**3. Energy detection**

The MIR-UWB architecture is based on energy detection. Thus the receiver detects *only* the energy of the received signal in a specified window. The disadvantage in performance is 2 Ultra-Wideband Radio Technologies for Communications, Localization and Sensor Applications MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture <sup>3</sup> MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture 3

**Figure 2.** MIR-UWB receiver

2 Will-be-set-by-IN-TECH

The idea of the MIR-UWB architecture is based on [45, 46]. The architecture proposed there comprises a transmitter using multiple bands and impulse radio within the bands to transmit data and a receiver, which detects only the energy of the transmitted impulses. The combination of energy detection receiver and multiband enables a flexible high data rate

The MIR-UWB transmitter is based on a multiband pulse generation followed by a modulator. The multiband pulse generator generates a pulse with a specified bandwidth for every subband. Subbands can be activated or deactivated using the bandplan. Different possibilities to generate theses pulses are shown in [30]. Each subband pulse will be modulated with different data, all subband pulses are summed up to a multiband pulse, amplified and

Bandplan Mapper

binary data

PAM/PPM Modulator1

PAM/PPM

Modulator2

PAM/PPM

Modulator*<sup>N</sup>*

The MIR-UWB receiver is based on *N* parallel energy detection receivers. A filter bank separates the individual subband pulses and an energy detector measures the energy in every subband. Based on the measured energy, the demodulator makes his decision. For pulse amplitude modulation (PAM) and its special case of on-off-keying (OOK) the demodulation process needs to know the SNR in each subband. This can be estimated using a preamble [46]. The channel state information can be used for Detect and Avoid (DAA) algorithms [34] and to increase the performance of the multiband system [28]. Pulse position modulation (PPM) and

The MIR-UWB architecture is based on energy detection. Thus the receiver detects *only* the energy of the received signal in a specified window. The disadvantage in performance is

transmitted. Figure 1 shows a transmitter based on an *oscillator bank* pulse generator.

Pulse- PA

*fc*,1

*fc*,2

*fc*,*<sup>N</sup>*

**Figure 1.** MIR-UWB Transmitter based on *oscillator bank* pulse generation

transmit reference (TR) do not need any channel state information.

**2. Multiband impulse radio**

system with low power consumption.

Generator

Multibandpulse generator

**2.1. Transmitter**

**2.2. Receiver**

**3. Energy detection**

accompanied by a very simple receiver design [36, 38, 54]. The performance measure is based on the average symbol error probability (SEP) or bit error probability (BEP) and will be derived in the following section.

### **3.1. Demodulation**

In the additive white gaussian noise (AWGN) channel the received signal *R* is the sum of the transmitted signal *s* and white Gaussian noise *W* with the power spectral density of *N*0/2:

$$R(t) = s(t) + \mathcal{W}(t) = a\_m p(t) + \mathcal{W}(t). \tag{1}$$

The transmitted signal *s* is a weighted pulse *p*. The amplitude of the pulse is *am*. Thus, the energy of a transmitted signal is:

$$\begin{aligned} E\_S &= \int\_{t\_0}^{t\_0 + T\_i} s^2(t) \, \mathbf{d}t = a\_m^2 \int\_{t\_0}^{t\_0 + T\_i} p^2(t) \, \mathbf{d}t \\ &= \sum\_{i=1}^{2D} s\_i = a\_m^2 \sum\_{i=1}^{2D} p\_{i\prime} \end{aligned}$$

where *si* := *s*(*i*/(2*B*)) and *pi* := *p*(*i*/(2*B*)) and *B* denoting the bandwidth. The received energy can be approximated by a finite sum of 2*D* = 2*TiB* samples [56]. The event {*A* = *am*} with the range A = {*a*0, ... , *aM*−1} describes a transmitted symbol with the amplitude *am*. Without loss of generality the integration starts at *t*<sup>0</sup> = 0. In order to measure the energy, the detector squares the received signal *R*(*t*) and integrates the result over the time interval *Ti*. The received energy, normalized by the power spectral density *N*0/2, is:

$$Y = \frac{2}{N\_0} \int\_0^{T\_i} R^2(t) \,\mathrm{d}t.\tag{2}$$

#### 4 Will-be-set-by-IN-TECH 4 Ultra-Wideband Radio Technologies for Communications, Localization and Sensor Applications MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture <sup>5</sup>

A time discrete representation of the received energy *Y* is:

$$Y = \frac{1}{N\_0 B} \sum\_{i=1}^{2D} \left( s\_i + W\_i \right)^2 \,\text{\,\,}\tag{3}$$

where *Ep* is the energy of an unmodulated pulse *p*. The demodulator has to decide, which symbol *m* with the energy *ESm* and the amplitude *am* has been transmitted, based on the observation of the random variable *Y*. The optimal receiver, i. e. the receiver with the lowest probability to make a wrong decision, makes the decision for the symbol that has been sent most likely, given a certain energy *y* at the receiver. Thus, the receiver makes the decision for

This is the maximum a posteriori probability (MAP) decision rule. If all transmitted symbols

*fY*|*A*(*y*|*am*) <sup>≥</sup> *fY*|*A*(*y*|*ak*), <sup>∀</sup> *<sup>m</sup>* �<sup>=</sup> *<sup>k</sup>*,

because **P**(*A* = *am*) = 1/*M* for all *m* ∈ {0, 1, . . . , *M* − 1} and *fY*(*y*) are independent of *m*. For the *M*-PAM modulated signal, we use the ML receiver with multiple hypothesis testing:

The SEP *Pe* for the energy detection receiver in the AWGN channel with *M*-PAM signals can

where *Pc* is the probability of a correct decision and **P**(*ρ<sup>m</sup>* < *Y* ≤ *ρm*+1|*A* = *am*) is the conditional probability, that the received energy *Y* is in the interval [*ρm*, *ρm*+1), with the optimal interval thresholds *ρ*. Thus, the decision has been made using the ML decision rule (8). **P**(*A* = *am*) is the a priori probability, that the symbol *m* has been sent and **P**(*A* = *am*) = 1/*M* for all *m* ∈ {0, 1, . . . , *M* − 1}. The conditional probability **P**(*ρ<sup>m</sup>* ≤ *Y* < *ρm*+1|*A* = *am*) is:

*ρ<sup>m</sup>*

*<sup>u</sup>D*−<sup>1</sup> exp

− *u* 2 

are equal probable, it can be reduced to the maximum-likelihood (ML) decision rule:

**<sup>P</sup>**(*<sup>A</sup>* <sup>=</sup> *am*|*<sup>Y</sup>* <sup>=</sup> *<sup>y</sup>*) = *fY*|*A*(*y*|*am*)**P**(*<sup>A</sup>* <sup>=</sup> *am*)

*m* = arg max *k*∈[0, *M*−1]

with the conditional probability density function *fY*|*<sup>A</sup>* based on (5) and (6).

*M*−1 ∑ *m*=0

**<sup>P</sup>**(*ρ<sup>m</sup>* <sup>&</sup>lt; *<sup>y</sup>* <sup>≤</sup> *<sup>ρ</sup>m*+1|*<sup>A</sup>* <sup>=</sup> *am*) = *<sup>ρ</sup>m*+<sup>1</sup>

0

*FY*|*A*(*y*|0) = *<sup>y</sup>*

The related distribution function *FY*|*A*(·|0) can be calculated in closed form:

1 2*D*Γ (*D*)

*Pe*(*γ*, **a**, ρ, *D*) = 1 − *Pc*(*γ*, **a**, ρ, *D*)

= 1 −

**P**{*A* = *am*|*Y* = *y*} ≥ **P**{*A* = *ak*|*Y* = *y*}, ∀ *m* �= *k*. (7)

MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture 5

*fY*(*y*) <sup>=</sup> *fY*|*A*(*y*|*am*)1/*<sup>M</sup>*

*fY*(*y*) ,

*fY*|*A*(*y*|*ak*), (8)

**P**(*ρ<sup>m</sup>* ≤ *Y* < *ρm*+1|*A* = *am*)**P**(*A* = *am*), (9)

<sup>=</sup> *FY*|*A*(*ρm*+1|*am*) <sup>−</sup> *FY*|*A*(*ρm*|*am*). (10)

 *D*, *<sup>y</sup>* 2 

<sup>Γ</sup> (*D*) , (11)

*fY*|*A*(*y*|*am*) <sup>d</sup>*<sup>y</sup>*

<sup>d</sup>*<sup>u</sup>* <sup>=</sup> <sup>Γ</sup>

the symbol *m* with the amplitude *am*, when [25]:

using the Bayes theorem:

be calculated as:

where *Wi* := *W*(*i*/(2*B*)). If the symbol energy is *ES* = 0, the received energy *Y* will be *χ*<sup>2</sup> distributed with the degree of freedom of 2*D*. If the symbol energy *ES* > 0, the received energy will be noncentral *χ*<sup>2</sup> distributed with the degree of freedom of 2*D* and the noncentrality parameter *μ*:

$$\mu = \sum\_{i=1}^{2D} s\_i^2 = \frac{2}{N\_0} \int\_0^T s^2(t) \, \text{d}t =: \frac{2E\_\text{s}}{N\_0} = 2\gamma\_\prime \tag{4}$$

where *γ* = *ES*/*N*<sup>0</sup> is the SNR at the receiver. Thus, the distribution of the received energy *Y* depends on the symbol energy *ES*:

$$Y \sim \begin{cases} \chi\_{2D}^2 & \text{for } E\_S = 0 \\ \chi\_{2D}^2(2\gamma) & \text{for } E\_S > 0 \end{cases}.$$

The conditional probability density function *fY*|*A*(·|*a*0) with *<sup>a</sup>*<sup>0</sup> <sup>=</sup> 0 and *ES* <sup>=</sup> 0 of the received energy *Y* is:

$$f\_{Y|A}(y|a\_0) = \frac{1}{2^D \Gamma(D)} y^{D-1} \exp\left(-\frac{y}{2}\right). \tag{5}$$

The conditional probability density function *fY*|*A*(·|*am*) with *am* <sup>&</sup>gt; 0 and *ES* <sup>&</sup>gt; 0 of the received energy *Y* is:

$$f\_{Y|A}(y|a\_m) = \frac{1}{2} \left(\frac{y}{2\gamma}\right)^{\frac{D-1}{2}} \exp\left(-\frac{2\gamma+y}{2}\right) \mathbf{I}\_{D-1}\left(\sqrt{2\gamma y}\right),\tag{6}$$

where Γ is the gamma function [18, eq. 8.310.1] and I*n* is the modified Bessel function of the first kind of order *n* [1, eq. 9.6.3].

