Fourier analysis? • There is a numerical techniques to determine frequency content of such data known as discrete Fourier transform.
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
22
Analysis of two-dimensional Signals and systems • Linearity is a common property of many physical phenomena
System Response
Stimuli
Addition
System
Many stimuli PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
Many Responses 23
Linear systems theory • Optical imaging operation is a linear mapping of object light distributions to image light distributions. • This mapping is done by the wave equation. • Linear systems theory gives us the ability to express the response to a complicated stimulus in terms of the responses to certain elementary stimuli. • Some mathematical tools are used in describing g linear phenomena
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
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Two dimensional Fourier transformation Goodman's notation: f x is equivalent to k x spatial frequency in x diretion Fourier transform or Fourier spectrum of a complex-valued function g ( x, y ) with two independent variables x, y (space) is : ∞
G ( f x , f y ) = F { g ( x, y )} = ∫ ∫ g ( x, y ) e
− j 2π ( f x x + f y y )
−∞
dxdy
The transform itself is a complex-valued function of two independent variables f X , fY (frequency) The inverse Fourier transform of G ( f X , fY ) F
−1
∞
{G} = ∫ ∫−∞ G ( f X , fY ) e j 2π ( f x + f y ) df x df y x
y
or Fourier integral representaion of function g ( x, y )
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
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Fourier transformation as a decomposition of g(x g(x,y) y) in 2D In dealing with linear systems we need to decompose a complicated input to a number of simple elements. Fourier transformation does this job. ∞
2π f X x g ( x) = F −1 {G} = ∫ G ( f X ) ej
df −∞
Elementary W i hti Weighting function factor or Amplitudes
This is expressing the space function g ( x) in terms of its frequency spectrum G ( f X ). This is linear combination of the elementry functions of the form e j 2π f X x The G ( f )s are the weighting g g factors or amplitudes p of each function.
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
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Decomposition of g(x,y) in 2D Goal: Finding orientation and frequency of the constant phase lines for exponential elementary functions. For 2 D Fourier transforms the elementary functions have the form: e j 2π ( f X x + fY y ) For each freq quency yp pair ( f X , fY ) the corresponding p g elementary y function has a zero or 2π m phase along the lines described by: f − X fY N
y=
x+
n where n is an integer. fY
y
Slope of the line perpendicular to the constant phase plane
So the elementary functions are being directed in ( x, y ) plane at an angle θ with respect to the x axis.
1/fY
⎛ fX ⎞ ⎟ f ⎝ Y ⎠
θ = tan -1 ⎜
1/fX
and spatial period of L = (Use cos θ =
θ
L
1 f +f 2 X
2 Y
x
.
L L & sin θ = to find L) 1/ f X 1/ fY
So we see the inverse Fourier transform as a form of decomposition with a periodic nature of the exponential elementary functions. PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
27
Fourier transformation and existence conditions Existence conditions (sufficient): 1) g mustt b be absolutely b l t l iintegrable t bl over the th fi finite it ((x,y)) plane l 2) g must have only finite number of discontinuitiesand a finite number of maxima o a aa and d minima. a 3) g must have no infinite discontinuities. If a function does not satisfy all of the above conditions and yet can be written as sum of the functions that satisfy the conditions, we can Fourier transform it by taking the Fourier transform of the pieces. Limit of this new sequence is called generalized Fourier transform of the function. If the transform exists, then we use it and don't worry about the existanace conditios. PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
28
Fourier Transforms (summary) 1D Hecht ) 2 D Hecht)
1 g ( x) = 2π f ( x, y ) =
2 D inverse Hecht) 2 D Goodman)
∞
∫
F ( k ) e − ikx dk
∞
