ω (rad/s)

Transcription

1 1. (a) From the figure we see that the signal has energy content in frequencies up to about 15rad/s. According to the sampling theorem, we must therefore sample with at least twice that frequency: 3rad/s or 3/(2π) = 477.5Hz. If we sample with too low frequency, aliasing occurs: energy appears in the DTFT of the sampled signal where the FT of the continuous signal did not have any. Due to a typo in the exam exercise, the task got more complicated than initially intended. Figure 3 shows the DTFT when sampled with 25rad/s, which was supposed to be the actual exam exercise. Y(ω) ω (rad/s) Figure 1: Solution to (1a). Black: FT of the continuous signal. Green: FT after folding at multiples of the Nyquist frequency rad/s. Y T (e iωt ) ω (rad/s) Figure 2: Solution to (1a). The DTFT of the sampled signal is the sum of the folded contributions. Obviously, it is severely distorted. Y(ω) ω (rad/s) Figure 3: Complement to (1a). Black: FT of the continuous signal. Green: DTFT of the signal when sampled with 25rad/s, as initially intended for the exercise. (b) The correct pairings are 1 A, 2 C, 3 B. The saw tooth s 1 (t) includes a constant offset which can be seen in the DTFT at Hz. Furthermore, s 1 (t) is a periodic signal and has peaks at multiples of its fundamental frequency 1Hz. 1

2 The sine s 2 (t) contains only a single frequency and gives one peak at 35Hz. The product s 3 (t) exhibits several peaks around 35Hz. It can be seen that the multiplication in time domain corresponds to a convolution in frequency domain: S 3 (e iωt ) = 1 2πT π/t π/t S 1 (e i(ω φ)t )S 2 (e iφt )dφ. (c) The state space model is time-invariant because the parameters a, b, q and r are constant for all t. Consequently, a Kalman filter for the system converges to the stationary Kalman filter and P (t + 1 t) = P (t t 1) = P as t. The resulting algebraic Riccati equation simplifies to lim P (t t 1) = P = A P A T A P C T ( C T ) 1 P C + R C P A T + BQB T t P = a 2 P a 2 P 2 P + r + b2 q P ( P + r) = a 2 P ( P + r) a 2 P 2 + b 2 q( P + r) = P 2 + P (r ra 2 b 2 q) b 2 qr. lim t P (t t 1) can be found as the positive root of the above polynomial. 2. (a) The following code snipped shows how to implement a high pass filter in matlab: fnyq = 441/2; % Nyquist frequency [bf, af] = butter(4, 1/fNyq, high ); % butterworth HP yhf = filtfilt(bf, af, y); % zero-phase filtering A plot of the filtered signal is shown in Figure signal magnitude time in milliseconds Figure 4: Solution to (2a). HF signal obtain after high pass filtering. The first 3msec. (b) The pump frequency can be found from the DFT of the signal. The code N = length(y); % number of samples Nfft = 2^16; % number of samples after zero-padding 2

3 Y = fft(y, Nfft); % DFT with zero-padding f = (:Nfft-1)/Nfft*441; % frequency vector in Hz figure(2);clf(); plot(f, abs(y)) % plot DFT magnitude xlim([ 6]); % zoom to interesting frequencies xlabel( Frequency in Hz ); ylabel( DFT magnitude ); produces Figure 5 which shows a distinct peak at the pump frequency of 1Hz. DFT magnitude Frequency in Hz Figure 5: Solution to (2b). The DFT magnitude of the recording shows a clear peak at the pump frequency. (c) First, we introduce ω P = 2πf P to simplify the notation. The low frequency signal can be written as linear equation a 1 y LF (t) = [ 1 cos(ω P t)... cos(nω P t) sin(ω P t)... sin(nω P t) ]. a n = ϕ T (t)θ. b 1. After combining the noise and HF signal to we obtain a linear regression problem e(t) = y HF (t) + n(t) y(t) = ϕ T (t)θ + e(t). With N measurements y(t) available, we can define y(1) ϕ T (1) Y =, Φ =. y(n) and write the problem in vector form as. ϕ T (N) Y = Φθ + E., E = e(1)... e(n) a b n 3

4 Under the (slightly violated) assumption that e(t) is a white noise sequence, the least squares solution for θ can be found by solving the normal equations The least squares solution is given by The code Φ T Y = Φ T Φθ. ˆθ = (Φ T Φ) 1 Φ T Y. fpmp = 5; % assumed pump frequency in Hz wpmp = 2*pi*fPmp; % pump frequency in rad/s n = 3; % order of the Fourier series t = (:N-1)/441; % time vector arg = wpmp*t *(1:n); % argument for the sin/cos functions phi = [ones(n,1), cos(arg), sin(arg)]; th = phi\y; % least squares solution in matlab yrecon = phi*th; % reconstruction of the LF signal gives the coefficients â.138 â 1 â ˆθ = â =.136 ˆb1.526 ˆb ˆb3.122 and Figure 6 shows the recording and LF reconstruction for the first 3msec. From (b) we know that the pump frequency is actually 1Hz. Accordingly, â 2 and ˆb 2 are bigger than the other coefficients. 5 recorded LF reconstruction signal magnitude time in milliseconds Figure 6: Solution to (2c). Recording and LF reconstruction for the first 3msec. 4