### **3.2. AWGN channel**

First we calculate the bit error probability of the energy detection receiver in the AWGN channel (1). This receiver detects only the energy of the received signal (2), (3).

### *3.2.1. Pulse amplitude modulation*

The *M*-PAM modulated signal is:

$$s(t) = \sum\_{k=-\infty}^{\infty} a\_{m,k} p(t - kT\_r).$$

and transmits log2(*M*) bit per symbol. The energy of the *<sup>m</sup>*th symbol is:

$$E\_{S\_m} = \underbrace{\int\_0^{T\_l} s^2(t) \, \mathbf{d}t}\_{=\sum\_{i=1}^{2D} s\_i^2} = a\_m^2 \underbrace{\int\_0^{T\_l} p^2(t) \, \mathbf{d}t}\_{=\sum\_{i=1}^{2D} p\_i^2} = a\_m^2 E\_{p^2}$$

where *Ep* is the energy of an unmodulated pulse *p*. The demodulator has to decide, which symbol *m* with the energy *ESm* and the amplitude *am* has been transmitted, based on the observation of the random variable *Y*. The optimal receiver, i. e. the receiver with the lowest probability to make a wrong decision, makes the decision for the symbol that has been sent most likely, given a certain energy *y* at the receiver. Thus, the receiver makes the decision for the symbol *m* with the amplitude *am*, when [25]:

$$\mathbb{P}\{A = a\_m | Y = y\} \ge \mathbb{P}\{A = a\_k | Y = y\}, \quad \forall \ m \ne k. \tag{7}$$

This is the maximum a posteriori probability (MAP) decision rule. If all transmitted symbols are equal probable, it can be reduced to the maximum-likelihood (ML) decision rule:

$$f\_{Y|A}(y|a\_m) \ge f\_{Y|A}(y|a\_k)\_\prime \quad \forall \ m \ne k\_\prime$$

using the Bayes theorem:

4 Will-be-set-by-IN-TECH

2*D* ∑ *i*=1

where *Wi* := *W*(*i*/(2*B*)). If the symbol energy is *ES* = 0, the received energy *Y* will be *χ*<sup>2</sup> distributed with the degree of freedom of 2*D*. If the symbol energy *ES* > 0, the received energy will be noncentral *χ*<sup>2</sup> distributed with the degree of freedom of 2*D* and the noncentrality

where *γ* = *ES*/*N*<sup>0</sup> is the SNR at the receiver. Thus, the distribution of the received energy *Y*

The conditional probability density function *fY*|*A*(·|*a*0) with *<sup>a</sup>*<sup>0</sup> <sup>=</sup> 0 and *ES* <sup>=</sup> 0 of the received

The conditional probability density function *fY*|*A*(·|*am*) with *am* <sup>&</sup>gt; 0 and *ES* <sup>&</sup>gt; 0 of the

where Γ is the gamma function [18, eq. 8.310.1] and I*n* is the modified Bessel function of the

First we calculate the bit error probability of the energy detection receiver in the AWGN

2*D*Γ (*D*)

 *T* 0 *s* (*si* + *Wi*)

<sup>2</sup>(*t*) d*t* =:

<sup>2</sup>*<sup>D</sup>* for *ES* = 0

<sup>2</sup>*D*(2*γ*) for *ES* > 0 .

*yD*−<sup>1</sup> exp

−2*<sup>γ</sup>* <sup>+</sup> *<sup>y</sup>* 2

*am*,*kp*(*t* − *kTr*)

*p*2(*t*) d*t*

= *a*<sup>2</sup> *mEp*,

 =<sup>2</sup>*<sup>D</sup>* ∑ *i*=1 *p*2 *i*

 − *y* 2 

 I*D*−<sup>1</sup>

2*γ<sup>y</sup>* 

2*Es N*0

<sup>2</sup> , (3)

= 2*γ*, (4)

. (5)

, (6)

*<sup>Y</sup>* <sup>=</sup> <sup>1</sup> *N*0*B*

A time discrete representation of the received energy *Y* is:

*μ* = 2*D* ∑ *i*=1 *s* 2 *<sup>i</sup>* <sup>=</sup> <sup>2</sup> *N*0

*fY*|*A*(*y*|*am*) = <sup>1</sup>

2 *y* 2*γ*

*Y* ∼ *χ*2

*fY*|*A*(*y*|*a*0) = <sup>1</sup>

*χ*2

 *<sup>D</sup>*−<sup>1</sup> 2 exp 

channel (1). This receiver detects only the energy of the received signal (2), (3).

∞ ∑ *k*=−∞

> = *a*<sup>2</sup> *m Ti* 0

*s*(*t*) =

and transmits log2(*M*) bit per symbol. The energy of the *<sup>m</sup>*th symbol is:

 =<sup>2</sup>*<sup>D</sup>* ∑ *i*=1 *s*2 *i*

 *Ti* 0 *s* <sup>2</sup>(*t*) d*t*

*ESm* =

depends on the symbol energy *ES*:

parameter *μ*:

energy *Y* is:

received energy *Y* is:

**3.2. AWGN channel**

first kind of order *n* [1, eq. 9.6.3].

*3.2.1. Pulse amplitude modulation* The *M*-PAM modulated signal is:

$$\mathbb{P}(A = a\_m | Y = y) = \frac{f\_{Y \mid A}(y | a\_m) \mathbb{P}(A = a\_m)}{f\_Y(y)} = \frac{f\_{Y \mid A}(y | a\_m) 1/M}{f \gamma(y)}.$$

because **P**(*A* = *am*) = 1/*M* for all *m* ∈ {0, 1, . . . , *M* − 1} and *fY*(*y*) are independent of *m*. For the *M*-PAM modulated signal, we use the ML receiver with multiple hypothesis testing:

$$m = \underset{k \in [0, M-1]}{\text{arg}\max} \ f\_{Y|A}(y|a\_k)\_\prime \tag{8}$$

with the conditional probability density function *fY*|*<sup>A</sup>* based on (5) and (6).

The SEP *Pe* for the energy detection receiver in the AWGN channel with *M*-PAM signals can be calculated as:

$$\begin{split} P\_{\varepsilon}(\overline{\gamma}, \mathbf{a}, \boldsymbol{\rho}, D) &= 1 - P\_{\varepsilon}(\overline{\gamma}, \mathbf{a}, \boldsymbol{\rho}, D) \\ &= 1 - \sum\_{m=0}^{M-1} \mathbb{P}(\rho\_{m} \le Y < \rho\_{m+1} | A = a\_{m}) \mathbb{P}(A = a\_{m}), \end{split} \tag{9}$$

where *Pc* is the probability of a correct decision and **P**(*ρ<sup>m</sup>* < *Y* ≤ *ρm*+1|*A* = *am*) is the conditional probability, that the received energy *Y* is in the interval [*ρm*, *ρm*+1), with the optimal interval thresholds *ρ*. Thus, the decision has been made using the ML decision rule (8). **P**(*A* = *am*) is the a priori probability, that the symbol *m* has been sent and **P**(*A* = *am*) = 1/*M* for all *m* ∈ {0, 1, . . . , *M* − 1}. The conditional probability **P**(*ρ<sup>m</sup>* ≤ *Y* < *ρm*+1|*A* = *am*) is:

$$\begin{split} \mathbb{P}(\rho\_{m} < y \le \rho\_{m+1} | A = a\_{m}) &= \int\_{\rho\_{m}}^{\rho\_{m+1}} f\_{Y|A}(y|a\_{m}) \, \mathrm{d}y \\ &= F\_{Y|A}(\rho\_{m+1}|a\_{m}) - F\_{Y|A}(\rho\_{m}|a\_{m}). \end{split} \tag{10}$$

The related distribution function *FY*|*A*(·|0) can be calculated in closed form:

$$F\_{Y|A}(y|0) = \int\_0^y \frac{1}{2^D \Gamma(D)} u^{D-1} \exp\left(-\frac{u}{2}\right) \,\mathrm{d}u = \frac{\Gamma\left(D, \frac{y}{2}\right)}{\Gamma\left(D\right)},\tag{11}$$

#### 6 Will-be-set-by-IN-TECH 6 Ultra-Wideband Radio Technologies for Communications, Localization and Sensor Applications MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture <sup>7</sup>

where Γ(·) is the Gamma function [18, eq. 8.310.1] and Γ(·, ·) is the incomplete Gamma function [18, eq. 8.350.1]. The distribution function (11) can be also displayed with the help of the Marcum-Q function:

$$\mathcal{Q}\_{m}(a,b) = \int\_{b}^{\infty} x \left(\frac{x}{a}\right)^{m-1} \exp\left(-\frac{x^{2}+a^{2}}{2}\right) \mathbf{I}\_{m-1}\left(a\mathbf{x}\right) \,\mathrm{d}x,\tag{12}$$

where I*n* is the modified Bessel function of the first kind of order *n* [1, eq. 9.6.3]. Thus, an alternative representation for the distribution function (11) is for *<sup>D</sup>* <sup>∈</sup> **<sup>Z</sup>**<sup>+</sup> with [49, eq. 2.1.124] and [52, eq. 4.71]:

$$F\_{Y|A}(y|0) = 1 - \mathcal{Q}\_D(0, \sqrt{y}).\tag{13}$$

*Eb*/*N*<sup>0</sup> in dB

MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture 7

*Eb*/*N*<sup>0</sup> in dB

*M*−1 ∑ *m*=0 *a*2 *<sup>m</sup>* = 1,

*Pe*(*γ*,a, ρ, *M*, *D*)

*fY*|*A*(*ρ*opt|*am*) = *fY*|*A*(*ρ*opt|*am*+1), (17)

0 5 10 15 20 <sup>10</sup>-6

*M*

Thus, the optimal interval threshold *ρ*opt between the two symbol amplitudes *am* and *am*+<sup>1</sup>

where *fY*|*<sup>A</sup>* are the conditional probability density functions based on (5) and (6). With these optimal interval thresholds, the symbol decision is based on the ML-criteria (8). Unfortunately, there is no closed form solution for determining the optimal interval thresholds. Thus, they have to be calculated numerically. Figure 5 shows the conditional probability density functions with equidistant symbol amplitudes *am* and optimal interval

