Estimation of power spectral density from laser Doppler data via linear interpolation and deconvolution

Abstract

Spectral estimation of irregularly sampled velocity data issued from Laser Doppler Anemometry measurements is considered in this paper. A new method is proposed based on linear interpolation followed by a deconvolution procedure. In this method, the analytic expression of the autocorrelation function of the interpolated data is expressed as a linear function of the autocorrelation function of the data to be estimated. For the analysis of both simulated and experimental data, the results of the proposed method is compared with the one of the reference methods in LDA: refinement of autocorrelation function of sample-and-hold interpolated signal method given by Nobach et al. (Exp Fluids 24:499–509, 1998), refinement of power spectral density of sample-and-hold interpolated signal method given by Simon and Fitzpatrick (Exp Fluids 37:272–280, 2004) and fuzzy slotting technique with local normalization and weighting algorithm given by Nobach (Exp Fluids 32:337–345, 2002). Based on these results, it is concluded that the performances of the proposed method are better than the one of the other methods, especially for what concerns bias and variance.

Appendix: A calculation of LIN interpolated signal autocorrelation function

In this appendix, an expression of R _y (k) as a function of R _r (k) for LIN interpolation scheme is setup. The calculation of the interpolated signal autocorrelation function R _y (k) can be carried out by identifying all the cases corresponding to different random sampling times with respect to l and l + k, as shown by Fig. 7:

• A. The calculation of R _y (0) given below can be carried out by identifying two exclusive cases denoted C ₁(0) and C ₂(0) illustrated in the first two subplot of Fig. 7.

  • 1. Case C ₁(0): one sampling time occurs exactly at l.

  • 2. Case C ₂(0): there is not any sampling time at l.

• B. The calculation of R _y (k), for k ≠ 0, can be carried out by identifying eight mutually exclusive cases denoted C _c (k), with c ∈ [1, 8].

  • 1. Case C ₁(k): Two sampling times occur exactly at l and l + k.

  • 2. Case C ₂(k): One sampling time occurs exactly at l and at least one between l and l + k.

  • 3. Case C ₃(k): One sampling time occurs exactly at l and none between l and l + k.

  • 4. Case C ₄(k): One sampling time occurs exactly at l + k and at least one between l and l + k.

  • 5. Case C ₅(k): One sampling time occurs exactly at l + k and none between l and l + k.

  • 6. Case C ₆(k): Only one sampling time occurs between l and l + k.

  • 7. Case C ₇(k): More than one sampling time occurs between l and l + k.

  • 8. Case C ₈(k): No sampling time occurs between l and l + k.

Fig. 7

Example of reconstructed signals with the LIN interpolated signal. Open circle measured missing data sequence, filled circle associated LIN interpolated sequence, Continues line associated LIN interpolated signal

1.1 A.1 calculation of R _y (0)

The two cases C ₁(0) and C ₂(0) that have been identified for the calculation of R _y (0) presented above yield two conditional ensemble average terms for R _y (0) named R _y|1(0) and R _y|2(0) respectively.

• 1. Case C ₁(0): one sampling time occurs exactly at l. This case happens with the probability

  $$ \hbox{Prob}[C_1(0)] = \alpha. $$

  (21)

  The LIN interpolation of x(l) gives y(l) = r(l). Therefore ∀l

  $$ \begin{aligned} R_{y|1}(0) &= \hbox{E}\left[y^2(l)|C_1(0)\right] = \alpha \hbox{E}\left[r^2(l)\right] \\ &= \alpha R_r(0). \end{aligned} $$

  (22)

• 2. Case C ₂(0): no sampling time occurs exactly at l. In this case, the nearest sampling times occur before and after l at l − m ₁ and l + n ₁ respectively, where m ₁ ∈ [1, l] and n ₁ ∈ [1, N − l]. This case happens with the probability

  $$ \hbox{Prob}[C_2(0)] = \sum_{m_1=1}^{l}\sum_{n_1=1}^{N-l}\alpha^2(1-\alpha)^{m_1+n_1-1}. $$

  (23)

  The LIN interpolation of x(l) gives

  $$ y(l) = \frac{n_1}{m_1+n_1} r(l-m_1) + \frac{m_1}{m_1+n_1} r(l+n_1). $$

  Therefore ∀l

  $$ \begin{aligned} R_{y|2}(0) &= \hbox{E}\left[y^2(l)|C_2(0)\right] \\ &= \sum_{m_1=1}^{l}\sum_{n_1=1}^{N-l}\alpha^2(1-\alpha)^{m_1+n_1-1} \hbox{E}\left[y^2(l)\right]\\ &= \sum_{m_1=1}^{l} \sum_{n_1=1}^{N-l} \left\{ \frac{\alpha^2(1-\alpha)^{m_1+n_1-1}}{(m_1+n_1)^2} \left[\left(m_1^2 +n_1^2\right) R_r(0)+2m_1n_1R_r(m_1+n_1) \right] \vphantom {\frac{\alpha^2(1-\alpha)^{m_1+n_1-1}}{(m_1+n_1)^2}} \right\}. \end{aligned} $$