F (k ) =
−∞
−∞
1 (2π )
∫
f ( x)eikx dx
2
∫∫
∞
−∞
F ( kx , k y ) e
F (k x , k y ) = ∫ ∫
∞
−∞
g ( x, y ) = F
−1
−i ( kx x + k y y )
f ( x, y ) e
dk x dk y
i ( kx x+ k y y )
dxdy
∞
{G} = ∫ ∫−∞ G ( f X , fY ) e j 2π ( f
X
x + fY y )
df X dfY
∞
2 D inverse Goodman) G ( f X , fY ) = F { g ( x, y )} = ∫ ∫ g ( x, y ) e − j 2π ( f X x + fY y ) dxdy −∞
Note: the reason Goodman's notation does not have the coefficients 1/ ( 2π )
2
is that he uses spatial frequency f x = 1/ λx insted of wavenumber k x = 2π / λx Here e
−i ( kx x + k y y )
or e j 2π ( f X x + fY y ) are the elementary components that the signal
is made up of and F ( k x , k y ) or G ( f X , fY ) is a complex function that containes information about the phase and amplitude of each elementary components. PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
29
Fourier transform theorems I 1 Linearity theorem: transform of sum is sum of transforms 1. F {α g + β h} = α F { g } + β F {h} 2. Similarity theorem: sreatch of coordinates in space domain results in contraction of coordinates in frequency q y domain and a change in amplitude of the spectrum 1 ⎛ f X fY ⎞ If F { g ( x, y )} = G ( f X , fY ) then F { g (ax, by )} = G⎜ , ⎟ | ab | ⎝ a b ⎠ 3 Shift theorem: 3. th translation t l ti is i space domain d i produces d a linear phase shift in frequency domain. - j 2 π ( f X a + fY b ) If F { g ( x, y )} = G ( f X , fY ) then F { g ( x − a, y − b)} = G ( f X , fY ) e
Phase shift Fourier transform is not affected
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
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Fourier transform theorems II 4. Rayleigh's enrgy theorem (parsewal's theorem): the sum (or integral) of the square of a function is equal to the sum (or integral) of the square of its transform. If F { g ( x, y )} = G ( f X , fY ) ∞
Then
∞
∫ ∫ | g ( x, y ) |
2
dxdy -∞
= ∫ ∫ | G ( f X , fY ) |2 df X dfY
- ∞ energy density in frequency domain
Energy contained in the waveform g ( x , y )
For the Fourier series the Parsewal's theorem takes the following form: A02 ∞ 1 L 2 f ( x)} dx = + ∑ Am2 + Bm2 { ∫ L −L 2 m=1
(
)
5. Convolution theorem: If F { g ( x, y )} = G ( f X , fY ) and F {h( x, y )} = H ( f X , fY ) then ∞ ⎪⎧ ⎪⎫ F ⎨ ∫ ∫ g (ξ ,η )h( x − ξ , y − η )d ξ dη ⎬ = G ( f X , fY ) H ( f X , fY ) ⎩⎪ - ∞ ⎭⎪ Convolution in space domain ⇔ multiplication in frequency domain.
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
31
Fourier transform theorems III 6. Cross-correlation is a measure of similarity of two waveforms as a function of a time lag applied to one of them time-lag them. It is commonly used to search a long duration signal for a shorter, known feature. It also has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology. F { g ( x, y )} = G ( f X , fY ) & F { f ( x, y )} = F ( f X , fY ) then cross-correlation of f , g is ∞
h( x, y ) = f * g = ∫ ∫ f (ξ ,η ) g * (ξ − x,η − y )dξ dη -∞
Autocorrelation is the cross-correlation of a function with itself. If F { g ( x, y )} = G ( f X , fY ) ⎧ ⎪⎧ ∞ 2 ⎪⎫ * ⎪F ⎨ ∫ ∫ g (ξ ,η ) g (ξ − x,η − y )d ξ dη ⎬ = G ( f X , fY ) ⎪ ⎪ -∞ ⎭⎪ then ⎨ ⎩ The theorem is a special case of ∞ ⎪ 2 * ⎪F g ( x, y ) = ∫ ∫ G (ξ ,η )G (ξ − f X ,η − fY )dξ dη -∞ ⎩ the convolution theorem in which we convolve g ( x, y ) with g * (− x, − y ) 7. Fourier integral theorem : at each point of continuity of g ,
{
}
FF −1 { g ( x, y )} = F −1F { g ( x, y )} = g ( x, y )
{
}
The two successive transforms yeild F F { g ( x, y )} = ag (− x, − y ) the successive transformation and inverse transformation of a function are not y the same although g for even functions they y differ within a constant. exactly 8. Fourier transform of a separable function can be written as: F { g ( x, y )} = FX { g X ( x)} FY { gY ( y )} if g ( x, y ) = g X ( x) gY ( y ) PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
32
Exercise
4 4) Fourier transform of the square pulse f ( x ) = {0 f 4.4) E
- L / 2≤ x ≤ L / 2 | x| > L / 2
is
F { f ( x )} = E f L sin c
kL . Use the Fourier transform theorems to find the Fourier 2 tansform of the following pulses. In each case mention the theorem used.
