EE 485, Winter 2004, Lih Y. Lin

Chapter 4 Fourier

- Based on harmonic analysis () and liner system (superposition). - An arbitrary function ∞ ∞ = ν ν − π ν +ν ν ν f (x, y) ∫∫F( x , y )exp[ j2 ( x x y y)]d xd y −∞−∞ → Superposition, or integral of harmonic functions of x and y. ν ν F( x , y ) : Complex amplitude ν ν x , y : Spatial (cycles/unit length)

- Compare this with plane = − + + U (x, y, z) Aexp[ j(kx x k y y kz z)]

= − π ν +ν U (x, y,0) Aexp[ j2 ( x x y y)] k k ν ↔ x , ν ↔ y x 2π y 2π - An arbitrary function can be analyzed as a superposition of harmonic functions. → An arbitrary traveling wave U (x, y, z) may be analyzed as a sum of plane !

4.1 Propagation of Light in Free Space A. Correspondence Between the Spatial Harmonic Function and the Plane Wave

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−1k  −1 θ = sin  x  = sin (λν ) x  k  x (4.1-1) − k  − θ = sin 1 y  = sin 1(λν ) y  k  y A physical way of picturing the spatial harmonic function is to project a plane wave on the x-y plane. Λ = 1 Λ = 1 x ν , y ν x y

−   −   θ = 1 λ  θ = 1 λ  → x sin Λ , y sin Λ  x   y  Paraxial approximation: θ = λ = λν θ = λ = λν x Λ x , y Λ y (4.1-2) x y

Spatial spectral analysis (Response of a plane wave after a thin optical element.) Consider a simple case:

= − π ν +ν t(x, y) exp[ j2 ( x x y y)] Λ = 1 Λ = 1 → Harmonic function on x-y plane with period x ν , y ν . x y = − π ν +ν − U (x, y, z) Aexp[ j2 ( x x y y)]exp( jkz z) θ = −1 λν θ = −1 λν → Output wave is bent with angles x sin ( x ), y sin ( y ). The harmonic function pattern works like a grating. Now consider a general case: = ν ν − π ν +ν ν ν t(x, y) ∫∫ F( x , y )exp[ j2 ( x x y y)]d xd y (4.1-4)

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= ν ν − π ν +ν − ν ν U (x, y, z) ∫∫ F( x , y )exp[ j2 ( x x y y)]exp( jkz z)d xd y 1 1 1 k = k 2 − k 2 − k 2 = 2π − − z x y λ2 λ2 λ2 x y → An incident plane wave is decomposed into many plane waves, each θ = −1 λν θ = −1 λν traveling at angles x sin ( x ), y sin ( y ), with a complex envelope ν ν F( x , y ) , the Fourier transform of f (x, y) .

Example 4.1-2, Imaging  πx2  t(x, y) = exp j  = exp()− j2πϕ(x, y)  λf 

x2 ϕ(x, y) = − 2λf ϕ ↔ν +ν Compare to earlier: (x, y) x x y y ∂ϕ(x, y) x Now ν varies with x ⇒ν = = − x x ∂x λf

−  x  ⇒ θ = sin 1−  x  f  → A cylindrical with focal length f

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B. Transfer Function of Free Space Since an arbitrary function can be analyzed as sum of harmonic functions, we consider a harmonic input function.

= = − π ν +ν f (x, y) U (x, y,0) Aexp[ j2 ( x x y y)] = = − + + Output g(x, y) U (x, y,d) Aexp[ j(kx x k y y kz d)] g(x, y) Η(ν ,ν ) = = exp(− jk d) x y f (x, y) z  1  (4.1-6)  1  2 = exp− j2π  −ν 2 −ν 2  d λ2 x y    

Fresnel approximation ν 2 +ν 2 << 1 x y λ2 → The plane-wave components of the propagating light make small angles θ λν θ λν x ~ x , y ~ y . → Paraxial waves: ν ν = − [ πλ ()ν 2 +ν 2 ] H ( x , y ) exp( jkd)exp j d x y (4.1-8) Validity of Fresnel approximation has the same expression as in Sec. 2.2.

