利用激光修饰有限样本模拟极端紫外激光脉冲的透射光谱(MATLAB版本)

Question

我目前正在参与瞬态吸收光谱和四波混频的研究。在实验设计中，极限紫外(XUV)激光脉冲和红外(IR)激光脉冲通过有限的气体样品发送，两者之间存在一定的时间延迟。我目前的任务是开发一个功能，给出系统的数据和输入脉冲，提供XUV脉冲电场在离开样品后的透射光谱。

过程

该方法的基本思想是取两个描述样本磁化率的矩阵，将它们组合成一个矩阵，对该矩阵进行对角化，对左、右特征向量矩阵进行归一化，对特征值矩阵应用指数函数，最后计算给定输入谱的现有谱。

代码

function E_out = FiniteSampleTransmissionXUV(w, X_0, X_nd, E_in, a, N, tau)
% Function takes as input:
% > w - A vector that describes the frequency spectrum range
% > X_0 - A vector that describes the susceptibility 
%   without the IR pulse
% > X_nd - A matrix that describes the susceptibility with the
%   IR pulse
% > E_in - A vector that describes the frequency spectrum of the 
%   incoming XUV pulse electric field
% > a - A constant that describes the intensity of the IR pulse
% > N - A constant that describes the optical depth of the
%   sample
% > tau - A constant that describes the time delay between the 
%   IR pulse and the XUV pulse
%
% Function provides output:
% > A vector that describes the frequency spectrum of the
%   XUV pulse electric field at the end of the sample


    % determines number of frequency steps from input frequency range
    n = size(w);    % constant
    
    % determines frequency step size
    delta_w = 1 / (n(2) - 1);   % constant
    
    % create matrix sqrt(w_i*w_j)
    sqrt_w = w.^(1/2).' * w.^(1/2);     % nxn matrix

    % combine X_0 and X_nd into total suscptibility matrix
    X_ij = (a * delta_w * sqrt_w .* X_nd) + ...
            diag(diag(sqrt_w).' .* (X_0));      % nxn matrix
    
    
    % diagonalize X_ij where sum(U_R_i^2) = 1
    [U_R, LAMBDA, U_L] = eig(X_ij);     % nxn matrices
    
    % attain the function that scales U_L'*U_R
    F = sum(U_L' * U_R, 1);     % row vector
    
    sqrt_F = F.^2;
    
    % scale U_R and U_L so that U_L'*U_R = 1
    U_R_bar = U_R ./ sqrt_F;    % nxn matrix
    U_L_bar = U_L ./ sqrt_F;    % nxn matrix

    
    % apply exponential transform to eigenvalues      % diagonal nxn matrix
    exp_LAMBDA = diag(exp(1i * (2*pi*N / (3*10^8)) * diag(LAMBDA)));
    
    % create phase shift due to the time delay
    tau_phase = exp(1i * w * tau);  % row vector
    
    % recombine transformed susceptibility matrix
    ULAMBDAU = U_R_bar * exp_LAMBDA * U_L_bar';     % nxn matrix
    
    % apply effects of finite propagation and pulse
    % time delay to input electric field spectrum
    E_out = ULAMBDAU * (tau_phase .* E_in).';   % vector
    
    
end

测试

下面是测试和演示该函数的脚本。

n = 100;    % number of frequency steps
w = linspace(0,1,n);    % linearly spaced 1D array

X_0 = (1 + 1i) * (sin(5*pi*w)).^2;

X_nd = (1 + 1i) * ((sin(1*pi*w')) * sin(1*pi*w)).^2;

E_in = GaussianSpectrum(w, 1, 0.5, 0.1);


a = 1;
N = 10^8;
tau = 0;
E_out = FiniteSampleTransmissionXUV(w, X_0, X_nd, E_in, a, N, tau);


figure(1)
plot(w, E_in,'b')
hold on
plot(w, abs(E_out),'r')
hold off


function E_w_x0 = GaussianSpectrum(w, E_0, w_0, sigma)
% creates a quasi-gaussian spectrum defined by the input
% frequency range w, amplitude E_0, center frequency w_0,
% and spectral width sigma

    E_w_x0 = E_0 * sin(pi*w) .* ...
             (exp(-((w - w_0) / (2^(1/2) * sigma)).^2));

end

在测试过程中，我们看到了XUV脉冲在进入蓝色样品之前的频谱和红色的现有光谱。人们看到XUV脉冲的频率已经被吸收，而新的频率已经产生。

请让我知道什么违反了最佳实践，以及如何重构该函数，使其更加健壮和干净。谢谢!

Answer 1

非常好的代码，很难找到任何评论。

3*10^8更好地写成3e8 (IEEE )，那么MATLAB就不需要做任何计算，尽管我怀疑这将是计算时间的一个重要部分。这取决于你觉得最易读的是什么。

在您的测试代码中，2^(1/2)更常见地编写为sqrt(2)。同样，什么是更易读的取决于你。