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HW03_in_class.jl
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HW03_in_class.jl
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### A Pluto.jl notebook ###
# v0.19.40
using Markdown
using InteractiveUtils
# This Pluto notebook uses @bind for interactivity. When running this notebook outside of Pluto, the following 'mock version' of @bind gives bound variables a default value (instead of an error).
macro bind(def, element)
quote
local iv = try Base.loaded_modules[Base.PkgId(Base.UUID("6e696c72-6542-2067-7265-42206c756150"), "AbstractPlutoDingetjes")].Bonds.initial_value catch; b -> missing; end
local el = $(esc(element))
global $(esc(def)) = Core.applicable(Base.get, el) ? Base.get(el) : iv(el)
el
end
end
# ╔═╡ d1980bd2-babf-11ee-1dbb-dfefbdbdb36d
using PlutoUI, Plots, ImageShow, TestImages, FFTW, NDTools, IndexFunArrays, FileIO, FourierTools, SpecialFunctions, UrlDownload, ImageMagick
# ╔═╡ 90c5b641-ac4d-410a-bb78-e5e1be0d1912
# ╔═╡ 17adc747-9822-42be-91c1-ad8a2e1532bc
md"# 0. Load packages
We had to add some new packages, so wait a couple of minutes to load the new packages
"
# ╔═╡ d2c6102c-7a62-4899-9c5c-b08ce9a5baa8
FFTW.set_num_threads(4)
# ╔═╡ 8e6ec43c-9135-4829-8146-a56b8624750a
const TODO = nothing
# ╔═╡ 7331d6a5-dd03-42c7-9ab3-c84a641296bc
TableOfContents()
# ╔═╡ f326cd36-9c05-4b1c-81c0-56c954cd51f3
gauss_R(z::T, z_R) where T = iszero(z) ? T(Inf) : (1 + (z_R / z)^2)
# ╔═╡ 7c6b4ddd-ca5a-4b23-96f7-e896fb1cda6d
gauss_ψ(z, z_R) = atan(z, z_R)
# ╔═╡ dd16735b-c1d6-4977-a396-a3e23823ee68
gauss_w(z, z_R, w_0) = w_0 * sqrt(1 + (z / z_R)^2)
# ╔═╡ 8e04ae26-5474-4c2c-9fda-dbb7efd77acd
"""
gauss_beam(y, x, z, λ, w_0)
Returns the eletrical field of a Gaussian beam at position `(y, x)` at optical axis position `z` with respect to the beam waist `w_0`.
Wavelength is `λ`.
"""
function gauss_beam(y, x, z, λ, w_0)
k = π / λ * 2
z_R = π * w_0^2 / λ
r² = x ^ 2 + y ^ 2
# don't put exp(i * k * z) into the same exp, it causes some strange wraps
return w_0 / gauss_w(z, z_R, w_0) * exp(-r² / gauss_w(z, z_R, w_0)^2) *
exp(1im * k * z) *
exp(1im * (k * r² / 2 / gauss_R(z, z_R) - gauss_ψ(z, z_R)))
end
# ╔═╡ 5f21a849-0272-4bca-9659-54cd38068e09
"""
bpm(field, λ0, Lx, Ly, z; window=true, paraxial=true, amplitude_array)
Propagates the array `field` with wavelength `λ0` and the filed size in meter size
`(Lx, Ly)`. The propagation distance `z` should be a vector of distances.
.
The returned array is a three dimensional array where `size(arr, 3) == size(z, 1)`.
If `window=true` we apply a Hann window function to dampen boundaries.
A keyword `amplitude_array` can be provided, which multiplies with the field at each point. This allows to include obstacles.
If `paraxial=true` the Fresnel approximation is applied.