5 3. (a) One oscillation can be modeled with an AR(2) model with poles on the unit circle. From a complex pole r i of an AR model, the frequency in Hz can be found from the angle f i = arg(r i )/(2πT ). For an estimated AR(n) model with n > 2, poles close to the unit circle correspond to dominant oscillations. (b) The AR order n can be selected by estimating models of different order on estimation data, and then comparing their performance on validation data. The following code snippet evaluates the mean squared prediction error ( the loss function ) on validation data. The result is shown in Figure 7 and suggests n = 5. sest = s(1:5); % split sest = detrend(sest); % detrend sval = s(51:1); % split sval = detrend(sval); % detrend lf = zeros(1,1); % initialize loss function for i=1:length(lf) m = ar(sest, i); % estimate on sest lf(i) = mean(pe(m, sval).^2); % validate on sval end.4.3 loss function AR model order Figure 7: Solution to (3b). Loss function on validation data for different AR model orders. (c) RLS and LMS adaptive filters can be implemented using matlab s rarx function. The function returns a vector with all estimated coefficients for each time step. In order to compute the estimated frequencies, we loop over the time steps. Note that we append a one to get [1 thrls(i,:)] before calling roots to get the correct coefficient vector. Picking r(1) gives the frequency that we try to recover, but we could as well have checked for the largest entry in r to be safe. The following code generates the estimates of Figure 8. Picking λ =.98 and µ = 1 λ =.2 gives the required convergence speed. n = 5; % from (b) % RLS la =.98; thrls = rarx(s, n, ff, la); frls = zeros(size(t)); 5

6 for i=1:length(frls) r = roots([1 thrls(i,:)]); frls(i) = angle(r(1))/(2*pi*t); end % LMS mu =.2; thlms = rarx(s, n, ug, mu); flms = zeros(size(t)); for i=1:length(frls) r = roots([1 thlms(i,:)]); flms(i) = angle(r(1))/(2*pi*t); end AR model order = 5 frequency estimates ftrue frls, λ=.98 flms, µ= time in sec Figure 8: Solution to (3c). Estimated frequencies for RLS and LMS. 4. (a) A difference equation for s(t) is s(t) = 2r cos(ω 2 )s(t 1) r 2 s(t 2) + v(t) 2r cos(ω 1 )v(t 1) + r 2 v(t 2). The spectrum can be obtained from the super formula Similarly, we can compute Φ ss (ω) = H s (e iω ) 2 Φ vv (ω) = H s (e iω ) 2 1. Φ nn (ω) = H n (e iω ) 2 Φ ww (ω) = H n (e iω ) 2 1. (b) In matlab, the spectra can be obtained with the function freqz from the signal processing toolbox. The following code produces Figure 9. Obviously, the signal and noise spectra overlap significantly. % specify the filters that generate s and n r =.9; om1 =.9*pi; bs = [1-2*r*cos(om1) r^2]; % b polynomial signal om2 =.6*pi; as = [1-2*r*cos(om2) r^2]; % a polynomial signal om3 =.2*pi; 6

7 bn = [1-2*r*cos(om3) r^2]; % b polynomial noise om4 =.5*pi; an = [1-2*r*cos(om4) r^2]; % a polynomial noise % compute and plot the signal spectra [hs, ws] = freqz(bs, as); % ws in [ pi]: divide by pi to normalize phis = hs.*conj(hs); % signal spectrum hn = freqz(bn, an); phin = hn.*conj(hn); % noise spectrum phiy = phis + phin; % measurement spectrum figure(1);clf();hold on; plot(ws/pi, phis); plot(ws/pi, phin, g ); plot(ws/pi, phiy, m ); xlabel( normalized frequency ); ylabel( spectra ); legend( signal, noise, measurement ); 8 6 signal noise measurement spectra normalized frequency Figure 9: Solution to (4b). Signal, noise and measurement spectrum as a function of normalized frequency. The non-causal Wiener filter can be obtained from the ratio H nc (e iω ) = Φ sy(ω) Φ yy (ω) = Φ ss (ω) Φ ss (ω) + Φ nn (ω), where the second equality follows from the independence of v(t) and w(t). By construction H nc (e iω ) 1. The following code computes the non-causal Wiener filter and plots the magnitude response of Figure 1. hwf = phis./phiy; % non-causal Wiener filter figure(2);clf();hold on; plot(ws/pi, hwf); xlabel( normalized frequency ); ylabel( magnitude ); 7

8 1 magnitude normalized frequency Figure 1: Solution to (4b). The non-causal Wiener filter as a function of normalized frequency. (c) Both s(t) and n(t) are generated by ARMA(2,2) models. A state space model for s(t), here in observer canonical form, is given by with A s = [ ] 2r cos(ω2 ) 1 r 2, B s = Similarly, we can construct with A n = [ ] 2r cos(ω4 ) 1 r 2, B n = Combining both models gives x s (t + 1) = A s x s (t) + B s v(t), s(t) = C s x s (t) + D s v(t) [ 2r cos(ω1 ) + 2r cos(ω 2 ) x n (t + 1) = A n x n (t) + B n w(t), n(t) = C n x n (t) + D n w(t) [ 2r cos(ω3 ) + 2r cos(ω 4 ) x(t) T = [ x s (t) T x n (t) T], u(t) T = [ v(t) T w(t) T], x(t + 1) = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t) ], C s = [ 1 ], D s = 1. ], C n = [ 1 ], D s = 1. with [ ] [ ] As Bs A =, B =, C = [ ] [ ] C A n B s C n, D = Ds D n. n For the combined model, process and measurement noise are correlated. 8