Detection of UWB signal using dirty template approach

Abstract

This paper handles the problem of detecting ultra-wideband signals in the presence of dense multipath channel and ambient noise. To design a low-complexity high-performance signal detection process, the dirty template approach is proposed. We first explain the dirty template technique and its implementation in UWB communication systems. Then, the Neyman–Pearson theorem is applied to derive the UWB signal detector and select the suitable detection threshold values. Finally, the performance of the proposed dirty template detector is evaluated in terms of the detection and false alarm probabilities for different threshold values, signal-to-noise ratio and number of data-aided symbols.

Appendix: Dirty template noise \({\check{\omega }}[n]\)

In this section, we derive the dirty template noise ${\check{\omega }}[n]$ model and calculate its statistic properties. Looking at the Eq. (5), it could be represented as follows:

$$\begin{aligned} x({t+nT_s})=w({t+nT_s })+f({t+nT_s ,\tau }), \end{aligned}$$

(20)

where

$$\begin{aligned} f({t+nT_s ,\tau })&= \left. {\sqrt{\varepsilon _s}s[{n-n_s -1}]p_R ({t+T_s -\tau })} \right| _{t=0}^{t=\tau }\nonumber \\&\quad +\left. {\sqrt{\varepsilon _s}s[{n-n_s}]p_R ({t-\tau })} \right| _{t=\tau }^{t=T_s}, \end{aligned}$$

(21)

and $w(t)$ represents the band-pass filtered zero-mean AGWN with PSD $N_0/2$ and with double-sided bandwidth B $(\gg 1/T_s)$ [14]. The autocorrelation function of $w(t)$ is given as follows [14]:

$$\begin{aligned} R_w (t)&= N_0 B\sin c({B\tau })\hbox {cos}({2\pi f_0 \tau })\nonumber \\&= TF^{-1}\left\{ {\frac{N_0}{2}} \right\} _{\pm f_0 -B/2}^{\pm f_0 +B/2}, \end{aligned}$$

(22)

where $f_0$ represents the center frequency of receiver bandwidth filter. To simplify $f({t+nT_s ,\tau })$, we could rewrite (21) as:

$$\begin{aligned} f({t+nT_s ,\tau })=\sqrt{\varepsilon _s}\sum _{m=0}^1 s[{n-n_s -m}]p_R ({t+mT_s -\tau })\nonumber \\ \end{aligned}$$

(23)

Fig. 14 represents the noise-free parts of the received signal. To simplify, $p_R(t)$ is plotted as a triangle.

Fig. 14

The received waveform

Note: The function $p_R \left( {t-\tau } \right)$ is given here by (Fig. 14):

$$\begin{aligned} p_R ({t-\tau })=\left\{ {{\begin{array}{ll} 0&{}:t\in \{0,\tau ) \\ p_R ({t-\tau })&{}:t\in \{\tau ,T_s ) \\ \end{array} }} \right. \end{aligned}$$

And the function $p_R ({t+T_s -\tau })$ is given by (Fig. 14):

$$\begin{aligned} p_R ({t+T_s -\tau })=\left\{ {{\begin{array}{ll} p_R ({t+T_s -\tau } )&{}:t\in \{0,\tau ) \\ 0&{}:t\in \{\tau ,T_s ) \\ \end{array} }} \right. \end{aligned}$$

• 1. Dirty template noise model ${\check{\omega }}[n]$ From (6), the sampled noise ${\check{\omega }}[n]$ is composed of three terms as follows:

  $$\begin{aligned} {\check{\omega }}[n]=\xi _1 [n]+\xi _2 [n]+\xi _3 [n] \end{aligned}$$

  (24)

• 1.1 $\xi _1 [n]$

  $$\begin{aligned}&\xi _1 [n] =\sqrt{\varepsilon _s}\sum \limits _{m=0}^1 s[{n-n_s -m}]\nonumber \\&\quad \times \int \limits _0^{T_s} p_R ({t+mT_s-\tau })w({t+({n+1})T_s})\hbox {d}t \end{aligned}$$

  (25)

