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1 11.2 Fourier transforms 11.2.1 One-dimensional transforms Complex exponential representation:  Fourier transform   Inverse Fourier transform  In.

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Presentation on theme: "1 11.2 Fourier transforms 11.2.1 One-dimensional transforms Complex exponential representation:  Fourier transform   Inverse Fourier transform  In."— Presentation transcript:

1 1 11.2 Fourier transforms 11.2.1 One-dimensional transforms Complex exponential representation:  Fourier transform   Inverse Fourier transform  In time domain: April 27 Fourier transform Chapter 11 Fourier optics Note: Other alternative definitions exist. Proof: Notations: The fancy “ F ” (script MT) is very hard to type.

2 2 Example: Fourier transform of the Gaussian function 1) The Fourier transform of a Gaussian function is again a Gaussian function (self Fourier transform functions). 2) Standard deviations (where f (x) drops to e -1/2 ):

3 3 11.2.2 Two-dimensional transforms x y a

4 4 J0(u)J0(u) J1(u)J1(u) u

5 5

6 6 Read: Ch11: 1-2 Homework: Ch11: 2,3 Due: May 8

7 7 April 29 Dirac delta function 11. 2.3 The Dirac delta function Dirac delta function: A sharp distribution function that satisfies: 0 1 x  (x) Sifting property of the Dirac delta function: If we shift the origin, then 0 1 x  (x- x 0 ) x0x0 A way to show delta function

8 8 Delta sequence: A sequence that approaches the delta function when the distribution is gradually narrowed. The 2D delta function: The Fourier representation of delta function:

9 9 Displacement and phase shift: The Fourier transform of a function displaced in space is the transform of the undisplaced function multiplied by a linear phase factor. Proof:

10 10 Fourier transform of some functions: (constants, delta functions, combs, sines and cosines): 0 22 k F(k)F(k) 0 1 x f(x)f(x) 0 1 x f(x)f(x) 0 1 k F(k)F(k) 0 x f(x)f(x) 0 k F(k)F(k) 0 x f(x)f(x) 0 k F(k)F(k) … …

11 11 0 x f(x)f(x) k A(k)A(k) 0 x f(x)f(x) k B(k)B(k) x f(x)f(x) 0 k A(k)A(k) x f(x)f(x) 0 k B(k)B(k)

12 12 Read: Ch11: 2 Homework: Ch11: 4,7,8,9,14 Due: May 8

13 13 May 1,4 Convolution theorem 11.3 Optical applications 11.3.1 Linear systems Linear system: Suppose an object f (y, z) passing through an optical system results in an image g(Y, Z), the system is linear if 1) af (y, z)  ag(Y, Z), 2) af 1 (y, z) + bf 2 (y, z)  ag 1 (Y, Z) + bg 2 (Y, Z). We now consider the case of 1) incoherent light (intensity addible), and 2) M T = +1. The flux density arriving at the image point (Y, Z) from dydz is Point-spread function y I 0  (y,z) z Y I i  (Y,Z) Z

14 14 Example:  The point-spread function is the irradiance produced by the system with an input point source. In the diffraction-limited case with no aberration, the point-spread function is the Airy distribution function.  The image is the superposition of the point-spread function, weighted by the source radiant fluxes. y I 0  (y,z) z Y I i  (Y,Z) Z y I 0  (y,z) z Y I i  (Y,Z) Z

15 15 Space invariance: Shifting the object will only cause the shift of the image:

16 16 11.3.2 The convolution integral Convolution integral: The convolution integral of two functions f (x) and h(x) is Symbol: g(X) = f(x)  h(x) Example 1: The convolution of a triangular function and a narrow Gaussian function. Question: What is f(x-a)  h(x-b)? Answer: x f(x)f(x) x h(x)h(x) x h(X-x)h(X-x) X X f(x)h(x)f(x)h(x)

17 17 Example 2: The convolution of two square functions. x f(x)f(x) x h(x)h(x) x h(X-x)h(X-x) X X f(x)h(x)f(x)h(x) The convolution theorem: Proof: Example: f (x) and h(x) are square functions.

18 18 Transfer functions: Optical transfer function T (OTF) Modulation transfer function M (MTF) Phase transfer function  (PTF) Frequency convolution theorem: Example: Transform of a Gaussian wave packet. Please prove it.

19 19 Read: Ch11: 3 Homework: Ch11: 15,19,20,21,22,26,27 Due: May 8

20 20 May 6 Fourier methods in diffraction theory 11.3.3 Fourier methods in diffraction theory Fraunhofer diffraction: Aperture function: The field distribution over the aperture: A(y, z) = A 0 (y, z) exp[i  (y, z)] dydz P(Y,Z) r R x y z Y Z X Y Z Each image point corresponds to a spatial frequency. The field distribution of the Fraunhofer diffraction pattern is the Fourier transform of the aperture function:

21 21 The single slit: Rectangular aperture: z A(z)A(z) b/2 -b/2 kZkZ E(kZ)E(kZ) 2/b2/b Fraunhofer-: The light interferes destructively here. Fourier-: The source has no spatial frequency here. The double slit (with finite width): z f(z)f(z) b/2 -b/2 kZkZ F(kZ)F(kZ) z h(z)h(z) a/2 -a/2 z g(z)g(z) a/2 -a/2 kZkZ H(kZ)H(kZ) kZkZ G(kZ)G(kZ)  = × =

22 22 Three slits: z f(z)f(z) a -a 0 kZkZ F(kZ)F(kZ) kZkZ |F  (k Z )| 2 Apodization: Removing the secondary maximum of a diffraction pattern. Rectangular aperture  sinc function  secondary maxima. Circular aperture  Bessel function (Airy pattern)  secondary maxima.  Gaussian aperture  Gaussian function  no secondary maxima. z f(z)f(z) kZkZ F(kZ)F(kZ)

23 23 Array theorem: The Fraunhofer diffraction pattern from an array of identical apertures = The Fourier transform of an individual aperture × The Fourier transform of a set of point sources arrayed in the same manner. Convolution theorem y z y z y =  Example: The double slit (with finite width).

24 24 Read: Ch11: 3 Homework: Ch11: 29,30,32 Due: May 11

25 25 Everything is a lost horse.


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