Technische Universität Kaiserslautern WS 2016/17 Fachbereich Mathematik Prof. Dr. J. Franke 19. January 2017

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1 Technische Universität Kaiserslautern WS 2016/17 Fachbereich Mathematik Prof. Dr. J. Franke 19. January 2017 Exercises for Computational Modelling with Statistics IV This exercise is about the spectral analysis of signals as a tool to discover hidden periodicities or quasiperiodicities in data. Exercise 1 (Spectral analysis of periodic signals) The original spectral analysis developped by Schuster (1898) was meant to discover hidden periodicities in signals which are observed with measurement noise. We develop this approach step by step. In practice one usually has larger sample sizes, but we focus on sample size N=128 such that the signals are clearly visible in the plots. Note that N = 2 7 ; we use it as the FFT-algorithm is most efficient for powers of 2. In practice, if one observes a continuous signal (like a EEG recording) and if one can choose the rate of discretization, then it is optimal for the spectral analysis, to choose it such that the resulting sample size is a power of 2. We ll look at various examples and store the relevant results in the following matrices whose size we now fix: FREQ=zeros(6,3); AMP=zeros(6,3); a) i) Generate a pure cosine signal with period l = 16, amplitude A = 1, angular frequency α = 2π l : N=128; alpha = 2*pi/16; t=1:1:n; X = cos(alpha*t); Have a look at the signal data plot(t,x, * ) and at the interpolated continuous signal ii) We use the fast Fourier transform (FFT) to calculate the discrete Fourier transform (DFT) of the 1 data. We use the scaling factor N as in the lecture. FX is complex-valued. To look at a plot, we calculate the periodogram which is the square of the absolute value of the DFT: Just have a look at it: plot(px) PX is a symmetric function. Therefore, we know it completely if we only look at the first half of it. Also, for interpretability, we plot it against angular frequency, more precisely against the equidistant Fourier frequencies ω j = 2πj N, j = 0,..., N 2. omega = 2*pi*(0:1:N/2)/N; ; To recover the frequency from the signal, we have to look where the periodogram assumes its maximum. We are also interested in the value of the maximum. The MATLAB command [M,I] = max(px) determines the value M of the maximum and the index I of the vector PX where the maximum is located. To relate I to angular frequency, frequency and period of the signal component we have to look at the corresponding value of the vector omega: angfreq = omega(i); freq = angfreq/(2*pi) If everything is ok, you should recover exactly the period l. This only works so nicely as the corresponding 1

2 angular frequency is a Fourier frequency, or, equivalently, N is an integer multiple of l, as N = 8l. iii) Please store M and freq: AMP(1,1) = M; FREQ(1,1) = freq; iv) To check the influence of a different amplitude A and a phase shift φ repeat step i) and ii) for the following signal: A=2; phi = pi/4; X = A*cos(alpha*t+phi); [M,I] = max(px) angfreq = omega(i); freq = angfreq/(2*pi) Please store M and freq: AMP(2,1)=M; FREQ(2,1)=freq; Note that the maximum value M of the periodogram is quadratic in A, and M and I are not influenced by the phase shift φ. The phase information is lost by squaring the DFT. You can play around with different values of A and φ (e.g. A=3; phi=0.123;). b) Now we look at a cosine function with an angular frequency α which is not a Fourier frequency, but lies between two of them alpha = 2*pi*7.5/128; X = cos(alpha*t); [M,I] = max(px) From the plot, the periodogram value at I-1 is almost as large, so we also store it: M2=PX(I-1); The true frequency lies somewhere between the two maxima. We just interpolate and set: angfreq = (omega(i)+omega(i-1))/2; freq = angfreq/(2*pi) Compare this with the true frequency f = α 2π alpha/(2*pi) Please store M, M2 and freq: AMP(3,1)=M2; AMP(3,2)=M; FREQ(3,1)=freq; c) We now look at a superposition of 3 cosine functions (the sine is just a cosine with phase shift): alpha1 = 2*pi/32; alpha2 = 2*pi/16; alpha3 = 2*pi/8; X = cos(alpha1*t + pi/4) + 2*cos(alpha2*t) + 0.5*sin(alpha3*t); plot(t,x, * ) [M1,I] = max(px) angfreq = omega(i); freq1 = angfreq/(2*pi) 1 To get the next largest maximum, we set the value of PX at I to 0 and repeat the procedure: P=PX; P(I)=0; [M2,I] = max(p) angfreq = omega(i); freq2 = angfreq/(2*pi) 2 And finally for the third local maximum: 2