  Substituting (23) into (25), the term $\xi _1 [\hbox {n}]$ could be simplified by

  $$\begin{aligned} \xi _1 [n]=\int \limits _0^{T_s } f(t+nT_s ,\tau )w({t+({n+1})T_s })\hbox {d}t \end{aligned}$$

  (26)

• 1.2 $\xi _2 [n]$

  $$\begin{aligned}&\xi _2 [n]=\sqrt{\varepsilon _s }\sum \limits _{m=0}^1 s[{n-n_s -m+1}]\nonumber \\&\quad \times \int \limits _0^{T_s } p_R ({t+mT_s -\tau })w({t+nT_s})\hbox {d}t \end{aligned}$$

  (27)

  Substituting (23) into (27), the term $\xi _2 [n]$ could be simplified by

  $$\begin{aligned} \xi _2 [n]=\int \limits _0^{T_s} f(t+({n+1})T_s ,\tau )w({t+nT_s})\hbox {d}t \end{aligned}$$

  (28)

• 1.3 $\xi _3 [n]$

  $$\begin{aligned} \xi _3 [n]=\int \limits _0^{T_s } w({t+nT_s})w({t+({n+1})T_s})\hbox {d}t \end{aligned}$$

  (29)

  Clearly, the statistic properties of the DT noise ${\check{\omega }}[n]$ depends on information-bearing symbols $\{{s[n]}\}$. Therefore, to simplify the calculus of the mean and the autocorrelation of ${\check{\omega }}[n]$, we suppose that the symbols $\{{s[n]}\}$ are i.i.d. PAM (binary pulse amplitude modulation) symbols with zero-mean and unit energy $E\{{s[k]s[l]}\}=\delta _{k,l}$ (see Table 2).

  Table 2 The autocorrelation function of $s[k]$

• 2. The mean of dirty template noise

  $$\begin{aligned} E\left\{ {\check{\omega }}[n]\right\} =E\left\{ {\xi _1 [n]}\right\} +E\left\{ {\xi _2 [n]}\right\} +E\left\{ {\xi _3[n]}\right\} \end{aligned}$$

• 2.1 $E\{{\xi _1 [n]}\}$ & $E\{{\xi _2 [n]}\}$

  $$\begin{aligned}&E\{{\xi _1 [n]}\}=\sqrt{\varepsilon _s }\sum \limits _{m=0}^1 s[{n-n_s-m}]\\&\quad \times \int \limits _0^{T_s} p_R ({t+mT_s-\tau })\underbrace{{E\{w({t+({n+1})T_s})\}}}_{{=0}} \hbox {d}t=0 \end{aligned}$$

  The mean of the first and the second terms ($E\{{\xi _2 [n]}\}=E\{{\xi _1 [n]}\}=0$) is zero, because the signal and the noise are uncorrelated.

• 2.2 $E\{{\xi _3 [n]}\}$

  $$\begin{aligned} E\{{\xi _3 [n]}\}=\int \limits _0^{T_s}\underbrace{E\left\{ {w({t+nT_s})w({t+({n+1})T_s})} \right\} }_{R_w \left( {T_s} \right) }\hbox {d}t \end{aligned}$$

  The pulse repetition time in UWB systems typically ranges from hundred to thousand times the pulse (monocycle) width ($T_s \gg T_p\approx 1/B$). In this situation, the noise auto-correlation proposed in (22) is approximately zero for time late $T_{s}$ second ($R_w ({T_s})\approx 0$). This leads to $E\{{\xi _3 [n]}\}=0$. So, the mean of DT noise $E\{{{\check{\omega }}[n]}\}$ is approximately zero.

• 3. The autocorrelation of dirty template noise

  $$\begin{aligned} E\{{\check{\omega }}[k]{\check{\omega }}[l]\}&= E\{{\xi _1 [k]\xi _1 [l]}\}+E\{{\xi _2 [k]\xi _2[l]}\}\\&+E\{{\xi _3[k]\xi _3 [l]} \}+2E\{{\xi _1 [k]\xi _2 [l]} \}\\&+2E\{{\xi _1 [k]\xi _3 [l]} \}+2E\{{\xi _2 [k]\xi _3 [l]}\} \end{aligned}$$