0 5 10 15 20 25 <sup>10</sup>-6

OOK, *D* = 2 OOK, *D* = 200 2-PPM, *D* = 2 2-PPM, *D* = 200

BEP

BEP

10-4

**Figure 4.** BEP for multilevel *M*-PAM and *M*-PPM for *D* = 2

thresholds *ρ*1, *ρ*<sup>2</sup> and *ρ*<sup>3</sup> with an SNR of *γ* = 10 dB.

has to fulfil the following equation:

OOK 4-PAM 2-PPM 4-PPM

minimise ρ

subject to <sup>1</sup>

10-2

100

10-4

**Figure 3.** BEP for OOK and 2-PPM with different degrees of freedom

10-2

10<sup>0</sup>

If a symbol with an amplitude *am* > 0 has been sent, the conditional probability density function *fY*|*A*(·|*am*) has the form (6). There does not exist a closed form for the distribution function in general. But for *<sup>D</sup>* <sup>∈</sup> **<sup>Z</sup>**<sup>+</sup> it can also be solved in closed form with the help of the Marcum-Q function (12):

$$\begin{split} F\_{Y|A}(y|a\_m) &= \int\_0^y \frac{1}{2} \left(\frac{u}{2a\_m^2 \gamma}\right)^{\frac{D-1}{2}} \exp\left(-\frac{2a\_m^2 \gamma + u}{2}\right) \mathbf{I}\_{D-1}\left(\sqrt{2a\_m^2 \gamma u}\right) \,\mathrm{d}u \\ &= 1 - \mathcal{Q}\_D\left(a\_m \sqrt{2\gamma\_\prime}, \sqrt{y}\right). \end{split} \tag{14}$$

Combining (13) and (14) with (9), the SEP *Pe* for an energy detection receiver with *M*-PAM for an SNR of *γ* = *Ep*/*N*<sup>0</sup> is [35]:

$$P\_{\ell}(\gamma\_{\prime}, \mathbf{a}, \rho, M, D) = 1 - \frac{1}{M} \left[ \sum\_{m=0}^{M-1} \mathcal{Q}\_{D} \left( a\_{m} \sqrt{2\gamma\_{\prime}} \sqrt{\rho\_{m+1}} \right) - \mathcal{Q}\_{D} \left( a\_{m} \sqrt{2\gamma\_{\prime}} \sqrt{\rho\_{m}} \right) \right] \tag{15}$$

with the *<sup>M</sup>* symbol amplitudes <sup>a</sup> = (*a*0, *<sup>a</sup>*1, ... , *aM*−2, *aM*−1), *<sup>M</sup>* + 1 interval thresholds <sup>ρ</sup> = (*ρ*0, *<sup>ρ</sup>*1, ... , *<sup>ρ</sup>M*−1, *<sup>ρ</sup>M*) and the degree of freedom 2*D*. Applying the interval thresholds *<sup>ρ</sup>*<sup>0</sup> = <sup>0</sup> und *ρ<sup>M</sup>* → ∞, we get Q*<sup>D</sup> am* <sup>√</sup>2*γ*, √*ρ*<sup>0</sup> <sup>=</sup> 0 and <sup>Q</sup>*<sup>D</sup> am* <sup>√</sup>2*γ*, √*ρ<sup>M</sup>* = 1. Combining this with (15), it reduces to:

$$P\_{\varepsilon}(\gamma, \mathbf{a}, \boldsymbol{\rho}, M, D) = \frac{1}{M} \left[ M - 1 + \sum\_{m=1}^{M-1} \mathcal{Q}\_{D} \left( a\_{\mathfrak{m}} \sqrt{2\gamma} \, \_{\prime} \sqrt{\boldsymbol{\rho}\_{m}} \right) - \sum\_{m=0}^{M-2} \mathcal{Q}\_{D} \left( a\_{\mathfrak{m}} \sqrt{2\gamma} \, \_{\prime} \sqrt{\boldsymbol{\rho}\_{m+1}} \right) \right]. \tag{16}$$

Figure 3 shows the influence of different degrees of freedom on the BEP of an energy detection receiver with OOK and 2-PPM. For large degrees of freedom, a higher SNR is necessary to achieve the same BEP. This is due to an increasing amount of noise at the detector. OOK shows a slightly better performance than 2-PPM. Figure 4 shows the influence of higher order modulation on the BEP.

### 3.2.1.1. Optimal interval thresholds

The optimal interval thresholds to minimise the SEP have to fulfil the following optimisation problem:

**Figure 3.** BEP for OOK and 2-PPM with different degrees of freedom

6 Will-be-set-by-IN-TECH

where Γ(·) is the Gamma function [18, eq. 8.310.1] and Γ(·, ·) is the incomplete Gamma function [18, eq. 8.350.1]. The distribution function (11) can be also displayed with the help of

> exp

where I*n* is the modified Bessel function of the first kind of order *n* [1, eq. 9.6.3]. Thus, an alternative representation for the distribution function (11) is for *<sup>D</sup>* <sup>∈</sup> **<sup>Z</sup>**<sup>+</sup> with [49, eq. 2.1.124]

If a symbol with an amplitude *am* > 0 has been sent, the conditional probability density function *fY*|*A*(·|*am*) has the form (6). There does not exist a closed form for the distribution function in general. But for *<sup>D</sup>* <sup>∈</sup> **<sup>Z</sup>**<sup>+</sup> it can also be solved in closed form with the help of the

Combining (13) and (14) with (9), the SEP *Pe* for an energy detection receiver with *M*-PAM for

with the *<sup>M</sup>* symbol amplitudes <sup>a</sup> = (*a*0, *<sup>a</sup>*1, ... , *aM*−2, *aM*−1), *<sup>M</sup>* + 1 interval thresholds <sup>ρ</sup> = (*ρ*0, *<sup>ρ</sup>*1, ... , *<sup>ρ</sup>M*−1, *<sup>ρ</sup>M*) and the degree of freedom 2*D*. Applying the interval thresholds *<sup>ρ</sup>*<sup>0</sup> = <sup>0</sup>

Figure 3 shows the influence of different degrees of freedom on the BEP of an energy detection receiver with OOK and 2-PPM. For large degrees of freedom, a higher SNR is necessary to achieve the same BEP. This is due to an increasing amount of noise at the detector. OOK shows a slightly better performance than 2-PPM. Figure 4 shows the influence of higher order

The optimal interval thresholds to minimise the SEP have to fulfil the following optimisation

<sup>=</sup> 0 and <sup>Q</sup>*<sup>D</sup>*

<sup>−</sup> *<sup>x</sup>*<sup>2</sup> <sup>+</sup> *<sup>a</sup>*<sup>2</sup> 2

> *<sup>m</sup>γ* + *u* 2

<sup>√</sup>*ρm*+<sup>1</sup> − Q*D am* 2*γ*,

> *am* <sup>√</sup>2*γ*,

√*ρ<sup>m</sup>* − *M*−2 ∑ *m*=0 Q*D am* 2*γ*,

 I*D*−<sup>1</sup>  2*a*<sup>2</sup> *<sup>m</sup>γu* d*u*

. (14)

√*ρ<sup>M</sup>*

<sup>√</sup>*ρ<sup>m</sup>* 

= 1. Combining this

<sup>√</sup>*ρm*+<sup>1</sup>

 .

(16)

(15)

*FY*|*A*(*y*|0) = <sup>1</sup> − Q*D*(0, <sup>√</sup>*y*). (13)

I*m*−<sup>1</sup> (*ax*) d*x*, (12)

the Marcum-Q function:

and [52, eq. 4.71]:

Marcum-Q function (12):

*FY*|*A*(*y*|*am*) =

an SNR of *γ* = *Ep*/*N*<sup>0</sup> is [35]:

und *ρ<sup>M</sup>* → ∞, we get Q*<sup>D</sup>*

*Pe*(*γ*, **<sup>a</sup>**, <sup>ρ</sup>, *<sup>M</sup>*, *<sup>D</sup>*) = <sup>1</sup>

modulation on the BEP.

problem:

3.2.1.1. Optimal interval thresholds

with (15), it reduces to:

*Pe*(*γ*, **<sup>a</sup>**, <sup>ρ</sup>, *<sup>M</sup>*, *<sup>D</sup>*) = <sup>1</sup> <sup>−</sup> <sup>1</sup>

Q*m*(*a*, *b*) =

= 1 − Q*<sup>D</sup>*

 *u* 2*a*<sup>2</sup> *mγ*

> *am* 2*γ*, √*y*

*M*−<sup>1</sup> ∑ *m*=0 Q*D am* 2*γ*,

√*ρ*<sup>0</sup>

*M*−1 ∑ *m*=1 Q*D am* 2*γ*,

*M*

*M* − 1 +

 *am* <sup>√</sup>2*γ*,

*M*   *<sup>D</sup>*−<sup>1</sup> 2 exp −2*a*<sup>2</sup>

 ∞ *b x x a <sup>m</sup>*−<sup>1</sup>

**Figure 4.** BEP for multilevel *M*-PAM and *M*-PPM for *D* = 2

$$\underset{\rho}{\text{minimise}}\quad P\_{\varepsilon}(\gamma, \mathfrak{a}, \mathfrak{p}, M, D)$$

$$\text{subject to}\quad \frac{1}{M}\sum\_{m=0}^{M-1} a\_{m}^{2} = 1,$$

Thus, the optimal interval threshold *ρ*opt between the two symbol amplitudes *am* and *am*+<sup>1</sup> has to fulfil the following equation:

$$f\_{Y|A}(\rho\_{\text{opt}}|a\_m) = f\_{Y|A}(\rho\_{\text{opt}}|a\_{m+1})\_\prime \tag{17}$$

where *fY*|*<sup>A</sup>* are the conditional probability density functions based on (5) and (6). With these optimal interval thresholds, the symbol decision is based on the ML-criteria (8). Unfortunately, there is no closed form solution for determining the optimal interval thresholds. Thus, they have to be calculated numerically. Figure 5 shows the conditional probability density functions with equidistant symbol amplitudes *am* and optimal interval thresholds *ρ*1, *ρ*<sup>2</sup> and *ρ*<sup>3</sup> with an SNR of *γ* = 10 dB.

