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The model is an FIR (MA) filter

$$x(t) = h_1 u(t-1) + h_2 u(t-2) + u(t) \tag{1}$$

$$ y(t) = h^T x(t) + v(t) \tag{2}$$

$u(t)$ is a pseudo-random binary signal (PRBS) that excites/ drives the process ; and $v(t) = N(0,\sigma^2)$ is the measurement noise.

Q1 : VEctor representation: Is the representation $$y(t) = \mathbf{h^T x(t)} + w(t) \tag{3}$$ correct? Or will it be $$y(t) = \mathbf{h^T u(t)} + w(t) \tag{4}$$

Q2: In Book : Fundamentals of Statistical Signal Processing by Steven Kay, Chapter 4 : Eq(4.21) fives the variance of the estimated coefficients as $var(\hat{h(I)} = \sigma^2/Nr_u[0]$ . Will the CRLB for the MA model be the inverse of this variance when the input $u$ is a PRN sequence? In that case how do I find out $r_u$ in order to implement the formula?

Please help in clearing these problems. Thank you

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  • $\begingroup$ Ok, few counterquestions first: 1. Why is there a $h^T$ on (2)?. Should not be $g_1$. 2. Could you paste an image of the paragraph of the book?. $\endgroup$ – Brethlosze May 30 '15 at 1:41
  • $\begingroup$ With that, i would complete the answer below... $\endgroup$ – Brethlosze May 30 '15 at 1:57
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The CRLB Estimator proposed, is just the standard estimator on the MSE conventional sense. There is nothing too strange to worry about. Just follow the cautions here indicated regarding the estimation of MA processes.

Answer 1

The proper representation is: $$x(t)=b_0 u(t) + b_1 u(t−1)+b_2 u(t−2)+v(t)$$

that is: $$x(t)=\mathbf{u(t)b} + w(t)$$ or: $$\mathbf{x=H \theta + w}$$

This is an ARMAX model with no autoregressive (AR) order, $n_a$=0, two moving average (MA) orders, $n_b=2+1$, and white noise $w=N(0,\sigma^2)$. $H$ is the matrix of observations from $u(0)...u(n)$.

Answer 2

The expression on the document on p.37 are somehow incorrect, on the sense the matrix $H(t)^TH(t)$ depends on time. For making the estimator MVU (Minimum Variance & Unbiased), in the sense of the slides presented,

  1. Ensure the measurement over a large $t$ in order to reduce the impact of the initial conditions on the Parameter Estimation: $$t -> \inf$$ so $$H(t)^TH(t)$$ is not triangular (just as $H(0)^TH(0)$ is), but tends to diagonal.

  2. Ensure an i.i.d. random white-noise (RN) input: $$u(t)=N(0,\sigma_u)$$ Thus by definition of i.i.d., the autocorrelation of the input will be: $$r_{uu}(0)=\sigma_u^2$$ $$r_{uu}(\tau)|_{\tau<>0}=0$$ Qualitatively and theoretically there is little difference between a perfect and a pseudo white noise, as the i.i.d. is a stronger abstraction than the perfection of the whiteness for these identically and independency.

  3. Build the matrix $H(t)^TH(t)$ and construct the estimators: $$\theta=(H(t)^TH(t))^{-1}H(t)^Tx(t)$$ $$C_{\theta}=\sigma^2(H(t)^TH(t))^{-1}$$

Cheers...

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  • $\begingroup$ Check the answer. $\endgroup$ – Brethlosze May 30 '15 at 3:02
  • $\begingroup$ I am a bit confused by your current background and the question itself. The EM Kalman, the CRLB Estimate and the slides document are three difficult level kind of objects. WHich one do you desire to be explained?. According the slides, the CRLB estimate is just.. the standard one. The EM estimate is another story. $\endgroup$ – Brethlosze May 30 '15 at 6:16
  • $\begingroup$ The matrix H(1000) is then built for example as: $H=[u(998) u(999) u(1000); u(997) u(998) u(999);...;u(98) u(99) u(100)]$... $\endgroup$ – Brethlosze May 30 '15 at 6:23
  • $\begingroup$ The u(t) for identification purposes is a white noise (Pseudo Random Noise, PRN) $u=randn(1000,1)$ $\endgroup$ – Brethlosze May 30 '15 at 6:25
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    $\begingroup$ The standard estimate $$θ=(H(t)^TH(t))^{−1}H(t)^Tx(t)$$ is built from H(t) rows as ^$$H(t,:)=[u(t−2)u(t−1)u(t)]$$ , with as many $t$s as data is available, provided that $t$ is not close to the initial condition. $\endgroup$ – Brethlosze May 30 '15 at 6:31

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