• 3.1 $E\{ {\xi _1 [k]\xi _1 [l]} \}$ & $E\{{\xi _2[k]\xi _2 [l]}\}$

  $$\begin{aligned}&=E\left\{ \left( {\int \limits _0^{T_s} f(t+kT_s ,\tau )w( {t+({k+1})T_s})\hbox {d}t} \right) \right. \\&\quad \left. \times \left( {\int \limits _0^{T_s} f(t+lT_s ,\tau )w({t+({l+1})T_s})\hbox {d}t} \right) \right\} \\&=\int \limits _0^{T_s } \int \limits _0^{T_s} E\left\{ {f(t_1 +kT_s ,\tau )f(t_2 +lT_s ,\tau )} \right\} \\&\quad \times E\left\{ {w\left( {t_1 +({k+1})T_s } \right) w\left( {t_2 +({l+1})T_s } \right) } \right\} \hbox {d}t_2 \hbox {d}t_1 \end{aligned}$$

  Let us find the value of $E\{{f(t_1+kT_s ,\tau )f(t_2 +lT_s,\tau )} \}$, and substituting it in the last equation,

  $$\begin{aligned}&E\left\{ {f(t_1 +kT_s ,\tau )f(t_2 +lT_s ,\tau )} \right\} \\&\quad =\varepsilon _s \sum \limits _{m_1 =0}^1 \sum \limits _{m_2 =0}^1 \underbrace{E\left\{ {s\left[ {k-n_s -m_1 } \right] s\left[ {l-n_s -m_2 } \right] } \right\} }_{=0\,\forall k-m_1 \ne l-m_2}\\&\quad \times \underbrace{p_R \left( {t_1 +m_1 T_s -\tau } \right) p_R \left( {t_2 +m_2 T_s -\tau } \right) }_{=0 : m_1 \ne m_2}\\&=\varepsilon _s \delta _{k,l} \sum \limits _{m=0}^1 p_R \left( {t_1 +mT_s -\tau } \right) p_R \left( {t_2 +mT_s -\tau } \right) \\&\Longrightarrow E\left\{ {f(t_1 +kT_s ,\tau )f(t_2 +lT_s ,\tau )} \right\} \\&\quad =\varepsilon _s \delta _{k,l} p_R \left( {t_1 } \right) p_R \left( {t_2 } \right) \end{aligned}$$

  So, $E\left\{ {\xi _1 \left[ k \right] \xi _1 \left[ l \right] } \right\}$ becomes:

  $$\begin{aligned}&E\left\{ {\xi _1 \left[ k \right] \xi _1 \left[ l \right] } \right\} \\&\quad =\varepsilon _s \delta _{k,l} \int \limits _0^{T_s } \int \limits _0^{T_s } p_R \left( {t_1 } \right) p_R \left( {t_2 } \right) R_w \left( {t_1 -t_2 } \right) \hbox {d}t_2 \hbox {d}t_1\\&=\varepsilon _s \delta _{k,l} \int \limits _0^{T_s } \int \limits _0^{T_s } p_R \left( {t_1 } \right) p_R \left( {t_2 } \right) N_0 B\sin c\left( {B\left( {t_1 -t_2 } \right) } \right) \\&\quad \times \cos \left( {2\pi f_0 \left( {t_1 -t_2 } \right) } \right) \hbox {d}t_2 \hbox {d}t_1\\&=\varepsilon _s \delta _{k,l} \int \limits _0^{T_s } \int \limits _0^{T_s } p_R \left( {t_1 } \right) p_R \left( {t_2 } \right) N_0 \!\left( {\int \limits _{-B/2}^{B/2} e^{j2\pi f\left( {t_1 -t_2 } \right) }\hbox {d}f}\!\right) \\&\quad \times \left( {\frac{e^{-j2\pi f_0 \left( {t_1 -t_2 } \right) }+e^{j2\pi f_0 \left( {t_1 -t_2 } \right) }}{2}} \right) \hbox {d}t_2 \hbox {d}t_1\\&=\varepsilon _s \delta _{k,l} \int \limits _0^{T_s } \int \limits _0^{T_s } p_R \left( {t_1 } \right) \\&\quad \times p_R \left( {t_2 } \right) \frac{N_0 }{2}\left( \int \limits _{-B/2}^{B/2} e^{j2\pi \left( {f-f_0 } \right) \left( {t_1 -t_2 } \right) }\hbox {d}f\right. \\&\quad \left. +\int \limits _{B/2}^{B/2} e^{j2\pi \left( {f+f_0 } \right) \left( {t_1 -t_2 } \right) }\hbox {d}f \right) \hbox {d}t_2 \hbox {d}t_1 \end{aligned}$$
  $$\begin{aligned}&=\varepsilon _s \delta _{k,l} \int \limits _0^{T_s } \int \limits _0^{T_s } p_R \left( {t_1 } \right) \\&\quad \times p_R \left( {t_2 } \right) \frac{N_0 }{2}\left( \int \limits _{-f_0 -B/2}^{-f_0 +B/2} e^{j2\pi f\left( {t_1 -t_2 } \right) }\hbox {d}f\right. \\&\quad \left. +\int \limits _{f_0 -B/2}^{f_0 +B/2} e^{j2\pi f\left( {t_1 -t_2 } \right) }\hbox {d}f \right) \hbox {d}t_2 \hbox {d}t_1 \end{aligned}$$