8 Will-be-set-by-IN-TECH 8 Ultra-Wideband Radio Technologies for Communications, Localization and Sensor Applications MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture <sup>9</sup>

*a*<sup>3</sup> = 1 are set. The minimal SEP has been reached for a = (0, 0.35, 0.67, 1). Figure 8 shows

*<sup>a</sup>*<sup>2</sup> *<sup>a</sup>*<sup>1</sup>

the gain for 4-PAM with optimal amplitudes for different degrees of freedom. The results show impressive gains for large degrees of freedom. Figure 9 shows the optimal amplitudes for different degrees of freedom. For *D* = 2 the amplitudes are almost equidistant but for

*Eb*/*N*<sup>0</sup> in dB

<sup>10</sup> <sup>15</sup> <sup>20</sup> <sup>25</sup> <sup>10</sup>-6

To analyse the performance of an energy detection receiver we need a channel model that enables a good approximation of the energy at the receiver. Investigations of the IEEE channel model (802.15.3a) show that the energy at the receiver can be approximated by a random

Figure 10 compares the channel's magnitude (denoted as CIR) to a moving average of width 100 MHz and 1 GHz of the energy in the IEEE channel model. Figure 11 shows the magnitude at the receiver for a detector with 100 MHz and 1 GHz bandwidth. Thus we can use the

equi., *D* = 2 opt., *D* = 2 equi., *D* = 100 opt., *D* = 100

0,5

MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture 9

1

0

0

0,5

1 10-4

*D* = 200 the amplitudes are adjusted and not equidistant any more.

10-2

10<sup>0</sup>

BEP

**Figure 7.** SEP for 4-PAM with different interval thresholds

BEP

10-4

**Figure 8.** BEP for 4-PAM with equidistant and optimal amplitudes

**3.3. Flat fading channel**

variable which is constant for one symbol.

10-2

10<sup>0</sup>

**Figure 5.** Optimal interval thresholds for 4-PAM (*γ* = 10 dB)

**Figure 6.** Sensitivity of the BEP related to the interval threshold *ρ*<sup>1</sup> for OOK

Figure 6 shows the influence of a non optimal interval threshold *ρ*<sup>1</sup> on the SEP for OOK (*M* = 2). In such a case, the SEP gets more sensitive for high SNR.

### 3.2.1.2. Optimal amplitudes

Now we try to minimise the SEP by optimising the symbol amplitudes a. The optimisation problem is now [33, 37]:

$$\begin{aligned} \underset{a}{\text{minimise}} \quad & P\_{\varepsilon}(\gamma, a, \rho, M, D), \\ \text{subject to} \quad & \frac{1}{M} \sum\_{m=0}^{M-1} a\_m^2 = 1, \end{aligned}$$

This optimisation problem can only be solved numerically, because the optimal interval thresholds ρ are based on the amplitudes a and there exists no closed form solution for the optimal interval thresholds ρ (17). For OOK (*M* = 2) the optimal amplitudes are aopt = (0, 2). In this case they are independent of the SNR *γ*. For *M* > 2 it is possible to calculate a set of optimal amplitudes aopt for every SNR *γ*. Figure 7 shows the SEP for 4-PAM for different symbol amplitudes *a*<sup>1</sup> and *a*<sup>2</sup> for a SNR of 16 dB. For figure 7 the amplitudes *a*<sup>0</sup> = 0 and *a*<sup>3</sup> = 1 are set. The minimal SEP has been reached for a = (0, 0.35, 0.67, 1). Figure 8 shows

**Figure 7.** SEP for 4-PAM with different interval thresholds

the gain for 4-PAM with optimal amplitudes for different degrees of freedom. The results show impressive gains for large degrees of freedom. Figure 9 shows the optimal amplitudes for different degrees of freedom. For *D* = 2 the amplitudes are almost equidistant but for *D* = 200 the amplitudes are adjusted and not equidistant any more.

**Figure 8.** BEP for 4-PAM with equidistant and optimal amplitudes

### **3.3. Flat fading channel**

8 Will-be-set-by-IN-TECH

*ρ*2

*fY*|*<sup>A</sup>*(*y*|*a*0) *fY*|*<sup>A</sup>*(*y*|*a*1) *fY*|*<sup>A</sup>*(*y*|*a*2)

*γ* = 5 dB *γ* = 10 dB *γ* = 15 dB

*fY*|*A*(*y*|*am*)

0

0,02

**Figure 5.** Optimal interval thresholds for 4-PAM (*γ* = 10 dB)

BEP/BEPopt

10<sup>1</sup>

**Figure 6.** Sensitivity of the BEP related to the interval threshold *ρ*<sup>1</sup> for OOK

2). In such a case, the SEP gets more sensitive for high SNR.

3.2.1.2. Optimal amplitudes

problem is now [33, 37]:

10<sup>2</sup>

10<sup>3</sup>

10<sup>4</sup>

0,04

0,06

0,08

0,10

0,12

*y*

*ρ*/*ρ*opt

0 0,5 1,0 1,5 2,0 <sup>10</sup><sup>0</sup>

Figure 6 shows the influence of a non optimal interval threshold *ρ*<sup>1</sup> on the SEP for OOK (*M* =

Now we try to minimise the SEP by optimising the symbol amplitudes a. The optimisation

minimise <sup>a</sup> *Pe*(*γ*,a, <sup>ρ</sup>, *<sup>M</sup>*, *<sup>D</sup>*)

*M*−1 ∑ *m*=0 *a*2 *<sup>m</sup>* = 1,

*M*

This optimisation problem can only be solved numerically, because the optimal interval thresholds ρ are based on the amplitudes a and there exists no closed form solution for the optimal interval thresholds ρ (17). For OOK (*M* = 2) the optimal amplitudes are aopt = (0, 2). In this case they are independent of the SNR *γ*. For *M* > 2 it is possible to calculate a set of optimal amplitudes aopt for every SNR *γ*. Figure 7 shows the SEP for 4-PAM for different symbol amplitudes *a*<sup>1</sup> and *a*<sup>2</sup> for a SNR of 16 dB. For figure 7 the amplitudes *a*<sup>0</sup> = 0 and

subject to <sup>1</sup>

0 50 100 150

*fY*|*<sup>A</sup>*(*y*|*a*3) *<sup>ρ</sup>*<sup>1</sup>

*ρ*3

To analyse the performance of an energy detection receiver we need a channel model that enables a good approximation of the energy at the receiver. Investigations of the IEEE channel model (802.15.3a) show that the energy at the receiver can be approximated by a random variable which is constant for one symbol.

Figure 10 compares the channel's magnitude (denoted as CIR) to a moving average of width 100 MHz and 1 GHz of the energy in the IEEE channel model. Figure 11 shows the magnitude at the receiver for a detector with 100 MHz and 1 GHz bandwidth. Thus we can use the

In a flat fading channel, the random path attenuation *H* is assumed to be constant for the

MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture 11

*R*(*t*) = *Hs*(*t*) + *W*(*t*),

where *W*(*t*) denotes the additive white Gaussian noise (AWGN) with a spectral power density

*<sup>γ</sup>* <sup>=</sup> *<sup>h</sup>*<sup>2</sup> *ES N*0 ,

where *h* denotes the instantaneous path attenuation and *ES* denotes the symbol energy. Since the SNR *γ* is random, *γ* is a realization of a a random variable Γ. The average SNR *γ* can be

with Ω = **E**(*H*2). Introducing a change of variable, the probability density function of the

2 *γγ* Ω

To calculate the average SEP in a flat fading channel, we have to solve the following integral

Using the SEP in the AWGN channel (16) and the probability density function of the random

*<sup>M</sup>* <sup>−</sup> <sup>1</sup> − Q*<sup>D</sup>* (0, <sup>√</sup>*ρ*1)

 ∞ 0 Q*D am* 2*γ*,

 ∞ 0 Q*D am* 2*γ*,

with the symbol amplitudes <sup>a</sup> = (*a*0, *<sup>a</sup>*1, ... , *aM*−2, *aM*−1) and the interval thresholds <sup>ρ</sup> = (*ρ*0, *ρ*1, ... , *ρM*−1, *ρM*). (20) is a general solution for the average SEP of an energy detection

*<sup>f</sup>*Γ(*γ*) = *fH*(

*γ f*Γ(*γ*) d*γ*

. (18)

*Pe*,*AWGN*(*γ*)*f*Γ(*γ*) d*γ*. (19)

<sup>√</sup>*ρ<sup>m</sup>* 

<sup>√</sup>*ρm*+<sup>1</sup> 

*f*Γ(*γ*) d*γ*

*f*Γ(*γ*) d*γ*

, (20)

<sup>Ω</sup>*<sup>γ</sup> <sup>γ</sup>* )

duration of a symbol. Thus, the received signal *R*(*t*) in the interval 0 ≤ *t* ≤ *TS* is:

The instantaneous SNR *γ* in the flat fading channel is:

calculated using the expectation of the random variable *H*2:

random variable Γ is [52, eq. 2.3]:

*3.3.1. Pulse amplitude modulation*

*Pe*(*γ*, <sup>a</sup>, <sup>ρ</sup>, *<sup>M</sup>*, *<sup>D</sup>*) = <sup>1</sup>

receiver in a flat fading channel with *M*-PAM.

*<sup>γ</sup>* <sup>=</sup> <sup>Ω</sup> *ES N*0 = ∞ 0

*Pe*(*γ*) = <sup>∞</sup>

SNR Γ in the flat fading channel, the average SEP for *M*-PAM is:

*M* 

> + *M*−1 ∑ *m*=1

> − *M*−2 ∑ *m*=1

0

of *N*0/2.

[52, eq. 1.8]:

**Figure 9.** Optimal amplitudes for different degrees of freedom (*D* = 2, *D* = 200)

flat fading channel model to model the energy at the receiver in a frequency selective fading channel.

**Figure 10.** Moving average of the energy at the receiver (IEEE 802.15.3a, CM1)

**Figure 11.** Energy at the receiver for different bandwidth (IEEE 802.15.3a, CM1)

In a flat fading channel, the random path attenuation *H* is assumed to be constant for the duration of a symbol. Thus, the received signal *R*(*t*) in the interval 0 ≤ *t* ≤ *TS* is:

$$R(t) = H\mathbf{s}(t) + W(t)\_{\prime\prime}$$

where *W*(*t*) denotes the additive white Gaussian noise (AWGN) with a spectral power density of *N*0/2.

The instantaneous SNR *γ* in the flat fading channel is:

10 Will-be-set-by-IN-TECH

*am*




**Figure 11.** Energy at the receiver for different bandwidth (IEEE 802.15.3a, CM1)







**Figure 10.** Moving average of the energy at the receiver (IEEE 802.15.3a, CM1)






channel.

