Solutions for examination in TSRT78 Digital Signal Processing,

Transcription

1 Solutions for examination in TSRT78 Digital Signal Processing (a) The forgetting factor balances the accuracy and the adaptation speed of the RLS algorithm. This as the forgetting factor controls the number of observations to take into account. So a large number of observations gives high static accuracy as it mitigates the noise but then fails to adapt to sudden parameter changes. (b) The circular convolution can e.g. be done in Matlab using the following command: %% Exercise 1b) >> x=[ 1 1 ; y=[ ; >> ifft(fft(x).*fft(y) >> real(ans) which returns the answer { }. (c) We have a system with a finite impulse response with five non-zero impulse coefficients this together with a straightforward application of the DTFT definition results in H(e iωt ) = T 2 k= 2 h[ke iωt k = T ( e 2iωT + e 2iωT ) + 2T (e iωt + e iωt ) + 3T = 3T + 2T cos(2ωt ) + 4T cos(ωt ). (d) To find the first overtone we use the DFT directly on the signal. The signal that corresponds to the first overtone is then extracted using a Butterworth filter of order 2 (note the normalized frequency which results from the fact that ω =.5 Hz.). %% Exercise 1d >> y1=kron((-1).^(1:2)ones(16)) ; figure(1); % plot the signal >> subplot(311); plot(y1); xlabel( time ); ylabel( signal ); >> axis([ ); % calculate the DFT >> Y1=fft(y1); =length(y1); T=1; f=((1:)-1)/(*t); >> subplot(312); plot(fabs(y1)); xlabel( freq. (Hz) ); ylabel( amplitude ); % filter the signal >> [BA=butter(2.25/.5*[.9 1.1); >> y1filtered=filtfilt(bay1); >> subplot(313); plot(y1filtered); xlabel( time ); ylabel( filtered signal ); Executing this code returns Figure 1 where the first overtone is obtained as the second peak in the middle plot as.25 Hertz. The bottom plot is the filter signal which is a sine which is as expected knowing that a square wave can be written as a Fourier series. 2. The non-parametric spectral analysis can for example by done by using spa or pwelch in Matlab. The command ar estimates an AR model of a given order and bode gives the spectra of the estimated model. The following code applies these commands and returns Figure 2: %% Exericse 2 >> =1; T=1; >> a1=[ ; >> a2=conv([ [ ); >> y2=filter(1a1zeros(1)[1;;;)+filter(1a2randn(1)); >> figure(2); >> subplot(411); plot(y2 k ); xlabel( time ); ylabel( signal ); % DFT analysis 1 Ver: 213/1/8

2 1 signal amplitude filtered signal time freq. (Hz) time Figur 1: The signal its DFT and the filtered signal from exercise 1d. >> Y2=fft(y2); f=2*pi*(:-1)/(*t); >> subplot(412); semilogx(fabs(y2) k ); xlabel( frequency (rad/s) ); ylabel( Amplitude ); >> axis([1e-1 1e1 1e-5 7*1e5); % non-parametric spectral analysis >> [Pw=pwelch(y2); >> subplot(413); loglog(wp k ); >> xlabel( freq. (rad/s) ); ylabel( Power (db) ); axis([1e-1 1e1 1e-5 1e6); % parameteric spectral analysis >> m=ar(y25); >> [magphasew=bode(m); >> subplot(414); loglog(wmag(:) k ); >> xlabel( freq. (rad/s) ); ylabel( Power (db) ); axis([1e-1 1e1 1e-5 1e7); (a) The DFT tells us that there are at least three (the third is small in the DFT) resonance peaks in the data. This is verified by the non-parametric spectral estimate shown as the third plot from the top in the figure where the third peak is more visible (together with a fourth which is corrupted by noise). (b) An AR model of order 5 is enough to find all three peaks. The fourth can be found with an AR model of order 15 but the result is difficult to interpret. (c) To determine the best model we apply the loss function criteria using validation data. The best choice for AR-modelling seems to be AR(4) this is were the loss function has its knee point. This gives a model with 99.15% model fit on validation data but with remaining correlation in the residuals. To remove these correlations a model of order p = 25 is needed. %% AR modelling >> y2e=y2(1:6666); % estimation data >> y2v=y2(6667:end); % validation data % Determine the optimal model order >> lossfunc=[; >> for p=1:2 m=arx(y2ep); 2 Ver: 213/1/8

3 5 signal time 6 x 1 5 Amplitude frequency (rad/s) Power (db) freq. (rad/s) Power (db) freq. (rad/s) Figur 2: The signal its DFT non-parametric spectral estimate and parametric spectral estimate. 3 Ver: 213/1/8