a) f a ( x ) = {0 f
- L≤ x≤ L | x| > L
d) f d ( x ) = {0 f
-α L ≤ x ≤ β L x β L
E
E
b) f b ( x ) = {0 f E
- aL ≤ x ≤ aL | x| > aL
e) f d ( x ) = {0 f E
c) f c ( x ) = {0 f E
- L / b≤ x≤ L / b | x| > L / b
5 L ≤ x ≤10 L x 10 L
{
}
{
}
f)) If g ( x) = Eg cos k g x find F { g ( x)} then F { f ( x) + g ( x)}; F f 2 ( x) ; F g 2 ( x) F { f ( x) * g ( x)} where * is the sign for convolution.
g) Calculate f ( x) * g ( x) directly without using any of the theorems. Then take the Fourier transform of the result to confirm the convolution theorem.
g(x)
f(x)
Eg
0
Ef
x
-L/2
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
0
+L/2
x 33
Delta function The delta dunction is a Generalized Function that is defined as the limit of a class of delta sequences. It is also called "Dirac's delta function" or "impulse symbole". A generalized function is generalization of the concept of a function and they are particularly useful f in making discontinuous functions f more like smooth functions. A delta sequence q is a sequence q of strongly g y peaked p functions for which
∫
lim
∞
−∞
n →∞
δ n ( x) f ( x)dx = f (0)
As n → ∞ the sequences become delta functions. A Fundamental property of the delta function: I)
∫
∞
−∞
f ( x)δ ( x − α )dx = f (α ) and in fact
α +ε
∫α ε −
f ( x)δ ( x − α )dx = f (α )
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
34
More identities with delta function II ) δ ( x - α ) = 0 for x ≠ α IV ) δ ( x 2 - α 2 ) = V ) δ [ g ( x)] = ∑ i
1 2α
III ) δ (α x) =
1
α
δ ( x)
[δ ( x + α ) + δ ( x - α )]
δ ( x − xi ) g '( xi )
where e e xi a are e the e roots oo s o of g
VI ) xδ '( x) = -1( x) or more generally x nδ ( n ) = (−1) n n !δ ( x) p to x where δ ( n ) ((x)) is nth derivative of the delta function with respect VII ) δ '(- x) = -δ '( x)
∫
VIII )
∞
−∞
f ( x)δ '( x - α )dx = - f '(α ) which is equivalent to convolution ∞
(δ '* f )(α ) = ∫ δ ' (α − x ) f ( x)dx = − f '(α ) ?? -∞
IX ) XII)
∫
∞
-∞
δ '( x) dx = ∞
X ) x δ '( x) = 0 2
∫ f ( x)δ ( x - x )dx = f ( x ) 0
0
⎛1⎞ XI ) ∫ δ ⎜ ⎟dx = 0 -1 1 ⎝x⎠ 1
sifting property
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
35
Delta function in higher dimensions In two dimension:
In three dimension:
⎧⎪0 2 δ ( x, y ) = ⎨ ⎪⎩∞
⎧⎪0 δ ( x, y , z ) = ⎨ ⎪⎩∞
∞
∫ ∫
∞
−∞ −∞
x2 + y 2 ≠ 0 x2 + y2 = 0
δ ( x, y )dxdy d d =1 2
1 δ (ax, by ) = δ 2 ( x, y ) ab δ 2 ( x, y ) = δ ( x)δ ( y ) In polar coordinates: 2
δ (r ) δ (r ,θ ) = π r 2
3
∞
∞
∫ ∫ ∫
∞
−∞ −∞ −∞
x2 + y2 + z 2 ≠ 0 x2 + y 2 + z 2 = 0
δ 3 ( x, y, z )dxdydz d d d =1
1 3 δ ( x, y , z ) abc δ 3 ( x, y, z ) = δ ( x)δ ( y )δ ( z )
δ 3 (ax, by, cz ) =
In cylinderical coordinates: δ (r )δ ( z ) δ 3 (r ,θ , z ) = πr In polar coordinates:
δ 3 (r ,θ , φ ) =
δ (r ) 2π r 2
PHYS 258 Spring 2010 SJSU Eradat Fouriet integrals
36
Some delta sequences* If we take limit of anyy of these sequences q and let n → ∞, we will have a delta function. 1 1 ⎧ < < n x ⎪⎪ 2n 2n 1) δ ( x) = lim ⎨ n →∞ 1 ⎪0