Input-output relation Given the input function f (x, y) , how to obtain the output g(x, y) : (1) Determine the complex envelopes of the plane-wave components in the input plane by Fourier transform. ∞ ∞ ν ν = π ν +ν F( x , y ) ∫∫f (x,y)exp[ j2 ( x x y y)]dxdy −∞−∞

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(2) Complex envelopes of the plane-wave components in the output plane = ν ν ν ν H( x , y )F( x , y ) = ν ν ν ν − π ν +ν ν ν (3) g(x, y) ∫∫ H ( x , y )F( x , y )exp[ j2 ( x x y y)]d xd y Under Fresnel approximation, g(x, y) = H ∫∫ F(ν ,ν )exp[ jπλd(ν 2 +ν 2 )]exp[− j2π (ν x +ν y)]dν dν 0 x y x y x y x y ≡ − H 0 exp( jkd)

Free-space propagation as a Each point generates a spherical wave. Under Fresnel approximation (observation point close to the propagation axis), spherical wave → parabolic wave.

 x2 + y 2  h(x, y) ≈ h exp− jk  (4.1-13) 0  2d  j h = exp(− jkd) 0 λd  (x − x')2 + (y − y')2  g(x, y) = h ∫∫ f (x', y')exp− jπ dx'dy' (4.1-14) 0  λd 

4.2 Optical Fourier Transform A plane wave transmitting through an optical element can be used to ν ν decompose the harmonic functions (Fourier components F( x , y ) ) that compose the pattern ( f (x, y) ) on the optical element.

A. Fourier Transform in the Far Field (Fraunhofer Approximation) If f (x, y) is confined to a small area of radius b, distance d to the observation 2 plane is sufficiently large, so that Fresnel number for f (x, y) , N '= b << 1. F λd

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 x2 + y 2  x y g(x, y) = h exp− jπ F( , ) (4.2-4) 0  λd  λd λd Furthermore, if we limit our interest to points at the output plane within a circle 2 of radius a centered about the z axis, so that N = a <<1 for g(x, y) . F λd x y g(x, y) = h F( , ) (4.2-1) 0 λd λd

→ The only plane wave that contributes to the complex amplitude at (x, y) at output plane is the wave making angles θ = x ,θ = y with the optical axis. x d y d   This is also the wave with wave-vector components k = ()x k, k =  y k and x d y  d  amplitude F(ν ,ν ) with ν = x ,ν = y . x y x λd y λd • Fraunhofer approximation is valid when both N F and N F ' are small.

B. Fourier Transform Using a Lens

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θ θ = λν λν Amplitude of the plane wave with direction ( x , y ) ( x , y ) is ν ν proportional to the Fourier transform F( x , y ) and is located at the point = θ θ = λ ν λ ν (x, y) ( x f , y f ) ( f x , f y ) . x y → g(x, y) ∝ F( , ) (4.2-5) λf λf

j  (x2 + y 2 )(d − f ) x y g(x, y) = exp[− jk(d + f )]exp jπ F( , ) (4.2-8) λf  λf 2  λf λf 2 1 x y I(x, y) = F( , ) (4.2-9) | λf |2 λf λf j x y If d = f , g(x, y) = exp[− j2kf ]F( , ) (4.2-10) λf λf λf Fourier transform using a lens is valid in Fresnel approximation (only radius at the output is limited). Without the lens, we need Fraunhofer approximation (radii at both output and input are limited).

4.3 of Light

Light not simply blocked by an opaque object, as in Ray Optics. It depends on the , the dimension of the object, and the distance between the object and the observation plane.

A. x y Aperture function p(x, y) , with Fourier components P(ν ,ν ) = P( , ). x y λd λd

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Assume the incident wave is a plane wave of intensity Ii in z-direction. Using Eq. (4.2-1), Fraunhofer approximation, we obtain: I x y 2 I(x, y) = i P( , ) (4.3-4) (λd)2 λd λd → Proportional to the squared magnitude of the Fourier transform of the x y aperture function p(x, y) evaluated at the ν = , ν = . x λd y λd

Example: Fraunhofer diffraction from a circular aperture 2 2 πD 2  2J (πDρ λd) I(ρ) =   I  1  (4.3-7)  4λd  i  πDρ λd  ρ = λ → Airy pattern. Center disk () has radius s 1.22 d / D , subtending an angle θ =1.22λ / D .

B.

