Matlab找到自然频率，interp1函数创建NaN值

Question

我创建了以下代码，以查找通过使用冲击锤激发并连接有加速度计的测试样本的固有频率。但是，我被困在interp_accelerance_dB_first。这种插值创建了一组 NaN 值，我不知道为什么。我觉得这很奇怪，因为interp_accelerance 工作正常。我希望有人可以帮助我！

N = 125000;
fs = 1/(x(2)-x(1));
ts = 1/fs;
f = -fs/2:fs/(N-1):fs/2;

% Set x-axis of graph
x_max = (N-1)*ts;
x_axis=0:ts:x_max;

% find the first natural frequency between these boundaries 
First_lower_boundary = 15; 
First_upper_boundary = 30; 


Input = abs(fft(y)); %FFT input force
Output = abs(fft(o)); %FFT output acceleration

Accelerance = Output./Input;

bin_vals = [0 : N-1]; 
fax_Hz = bin_vals*fs/N; 
N_2 = ceil(N/2);

% Interpolate accelerance function in order to be able to average all accelerance functions 
Interp_accelerance = interp1(fax_Hz(1:N_2),Accelerance(1:N_2),x_axis); 


% --- Find damping ratio of first natural frequency 

% Determine the x-axis (from the boundries at the beginning of this script) 
x_axis_first_peak = First_lower_boundary:ts:First_upper_boundary;

% Accelerance function with a logarithmic scale [dB] 
Accelerance_dB_first =  20*log10(Accelerance(First_lower_boundary:First_upper_boundary));


% Interpolate the accelerance function [dB] 
Interp_accelerance_dB_first = interp1(fax_Hz(First_lower_boundary:First_upper_boundary),Accelerance_dB_first,x_axis_first_peak);

Answer 1

在不知道x,y,o 是什么的情况下很难确定，但通常interp1 在您尝试在数据x 轴边界之外进行插值时返回NaN。在代码末尾附加以下内容：

[min(fax_Hz(First_lower_boundary:First_upper_boundary)),max(fax_Hz(First_lower_boundary:First_upper_boundary))]
[min(x_axis_first_peak),max(x_axis_first_peak)]

如果第二个片段不在第一个片段之内，那么您就发现了问题。

顺便说一下，我认为interp_accelerance 可能会出现同样的错误，这又取决于输入参数的确切性质。