"""
function bpm(field, λ0, Lx, Ly, z, n=1; window=true, amplitude_array=ones(size(field)..., length(z)), paraxial=true)
# free space wavenumber in m-1
k0 = 2 * π / λ0
# medium wavenumber m-1
k = n * k0
# medium in m
dz = z[2] - z[1]
# field parameters
Nx = size(field, 1)
dx = Lx / Nx
x = range(-Lx/2, Lx/2, Nx)
fx = reshape(fftfreq(Nx, 1 / dx), (1, Nx))
Ny = size(field, 1)
dy = Ly / Ny
y = range(-Ly/2, Ly/2, Ny)
fy = fftfreq(Ny, 1 / dy)
if paraxial
# important step, this calculates the Fourier space kernel
H = exp.(-1im .* k .* λ0^2 .* (fx.^2 .+ fy.^2) ./ (2) * dz)
else
H = exp.(1im .* sqrt.(1 .+ 0im .- λ0^2 .* fx.^2 .- λ0^2 .* fy.^2) .* k .* dz) .* ((λ0^2 .* fx.^2 .+ λ0^2 .* fy.^2) .< 1)
end
# 3d output fields we save
# third dimensions stores the different z propagation distances
out_field = zeros(ComplexF64, (Ny, Nx, size(z, 1)))
# first entry corresponds to z[1] = 0
out_field[:, :, 1] = field
# FFT plan for calculating FFTs
# It's a more efficient sytanx: p * x == fft(x)
p = plan_fft(field, (1,2))
window_f = window ? IndexFunArrays.window_hanning(size(out_field)[1:2], border_in=0.8) : 1
# inverse FFT
invp = inv(p)
for z_index in 2:size(out_field, 3)
u0 = out_field[:, :, z_index - 1] .* window_f .* amplitude_array[:, :, z_index - 1]
u1 = invp * ((p * u0) .* H)
out_field[:, :, z_index] .= u1
end
return out_field
end
# ╔═╡ aaf7cf09-1244-461b-aaa6-731a65fc3d57
md"# 1. Bessel Beam with an Axicon
A Bessel beam is a field whose amplitude is described by a Bessel function of the first kind. As we discussed in the class the Bessel beam can propagate without diverging for much longer distances than the Gaussian beam. Bessel beam can be obtained by illuminating an axicon with a plane wave. The axicon is an optical component that looks like a lens but it has a conical shape. The angle θ between the side of the axicon and its base determines the shape and length of the Bessel beam.
In addition, Bessel beams are characterized by their self-healing property. This characteristic is referred to their ability to reconstruct their field when they are disturbed by an obstacle.
"
# ╔═╡ 15674d2a-e0f1-45b4-8471-886c6e879a12
urldownload("https://upload.wikimedia.org/wikipedia/commons/thumb/b/b8/Bessel_beam.svg/640px-Bessel_beam.svg.png")
# ╔═╡ 955baec1-a632-4ea0-8e6f-545057316a1d
md"**Figure 1**: Axicon"
# ╔═╡ 233260af-965e-47af-aa27-183bf34b6560
urldownload("https://upload.wikimedia.org/wikipedia/commons/thumb/c/c7/Bessel_beam_reform.svg/640px-Bessel_beam_reform.svg.png")
# ╔═╡ 6f3aadd7-9ce5-47ca-b3aa-e54001aca4e8
md"**Figure 2**: Axicon with Obstacle"
# ╔═╡ ac3147f3-1831-4ca8-b9d9-da78b3a9b6c3
md"## 1.1 Bessel Beam
A bessel beam is given by
$$A(x,y) = J_m(k_t \cdot \rho)$$
where $J_m$ is the m-th bessel function of the first kind.
$\rho = \sqrt{x^2+y^2}$ and $k_t = \sqrt{k^2 (1 - \sin^2\theta)}$ and $\theta$ is the angle of the axicon. $k_t$ is called the transversal wave number.