  (24)

The relations Eqs. 22 and 24 give, ∀l

$$ R_y(l,l) =R_{y|1}(0) + R_{y|2}(0) = R_r(0). $$

(25)

Elementary arithmetic allows to verify that

$$ \hbox{Prob}[C_1(0)] + \hbox{Prob}[C_2(0)] = 1. $$

(26)

1.2 A.2 calculation of R _y (k), for k ≠ 0

The eight cases that have been identified for the calculation of R _y (k) presented above yield eight conditional ensemble average terms for R _y (k), named R _y|c(k) with c ∈ [1, 8].

• 1. Case C ₁(k): Two sampling times occur exactly at l and l + k. This case happens with the probability

  $$ \hbox{Prob}[C_1(k)] = \alpha^2. $$

  (27)

  The LIN interpolation of x(l) gives y(l) = r(l) and y(l + k) = r(l + k). Therefore ∀l

  $$ \begin{aligned} R_{y|1}(k) & = \hbox{E}[y(l)y(l+k)|C_1(k)] \\ & = \alpha^2 \hbox{E}[r(l)r(l+k)] \\ & = \alpha^2 R_r(k). \end{aligned} $$

  (28)

• 2. Case C ₂(k): One sampling time occurs exactly at l and at least one between l and l + k. In this case, the nearest sampling times occur before and after l + k at l + k − m ₂ and l + k + n ₂ respectively, where m ₂ ∈ [1, k − 1] and n ₂ ∈ [1, N − l − k]. This case happens with the probability

  $$ \hbox{Prob}[C_2(k)] = \sum_{m_2=1}^{k-1} \sum_{n_2=1}^{N-l-k} \alpha^3(1-\alpha)^{m_2+n_2-1}. $$

  (29)

  The LIN interpolation of x(l) gives y(l) = r(l) and

  $$ y(l+k) = \frac{n_2}{m_2+n_2} r(l+k-m_2)+ \frac{m_2}{m_2+n_2}r(l+k+n_2). $$

  Therefore ∀l

  $$ \begin{aligned} R_{y|2}(k) & = \hbox{E}[y(l)y(l+k)|C_2(k)] \\ & = \sum_{m_2=1}^{k-1} \sum_{n_2=1}^{N-l-k} \alpha^3(1-\alpha)^{m_2+n_2-1} \hbox{E} [y(l)y(l+k)] \\ & = \sum_{m_2=1}^{k-1} \sum_{n_2=1}^{N-l-k} \left\{ \frac{\alpha^3(1-\alpha)^{m_2+n_2-1}}{m_2+n_2} [n_2 R_r(k-m_2)+m_2 R_r(k+n_2)] \vphantom{\frac{\alpha^3(1-\alpha)^{m_2+n_2-1}}{m_2+n_2}}\right\} \end{aligned} $$

  (30)

• 3. Case C ₃(k): One sampling time occurs exactly at l and none between l and l + k. In this case, the nearest sampling time occurs after l + k at l + k + n ₂, where n ₂ ∈ [1, N − l − k]. This case happens with the probability

  $$ \hbox{Prob}[C_3(k)] = \sum_{n_2=1}^{N-l-k}\alpha^2(1-\alpha)^{k+n_2-1}. $$

  (31)

  The LIN interpolation of x(l) gives y(l) = r(l) and

  $$ y(l+k) = \frac{n_2}{k+n_2}r(l) + \frac{k}{k+n_2}r(l+k+n_2). $$

  Therefore ∀l

  $$ \begin{aligned} R_{y|3}(k) & = \hbox{E}[y(l)y(l+k)|C_3(k)] \\ & = \sum_{n_2=1}^{N-l-k}\alpha^2(1-\alpha)^{k+n_2-1} \hbox{E}[y(l)y(l+k)] \\ & = \sum_{n_2=1}^{N-l-k} \left\{\frac{\alpha^2(1-\alpha)^{k+n_2-1}}{k+n_2} [k R_r(k+n_2)+ n_2 R_r(0)] \right\}. \end{aligned} $$

  (32)

• 4. Case C ₄(k): One sampling time occurs exactly at l + k and at least one between l and l + k. In this case, the nearest sampling times occur before and after l at l − m ₁ and l + n ₁ respectively, where m ₁ ∈ [1, l] and n ₁ ∈ [1, k − 1]. This case happens with the probability

  $$ \hbox{Prob}[C_4(k)] = \sum_{m_1=1}^{l}\sum_{n_1=1}^{k-1}\alpha^3(1-\alpha)^{m_1+n_1-1}. $$

