Introduction to Signal Analysis Parts I and II

Transcription

1 41614 Dynamics of Machinery 23/03/2005 IFS Introduction to Signal Analysis Parts I and II Contents 1 Topics of the Lecture 11/03/2005 (Part I) 2 2 Fourier Analysis Fourier Series, Integral and Complex Form 3 3 Fourier Analysis Example with a Square Wave Function 8 4 Fourier Analysis Example with a Triangular Wave Function 11 5 Discrete Form of Fourier Transform 15 6 Discrete Fourier Analysis Examples 18 7 Matlab Routines 13 Examples with 5 Different Signals 20 8 Analyzers and Signal Processing 24 9 Topics of the Lecture 18/03/2005 (Part II) Impulse Response Response to an Arbitrary Input Response to Random Input Estimators H1, H2 and Coherence Function Matlab Routine FRF-general.m Extracting the Modal Parameters 53 1

2 41614 Dynamics of Machinery 18/03/2005 IFS Introduction to Signal Analysis Part I 1 1 Topics of the Lecture 11/03/2005 (Part I) Fourier Analysis Fourier Series Fourier Integral Complex Form of Fourier Transform Examples: Square Wave and Triangular Wave Discrete Form of Fourier Transform Example: Sinus with 16 points Frequency Response Analyzers Digital Signal Processing Aliasing Anti-Aliasing Filtering Leakage Windowing Averaging Overlapping 1 References Bendat, J. S. (1993) Engineering Applications of Correlation and Spectral Analysis, John Wiley & Sons, Inc., New York. Carlson, G. E. (1998) Signal and Linear System Analysis with Matlab, John Wiley & Sons, Inc., New York. Randall, R. B. (1987) Frequency Analysis, Brüel & Kjaer, Naerum, Denmark. Ewins, D. J. (1984) Modal Testing: theory and practice, Letchworth: Research Studies Press Ltd., England. 2

3 2 Fourier Analysis Fourier Series, Integral and Complex Form 3

4 4

5 5

6 6

7 7

8 3 Fourier Analysis Example with a Square Wave Function 8

9 9

10 10

11 4 Fourier Analysis Example with a Triangular Wave Function 11

12 12

13 13

14 14

15 5 Discrete Form of Fourier Transform 15

16 16

17 17

18 6 Discrete Fourier Analysis Examples EXAMPLE 1: function x(t) = cos4πt It is important to see in table 1 that the spectral coefficients a k and b k provides the information about the frequency of the function, i.e 4π rad/s or 2 Hz. Moreover, all coefficients b k are zero. Analyzing equations (??) and (??) you can easily see that the spectral coefficients a k are related to the real part of X k or cos(k ω t) and b k to the imaginary part of X k or sin(k ω t). Obviously, all coefficients b k shall be zero. t k [s] x k f k [Hz] a k b k x(t) x r t r [s] 10 a k b k f k =ω k /2π [Hz] Table 1: Data of a digitalized cosine function x(t) = cos4πt composed of N = 16 points in the time domain, where T = 1 s, T = T/N = s, ω k = 2π/T = 2π [rad/s] = 1 [Hz], f k = k ω/2π = k 1 [Hz] (k = 0,1,2,...,(N 1)) and spectral coefficients X k = a k j b k. 18

19 EXAMPLE 2: function x(t) = sin4πt It is important to see in table 2 that the spectral coefficients a k and b k provides the information about the frequency of the function, i.e 4π rad/s or 2 Hz. Moreover, all coefficients a k are zero. Analyzing equations (??) and (??) you can easily see that the spectral coefficients a k are related to the real part of X k or cos(k ω t) and b k to the imaginary part of X k or sin(k ω t). Obviously, all coefficients a k shall be zero. t k [s] x k f k [Hz] a k b k x(t) x r t r [s] 10 a k b k f k =ω k /2π [Hz] Table 2: Data of a digitalized cosine function x(t) = cos4πt composed of N = 16 points in the time domain, where T = 1 s, T = T/N = s, ω k = 2π/T = 2π [rad/s] = 1 [Hz], f k = k ω/2π = k 1 [Hz] (k = 0,1,2,...,(N 1)) and spectral coefficients X k = a k j b k. 19