"
# ╔═╡ 429a9db1-f696-40da-8aba-ee49e4680c47
λ = 633e-9
# ╔═╡ 78ef08a1-1486-4f0f-aee5-d56c91e53561
k = 2π / λ
# ╔═╡ aa2bb7f0-24b6-49e5-bb49-14be4936fb9e
md"θ in degree = $(@bind θ PlutoUI.Slider(0:88, show_value=true, default=88))"
# ╔═╡ 200cf84b-2f99-4ce9-aee0-c9a823ede21b
# define the transversal wave vector
kt = sqrt(k^2 * (1 - sind(θ)^2))
# ╔═╡ 6e737c1e-ca25-4d21-a66b-769677335f1d
sz = (256, 256)
# ╔═╡ abbc7471-0d30-41aa-b42b-248628175846
L = 100f-6
# ╔═╡ 6935acba-6e78-4134-ba72-e21ec0261355
dx = L ./ sz
# ╔═╡ 368881fa-4834-4ead-9322-22c1523a6d03
bb = SpecialFunctions.besselj.(0, kt .* rr(sz, scale=dx));
# ╔═╡ 1455fff6-dcc3-46d2-95ef-be4e0d2e5b39
y = range(- L / 2, L / 2, sz[2])
# ╔═╡ 0b428a2d-37e6-4fa4-91be-f6888ec49d22
x = y';
# ╔═╡ cecd2ae7-b6c9-4d8b-b262-79bc4b5e5cda
gb = gauss_beam.(y, x, 0.0, λ, 5f-6);
# ╔═╡ ef35926c-5acb-4e7b-aa97-579ea2eec22e
heatmap(y, y, abs.(bb), xlabel="x in m", ylabel="y in m",cmap=:turbo, aspect_ratio=:equal)
# ╔═╡ eb683da7-cf08-4505-925f-a2aa4dd59b12
md"## 1.2 Propagate Bessel Beam
Propagate the Bessel and the Gaussian beam each over 1.5 millimeters
"
# ╔═╡ 906533cc-1c47-4d95-9a37-54a22cd55048
z = range(0, 1.5f-3, 100)
# ╔═╡ 7d0afb02-82b9-4af9-b04e-7680d4649d2a
bb_p =bpm(bb, λ, L, L, z, window=true);
# ╔═╡ 45171f64-bacc-4278-81c5-c2faf730f7fb
gb_p = bpm(gb, λ, L, L, z, window=true);
# ╔═╡ e37276c3-de6e-4e01-859f-c8815a89685c
heatmap(z, y, abs2.(bb_p[sz[1]÷2 + 1, :, :]), xlabel="z in m", ylabel="y in m",cmap=:turbo)
# ╔═╡ a91e91f2-008c-4462-a1a1-7e63b34612bb
heatmap(z, y, abs2.(gb_p[sz[1]÷2 + 1, :, :]), xlabel="z in m", ylabel="y in m", cmap=:turbo)
# ╔═╡ 4709e734-9d44-488c-9380-804758a6b96a
md"### 1.2 Question
Explain in this box, why is the bessel beam still diverging and not further propagating (it seems to diverge after some propagation)?
Shortly compare Gaussian and Bessel beam.
### 1.2 Answer
A true Bessel beam is non-diffractive. This means that as it propagates, it does not diffract and spread out. However, as with a plane wave, a true Bessel beam cannot be created, as it requires an infinite amount of energy. That's why the bessel beam you defined eventially diverges.
Although an ideal Gaussian beam stays bounded over a certain propagation range after which they diverge, an ideal Bessel beam is among a class of solutions to the wave equation that are ideally diffraction-free and do not diverge when they propagate.
In reality, the gaussian beam has an ellipsoidal focus, while the bessel beam has a line shape focus.