3 P(I)=0; [M3,I] = max(p) angfreq = omega(i); freq3 = angfreq/(2*pi) 3 As all 3 angular frequencies were Fourier frequencies you should recover the periods exactly in the order of amplitude size, i.e. 16, 32 and 8. Please store M1, M2, M3 and freq1, freq2, freq3: AMP(4,1)=M1; AMP(4,2)=M2; AMP(4,3)=M3; FREQ(4,1)=freq1; FREQ(4,2)=freq2; FREQ(4,3)=freq3; d) We add some Gaussian noise to the previous function: X = X + normrnd(0,1,1,n); plot(t,x, * ) We now repeat the steps from c): [M1,I] = max(px) angfreq = omega(i); freq1 = angfreq/(2*pi) 1 P=PX; P(I)=0; [M2,I] = max(p) angfreq = omega(i); freq2 = angfreq/(2*pi) 2 P(I)=0; [M3,I] = max(p) angfreq = omega(i); freq3 = angfreq/(2*pi) 3 We still see the two dominating frequencies corresponding to periods 16 and 32. The periodogram at the frequency corresponding to period 8 is still larger then the rest, but in another simulation it was only marginally larger than the next largest periodogram value at a completely different frequency. So it is not guaranteed that we discover a periodic component with such a small amplitude from only 128 observations. The chance to discover a particular periodic component is related to its amplitude, the noise standard deviation (in our case 1) and the sample size N. Please store M1, M2, M3 and freq1, freq2, freq3: AMP(5,1)=M1; AMP(5,2)=M2; AMP(5,3)=M3; FREQ(5,1)=freq1; FREQ(5,2)=freq2; FREQ(5,3)=freq3; e) Finally, we look at the function with non-fourier frequency α from b) with added random noise: alpha = 2*pi*7.5/128; X = cos(alpha*t) + normrnd(0,1,1,n); [M1,I1] = max(px); From the plot, the periodogram value at a neighbouring frequency is almost as large: P=PX; P(I)=0; [M2,I2] = max(p); The true maximum lies somewhere between I1 and I2, so we interpolate again: angfreq = (omega(i1)+omega(i2))/2; freq = angfreq/(2*pi) Please store M1, M2 and freq: AMP(6,1)=M1; AMP(6,2)=M2; FREQ(6,1)=freq; 3

4 Exercise 2 (The effect of the mean on spectral analysis) The signals of Exercise 1 have approximately mean 0 if we average over time t = 1,..., N. This is due to the symmetric fluctuation of the cosine around 0 and due to the law of large numbers for the noise component which we have chosen with expectation 0. In practice, signals frequently have a nonvanishing mean (compare the sunspot numbers example in the lecture). a) As in Exercise 1 a), generate a pure cosine signal with period l = 16, amplitude A = 1, angular frequency α = 2π l, but with a nonvanishing mean 1: N=128; alpha = 2*pi/16; t=1:1:n; X = 1 + cos(alpha*t); Then, calculate the periodogram: omega = 2*pi*(0:1:N/2)/N; Now the same with the mean-corrected data: X=X-mean(X); b) We now generate Gaussian white noise, i.e. a sequence of i.i.d. normal random variables, first with mean 0, and calculate the periodogram: X=normrnd(0,1,1,N); The periodogram fluctuates around the true spectral density which, in case of white noise, is a constant function. Now we change the mean from 0 to 1: X=1+X; and the same with mean 10: X=normrnd(10,1,1,N); Even for moderately large mean, the peak of the periodogram is so extreme that due to the scaling of the figure we cannot see the structure of the periodogram at the other frequencies. Therefore, mean correction is necessary: X=X-mean(X); c) We now consider a so-called autoregressive process of order 1 or AR(1)-process with Gaussian innovations, i.e. a signal satisfying X t = ax t 1 + Z t, t = 2,..., N, where Z 1,..., Z N are i.i.d. N (0, σ 2 ) random variables. To get a stationary signal, we need a < 1. To get a positive correlation between neighbouring values, we need a > 0. We choose a = 0.95, σ = 0.1. We first generate a realization of the AR(1)-process with mean 0: X=zeros(1,N); Z=normrnd(0,0.1,1,N); for j=2:n; X(j) = 0.95*X(j-1) + Z(j); end; Then, we change the mean to 1: X=1+X; We calculate the periodogram of the raw signal: 4