  The signal $p_R(t)$ has the duration $T_s$, so $\int _0^{T_s } p_R (t) e^{-j2\pi ft} \hbox {d}t=\int _{-\infty }^\infty p_R (t)e^{-j2\pi ft}\hbox {d}t=TF\left\{ {p_R \left( t \right) } \right\}$, and the bandwidth of $p_R (t)$ is $1/T_p$. Recalling that the bandwidth of the receiver band-pass front end $B\approx 1/T_p$. Hence, $E\left\{ {\xi _1 [k]\xi _1 [l]} \right\}$ is developed as follows:

  $$\begin{aligned}&=\frac{N_0 }{2}\varepsilon _s \delta _{k,l} \int \limits _0^{T_s } p_R \left( {t_1 } \right) \left( \int \limits _{-f_0 -B/2}^{-f_0 +B/2} \underbrace{\int _0^{T_s } p_R \left( {t_2 } \right) e^{j2\pi f\left( {t_1 -t_2 } \right) }\hbox {d}t_2}_{TF\left\{ {p_R \left( t \right) } \right\} e^{j2\pi ft_1}} \hbox {d}f\right. \\&\quad \left. +\int \limits _{f_0 -B/2}^{f_0 +B/2} \underbrace{\int _0^{T_s } p_R \left( {t_2 } \right) e^{j2\pi f\left( {t_1 -t_2 } \right) }\hbox {d}t_2}_{TF\left\{ {p_R \left( t \right) } \right\} e^{j2\pi ft_1 }} \hbox {d}f \right) \hbox {d}t_1\\&=\frac{N_0 }{2}\varepsilon _s \delta _{k,l}\int \limits _0^{T_s }p_R \left( t \right) \\&\quad \times \!\left( \!\! \underbrace{ \int _{-f_0 -B/2}^{-f_0 +B/2} TF[ {p_R ( t)} ]e^{j2\pi ft}\hbox {d}f{+}\!\int _{f_0 -B/2}^{f_0 +B/2} TF\!\left[ {p_R ( t)} \right] e^{j2\pi ft}\hbox {d}f}_{TF^{-1}\left\{ {TF\left\{ {p_R (t)} \right\} } \right\} =p_R( t )}\! \right) \!\!\hbox {d}t\\&=\frac{N_0 }{2}\varepsilon _s \delta _{k,l} \underbrace{\int _0^{T_s } p_R^2 \left( t \right) \hbox {d}t}_{\varepsilon _{\max }} \end{aligned}$$

  Hence,

  $$\begin{aligned} E\left\{ {\xi _1 [k]\xi _1 [l]} \right\} =E\left\{ {\xi _2 [k]\xi _2 [l]} \right\} =\frac{N_0 }{2}\varepsilon _s \varepsilon _{\max } \delta _{k,l}, \end{aligned}$$

  (30)

  where $\varepsilon _{max}$ is an unknown constant received symbol energy, which depends on the transmitted channel.