4 yp=predict(y2vm1); lossfunc(p)=mean((y2v-yp).^2); >> end >> figure(3) >> plot(lossfunc); xlabel( model order (p) ); ylabel( loss function ); % estimate a model of order p=4 >> p=4; mar=arx(y2ep); % quick model checks >> compare(y2vmar1); % model fit on validation data >> resid(mary2v); % some patterns left in the data In estimating the ARMA model we start off using varying p and q between and 1 we find that all models with p geq4 and q 1 have almost the same loss function. Using for example p = 1 and q = 2 gives a model fit of 99.18% on validation data with no significant correlation in the residuals. As ARMA(12) is smaller than AR(25) it is the natural choice for a small model with good quality satisfying the model assumptions. %% ARMA modelling % Use try different p and q to optimize the model fit >> lossfunc=zeros(11); >> for p=1:1 >> for q=1:1 m=armax(y2e[p q); yp=predict(y2vm1); lossfunc(pq)=mean((y2v-yp).^2); >> end >> suef(lossfunc) % use e.g. [1 2 >> m=armax(y2e[1 2); >> compare(y2vm1); % model fit on validation data >> resid(my2v) 3. (a) First the signal model must be rewritten on the form of a linear regression. This is done using simple trigonometric identities realising that A sin(ωt + φ) = A sin(ωt) cos(φ) + A cos(ωt) sin(φ) = [ sin(ωt) cos(ωt) [ A cos(φ) }{{} A sin(φ) }{{} ψ (t) so that the model is written as (note that ψ is used instead of the standard notation in the course using ϕ as this character is used for the phase of the sine function in this exercise) y(t) = ψ (t)θ + e(t) i.e. the standard linear regression form. The least square estimate is defined as 1 θ = arg min θ θ ( y(t) ψ (t)θ ) 2 taking the derivative of the right hand side and set it to zero yields = 2 ψ(t) ( y(t) ψ (t)θ ) = 2 ψ(t)y(t) + 2 θ ψ(t)ψ (t) θ = ψ(t)y(t) ψ(t)ψ (t) 4 Ver: 213/1/8

5 where we have θ = [ θ(1) θ (2) = [ A cos(φ). A sin(φ) We see that A can be estimated using a simple application of the Pythagorean trigonometric identity as follows  = ( θ (1) )2 + ( θ (2) )2 and the phase can be estimated using arctan ϕ = arctan arctan (2) [ θ θ (1) (2) [ θ θ (1) (2) [ θ + π π θ (1) which is implemented in the atan2 command in Matlab. θ(1) > θ(2) θ(2) (b) The noise variance can be estimated using the following relation σ 2 e = 1 ɛ 2 (t) = 1 θ (1) < < θ (1) < ( y(t) ϕ (t) θ ) 2 which follows directly from the standard expression for the estimation of the sample variance and the model. 4. (a) According to the assignment the one-step FIR predictor is ŝ(t + 1) = H(q)y(t) = h()y(t) + h(1)y(t 1) The Wiener predictor is found by minimizing the cost function V (h) = 1 2 E(s(t + 1) ŝ(t + 1))2 = 1 2 E(s(t + 1) h()y(t) h(1)y(t 1))2. Since V (h) is a quadratic function we can easily find the unique minimizer simply by computing the dv/dh and solve dv/dh =. = E (y(t) (s(t + 1) h()y(t) h(1)y(t 1))) = h() = E (y(t 1) (s(t + 1) h()y(t) h(1)y(t 1))) =. h(1) The resulting Wiener-Hopf equations are ( ) ( ) Ryy () R yy (1) h() = R yy ( 1) R yy () h(1) ( ) E (s(t + 1)y(t)) E (s(t + 1)y(t 1)) ote that E (s(t + 1)y(t k)) = E (s(t + 1)(s(t k)) + e(t k)) = E(s(t + 1)s(t k)) = R ss (k + 1) where the last equality follows from the fact that s(t + 1) and e(t k) are independent. In order to be able to solve for h = (h() h(1)) we need R yy ( 1) R yy () R yy (1) R ss (1) and R ss (2). It is straightforward to show that R yy ( 1) = R yy (1) = R yy () = R ss(1) = R ss (2) = 4 15 which inserted into (1) results in The MSE for the predictions is defined as h() = h(1) = var(s(t + 1) ŝ(t + 1)) = E(s(t + 1) ŝ(t + 1)) 2 = E((s(t + 1) ŝ(t + 1))s(t + 1)) where the last equality follows from the fact that (s(t + 1) ŝ(t + 1)) is independent of ŝ(t + 1). Hence var(s(t + 1) ŝ(t + 1)) = E(((s(t + 1) h(1)y(t 1))s(t + 1)) = R ss () h(1)r ss (2) = (1) 5 Ver: 213/1/8

6 (b) Based on the model for s(t) it is natural to try a FIR predictor with H(q) = h()q 1 + h(1)q 3 in other words ŝ(t + 1) = h()y(t 1) + h(1)y(t 3). Analogously to the previous assignment using the cost function V (h) = E(s(t + 1) h()y(t 1) h(1)y(t 3)) 2 results in the following Wiener-Hopf equations to be solved = E(y(t 1)(s(t + 1) h()y(t 1) h(1)y(t 3))) = h() = E(y(t 3)(s(t + 1) h()y(t 1) h(1)y(t 3))) =. h(1) The corresponding Wiener-Hopf equations are given by ( ) ( ) Ryy () R yy (2) h() = R yy ( 2) R yy () h(1) resulting in h().127 h(1).159 ( ) Rss (2) (2) R ss (4) The MSE of the prediction error is given by (analogously to the previous assignment) var(ŝ(t + 1) s(t + 1)) = R ss () h()r ss (2) h(1)r ss (4) This error is smaller than the error in the previous assignment. 6 Ver: 213/1/8