*a*1

**Figure 9.** Optimal amplitudes for different degrees of freedom (*D* = 2, *D* = 200)

*a*2

*a*3

*Eb*/*N*<sup>0</sup> in dB

*f* in GHz

*f* in GHz

2 4 6 8 10 12

CIR

ED – 100 MHz ED – 1 GHz

2 4 6 8 10 12

CIR

CIR – 100 MHz CIR – 1 GHz

<sup>10</sup> <sup>15</sup> <sup>20</sup> <sup>25</sup> *<sup>a</sup>*<sup>0</sup>

flat fading channel model to model the energy at the receiver in a frequency selective fading

$$\gamma = h^2 \frac{E\_S}{N\_0} \prime$$

where *h* denotes the instantaneous path attenuation and *ES* denotes the symbol energy. Since the SNR *γ* is random, *γ* is a realization of a a random variable Γ. The average SNR *γ* can be calculated using the expectation of the random variable *H*2:

$$\overline{\gamma} = \Omega \frac{\overline{E\_S}}{N\_0} = \int\_0^\infty \gamma f\_\Gamma(\gamma) \,\mathrm{d}\gamma$$

with Ω = **E**(*H*2). Introducing a change of variable, the probability density function of the random variable Γ is [52, eq. 2.3]:

$$f\_{\Gamma}(\gamma) = \frac{f\_H(\sqrt{\frac{\Omega \gamma}{\gamma}})}{2\sqrt{\frac{\gamma \Gamma}{\Omega}}}.\tag{18}$$

To calculate the average SEP in a flat fading channel, we have to solve the following integral [52, eq. 1.8]:

$$\overline{P}\_{\varepsilon}(\overline{\gamma}) = \int\_{0}^{\infty} P\_{\varepsilon, AWGN}(\gamma) f \mathbf{r}(\gamma) \, \mathbf{d}\gamma. \tag{19}$$

### *3.3.1. Pulse amplitude modulation*

Using the SEP in the AWGN channel (16) and the probability density function of the random SNR Γ in the flat fading channel, the average SEP for *M*-PAM is:

$$\begin{split} \overline{P}\_{\varepsilon}(\overline{\gamma}, a, \rho, M, D) &= \frac{1}{M} \Big[ M - 1 - \mathcal{Q}\_{D} \left( 0, \sqrt{\rho\_{1}} \right) \\ &\quad + \sum\_{m=1}^{M-1} \int\_{0}^{\infty} \mathcal{Q}\_{D} \left( a\_{m} \sqrt{2\gamma}, \sqrt{\rho\_{m}} \right) f\_{\Gamma}(\gamma) \, \mathrm{d}\gamma \\ &\quad - \sum\_{m=1}^{M-2} \int\_{0}^{\infty} \mathcal{Q}\_{D} \left( a\_{m} \sqrt{2\gamma}, \sqrt{\rho\_{m+1}} \right) f\_{\Gamma}(\gamma) \, \mathrm{d}\gamma \Big], \end{split} \tag{20}$$

with the symbol amplitudes <sup>a</sup> = (*a*0, *<sup>a</sup>*1, ... , *aM*−2, *aM*−1) and the interval thresholds <sup>ρ</sup> = (*ρ*0, *ρ*1, ... , *ρM*−1, *ρM*). (20) is a general solution for the average SEP of an energy detection receiver in a flat fading channel with *M*-PAM.

#### 12 Will-be-set-by-IN-TECH 12 Ultra-Wideband Radio Technologies for Communications, Localization and Sensor Applications MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture <sup>13</sup>

### *3.3.2. Rayleigh fading*

*Rayleigh* distributed path gains are used to model fading channels with no line-of-sight (NLOS) [49, 50, 55]. Thus, the random variable *H* is *Rayleigh* distributed:

$$f\_H(h) = \frac{2h}{\Omega} \exp\left(-\frac{h^2}{\Omega}\right), \quad h \ge 0. \tag{21}$$

where I0 denotes the modified Bessel function of the first kind of order zero. The *Rician*-k-factor is the ratio between the power in the direct path and the power in the scattered paths. For *k* = 0 the *Rice* distribution is equal to the *Rayleigh* distribution. For *k* → ∞ the *Rician* fading channel converges to the AWGN channel. Different UWB measurement campaigns show a good fit with the distribution of the path gains with a *Rice* distribution [20, 26, 43].

> � I0 � 2 �

Q*<sup>D</sup>* (*amu*,

2*ka*<sup>2</sup> *mγ* (*k* + 1) + *a*<sup>2</sup>

Φ<sup>1</sup> (*γ*, *am*, *ρm*, *k*) −

A closed form solution for the integral in (28) is not known for *D* > 1. In this case, the integral

The probability density function of the *Nakagami*-*m* distribution of the random path gains is

where *m* denotes the *Nakagami*-*m* fading parameter with *m* ∈ [1/2, ∞) and Γ denotes the Gamma function. The *Nakagami*-*m* distribution includes as special cases the one-sided normal distribution (*m* = 1/2) and the *Rayleigh*-distribution (*m* = 1). For *m* → ∞ the *Nakagami*-*m* fading channel converges to the AWGN channel. Different UWB measurement campaigns show a good fit to the *Nakagami*-*m* distribution [4, 19]. *Nakagami*-*m* distribution is also used

− *m* Ω *h*2 �

� *m* Ω �*<sup>m</sup>* exp �

· I0 ⎛ ⎝2 �

Thus, the average SEP in a *Rician* fading channel is with a degree of freedom of 2*D* = 2:

*<sup>M</sup>* <sup>−</sup> <sup>1</sup> − Q<sup>1</sup> (0, <sup>√</sup>*ρ*1)

There exists only a closed form solution for *D* = 1 [39, eq. 45]. In this case, we get:

��

*k* (*k* + 1) *γ γ*

⎞

(*k* + 1)*ρ<sup>m</sup>* (*k* + 1) + *a*<sup>2</sup>

*M*−2 ∑ *m*=1

*mγ*

Φ<sup>1</sup> (*γ*, *am*, *ρm*+1, *k*)

, *h* ≥ 0, (31)

�

<sup>√</sup>*ρm*) exp �

MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture 13

*K* (*K* + 1) *u*<sup>2</sup> *γ*

> *mγ* , �

�

<sup>−</sup>*<sup>K</sup>* <sup>−</sup> (*<sup>K</sup>* <sup>+</sup> <sup>1</sup>)*u*<sup>2</sup> *γ*

, *γ* ≥ 0. (27)

�

. (29)

�

. (30)

⎠ *u* d*u*. (28)

Using (18), the probability density function of the random SNR is:

*γ*

<sup>−</sup>*<sup>k</sup>* <sup>−</sup> (*<sup>k</sup>* <sup>+</sup> <sup>1</sup>) *<sup>γ</sup> γ*

> � ∞ 0

*<sup>γ</sup>* exp �

Combining (20) and (27) we get an integral of the form:

Φ1(*γ*, *am*, *ρm*, *k*) = Q<sup>1</sup>

*M* �

*fH*(*h*) = <sup>2</sup>*h*2*m*−<sup>1</sup>

Γ(*m*)

+ *M*−1 ∑ *m*=1

*Pe*,*ric*(*γ*, **<sup>a</sup>**, <sup>ρ</sup>, *<sup>k</sup>*, *<sup>M</sup>*) = <sup>1</sup>

has to be calculated numerically.

*3.3.4. Nakagami-m fading*

related to the *χ*<sup>2</sup> distribution:

<sup>Φ</sup>*D*(*γ*, *am*, *<sup>ρ</sup>m*, *<sup>k</sup>*) = *<sup>k</sup>* <sup>+</sup> <sup>1</sup>

*<sup>f</sup>*<sup>Γ</sup> (*γ*) <sup>=</sup> *<sup>k</sup>* <sup>+</sup> <sup>1</sup>

In UWB channels with a large bandwidth and a corresponding high temporal resolution, it is questionable, if the central limit theorem is applicable [4, 32, 59]. Nevertheless, some UWB channel measurements show a good fit to the *Rayleigh* distribution [17, 24, 51]. Using (18), the probability density function of the random SNR is:

$$f\_{\Gamma}(\gamma) = \frac{1}{\overline{\gamma}} \exp\left(-\frac{\gamma}{\overline{\gamma}}\right), \quad \gamma \ge 0. \tag{22}$$

Combining (20) and (22) we get an integral of the form:

$$\mathcal{Y}\_{D}(\overline{\gamma}, a\_{m}, \rho\_{m}) = \frac{1}{\overline{\gamma}} \int\_{0}^{\infty} \mathcal{Q}\_{D} \left( a\_{m} \sqrt{2\gamma}, \sqrt{\rho\_{m}} \right) \exp \left( -\frac{\gamma}{\overline{\gamma}} \right) \,\mathrm{d}\gamma. \tag{23}$$

The integral in (23) can be solved using [40, eq. 12], given that the interval thresholds do not depend on the instantaneous SNR *γ* but on the average SNR *γ*. This is the case for an energy detection receiver with limited channel state information (only knowledge of the average SNR). The closed form solution for Υ*<sup>D</sup>* is then:

$$\begin{split} \mathbf{Y}\_{\mathcal{D}}(\overline{\gamma}\_{\prime}a\_{m},\rho\_{m}) &= \exp\left(-\frac{\rho\_{m}}{2}\right) \left\{ \left(\frac{1/\overline{\gamma} + a\_{m}^{2}}{a\_{m}^{2}}\right)^{D-1} \\ & \cdot \left[ \exp\left(\frac{\rho\_{m}}{2} \frac{a\_{m}^{2}}{a\_{m}^{2} + 1/\overline{\gamma}}\right) - \sum\_{d=0}^{D-2} \frac{1}{d!} \left(\frac{\rho\_{m}}{2} \frac{a\_{m}^{2}}{a\_{m}^{2} + 1/\overline{\gamma}}\right)^{d} \right] + \sum\_{d=0}^{D-2} \frac{1}{d!} \left(\frac{\rho\_{m}}{2}\right)^{d} \right\}. \end{split} \tag{24}$$

Combining (20) and (24) yields to the closed form solution for the energy detection receiver in a *Rayleigh* fading channel with *M*-PAM:

$$\begin{split} \mathbf{T}\_{\varepsilon,\text{ray}}(\overline{\gamma}\_{\prime}\mathbf{a},\rho,M,D) &= \frac{1}{M} \Big[ M - 1 - \mathcal{Q}\_{D}\left(0, \sqrt{\rho\_{1}}\right) \\ &+ \sum\_{m=1}^{M-1} \mathbf{Y}\_{D}\left(\overline{\gamma}\_{\prime}a\_{m\prime}\rho\_{m}\right) - \sum\_{m=1}^{M-2} \mathbf{Y}\_{D}\left(\overline{\gamma}\_{\prime}a\_{m\prime}\rho\_{m+1}\right) \Big]. \end{split} \tag{25}$$