20 7 Matlab Routines 13 Examples with 5 Different Signals %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % MACHINERY DYNAMICS LECTURES (41614) % % IKS - DEPARTMENT OF CONTROL ENGINEERING DESIGN % % DTU - TECHNICAL UNIVERSITY OF DENMARK % % % % Copenhagen, February 20th, 2002 % % IFS % % % % LECTURE ABOUT SIGNAL ANALYSIS & SIGNAL PROCESSING % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all close all % * SIGNAL WITH DIFFERENT FREQUENCIES % (1) a perfect sinus with f= 10 Hz with 2048 points % (2) a perfect sinus with f= 100 Hz with 2048 points % % * DECREASING THE RATE FREQUENCY -> ALIASING % (3) a perfect sinus f=100.0 Hz with 1024 points % (4) a perfect sinus f=100.0 Hz with 512 points % (5) a perfect sinus f=100.0 Hz with 256 points % (6) a perfect sinus f=100.0 Hz with 128 points % (7) a perfect sinus f=100.0 Hz with 64 points % % * LEAKAGE % (8) an incomplete sinus leak= 1.0 f=1.0 Hz with 128 points % (9) an incomplete sinus leak= 1.5 f=1.0 Hz with 128 points % % * TRANSIENT SIGNAL & WINDOWING % (10) xi=0.1 with a sinus f=100 Hz with 512 points % (11) xi=0.5 with a sinus f=100 Hz with 512 points % (12) xi=0.9 with a sinus f=100 Hz with 512 points % (13) xi=0.1 with a sinus f=100 Hz with 128 points (aliasing) % % * SIGNAL WITH MULTIPLE FREQUENCY COMPONENTS - 2 COMPONENTS % % * SIGNAL WITH MULTIPLE FREQUENCY COMPONENTS - 4 COMPONENTS % % * SIGNAL CONTAMINATED BY NOISE % % * STEP SIGNAL % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % * SIGNAL WITH DIFFERENT FREQUENCIES % (Case 1) n = 2048; % number of points in the range of time freq = 10.0; % [Hz] - frequency of the simulated signal xi = 0.0; % - damping factor leak = 1.0; % - leakage factor % (Case 2) % n = 2048; % number of points in the range of time % freq =100.0; % [Hz] - frequency of the simulated signal 20

21 % xi = 0.0; % - damping factor % leak = 1.0; % - leakage factor % * DECREASING THE RATE FREQUENCY -> ALIASING % (Case 3) % n = 1024; % number of points in the range of time % freq =100.0; % [Hz] - frequency of the simulated signal % xi = 0.0; % - damping factor % leak = 1.0; % - leakage factor % (Case 4) % n = 512; % number of points in the range of time % freq =100.0; % [Hz] - frequency of the simulated signal % xi = 0.0; % - damping factor % leak = 1.0; % - leakage factor % (Case 5) % n = 256; % number of points in the range of time % freq =100.0; % [Hz] - frequency of the simulated signal % xi = 0.0; % - damping factor % leak = 1.0; % - leakage factor % (Case 6) % n = 128; % number of points in the range of time % freq =100.0; % [Hz] - frequency of the simulated signal % xi = 0.0; % - damping factor % leak = 1.0; % - leakage factor % (Case 7) % n = 64; % number of points in the range of time % freq =100.0; % [Hz] - frequency of the simulated signal % xi = 0.0; % - damping factor % leak = 1.0; % - leakage factor % * LEAKAGE % (Case 8) % n = 128; % number of points in the range of time % freq = 1.0; % [Hz] - frequency of the simulated signal % xi = 0.0; % - damping factor % leak = 1.0; % - leakage factor % (Case 9) % n = 128; % number of points in the range of time % freq = 1.0; % [Hz] - frequency of the simulated signal % xi = 0.0; % - damping factor % leak = 1.5; % - leakage factor % * TRANSIENT SIGNAL & WINDOWING % (Case 10) % n = 1024; % number of points in the range of time % freq =100.0; % [Hz] - frequency of the simulated signal % xi = 0.1; % - damping factor 21