• 3.2 $E\left\{ {\xi _3 [k]\xi _3 [l]}\right\}$

  $$\begin{aligned}&E\left\{ {\xi _3 [k]\xi _3 [l]} \right\} \\&\quad =E\left\{ \left( { \int \limits _0^{T_s } w\left( {t_1 +kT_s } \right) w\left( {t_1 +\left( {k+1} \right) T_s } \right) \hbox {d}t_1 } \right) \right. \\&\quad \times \left. \left( {\int \limits _0^{T_s } w\left( {t_2 +lT_s } \right) w\left( {t_2 +\left( {l+1} \right) T_s } \right) \hbox {d}t_2 } \right) \right\} \\&=\int \limits _0^{T_s } \int \limits _0^{T_s } E\left\{ w\left( {t_1 +kT_s } \right) \cdot w\left( {t_1 +\left( {k+1} \right) T_s } \right) \right. \\&\quad \times w\left( {t_2 +lT_s } \right) \left. w\left( {t_2 +\left( {l+1} \right) T_s } \right) \right\} \hbox {d}t_2 \hbox {d}t_1 \end{aligned}$$

  For finding $E\left\{ {\xi _3 [k]\xi _3 [l]} \right\}$, we have employed the fact that the expectation of the product of zero mean, jointly Gaussian random variables $\{ n\left( {t_1} \right) ,n\left( {t_2} \right) ,n\left( {t_3} \right) , n \left( {t_4} \right) \}$, is given as follows:

  $$\begin{aligned} E\left\{ {n({t_1})n({t_2 }),n({t_3 })n({t_4})} \right\}&= R_w ({t_1-t_2 })R_w ({t_3 -t_4})\\&\quad +R_w ({t_1 -t_3})R_w ({t_2 -t_4})\\&\quad +R_w ({t_1 -t_4})R_w ({t_2 -t_3}), \end{aligned}$$

  where $n({t_1})=w({t_1 +kT_s}),\,n({t_2})=w({t_1 +({k+1})T_s}), n({t_3})=w({t_2+lT_s})$, and $n\left( {t_4} \right) =w\left( {t_2+\left( {l+1} \right) T_s} \right)$. Thus, we can represent $E\left\{ {\xi _3 \left[ k \right] \xi _3 \left[ l \right] } \right\}$ as follows:

  $$\begin{aligned}&E\left\{ {\xi _3 [k]\xi _3 [l]} \right\} =\int \limits _0^{T_s } \int \limits _0^{T_s } \left\{ \underbrace{R_w^2 \left( {T_s } \right) }_{\approx 0}+\underbrace{R_w^2 \left( {t_1 -t_2 +(k-l)T_s}\right) }_{R_w^2 \left( {t_1 -t_2 } \right) \delta _{k,l} }\right. \\&\quad \left. +\underbrace{R_w \left( {t_1 -t_2 +(k-l-1)T_s } \right) R_w \left( {t_1 -t_2 +(k-l+1)T_s } \right) }_{\approx 0:\hbox {one of these two correlations is approximately zero}} \right\} \hbox {d}t_2 \hbox {d}t_1\\&=\delta _{k,l} \int \limits _0^{T_s } \int \limits _0^{T_s } R_w^2 \left( {t_1 -t_2 } \right) \hbox {d}t_2 \hbox {d}t_1\\&=\delta _{k,l} \int \limits _0^{T_s } \int \limits _0^{T_s } \left[ {N_0 B\sin c\left( {B\left( {t_1 -t_2 } \right) } \right) \cos \left( {2\pi f_0 \left( {t_1 -t_2 } \right) } \right) } \right] ^{2}\hbox {d}t_2 \hbox {d}t_1 \end{aligned}$$

  Since the symbol time $T_s \gg 1/B$ is large enough so that most of the energy of the pulse concentrates within $T_s$, we can express $E\{{\xi _3 [k]\xi _3 [l]} \}$ in the frequency domain by applying Parseval’s theorem $\int _{-\infty }^\infty |{x(t)}|^{2}\hbox {d}t=\int _{-\infty }^\infty |{X(f)}|^{2}\hbox {d}f$, and using (22) yields

  $$\begin{aligned} E\left\{ {\xi _3 [k]\xi _3 [l]} \right\} =\delta _{k,l} \int \limits _0^{T_s } \int \limits _{-B/2}^{B/2} 2({N_0 /2})^{2}\hbox {d}f\hbox {d}t=\delta _{k,l} \frac{N_0 ^{2}}{2}\hbox {BT}_s \end{aligned}$$