"
# ╔═╡ deb092ef-4341-48c6-bea8-2887fe2ae0d8
md"## 1.3 Propagate after an Obstacle"
# ╔═╡ 6f519ac0-cc1b-4b76-a92a-9ee640f119af
begin
obstacle = ones(size(gb_p))
obstacle[120:140, 120:140, 5] .= 0
end;
# ╔═╡ 7509c4c6-ce81-40e3-92ca-251a17359729
heatmap(z, y, obstacle[sz[1]÷2 + 1, :, :], xlabel="z in m", ylabel="y in m", cmap=:turbo)
# ╔═╡ ae7595ed-7673-461a-a48c-c9c35e4aefe0
bb_obstacle = bpm(bb, λ, L, L, z, window=true, amplitude_array=obstacle);
# ╔═╡ 961f1340-7e6c-4be9-9f0a-fe5b811f0331
heatmap(z, y, abs2.(bb_obstacle[sz[1]÷2 + 1, :, :]), xlabel="z in m", ylabel="y in m", cmap=:turbo)
# ╔═╡ ce9cdcdb-72da-4ee1-841a-2c1abe099df3
md"### 1.3 Question
Why can the beam recover behind the obstacle?
### 1.3 Answer
The Bessel beam is composed of many plane waves with the same spatial frequency but different orientations. Note that Fourier transform of a Bessel beam is a ring. When it faces with an obstacle, all the plane waves are partially blocked but they interfere after the obstacle to reconstruct the Bessel beam. This scenario is similar to its creation by an axicon lens in a geometric point of view.
"
# ╔═╡ bf30dceb-a00e-445f-9eb9-bad573c9551b
md"# 2. Scalable or Not?
In the paraxial beam propagation equation derived in class, λz appear always together implying a scalability between λ, the wavelength and z, the propagation distance.
This means that for the same initial optical field a change of the wavelength will reproduce the same diffraction pattern at a different z distance.
Verify that using the code provided in the moodle. Use as an input,
$$A(x, y, z = 0) = \frac12 \left(1 + \cos\frac{2\pi}{\Lambda}x \right)$$
with $\Lambda = 4μm$.
What happens if you use the non-paraxial code?
(Use `bpm` with the keyword `paraxial=false`)
## Question
* Define the field and propagate it
* Propagate the sam field with different parameters to show that the field scales with `λz`
* Propagate it with the non-paraxial BPM
* Does the scaling still apply?
"
# ╔═╡ 67f5c32e-b035-4dbe-a5a1-b06e1544c125
md" ## Answers
TODO
"
# ╔═╡ 2e05228d-084e-40e2-bbf6-64d04961f8ce
L3 = 1e-3
# ╔═╡ 647c72e5-17d8-445f-8375-960799fd54ed
N3 = 256
# ╔═╡ 6a12475c-20d2-4096-8ea5-27d7055a074a
y3 = range(-L3 / 2, L3 / 2, N3)
# ╔═╡ deafbee8-90ca-47b0-a572-19a1727a3780
x3 = y3';
# ╔═╡ 3fcb9bfb-f544-4aa7-a2da-09745becc1be
Λ = 20e-6
# ╔═╡ 050a9677-f91e-4795-bd08-036116127c31
A3 = 0.5 * (1 .+ cos.(2π./Λ .* x3)) .+ 0 .* y3
# ╔═╡ c5925699-2de3-46a1-aaf8-dbce77c80acf
simshow(A3)
# ╔═╡ 2a837cf8-c51c-4466-acee-e228ce44b62b
λ3 = 400e-9
# ╔═╡ acaf8c69-2c6b-4047-8c49-2824f267144b
z3 = range(0, 10e-3, 50)
# ╔═╡ 4fb25881-7e54-4ffa-99e7-6e8e679850fe
simshow(bpm(A3, λ3, L3, L3, z3)[:, :, end])
# ╔═╡ 71b38f92-22d6-45ed-80c8-7498ea5e16de
simshow(bpm(A3, λ3 .* 100, L3, L3, z3 ./ 100)[:, :, end])
# ╔═╡ 82156543-5c4e-4d1c-a52f-4e5ce55d9ef3
simshow(bpm(A3, λ3 .* 100, L3, L3, z3 ./ 100, paraxial=false)[:, :, end])
# ╔═╡ 855d09b4-87a3-4050-baf3-bcd189515e0a
# ╔═╡ 30b109fc-0a40-4724-ab4a-81a530d36d35
# ╔═╡ b3117642-1547-41dd-b241-e81e04a86450
md"# 3. Convolution Theorem
The convolution theorem states that
$$\mathcal{F} [(h * g)(x)] = \mathcal{F}[h] \cdot \mathcal{F}[g]$$
where $*$ is the convolution operator and $\cdot$ is the elementwise multiplication.