### *3.3.3. Rician fading*

*Rice* distributed path gains are used to model line-of-sight (LOS) fading channels [49, 50, 55]. Thus, the random variable *H* is *Rice* distributed:

$$f\_H(h) = \frac{2h(k+1)}{\Omega} \exp\left(-k - \frac{(k+1)h^2}{\Omega}\right) \mathbf{I}\_0\left(2h\sqrt{\frac{k(k+1)}{\Omega}}\right), \quad h \ge 0,\tag{26}$$

where I0 denotes the modified Bessel function of the first kind of order zero. The *Rician*-k-factor is the ratio between the power in the direct path and the power in the scattered paths. For *k* = 0 the *Rice* distribution is equal to the *Rayleigh* distribution. For *k* → ∞ the *Rician* fading channel converges to the AWGN channel. Different UWB measurement campaigns show a good fit with the distribution of the path gains with a *Rice* distribution [20, 26, 43]. Using (18), the probability density function of the random SNR is:

$$f\_{\gamma}(\gamma) = \frac{k+1}{\overline{\gamma}} \exp\left(-k - \frac{(k+1)\,\gamma}{\overline{\gamma}}\right) \mathbf{I}\_0\left(2\sqrt{\frac{k\left(k+1\right)\,\gamma}{\overline{\gamma}}}\right), \quad \gamma \ge 0. \tag{27}$$

Combining (20) and (27) we get an integral of the form:

$$\Phi\_{\rm D}(\overline{\gamma}, a\_{m}, \rho\_{m}, k) = \frac{k+1}{\overline{\gamma}} \int\_{0}^{\infty} \mathcal{Q}\_{\rm D} \left( a\_{m} u\_{\prime} \sqrt{\rho\_{m}} \right) \exp \left( -\mathcal{K} - \frac{(K+1)u^{2}}{\overline{\gamma}} \right)$$

$$\cdot \operatorname{I}\_{0} \left( 2 \sqrt{\frac{\mathcal{K} \left( K+1\right) u^{2}}{\overline{\gamma}}} \right) u \operatorname{d}u. \tag{28}$$

There exists only a closed form solution for *D* = 1 [39, eq. 45]. In this case, we get:

$$\Phi\_1(\overline{\gamma}, a\_{m\prime} \rho\_{m\prime} k) = \mathcal{Q}\_1 \left( \sqrt{\frac{2ka\_m^2 \gamma}{(k+1) + a\_m^2 \gamma}}, \sqrt{\frac{(k+1)\rho\_m}{(k+1) + a\_m^2 \gamma}} \right). \tag{29}$$

Thus, the average SEP in a *Rician* fading channel is with a degree of freedom of 2*D* = 2:

$$\begin{split} \overline{P}\_{\boldsymbol{\sigma},\text{ric}}(\overline{\gamma}\_{\prime}\mathbf{a}, \boldsymbol{\sigma}, k, M) &= \frac{1}{M} \Bigg[ M - 1 - Q\_1\left(0, \sqrt{\rho\_1}\right) \\ &+ \sum\_{m=1}^{M-1} \Phi\_1\left(\overline{\gamma}\_{\prime}a\_{m\prime}\rho\_{m\prime}k\right) - \sum\_{m=1}^{M-2} \Phi\_1\left(\overline{\gamma}\_{\prime}a\_{m\prime}\rho\_{m+1\prime}k\right) \Bigg]. \end{split} \tag{30}$$

A closed form solution for the integral in (28) is not known for *D* > 1. In this case, the integral has to be calculated numerically.

### *3.3.4. Nakagami-m fading*

12 Will-be-set-by-IN-TECH

*Rayleigh* distributed path gains are used to model fading channels with no line-of-sight

 − *h*2 Ω 

In UWB channels with a large bandwidth and a corresponding high temporal resolution, it is questionable, if the central limit theorem is applicable [4, 32, 59]. Nevertheless, some UWB channel measurements show a good fit to the *Rayleigh* distribution [17, 24, 51]. Using (18), the

> − *γ γ*

The integral in (23) can be solved using [40, eq. 12], given that the interval thresholds do not depend on the instantaneous SNR *γ* but on the average SNR *γ*. This is the case for an energy detection receiver with limited channel state information (only knowledge of the

*m*

1 *d*!

Combining (20) and (24) yields to the closed form solution for the energy detection receiver in

*<sup>M</sup>* <sup>−</sup> <sup>1</sup> − Q*<sup>D</sup>* (0, <sup>√</sup>*ρ*1)

*Rice* distributed path gains are used to model line-of-sight (LOS) fading channels [49, 50, 55].

<sup>−</sup>*<sup>k</sup>* <sup>−</sup> (*<sup>k</sup>* <sup>+</sup> <sup>1</sup>) *<sup>h</sup>*<sup>2</sup> Ω

Υ*<sup>D</sup>* (*γ*, *am*, *ρm*) −

 I0 2*h*

*<sup>D</sup>*−<sup>1</sup>

 *ρ<sup>m</sup>* 2

√*ρ<sup>m</sup>* exp − *γ γ* 

> *a*2 *m*

*<sup>d</sup>* + *D*−2 ∑ *d*=0

*M*−2 ∑ *m*=1

 *<sup>k</sup>* (*<sup>k</sup>* + <sup>1</sup>) Ω

1 *d*!

Υ*<sup>D</sup>* (*γ*, *am*, *ρm*+1)

 *ρm* 2 *d* 

, *h* ≥ 0, (26)

. (25)

. (24)

*a*2 *<sup>m</sup>* + 1/*γ*

, *h* ≥ 0. (21)

, *γ* ≥ 0. (22)

d*γ*. (23)

<sup>Ω</sup> exp

*<sup>γ</sup>* exp

(NLOS) [49, 50, 55]. Thus, the random variable *H* is *Rayleigh* distributed:

*fH*(*h*) = <sup>2</sup>*<sup>h</sup>*

*<sup>f</sup>*Γ(*γ*) = <sup>1</sup>

*γ* ∞ 0 Q*D am* 2*γ*,

1/*<sup>γ</sup>* <sup>+</sup> *<sup>a</sup>*<sup>2</sup>

 − *D*−2 ∑ *d*=0

*a*2 *m*

probability density function of the random SNR is:

Combining (20) and (22) we get an integral of the form:

<sup>Υ</sup>*D*(*γ*, *am*, *<sup>ρ</sup>m*) = <sup>1</sup>

average SNR). The closed form solution for Υ*<sup>D</sup>* is then:

*a*2 *<sup>m</sup>* + 1/*γ*

*a*2 *m*

> *M*

> > + *M*−1 ∑ *m*=1

 <sup>−</sup> *<sup>ρ</sup><sup>m</sup>* 2

*ρ<sup>m</sup>* 2

a *Rayleigh* fading channel with *M*-PAM:

*Pe*,*ray*(*γ*, **<sup>a</sup>**, <sup>ρ</sup>, *<sup>M</sup>*, *<sup>D</sup>*) = <sup>1</sup>

Thus, the random variable *H* is *Rice* distributed:

<sup>Ω</sup> exp

*fH* (*h*) <sup>=</sup> <sup>2</sup>*h*(*<sup>k</sup>* <sup>+</sup> <sup>1</sup>)

Υ*D*(*γ*, *am*, *ρm*) = exp

*3.3.3. Rician fading*

· exp

*3.3.2. Rayleigh fading*

The probability density function of the *Nakagami*-*m* distribution of the random path gains is related to the *χ*<sup>2</sup> distribution:

$$f\_H(h) = \frac{2h^{2m-1}}{\Gamma(m)} \left(\frac{m}{\Omega}\right)^m \exp\left(-\frac{m}{\Omega}h^2\right), \quad h \ge 0,\tag{31}$$

where *m* denotes the *Nakagami*-*m* fading parameter with *m* ∈ [1/2, ∞) and Γ denotes the Gamma function. The *Nakagami*-*m* distribution includes as special cases the one-sided normal distribution (*m* = 1/2) and the *Rayleigh*-distribution (*m* = 1). For *m* → ∞ the *Nakagami*-*m* fading channel converges to the AWGN channel. Different UWB measurement campaigns show a good fit to the *Nakagami*-*m* distribution [4, 19]. *Nakagami*-*m* distribution is also used

#### 14 Will-be-set-by-IN-TECH 14 Ultra-Wideband Radio Technologies for Communications, Localization and Sensor Applications MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture <sup>15</sup>

in the IEEE channel model 802.15.4a to model the path gains [31]. The probability density function of the random SNR Γ is with (18) and (31):

$$f\_{\Gamma}(\gamma) = \frac{\gamma^{m-1}}{\Gamma(m)} \left(\frac{m}{\overline{\gamma}}\right)^{m} \exp\left(-\frac{m}{\overline{\gamma}}\gamma\right), \quad \gamma \ge 0. \tag{32}$$

In this case we have to solve the following integral by substituting *γ* = *u*2/2:

$$\int\_0^\infty u^{2m-1} \exp\left(-\frac{mu^2}{a\overline{\gamma}}\right) \mathcal{Q}\_D\left(au, \sqrt{\rho\_m}\right) \,\mathrm{d}u.\tag{33}$$

*Eb*/*N*<sup>0</sup> in dB

MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture 15

*Eb*/*N*<sup>0</sup> in dB

0 5 10 15 20 25 10-6

The channel model used here is based on flat fading with independent and correlated fading

*γ<sup>l</sup>* = Ω*lES*/*N*0,*<sup>l</sup>*

At the SLC receiver we have a new SNR *Y*SLC at the receiver output based on the sum of the

*<sup>l</sup> ES*/*N*0,*<sup>l</sup>*

**Figure 13.** BEP in a flat fading channel with *Nakagami*-*m* distributed channel gains (OOK, *D* = 2)

gains *Hl* for all *l* diversity paths. The instantaneous SNR at the energy detector *l* is:

th detector is:

*γ<sup>l</sup>* = *h*<sup>2</sup>

*<sup>l</sup>* ). Figure 14 shows the model of multichannel receiver.