5 As before, the large peak at 0 obscures the remaining structure of the periodogram such that we have to subtract the mean Y=X-mean(X); FY = fft(y)/sqrt(n); PY = abs(fy). 2; PY = PY(1:(N/2+1));. plot(omega,py, * ) We see that the spectrum is large for small frequencies and then decreases for larger frequencies. This is typical for AR(1)-processes with a > 0, where the true unknown spectral density has a maximum at 0 and then decreases monotonically. Note also, that due to the mean correction, the periodogram at frequency 0 becomes 0. Otherwise, the periodogram is unchanged. To see this, plot the periodogram of the original signal X and that of the mean-corrected signal Y together omitting the first values at frequency 0: plot(omega(2:end),px(2:end), *,omega(2:end),py(2:end), o ) Save this figure, but not - as in the previous exercises - as MATLAB.fig file, but in the JPEG format which you can insert easily into text files (e.g. MS-Word, LaTeX, etc.). For this, in the plot window, click on File, then Save As (not Save!). The default option is MATLAB Figure (*.fig), but you can click on it, and then you get the other options for the figure format. Choose JPEG image (*.jpg). Then save the figure as AR1pgram.jpg. Exercise 3 (Smoothing the periodogram) On the average, if there are no strictly periodic components in the signal like in Exercise 1, the periodogram is equal to the spectral density. However, it is wildly fluctuating, and it becomes worse with increasing sample size. Therefore, to get a clear picture of the spectral structure of the signal, we have to smooth the periodogram, i.e. to calculate weighted averages of neighbouring periodogram values. We first generate data from a Gaussian autoregressive process of order 2 or AR(2)-process: X t = a 1 X t 1 + a 2 X t 2 + Z t, t = 2,..., N, with Z 1,..., Z N as in Exercise 1c). We choose a 1 = 0.95, a 2 = 0.8, σ = 0.1 and now N = 256. N=256; t=1:1:n; X=zeros(1,N); Z=normrnd(0,0.1,1,N); X(1)=Z(1); X(2)=0.95*X(1)+Z(2); for j=3:n; X(j) = 0.95*X(j-1) - 0.8*X(j-2) + Z(j); end; a) Now, we look at the periodogram. We can skip the mean correction as the simulated data have mean 0, but for real data a mean correction usually is necessary (compare Exercise 2). omega = 2*pi*(0:1:N/2)/N; We see that this has a peak at an angular frequency slightly larger than 1, i.e. a frequency slightly larger than 1 2π 0.16 and a period slightly smaller than 2π, i.e. approximately 6. However, the periodogram is rather random. b) To get a better impression about the form of the true spectrum, we smooth the periodogram P : SP (ω k ) = b w j P (ω k+j ), (1) j= b where w b,..., w b 0 are weights which sum up to 1 and are symmetric: b w j = 1, w j = w j, j = 0,..., b. j= b 5

6 The larger the b, the smoother the smoothed periodogram will be (depending also, but not so strongly, on the particular form of the weights). The signal analysis toolbox of MATLAB has fixed routines for all of this, but as we do not have it in the students licence, we have to do it in the pedestrian way. From (1) we see, that for e.g. calculating SP at frequency 0, we also need values P (ω j ), j = b,..., 1. The same problem appears at the right-hand boundary. So, we need an extended version of the periodogram, making use of the symmetry and periodicity of this function. P = abs(fx). 2; P=[P((N/2+1):N) P]; plot(p, * ) i) We define the vector (w b,..., w b ) of weights, where b = 4: W = [ ]; W=W/sum(W); The last command guarantees that the weights sum to 1. The operation (1) is a so-called convolution of the vector P and the vector W. In MATLAB, we can easily do it using the function conv: SP=conv(P,W, same ); We are only interested in the central part (as it already contains all the relevant information by symmetry and periodicity): SP=SP((N/2+1):(N+1)); This is the smoothed periodogram, and we can look at it: plot(omega,sp) This looks quite reasonable as it looks smooth, but not too smooth (showing yet a bit of small variability, such that we do not run into the danger of obscuring essential features of the spectrum by smoothing). We can also plot the smoothed periodogram and the raw periodogram together: plot(omega,sp,omega,px, * ) Please, save this figure as Spgram.jpg. ii) To check the influence of the weights, repeat the previous calculations with constant weights, but the same b: W = ones(1,9); W=W/sum(W); SP2=conv(P,W, same ); SP2=SP2((N/2+1):(N+1)); We compare the two smoothed periodograms: plot(omega,sp,omega,sp2) Please, save this figure as Spgram2.jpg. SP2 is a bit rougher. Weight vector with a maximum in the center and then decreasing to the left and right are preferable. Usually, they are generated as values of specific functions, the so-called kernels. Popular and also good kernels are the Gaussian kernel (= density of the standard normal distribution) or the Bartlett-Priestley kernel (quadratic function with a maximum at 0). ii) Finally, let us consider the effect of the choice of b by choosing a weight vector with b = 2: W = [ ]; W=W/sum(W); SP3=conv(P,W, same ); SP3=SP3((N/2+1):(N+1)); plot(omega,sp,omega,sp3) Please, save this figure as Spgram3.jpg. Note that this estimate of the true spectrum is still rather rough. Send me a mail (franke@mathematik.uni-kl.de) with the header Exercise 4. In the mail, you should list the names of the students who are members of the group handing in the solutions. Then, attach to the mail the file ex4work.mat and the 2 figure files AR1pgram.jpg, Spgram.jpg, Spgram2.jpg, Spgram3.jpg. Deadline: Sunday, February 5,