  Thus,

  $$\begin{aligned} E\left\{ {\xi _3 [k]\xi _3 [l]} \right\} =\frac{N_0 ^{2}}{2}\hbox {BT}_s \delta _{k,l} \end{aligned}$$

  (31)

• 3.3 $E\left\{ {\xi _1 [k]\xi _2 [l]}\right\}$

  $$\begin{aligned}&=E\left\{ \left( {\int \limits _0^{T_s } f\left( {t+kT_s ,\tau } \right) w\left( {t+\left( {k+1} \right) T_s } \right) \hbox {d}t} \right) \right. \\&\quad \times \left. \left( {\int \limits _0^{T_s } f(t+\left( {l+1} \right) T_s ,\tau )w\left( {t+lT_s } \right) \hbox {d}t} \right) \right\} \\&=\int \limits _0^{T_s } \int \limits _0^{T_s } E\left\{ {f(t_1 +kT_s ,\tau )f(t_2 +\left( {l+1} \right) T_s ,\tau )} \right\} \\&\quad \times \underbrace{E\left\{ {w\left( {t_1 +\left( {k+1} \right) T_s } \right) w\left( {t_2 +lT_s } \right) } \right\} }_{R_w \left( {t_1 -t_2 +\left( {k-l+1} \right) T_s } \right) } \hbox {d}t_2 \hbox {d}t_1 \end{aligned}$$

  The first product is given as:

  $$\begin{aligned}&E\left\{ {f(t_1 +kT_s ,\tau )f(t_2 +\left( {l+1} \right) T_s ,\tau )} \right\} \\&\quad =\varepsilon _s \sum _{m_1 =0}^1 \sum _{m_2 =0}^1 \underbrace{E\left\{ {s[ {k{-}n_s {-}m_1 } ]s[ {l+1-n_s -m_2 } ]} \right\} }_{=0 \,\forall k-m_1 \ne l+1-m_2}\\&\qquad \times \underbrace{E\left\{ {p_R \left( {t_1 +m_1 T_s -\tau } \right) p_R \left( {t_2 +m_2 T_s -\tau } \right) } \right\} }_{=0 :m_1 \ne m_2}\\&=\varepsilon _s \delta _{k,l+1} \sum _{m=0}^1 p_R \left( {t_1 +mT_s -\tau } \right) p_R \left( {t_2 +mT_s -\tau } \right) \\&=\varepsilon _s \delta _{k,l+1} p_R \left( {t_1 } \right) p_R \left( {t_2 } \right) \end{aligned}$$

  So, $E\left\{ {\xi _1 \left[ k \right] \xi _2 [l]} \right\}$ is :

  $$ \begin{aligned}&E\left\{ {\xi _1 \left[ k \right] \xi _2 \left[ l \right] } \right\} =\varepsilon _s \delta _{k,l+1} \int \limits _0^{T_s } \int \limits _0^{T_s }\\&\quad \times p_R \left( {t_1 } \right) p_R \left( {t_2 } \right) \underbrace{R_w \left( {t_1 -t_2 +2T_s } \right) }_{\approx 0 :t_1 \& t_2 \in \left[ {0,} \right. \left. {T_s } \right) } \hbox {d}t_2 \hbox {d}t_1 =0 \end{aligned}$$