$$\mathcal{F}^{-1}\mathcal{F} [(h * g)(x)] = \mathcal{F}^{-1}\mathcal{F}[h] \cdot \mathcal{F}[g]$$
$$[(h * g)(x)] = \mathcal{F}^{-1}\left[\mathcal{F}[h] \cdot \mathcal{F}[g] \right]$$
## Question
* Implement the `my_conv` function naively with `fft` and `ifft`.
* What is the problem? How can you fix it?
* Implement `my_conv_fixed` which does it correctly.
"
# ╔═╡ 5039a62f-ddd3-46ec-bbb8-a98890794ea4
md" ## Answers
By taking a look at the Fourier transformation integral. We remember that
$$F\{g(x)\}=\int_{-\infty}^{\infty}g(x)e^{-i\omega x}dx=\left\{\begin{array}{rcl}
2\int_{0}^{\infty}g(x)cos(\omega x),& \;if\;g(x)\;is\;even\\
-2i\int_{0}^{\infty}g(x)sin(\omega x),& \;if\;g(x)\;is\;odd
\end{array}\right.$$
which means the fourier transform of an even function is purly real and the fourier transfrom of an odd funtion is purely imaginary. Now, imagine a real signal, that starts from $$x=0$$. It means that you can write down your real signal as $$f(x)=u(x)g(x)$$, where $$g(x)=f(x)+f(-x)$$, is an even function. This will lead to
$$F(\omega)=G(\omega)*(\pi\delta(\omega)+\frac{1}{i\omega})$$
which means the Fourier transform of your signal, have an imaginary part! An experimentalist, never likes this because e.g., there are usually no physical interpretation for the imaginary part of an spectrum. fft shift, is actually creating the symmetric function $$g(x)$$ to have purely real signal int the spectrum.
"
# ╔═╡ a1430eb0-17dc-4751-ba3e-75e33d19ee8a
md"""## Difference of FFT and FT
`fft` interpretes the signal to be centered around the first index (`(1,1,...)`) and outputs a Fourier signal where the DC (zero frequency) is also centered at the first index (`(1,1,...)`).
`ft(x) = fftshift(fft(ifftshift(x)))` interpretes the signal to be centered around the center index (`(N ÷ 2 + 1, N ÷ 2 + 1,...)`) and outputs a Fourier signal where the DC (zero frequency) is also centered at the first index (`(N ÷ 2 + 1, N ÷ 2 + 1,...)`).
"""
# ╔═╡ 7c5ead00-6fec-4e94-924a-37bb6f96560f
img = Float32.(testimage("resolution_test_512"));
# ╔═╡ 97d54710-65ca-4b6c-9237-b645156be82c
@bind sigma Slider(0.1:0.1:3, show_value=true)
# ╔═╡ af791ce8-859d-4fd0-887e-004870ab5b04
psf = gaussian(size(img), sigma=sigma);
# ╔═╡ be8e8728-54ae-4c87-938c-d54d02504e16
[simshow(img) simshow(psf)]
# ╔═╡ 3da1f0c4-d2e5-4f7c-a2be-78a4c3b0a4dd
function my_conv(img, psf)
return ifft(fft(img) .* fft(psf))
end