  (33)

  The LIN interpolation of x(l) gives

  $$ y(l) = \frac{n_1}{m_1+n_1} r(l-m_1) + \frac{m_1}{m_1+n_1} r(l+n_1) $$

  and y(l + k) = r(l + k). Therefore ∀l

  $$ \begin{aligned} R_{y|4}(k) & = \hbox{E}[y(l)y(l+k)|C_4(k)] \\ & = \sum_{m_1=1}^{l}\sum_{n_1=1}^{k-1} \alpha^3(1-\alpha)^{m_1+n_1-1} \hbox{E}[y(l)y(l+k)] \\ & = \sum_{m_1=1}^{l}\sum_{n_1=1}^{k-1} \left\{ \frac{\alpha^3(1-\alpha)^{m_1+n_1-1}}{m_1+n_1} [n_1 R_r(k+m_1) + m_1 R_r(k-n_1)] \vphantom { \frac{\alpha^3(1-\alpha)^{m_1+n_1-1}}{m_1+n_1}}\right\}. \end{aligned} $$

  (34)

• 5. Case C ₅(k): One sampling time occurs exactly at l + k and none between l and l + k. In this case, the nearest sampling time occurs before l at l − m ₁, where m ₁ ∈ [1, l]. This case happens with the probability

  $$ \hbox{Prob}[C_5(k)] = \sum_{m_1=1}^{l}\alpha^2(1-\alpha)^{k+m_1-1}. $$

  (35)

  The LIN interpolation of x(l) gives

  $$ y(l) = \frac{m_1}{m_1+k}r(l+k)+ \frac{k}{m_1+k}r(l-m_1) $$

  and y(l + k) = r(l + k). Therefore ∀l

  $$ \begin{aligned} R_{y|5}(k) & = \hbox{E}[y(l)y(l+k)|C_5(k)] \\ & = \sum_{m_1=1}^{l} \alpha^2(1-\alpha)^{k+m_1-1} \hbox{E}[y(l)y(l+k)] \\ & = \sum_{m_1=1}^{l} \left\{ \frac{\alpha^2(1-\alpha)^{k+m_1-1}}{k+m_1} [k R_r(k+m_1) + m_1 R_r(0)]\right\}. \end{aligned} $$

  (36)

• 6. Case C ₆(k): Only one sampling time occurs between l and l + k. In this case, the nearest sampling time occurs before l at l − m ₁, where m ₁ ∈ [1, l], one sampling time occurs at l + n ₁, where n ₁ ∈ [1, k − 1] and the nearest sampling time occurs after l + k at l + k + n ₂, where n ₂ ∈ [1, N − l − k]. This case happens with the probability

  $$ \hbox{Prob}[C_6(k)] = \sum_{m_1=1}^{l}\sum_{n_1=1}^{k-1} \sum_{n_2=1}^{N-l-k} \alpha^3(1-\alpha)^{k+m_1+n_2-2}. $$

  (37)

  The LIN interpolation of x(l) gives

  $$ y(l) = \frac{n_1}{m_1+n_1} r(l-m_1) + \frac{m_1}{m_1+n_1} r(l+n_1) $$

  and

  $$ y(l+k) = \frac{n_2}{k+n_1+n_2} r(l+n_1) + \frac{k-n_1}{k+n_1+n_2} r(l+k+n_2). $$

  Therefore ∀l

  $$ \begin{aligned} R_{y|6}(k) & = \hbox{E}[y(l)y(l+k)|C_6(k)] \\ & = \sum_{m_1=1}^{l}\sum_{n_1=1}^{k-1}\sum_{n_2=1}^{N-l-k} \left\{ \alpha^3(1-\alpha)^{k+m_1+n_2-2} \hbox{E} [y(l)y(l+k)] \right\} \\ & = \sum_{m_1=1}^{l}\sum_{n_1=1}^{k-1}\sum_{n_2=1}^{N-l-k} \left\{ \frac{\alpha^3(1-\alpha)^{k+m_1+n_2-2}}{(m_1+n_1)(k+n_1+n_2)} [ m_1(k-n_1)R_r(k-n_1+n_2)\right. \\ & +n_1(k-n_1)R_r(k+m_1+n_2)\\ & \left. +n_1n_2R_r(m_1+n_1)+m_1n_2R_r(0) ] \vphantom { \frac{\alpha^3(1-\alpha)^{k+m_1+n_2-2}}{(m_1+n_1)(k+n_1+n_2)}} \right\}. \end{aligned} $$

  (38)