0 5 10 15 20 25 10-6

**Figure 12.** BEP in a flat fading channel with *Rayleigh* and *Rice* distributed channel gains (OOK, *D* = 2)

AWGN Rayleigh Rice, *k* = 2 Rice, *k* = 20 Rice, *k* = 200

AWGN Rayleigh Nak, *m* = 2 Nak, *m* = 20 Nak, *m* = 200

BEP

BEP

and the average SNR at the *l*

*3.4.1. Square law combining*

SNR *Yl* at the *l* detectors:

with Ω*<sup>l</sup>* = **E**(*H*<sup>2</sup>

10-4

10-2

10<sup>0</sup>

10-4

10-2

10<sup>0</sup>

(33) can be solved recursively [13]. The average symbol error rate in a fading channel with *Nakagami*-*m* distributed fading gains is:

$$\begin{split} \overline{P}\_{\varepsilon,\text{nak}}(\overline{\gamma}\_{\prime}m\_{\prime}\mathbf{a}\_{\prime}\boldsymbol{\rho}\_{\prime}\boldsymbol{M}\_{\prime}\boldsymbol{D}) &= \frac{1}{M} \Big[ \boldsymbol{M} - 1 - \mathcal{Q}\_{D}\left(0, \sqrt{\rho\_{1}}\right) \\ &+ \sum\_{\nu=1}^{M-1} \left( A\left(\overline{\gamma}\_{\prime}m\_{\prime}a\_{\nu\prime}\rho\_{\nu}\right) + \beta\_{\nu}^{\prime\prime}B\left(\overline{\gamma}\_{\prime}m\_{\prime}a\_{\nu\prime}\rho\_{\nu}\right) \right) \\ &- \sum\_{\nu=1}^{M-2} \left( A\left(\overline{\gamma}\_{\prime}m\_{\prime}a\_{\nu\prime}\rho\_{\nu-1}\right) + \beta\_{\nu}^{\prime\prime}B\left(\overline{\gamma}\_{\prime}m\_{\prime}a\_{\nu\prime}\rho\_{\nu+1}\right) \right) \Big], \end{split} \tag{34}$$

where

$$A\left(\overline{\gamma}\_{\prime}m,a,\rho\right) = \exp\left(-\frac{\beta\rho}{2}\right)\left[\beta^{m-1}L\_{m-1}\left(-\frac{\left(1-\beta\right)\rho}{2}\right) + \left(1-\beta\right)\sum\_{i=0}^{m-2}\beta^{i}L\_{i}\left(-\frac{\left(1-\beta\right)\rho}{2}\right)\right]$$

and

$$B\left(\overline{\gamma}, m, a\_{\prime}\rho\right) = \exp\left(-\frac{\rho}{2}\right) \sum\_{n=1}^{D-1} \frac{\rho\_w^n}{2^n n!} \,\_1F\_1\left(m; n+1; \frac{(1-\beta)\,\rho}{2}\right) \text{ with } \beta\_V = \frac{2m}{2m + a\_V\overline{\gamma}}.$$

*Li* is the *Laguerre* polynomial of degree *i* [18, eq. 8.970] and <sup>1</sup>*F*<sup>1</sup> is the confluent hypergeometric function [18, eq. 9.210.1].

Figure 12 shows the bit error probability in a flat fading channel with *Rayleigh* and *Rice* distributed channel gains. Figure 13 shows the bit error probability in a flat fading channel with *Nakagami*-*m* distributed channel gains.

### **3.4. Diversity reception**

Now we analyse the SEP of an energy detection receiver with diversity reception. The goal is to increase the SNR to improve its performance. Because of the architecture of the receiver, detecting only the energy of the received signal, the possibilities to improve its performance are limited and many combining techniques like maximum ratio combining (MRC) or equal gain combining (EGC) are not feasible. Thus we concentrate on square law combining (SLC) and square law selection (SLS) [29].

**Figure 12.** BEP in a flat fading channel with *Rayleigh* and *Rice* distributed channel gains (OOK, *D* = 2)

**Figure 13.** BEP in a flat fading channel with *Nakagami*-*m* distributed channel gains (OOK, *D* = 2)

The channel model used here is based on flat fading with independent and correlated fading gains *Hl* for all *l* diversity paths. The instantaneous SNR at the energy detector *l* is:

$$\gamma\_l = h\_l^2 E\_S / N\_{0J}$$

and the average SNR at the *l* th detector is:

$$\overline{\gamma}\_l = \Omega\_l E\_S / N\_{0,l}$$

with Ω*<sup>l</sup>* = **E**(*H*<sup>2</sup> *<sup>l</sup>* ). Figure 14 shows the model of multichannel receiver.

### *3.4.1. Square law combining*

14 Will-be-set-by-IN-TECH

in the IEEE channel model 802.15.4a to model the path gains [31]. The probability density

(33) can be solved recursively [13]. The average symbol error rate in a fading channel with

*<sup>M</sup>* <sup>−</sup> <sup>1</sup> − Q*<sup>D</sup>* (0, <sup>√</sup>*ρ*1)

*Li* is the *Laguerre* polynomial of degree *i* [18, eq. 8.970] and <sup>1</sup>*F*<sup>1</sup> is the confluent hypergeometric

Figure 12 shows the bit error probability in a flat fading channel with *Rayleigh* and *Rice* distributed channel gains. Figure 13 shows the bit error probability in a flat fading channel

Now we analyse the SEP of an energy detection receiver with diversity reception. The goal is to increase the SNR to improve its performance. Because of the architecture of the receiver, detecting only the energy of the received signal, the possibilities to improve its performance are limited and many combining techniques like maximum ratio combining (MRC) or equal gain combining (EGC) are not feasible. Thus we concentrate on square law combining (SLC)

Q*<sup>D</sup>* (*au*,

(*A* (*γ*, *m*, *aν*, *ρν*) + *β<sup>m</sup>*

(*<sup>A</sup>* (*γ*, *<sup>m</sup>*, *<sup>a</sup>ν*, *ρν*−1) <sup>+</sup> *<sup>β</sup><sup>m</sup>*

*<sup>m</sup>*; *<sup>n</sup>* <sup>+</sup> 1; (<sup>1</sup> <sup>−</sup> *<sup>β</sup>*) *<sup>ρ</sup>*

2

+ (1 − *β*)

<sup>−</sup> (<sup>1</sup> <sup>−</sup> *<sup>β</sup>*) *<sup>ρ</sup>* 2

, *γ* ≥ 0. (32)

<sup>√</sup>*ρm*) <sup>d</sup>*u*. (33)

*<sup>ν</sup> B* (*γ*, *m*, *aν*, *ρν*))

*m*−2 ∑ *i*=0 *βi Li* 

*<sup>ν</sup> B* (*γ*, *m*, *aν*, *ρν*+1))

with *βν* <sup>=</sup> <sup>2</sup>*<sup>m</sup>*

<sup>−</sup> (<sup>1</sup> <sup>−</sup> *<sup>β</sup>*) *<sup>ρ</sup>* 2

<sup>2</sup>*<sup>m</sup>* <sup>+</sup> *<sup>a</sup>νγ*.

, (34)

function of the random SNR Γ is with (18) and (31):

 ∞ 0

*Nakagami*-*m* distributed fading gains is:

*Pe*,*nak*(*γ*, *<sup>m</sup>*, **<sup>a</sup>**, <sup>ρ</sup>, *<sup>M</sup>*, *<sup>D</sup>*) = <sup>1</sup>

 <sup>−</sup> *βρ* 2

> − *ρ* 2

with *Nakagami*-*m* distributed channel gains.

where

and

*A* (*γ*, *m*, *a*, *ρ*) = exp

*B* (*γ*, *m*, *a*, *ρ*) = exp

function [18, eq. 9.210.1].

**3.4. Diversity reception**

and square law selection (SLS) [29].

*<sup>f</sup>*<sup>Γ</sup> (*γ*) <sup>=</sup> *<sup>γ</sup>m*−<sup>1</sup>

Γ(*m*)

*u*2*m*−<sup>1</sup> exp

*M* 

 *<sup>D</sup>*−<sup>1</sup> ∑ *n*=1

+ *M*−1 ∑ *ν*=1

− *M*−2 ∑ *ν*=1

*<sup>β</sup>m*−<sup>1</sup>*Lm*−<sup>1</sup>

*ρn w* 2*nn*! <sup>1</sup>*F*<sup>1</sup> 

In this case we have to solve the following integral by substituting *γ* = *u*2/2:

 *m γ <sup>m</sup>* exp − *m γ γ* 

 <sup>−</sup> *mu*<sup>2</sup> *aγ*

> At the SLC receiver we have a new SNR *Y*SLC at the receiver output based on the sum of the SNR *Yl* at the *l* detectors:

**Figure 14.** Channel model for multichannel receiver

$$\mathbf{Y\_{SLC}} = \sum\_{l=1}^{L} \mathbf{Y\_{l}}.$$

The new random variable *Y*SLC has a central *χ*<sup>2</sup> distribution for *a*<sup>0</sup> = 0 and a noncentral *χ*<sup>2</sup> distribution for *am* > 0 with the degree of freedom of 2*LD*. The conditional distribution function for SLC is:

$$F\_{Y|A}(y|a\_m) = \mathcal{Q}\_{LD}(a\_m\sqrt{\gamma\_{\rm SLC}}\sqrt{y})\tag{35}$$

3.4.1.1. Independent and identically distributed Rayleigh distributed channel gains

*<sup>f</sup>*ΓSLC (*γ*SLC) = *<sup>γ</sup>L*−<sup>1</sup>

channel gains we get the average SEP in I.I.D. *Rayleigh* fading channels:

*f*ΓSLC (*γ*SLC) = *c*<sup>1</sup>

*M*

+ *c*<sup>1</sup>

− *c*<sup>1</sup>

The receiver based on SLS chooses the detector with the highest received energy *Y*max:

*M*−1 ∑ *m*=1

*M*−2 ∑ *m*=1

*<sup>c</sup>*<sup>1</sup> <sup>=</sup> <sup>1</sup> Π*<sup>L</sup> <sup>l</sup>*=1*λ<sup>l</sup>*

distributed fading gains can be written as:

*3.4.2. Square law selection*

*Pe*,SLC(*γ*, **<sup>a</sup>**, ˜ρ, *<sup>M</sup>*, *<sup>D</sup>*, *<sup>L</sup>*) = <sup>1</sup>

3.4.1.2. Correlated Rayleigh distributed channel gains

density function *<sup>f</sup>*ΓSLC of <sup>Γ</sup>SLC = <sup>∑</sup>*<sup>L</sup>*

*Lγ*.