At small distance (d → 0 ), the diffraction pattern is the shadow of the aperture. At medium distance (Fresnel diffraction), the diffraction pattern is the convolution of the aperture. Using Eq. (4.1-14), free-space propagation as a convolution, we obtain: 2 I  (x − x')2 + (y − y')2  I(x, y) = i ∫∫ p(x', y')exp− jπ dx'dy' (4.3-11) (λd)2  λd  At large d, the diffraction pattern becomes Fraunhofer diffraction pattern. The far field has an angular divergence proportional to λ / D , where D is the diameter of the aperture.

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4.4 Image Formation Spatial filtering Two-lens imaging system (4-f system). Unity maginification.

4-f imaging system for Fourier transform. The Fourier components of f (x, y) are separated by the lens. Each point in the Fourier plane corresponds to a single spatial frequency (Recall Fig. 4.2-2). The second lens reconstructs the image.

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Spatial filtering: Add a mask at the Fourier plane to block unwanted Fourier components of f (x, y) .

Transfer function of the mask for the Fourier components: ν ν = λ ν λ ν H ( x , y ) p( f x , f y ) (4.4-4) ν ν = ν ν ν ν Output: G( x , y ) H ( x , y )F( x , y )

Example: (a) Low-pass filter ν ν = ν 2 +ν 2 <ν 2 ν H ( x , y ) 1 for x y s , s : cutoff frequency ν ν = H( x , y ) 0 othewise

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A low-pass filter for spatial frequency is a circular aperture of diameter = ν λ D 2 s f . (b) High-pass filter Complement of low-pass filter. Output is high at regions of large rate of change, small at regions of smooth or slow variation of the object. Application: Edge enhancement in image-processing. (c) Vertical-pass filter Blocks horizontal frequency and transmits vertical frequency.

4.5 Holography Recording and reconstruction of optical waves. Consider an arbitrary monochromatic optical wave. At z = 0 plane, = U U 0 (x, y) . If a thin optical element (transparency) has complex amplitude = transmittance t(x, y) U 0 (x, y). Illuminate the transparency with a uniform plane wave in z-direction, the optical wave U (x, y) can be reconstructed. Transparency → Hologram

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How to get t(x, y) from U 0 (x, y)? information is very important. Need some kind of coding to transform phase into intensity.

Holographic code and off-axis holography

Mixing the original wave (object wave) U 0 with a known reference wave U r , and recording their interference pattern in z = 0 plane. ∝ + + * + * t(x, y) I r I0 U rU 0 U rU 0 (4.5-1)

Decoding: Illuminate the hologram with U r , = ∝ + + + 2 * U tU r U r I r U r I0 I rU 0 U r U 0 (4.5-2) First and second terms → Reference wave Third term → Original wave Fourth term → Conjugate of the original wave

Example 4.5-1: Hologram of an oblique plane wave = − θ U 0 (x, y) I0 exp( jkxsin ) ∝ + + − θ + θ U (x,u) I r I0 I r I0 exp( jkxsin ) I r I0 exp( jkxsin ) → Diffraction grating

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How to make sure the object wave can be well separated? Consider an arbitrary object wave whose propagation direction centers about θ . At = = − θ z 0 plane, U 0 (x, y) f (x, y)exp( jkxsin ) . ν Assume f (x, y) varies slowly so that its maximum spatial frequency s , θ = −1 λν << θ corresponding to s sin ( s ) . ∝ + 2 + − θ + + θ U (x, y) I r f (x, y) I r f (x, y)exp( jkxsin ) I r f *(x, y)exp( jkxsin ) f (x, y) 2 : Ambiguity term → Non-uniform plane wave in directions with a cone of θ 2 s around the z-direction. θ > θ >> 2 If (1) 3 x , or (2) Ir f (x, y) , the original wave can be resolved unambiguously.

Fourier-transform holography ν ν Fourier transform F( x , y ) of a function f (x, y) can be obtained by a lens (see Sec. 4.2). F(ν ,ν ) = F( x , y ) = U (x, y). x y λf λf 0

Illuminate the hologram with U r reconstruct F. The original function f (x, y) is reconstructed at the focal plane using a lens.

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The holographic apparatus One essential requirement for holography: A monochromatic light source with minimum phase fluctuations. → A coherent light source, usually a laser.

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