# ╔═╡ f52e1eab-f038-4417-9e59-f3be9b301350
simshow(my_conv(img, psf))
# ╔═╡ 3176c625-adb0-4fb0-9516-ec72ce58ba9b
sum(imag, my_conv(img, psf))
# ╔═╡ a9d31f41-d3a7-494e-96ab-0973016907d4
eltype(img)
# ╔═╡ e7105ca2-46c8-4274-a971-1d1d24fe29d1
eltype(psf)
# ╔═╡ 5b78a39a-d2cc-489d-9ec7-63d89757aec7
# ╔═╡ 4c74598b-de48-4100-9de1-3da60c0f2b59
simshow(psf, γ=0.1)
# ╔═╡ 9b9616e0-e0c9-4980-96b4-108c34b79a9b
function my_conv_fixed(img, psf)
return real.(ifft(fft(img) .* fft(ifftshift(psf))))
end
# ╔═╡ 82ec2319-15bc-4e92-9b55-5e13a68e4980
function my_conv_fixed_ft(img, psf)
return real.(ift(ft(img) .* ft(psf)))
end
# ╔═╡ eef84a2a-6395-48e1-ab9e-2ce31032f47f
function my_conv_fixed_rft(img, psf)
return (irft(rft(img) .* rft(psf), size(img, 1)))
end
# ╔═╡ f3ede0a4-4219-455d-9c38-d10181584650
simshow(ft(img), γ=0.1)
# ╔═╡ dd312559-a1fa-4bbe-a32b-1d9e1a9744bf
simshow(ft(img), γ=0.1)
# ╔═╡ 24da8e1c-205d-4ffe-b8c1-3904565a1fa3
[simshow(my_conv_fixed(img, psf)) simshow(my_conv_fixed_ft(img, psf)) simshow(my_conv_fixed_rft(img, psf))]
# ╔═╡ 55243ccc-f66a-446d-a4bd-0b2df4a86610
simshow(fft(img), γ=0.3)
# ╔═╡ 1745f5b8-4314-43f2-8a35-a7df6382391f
simshow(fftshift(fft(ifftshift(img))), γ=0.1)
# ╔═╡ 1b5c00c7-3de5-4db1-8f6e-5ceecc397d19
simshow(ft(img),γ=0.1)
# ╔═╡ dce6a8dd-3855-462c-80b2-71fdc814bdc4
fft(img)[1,1]
# ╔═╡ 99594a61-997f-4a11-bb0b-d8c7fb879bc3
sum(img)
# ╔═╡ cfe9da9d-4eec-447e-ac62-c6dc4fc196fc
fft(ifftshift([0, 0, 1, 0, 0]))
# ╔═╡ 4fd1eef3-f646-4622-87ed-d7f0311b4f03
fft(ifftshift([0, 0, 0, 1, 0, 0]))
# ╔═╡ 02892ba9-ae09-4ad2-888a-c10f9c8bfa0f
fftshift(ifftshift([0,0,1,0,0]))
# ╔═╡ 0bd892b8-ae24-4d21-9198-800c96c62f1c
fftshift(ifftshift([0,0,0,1, 0,0]))
# ╔═╡ 39c0efad-b7ea-4363-ad5e-ee10c319b8a4
ifftshift([0,0,1, 0,0])
# ╔═╡ 1611a48a-0c00-457a-a50e-dfad91ff0722
fftshift([0,0,1, 0,0])
# ╔═╡ 5e8c485c-71ea-4732-a577-350ad15a3802
begin
arr = zeros((10,10))
arr[1,3] = 1
end
# ╔═╡ d6ecc41b-a91b-4937-a486-4dd0c4a0ec04
FourierTools.ft(arr)
# ╔═╡ a471ba83-6dbb-403a-b0bd-8691f2edb4ef
fftshift(ft(ifftshift(arr)))
# ╔═╡ d7a46f07-086f-49f4-9a35-f90c97999f05
simshow(fft(arr))
# ╔═╡ 3e7c4453-e4b2-4ec0-8131-82cd29af8742
fft(arr);
# ╔═╡ 19d008fe-f9a5-4ab4-9147-2d9988e459e4
abs.(fft(arr));
# ╔═╡ 23ae6225-2e02-4b36-9894-de49f1cf0a0c
angle.(fft(arr))
# ╔═╡ 6548955f-a27a-4a91-b4a9-ce544a831b7b
abs.(fft([0, 0, 1, 0, 0]))
# ╔═╡ 00000000-0000-0000-0000-000000000001
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