fft of impulse response does not equal bode of transfer function

20 次查看（过去 30 天）
Lukas
Lukas 2018-9-11
回答： Lukas 2018-11-2
Hello
I am struggling with the problem, that the fourier transform of the impulse response of my system does not equal the bode plot of my transfer function. It would be great if somebody could help me to find the misstake.
clear all, clc, close all
%.........
% System
%.........
s=tf('s');
stoerung_G = 75 / (s + 25);
figure; bode(stoerung_G); grid;
stoerund_dt = 0.00001;
stoerung_t = [0 : stoerund_dt : 0.4]';
[stoerung_impulse] = impulse(stoerung_G, stoerung_t);
%.........
% Fourieranalyse
%.........
stoerung_y = stoerung_impulse;
stoerung_Fs = 1/stoerund_dt; % Hz
stoerung_L = size(stoerung_y,1);
stoerung_Y_ft = fft(stoerung_y);
P2 = abs(stoerung_Y_ft/stoerung_L);
P1 = P2(1:stoerung_L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
P1_log = 20 * log10(P1);
Phase2 = angle(stoerung_Y_ft);
Phase1 = Phase2(1:stoerung_L/2+1);
Phase1(2:end-1) = 2*Phase1(2:end-1);
Phase1 = Phase1*180/pi();
stoerung_f = stoerung_Fs*(0:(stoerung_L/2))/stoerung_L;
stoerung_w = 2*pi()*stoerung_f;
figure
subplot(211)
semilogx(stoerung_w,P1_log), grid, xlim([1 3000]);
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('w (rad/s)')
ylabel('|Mag| (dB)')
subplot(212)
semilogx(stoerung_w,Phase1), grid, xlim([1 3000]);
xlabel('w (rad/s)')
ylabel('Phase (degree)')

请先登录，再回答此问题。

采纳的回答

Lukas
Lukas 2018-11-2
Hello
Following link helped me to solve my issue: https://www.12000.org/my_notes/on_scaling_factor_for_ftt_in_matlab/
I have divided the fft by the sampling time. Then I have also deleted the factor 2 at the magnitude and at the phase calculation. It is still not reasonable to me why, because I am plotting just one half (as I understood: frequencies from 0 to fs/2 and not -fs/2 to fs/2) of the fft. In order to keep the "signal energy" constant I should multiply by 2. Anyways, it works now without the "2".
clear all, clc, close all
%.........
% System
%.........
s=tf('s');
stoerung_G = 75 / (s + 25);
figure; bode(stoerung_G); grid;
stoerung_dt = 0.00001;
stoerung_t = [0 : stoerung_dt : 0.4]';
[stoerung_impulse] = impulse(stoerung_G, stoerung_t);
%.........
% Fourieranalyse
%.........
stoerung_y = stoerung_impulse;
stoerung_Fs = 1/stoerung_dt; % Hz
stoerung_L = size(stoerung_y,1);
stoerung_Y_ft = fft(stoerung_y)*stoerung_dt;
P2 = abs(stoerung_Y_ft);
P1 = P2(1:stoerung_L/2+1);
P1_log = 20 * log10(P1);
Phase2 = angle(stoerung_Y_ft);
Phase1 = Phase2(1:stoerung_L/2+1);
Phase1 = Phase1*180/pi();
stoerung_f = stoerung_Fs*(0:(stoerung_L/2))/stoerung_L;
stoerung_w = 2*pi()*stoerung_f;
figure
subplot(211)
semilogx(stoerung_w,P1_log), grid, xlim([1 3000]);
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('w (rad/s)')
ylabel('|Mag| (dB)')
subplot(212)
semilogx(stoerung_w,Phase1), grid, xlim([1 3000]);
xlabel('w (rad/s)')
ylabel('Phase (degree)')

更多回答（1 个）

Dimitris Kalogiros
Dimitris Kalogiros 2018-9-11
You have to change the time interval that you used to define stoerung_t
Try a larger one:
stoerung_t = [0 : stoerund_dt : 40-stoerund_dt]';
Now, two bode diagrams will become identical
  2 个评论
Lukas
Lukas 2018-9-18
Unfortunately this is not solving my problem. It is also strange, that the fft leads to an offset in the magnitude plot of -26dB and that the phase shift is -180° instead of -90° for a PT1 system.
Dimitris Kalogiros
Dimitris Kalogiros 2018-9-18
These artifacts stems from the fact that you compare a continous time system, discribed by laplace transform (stoerung_G), and a discrete time system, described by z-transform (stoerung_Y_ft).
For more details, have a look here: bilinear transformation

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Spectral Measurements 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by