• 3.4 $E\left\{ {\xi _1 \left[ k \right] \xi _3 \left[ l \right] } \right\}$ & $E\left\{ {\xi _2 \left[ k \right] \xi _3 \left[ l \right] } \right\}$

  $$\begin{aligned}&=E\left\{ \left( {\int \limits _0^{T_s } f(t_1 +kT_s ,\tau )w\left( {t_1 +\left( {k+1} \right) T_s } \right) \hbox {d}t_1 } \right) \right. \\&\quad \times \left. \left( {\int \limits _0^{T_s } w\left( {t_2 +lT_s } \right) w\left( {t_2 +\left( {l+1} \right) T_s } \right) \hbox {d}t_2 } \right) \right\} \\&=\int \limits _0^{T_s } \int \limits _0^{T_s } E\left\{ f(t_1 +kT_s ,\tau ) w\left( {t_1 +\left( {k+1} \right) T_s } \right) \right. \\&\quad \times \left. w\left( {t_2 +lT_s } \right) w\left( {t_2 +\left( {l+1} \right) T_s } \right) \right\} \hbox {d}t_2 \hbox {d}t_1\\&E\left\{ {\xi _{1} [k]\xi _{3} [l]} \right\} \\&\quad = \int \limits _{0}^{{T_{s} }} \int \limits _{0}^{{T_{s} }} \underbrace{E\left\{ {f(t_{1} + kT_{s} ,~\tau )w\left( {t_{1}+\left( {k + 1} \right) T_{s} } \right) } \right\} }_{{= 0}}\\&\quad \times \underbrace{E\left\{ {w\left( {t_{2} + lT_{s} } \right) w\left( {t_{2}+\left( {l + 1} \right) T_{s} } \right) } \right\} }_{{R_{w} \left( {T_{s}} \right) \approx 0}} \hbox {d}t_{2} \hbox {d}t_{1}\\&+\int \limits _{0}^{{T_{s} }} \int \limits _{0}^{{T_{s}}} \underbrace{E\left\{ {f(t_{1}+ kT_{s} ,\tau )w\left( {t_{2} + lT_{s} } \right) } \right\} }_{{= 0}}\\&\quad \times \underbrace{E\left\{ {w\left( {t_{1}+\left( {k + 1} \right) T_{s} } \right) w\left( {t_{2}+\left( {l + 1} \right) T_{s} } \right) } \right\} }_{{R_{w} \left( {t_{1}- t_{2} } \right) \delta _{{k,l}}}} \hbox {d}t_{2} \hbox {d}t_{1} \\&+\int \limits _{0}^{{T_{s}}} \int \limits _{0}^{{T_{s} }} \underbrace{E\left\{ {f(t_{1} + kT_{s} ,~\tau )w\left( {t_{2}+\left( {l + 1} \right) T_{s} } \right) } \right\} }_{{= 0}}\\&\times \quad \underbrace{E\left\{ {w\left( {t_{1} + \left( {k + 1} \right) T_{s} } \right) w\left( {t_{2}+ lT_{s} } \right) } \right\} }_{{R_{w} \left( {t_{1}-t_{2} } \right) \delta _{{k + 1,l}} }} \hbox {d}t_{2} \hbox {d}t_{1}=0 \end{aligned}$$

  In the UWB system, using the dirty template approach, the sampled noise $w\left( {t+kT_s } \right)$ is only weakly correlated with its adjacent $w\left( {t+\left( {k+1} \right) T_s } \right)$ since $(B\gg 1/T_s )$. We have also found that the three noise terms {$\xi _1 \left[ n \right] ,\xi _2 \left[ n \right] ,\xi _3 \left[ n \right]$} could be treated approximately as mutually uncorrelated zero mean noises. Furthermore, $\xi _3 \left[ n \right]$ contains the product of two uncorrelated Gaussian noise components, and the central limit theorem (CLT) asserts that it will also be approximately Gaussian as a consequence of the integration. In other word, $\xi _3 \left[ n \right]$ is also approximately white Gaussian noise $\left( {\xi _3 \left[ n \right] \sim \mathcal{N}(0,\frac{N_0 ^{2}}{2}\hbox {BT}_s )} \right)$.

Finally, ${\check{\omega }}[n]$ is the zero-mean Gaussian $\bigg ({\check{\omega }}[n]\sim {\mathcal {N}}\bigg (0,\frac{N_0 ^{2}}{2} \hbox {BT}_s +N_0 \varepsilon _s \varepsilon _{\max }\bigg )\bigg )$ with correlation function $\bigg ( \frac{N_0 ^{2}}{2}\hbox {BT}_s + N_0 \varepsilon _s \varepsilon _{\max } \bigg )\delta _{k,l}$.