22 % leak = 1.0; % - leakage factor % (Case 11) % n = 1024; % number of points in the range of time % freq =100.0; % [Hz] - frequency of the simulated signal % xi = 0.5; % - damping factor % leak = 1.0; % - leakage factor % (Case 12) % n = 1024; % number of points in the range of time % freq =100.0; % [Hz] - frequency of the simulated signal % xi = 0.9; % - damping factor % leak = 1.0; % - leakage factor % (Case 13) % n = 512 % number of points in the range of time % freq =100.0 % [Hz] - frequency of the simulated signal % xi = 0.1; % - damping factor % leak = 1.0; % - leakage factor % (Case 14) % n = 512; % number of points in the range of time % freq =10.0; % [Hz] - frequency of the simulated signal % xi = 0.0; % - damping factor % leak = 1.0; % - leakage factor tmax=1.0; % [s] - time in seconds xo=1.0e-0; % [m] - signal amplitude freqa = freq*sqrt(1-xi^2); % [Hz] - damped frequency fmax=n/(2*tmax); % [Hz] - max. frequency for i=1:n, t(i)=(i-1)/n*tmax; f(i)=(i-1)/tmax; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % FIVE DIFFERENT TYPES OF SIGNAL %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % (1) SINUS SIGNAL WITH 1 FREQUENCY COMPONENT x(i)=xo*exp(-xi*freqa*t(i))*sin(2*pi*freqa*t(i)*leak); % % (2) SINUS SIGNAL WITH 2 FREQUENCY COMPONENT % x(i)=xo*exp(-xi*freqa*t(i))*sin(2*pi*freqa*t(i)*leak) +... % xo*exp(-xi*freqa*t(i))*sin( 4*2*pi*freqa*t(i)*leak); % % (3) SINUS SIGNAL WITH 4 FREQUENCY COMPONENT % x(i)=xo*exp(-xi*freqa*t(i))*sin(0*2*pi*freqa*t(i)*leak) +... % xo/2*exp(-xi*freqa*t(i))*sin(2*2*pi*freqa*t(i)*leak) +... % xo/4*exp(-xi*freqa*t(i))*sin(4*2*pi*freqa*t(i)*leak) +... % xo/6*exp(-xi*freqa*t(i))*sin(6*2*pi*freqa*t(i)*leak); % % (4) SINUS SIGNAL + NOISE % const = 0.5; % ruido(i) = randn; 22

23 % x(i)=xo*exp(-xi*freqa*t(i))*sin(2*pi*freqa*t(i)*leak)+const*ruido(i); % % (5) STEP WAVE % x(i)=xo*exp(-xi*freqa*t(i))*sign(sin(2*pi*freqa*t(i)*leak)); end y=fft(x); figure(1) title( Signal Analysis in Time and Frequency Domain, FontSize,18) subplot(3,1,1), plot(t,x, r- ) title( (a) Time Domain - (b) and (c) Frequency Domain with N and N/2 points, FontSize,18) xlabel( time [s], FontSize,18) ylabel( (a), FontSize,18) grid subplot(3,1,2), plot(f(1:n),abs(y(1:n)), r- ) xlabel( frequency [Hz], FontSize,18) ylabel( (b), FontSize,18) grid subplot(3,1,3), plot(f(1:n/2),abs(y(1:n/2)), r- ) xlabel( frequency [Hz], FontSize,18) ylabel( (c), FontSize,18) grid 23

24 8 Analyzers and Signal Processing 24

25 25

26 26

27 27

28 28

29 29