• 7. Case C ₇(k): More than one sampling time occurs between l and l + k. In this case, the nearest sampling times occur before and after l at l − m ₁ and l + n ₁ respectively, where m ₁ ∈ [1, l] and n ₁ ∈ [1, k − 2]. Furthermore, the nearest sampling times occur before and after l + k at l + k − m ₂ and l + k + n ₂ respectively, where m ₂ ∈ [1, k − n ₁ − 1] and n ₂ ∈ [1, N − l − k]. This case happens with the probability

  $$ \hbox{Prob}[C_7(k)] = \sum_{m_1=1}^{l}\sum_{n_1=1}^{k-2} \sum_{m_2=1}^{k-n_1-1} \sum_{n_2=1}^{N-l-k}\alpha^4\\ \times (1-\alpha)^{m_1+n_1+m_2+n_2-2}. $$

  (39)

  The LIN interpolation of x(l) gives

  $$ y(l) = \frac{n_1}{m_1+n_1} r(l-m_1) + \frac{m_1}{m_1+n_1} r(l+n_1) $$

  and

  $$ y(l+k) = \frac{n_2}{m_2+n_2} r(l+k-m_2) + \frac{m_2}{m_2+n_2} r(l+k+n_2). $$

  Therefore ∀l

  $$ \begin{aligned} R_{y|7}(k) & = \hbox{E}[y(l)y(l+k)|C_7(k)] \\ & = \sum_{m_1=1}^{l} \sum_{n_1=1}^{k-2} \sum_{m_2=1}^{k-n_1-1} \sum_{n_2=1}^{N-l-k} \left\{ \alpha^4 (1-\alpha)^{m_1+n_1+m_2+n_2-2} \hbox{E}[y(l)y(l+k)] \right\} \\ & = \sum_{m_1=1}^{l} \sum_{n_1=1}^{k-2} \sum_{m_2=1}^{k-n_1-1} \sum_{n_2=1}^{N-l-k} \left\{\frac{\alpha^4(1-\alpha)^{m_1+n_1+m_2+n_2-2}} {(m_1+n_1)(m_2+n_2)} [ m_1m_2R_r(k-n_1+n_2) \right. \\ & + m_1n_2R_r(k-n_1-m_2) + n_1m_2R_r(k-m_1+n_2) \\ & \left. + n_1n_2R_r(k+m_1-m_2) ] \vphantom {frac{\alpha^4(1-\alpha)^{m_1+n_1+m_2+n_2-2}}} \right\}. \end{aligned} $$

  (40)

• 8. Case C ₈(k): No sampling time occurs between l and l + k. In this case, the nearest sampling times occur before l at l − m ₁, where m ₁ ∈ [1, l], and the nearest sampling time occurs after l + k at l + k + n ₂, where n ₂ ∈ [1, N − l − k]. This case happens with the probability

  $$ \hbox{Prob}[C_8(k)] = \sum_{m_1=1}^{l}\sum_{n_2=1}^{N-l-k}\alpha^2(1-\alpha)^{k+m_1+n_2-1}. $$

  (41)

  The LIN interpolation of x(l) gives

  $$ y(l) = \frac{k+n_2}{k+m_1+n_2} r(l-m_1) + \frac{m_1}{k+m_1+n_2} r(l+k+n_2) $$

  and

  $$ y(l+k) = \frac{n_2}{k+m_1+n_2} r(l-m_1) + \frac{k+m_1} {k+m_1+n_2} r(l+k+n_2). $$

  Therefore ∀l

  $$ \begin{aligned} R_{y|8}(k) & = \hbox{E}[y(l)y(l+k)|C_8(k)] \\ & = \sum_{m_1=1}^{l} \sum_{n_2=1}^{N-l-k} \alpha^2 (1-\alpha)^{k+m_1+n_2-1} \hbox{E}[y(l)y(l+k)] \\ & = \sum_{m_1=1}^l \sum_{n_2=1}^{N-l-k} \left\{ \frac{\alpha^2(1-\alpha)^{k+m_1+n_2-1}}{(k+m_1+n_2)^2} [ (m_1n_2+(k+m_1)(k+n_2))R_r(k+m_1+n_2) \right.\\ &\left. + (m_1(k+m_1)+n_2(k+n_2))R_r(0)] \right\}. \end{aligned} $$

  (42)

Elementary arithmetic allows to verify that

$$ \sum_{c=1}^8 \hbox{Prob}[C_c(k)] = 1. $$

(43)

R _y (k) is given as a linear combination of the different autocorrelation samples R _r (k)

$$ R_y(k) = \sum_{c=1}^8 R_{y|c}(k) $$

(44)

where the terms R_y|c(k) are given by Eqs. 28, 30, 32, 34, 36, 38, 40 and 42. Note that for $\alpha=1, R_y(k)=R_r(k), \forall k \in {\sf Z\!\!Z}.$