当啁啾信号应用于其输入时，我试图通过计算系统的频率响应函数 (FRF) 来识别振动系统。在比较计算的 FRF 和传递函数的波特图后，我发现共振已经被很好地检测到，但反共振没有。

该系统具有以下形式：

$$\textrm{sys} = \frac{s^2+\omega_{n_{z_1}}^2}{s^2+\omega_{n_{p_1}}^2}, \quad \textrm{where } \omega_{n_z} = 10\textrm{ Hz} \textrm{ and } \omega_{n_p} = 100\textrm{ Hz}.\\$$

我有一个假设：

线性系统对正弦信号的稳态响应是另一个具有相同频率的正弦信号，其幅值和相位取决于系统。当我施加啁啾信号时，系统从未达到稳定状态。因此，虽然系统对 \sin(\omega_{n_{z_1}}\cdot 2\pi) 的输出，但由于刺激 (弱）具有瞬态响应的频谱部分。$\sin(\omega_{n_{z_1}}\cdot 2\pi)$$100\textrm{ Hz}$

在我看来，正确的计算应该只测量与特定时刻输入频率相对应的输出分量。

• 你知道如何纠正这个问题吗？
• 是否可以应用移动过滤器或窗口来纠正此问题？

这是我的代码：

% Identification of the Frequency Response of a Transfer Function with
% Resonance and Antiresonance peaks
%% Cleaning the house
clc;
clear;
close all;
%% Definition of the System 
wn_z1 = 10*2*pi;           % Anti-Resonance Natural frequency 10 Hz
wn_p1 = 80*2*pi;           % Resonance Natural frequency 80 Hz
chi_z = 0;                 % Damping factor of the zeros 
chi_p = 0;                % Damping factor of the poles
s     = tf('s');
sys_1 = (s^2+2*chi_z*wn_z1*s+ wn_z1^2)/(s^2+2*chi_p*wn_p1*s+wn_p1^2);             
zpk(sys_1)
%% Definition of the Chirp Input
h                    = 0.001;
time                 = 0:0.001:60;
freq_ini             = 0.1*2*pi; % Initial frequency considered (0.1 Hz)
freq_final           = 240*2*pi; % Final frequency considered (240 Hz)
chirp_input          = 1.5*chirp(time,freq_ini,time(end),freq_final,'logarithmic');

frequency_evolution  = logspace(log10(freq_ini),log10(freq_final),length(time));
figure(1);
subplot(2,1,1);plot(time,chirp_input); xlabel('time (s)'); ylabel('Chirp    amplitude')
subplot(2,1,2);plot(time,frequency_evolution/(2*pi)); xlabel('time (s)'); ylabel('Frequency (Hz)');
%% Execution of the simulation
[sys_output,~]       = lsim(sys_1,chirp_input,time);
%% Computation of the FRF
fft_input   = fft(chirp_input,2^12);                                           %% Considering zero padding
fft_output  = fft(sys_output ,2^12);                                           %% Considering zero padding
fft_input   = fft_input(1:length(fft_input)/2);
fft_output  = fft_output(1:length(fft_output)/2);
fft_output  = fft_output(:);
fft_input   = fft_input(:);
FRF              = abs(fft_output)./abs(fft_input);
frequency_vector = linspace(0,1/(2*h),length(FRF));

%% Comparing the Bode Plot (Matlab) with the FRF (own implementation)
opts = bodeoptions('cstprefs');
opts.FreqUnits = 'Hz';
close all;
bodemag(sys_1,opts,'r'); hold on;
semilogx(frequency_vector, 20*log10(FRF ),'b')
xlim([0.1,frequency_vector(end)])
legend('Matlab Bode','FRF Identification')