eq. 10.60]:

with

Now we derive the SEP for SLC with independent and identical distributed (i.i.d.) Rayleigh distributed channel gains *Hl*. The SNR Γ*<sup>l</sup>* is exponential distributed for all *l*. The probability

SLC

Comparing (36) with the probability density function of the SNR with *Nakagami*-*m* distributed

*Pe*,*SLC*(*γ*, **a**, ˜ρ, *M*, *D*, *L*) = *Pe*,*nak*(*Lγ*, *L*, **a**, ˜ρ, *M*, *LD*),

substituting *γ* by *Lγ*, *m* by *L*, ρ by ˜ρ and 2*D* by 2*LD*. The intervall thresholds ˜ρ are based on

In the next step, we assume correlated *Rayleigh* distributed channel gains *Hl*. In this case the probability density function of the SNR Γ is a sum of weighted exponential distributions [27,

> *L* ∑ *l*=1

, *<sup>c</sup>*2,*<sup>l</sup>* <sup>=</sup> <sup>1</sup> Π*<sup>L</sup>*

where *λ<sup>l</sup>* are the eigenvalues of the *L* × *L* covariance matrix **Σ** of the normalized received signal. Using the function Υ (24) the SEP in a fading channel with correlated *Rayleigh*

*<sup>c</sup>*2,*<sup>l</sup>* exp *<sup>γ</sup>*SLC

*γ*

*<sup>l</sup>*�=*<sup>k</sup>* (1/*λ<sup>k</sup>* <sup>−</sup> 1/*λl*)

*<sup>M</sup>* <sup>−</sup> <sup>1</sup> − Q*LD* (0, <sup>√</sup>*ρ*1)

*L* ∑ *l*=1

*L* ∑ *l*=1 ,

*λ<sup>l</sup> c*2,*l*Υ*LD* (*λl*, *am*, *ρ*˜*m*)

*λ<sup>l</sup> c*2,*l*Υ*LD* (*λl*, *am*, *ρ*˜*m*+1)

*Y*max = max(*Y*1,*Y*2, ... ,*YL*). (38)

 .

*<sup>l</sup>*=<sup>1</sup> Γ*<sup>l</sup>* with *γ* = **E**(Γ*l*) is [27, eq. 10.61]:

(*<sup>L</sup>* <sup>−</sup> <sup>1</sup>)!*γ<sup>L</sup>* exp *<sup>γ</sup>*SLC

*γ*

MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture 17

. (36)

(37)

with the non centrality parameter:

$$
\mu\_{\rm SLC} = 2\gamma\_{\rm SLC} = 2\sum\_{I=1}^{L} \gamma\_I.
$$

Thus the SEP in *L* parallel AWGN channels with different SNR *γ<sup>l</sup>* using SLC is:

$$\begin{aligned} P\_{\boldsymbol{\varepsilon}}(\gamma\_{\text{SLC}'} \mathbf{a}, \tilde{\rho}, M, D, L) &= \frac{1}{M} \bigg[ M - 1 + \sum\_{m=1}^{M-1} \mathcal{Q}\_{LD} \left( \sqrt{a\_m^2 \gamma\_{\text{SLC}'}} \sqrt{\tilde{\rho}\_m} \right) \\ &- \sum\_{m=0}^{M-2} \mathcal{Q}\_{LD} \left( \sqrt{a\_m^2 \gamma\_{\text{SLC}}} \sqrt{\tilde{\rho}\_{m+1}} \right) \bigg], \end{aligned}$$

where *ρ*˜ are the optimal interval thresholds (section 3.2.1.1), based on the new random variable *Y*SLC.

### 3.4.1.1. Independent and identically distributed Rayleigh distributed channel gains

Now we derive the SEP for SLC with independent and identical distributed (i.i.d.) Rayleigh distributed channel gains *Hl*. The SNR Γ*<sup>l</sup>* is exponential distributed for all *l*. The probability density function *<sup>f</sup>*ΓSLC of <sup>Γ</sup>SLC = <sup>∑</sup>*<sup>L</sup> <sup>l</sup>*=<sup>1</sup> Γ*<sup>l</sup>* with *γ* = **E**(Γ*l*) is [27, eq. 10.61]:

$$f\_{\Gamma \text{sLC}}(\gamma\_{\text{SLC}}) = \frac{\gamma\_{\text{SLC}}^{L-1}}{(L-1)!\overline{\gamma}^L} \exp\left(\frac{\gamma\_{\text{SLC}}}{\overline{\gamma}}\right) \,. \tag{36}$$

Comparing (36) with the probability density function of the SNR with *Nakagami*-*m* distributed channel gains we get the average SEP in I.I.D. *Rayleigh* fading channels:

$$\overline{\mathcal{P}}\_{\mathfrak{e},\operatorname{SLC}}(\overline{\gamma}\_{\prime}\mathtt{a}, \widetilde{\mathfrak{p}}, \operatorname{M}, \operatorname{D}, \operatorname{L}) = \overline{\mathcal{P}}\_{\mathfrak{e},\operatorname{nak}}(L\overline{\gamma}\_{\prime}\mathtt{L}, \operatorname{a}, \widetilde{\mathfrak{p}}, \operatorname{M}, \operatorname{L}D)\_{\prime}$$

substituting *γ* by *Lγ*, *m* by *L*, ρ by ˜ρ and 2*D* by 2*LD*. The intervall thresholds ˜ρ are based on *Lγ*.

### 3.4.1.2. Correlated Rayleigh distributed channel gains

In the next step, we assume correlated *Rayleigh* distributed channel gains *Hl*. In this case the probability density function of the SNR Γ is a sum of weighted exponential distributions [27, eq. 10.60]:

$$f\_{\rm{TLC}}(\gamma\_{\rm{SLC}}) = c\_1 \sum\_{l=1}^{L} c\_{2,l} \exp\left(\frac{\gamma\_{\rm{SLC}}}{\overline{\gamma}}\right) \tag{37}$$

with

16 Will-be-set-by-IN-TECH

*h*1

*h*2

*h*3

*hL*

*Y*SLC =

*FY*|*A*(*y*|*am*) = <sup>Q</sup>*LD*(*am*

Thus the SEP in *L* parallel AWGN channels with different SNR *γ<sup>l</sup>* using SLC is:

*M* 

> − *M*−2 ∑ *m*=0

*μ*SLC = 2*γ*SLC = 2

*M* − 1 +

where *ρ*˜ are the optimal interval thresholds (section 3.2.1.1), based on the new random variable

**Figure 14.** Channel model for multichannel receiver

*Pe*(*γ*SLC, **<sup>a</sup>**, ˜ρ, *<sup>M</sup>*, *<sup>D</sup>*, *<sup>L</sup>*) = <sup>1</sup>

function for SLC is:

*Y*SLC.

with the non centrality parameter:

*s*(*t*)

*τ*2

*n*1

*n*2

*n*3

multichannel

√*y*) (35)

*ρ*˜*<sup>m</sup>* 

 ,

*<sup>ρ</sup>*˜*m*+<sup>1</sup>

 receiver

*nL*

<sup>√</sup>*γ*SLC,

*L* ∑ *l*=1 *γl*.

*M*−1 ∑ *m*=1

Q*LD*

Q*LD*

 *a*2 *<sup>m</sup>γ*SLC,

 *a*2 *<sup>m</sup>γ*SLC,

*τ*1

*τ*3

*τL*

*L* ∑ *l*=1 *Yl*.

The new random variable *Y*SLC has a central *χ*<sup>2</sup> distribution for *a*<sup>0</sup> = 0 and a noncentral *χ*<sup>2</sup> distribution for *am* > 0 with the degree of freedom of 2*LD*. The conditional distribution

$$c\_1 = \frac{1}{\Pi\_{l=1}^L \lambda\_l}, \ c\_{2,l} = \frac{1}{\Pi\_{l \neq k}^L \left(1/\lambda\_k - 1/\lambda\_l\right)}.$$

where *λ<sup>l</sup>* are the eigenvalues of the *L* × *L* covariance matrix **Σ** of the normalized received signal. Using the function Υ (24) the SEP in a fading channel with correlated *Rayleigh* distributed fading gains can be written as:

$$\begin{split} \overline{P}\_{c,\text{SLC}}(\overline{\gamma}, \mathbf{a}, \bar{\rho}, M\_{\prime}D\_{\prime}L) &= \frac{1}{M} \Big[ M - 1 - \mathcal{Q}\_{LD} \left( 0, \sqrt{\rho\_{1}} \right) \\ &+ c\_{1} \sum\_{m=1}^{M-1} \sum\_{l=1}^{L} \lambda\_{l} c\_{2,l} \mathbf{Y}\_{LD} \left( \lambda\_{l}, a\_{m\prime} \bar{\rho}\_{m} \right) \\ &- c\_{1} \sum\_{m=1}^{M-2} \sum\_{l=1}^{L} \lambda\_{l} c\_{2,l} \mathbf{Y}\_{LD} \left( \lambda\_{l}, a\_{m\prime} \bar{\rho}\_{m+1} \right) \Big]. \end{split}$$

### *3.4.2. Square law selection*

The receiver based on SLS chooses the detector with the highest received energy *Y*max:

$$Y\_{\text{max}} = \max(Y\_1, Y\_2, \dots, Y\_L). \tag{38}$$

#### 18 Will-be-set-by-IN-TECH 18 Ultra-Wideband Radio Technologies for Communications, Localization and Sensor Applications MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture <sup>19</sup>

Thus, this receiver collects only a fraction of the total received energy. Using (9) and (38), the SEP with SLS in the AWGN channel is:

$$P\_{\mathbf{f}}(\gamma, \mathbf{a}, \boldsymbol{\rho}, D, L) = 1 - P\_{\mathbf{f}}(\gamma, \mathbf{a}, \boldsymbol{\rho}, D)$$

$$= 1 - \sum\_{m=0}^{M-1} \mathbb{P}(\rho\_m < Y\_{\text{max}} \le \rho\_{m+1} | A = a\_m) \mathbb{P}(A = a\_m). \tag{39}$$

function Υ (24):

*Pe*(*γ*, **<sup>a</sup>**, <sup>ρ</sup>, *<sup>D</sup>*, *<sup>L</sup>*) = <sup>1</sup>

BEP

**Figure 15.** BEP in the *Rayleigh* channel with SLC

BEP

**Figure 16.** BEP in the *Rayleigh* channel with SLS

10-4

10-2

10<sup>0</sup>

10-4

10-2

10<sup>0</sup>

*M* 

> + *M*−2 ∑ *m*=1

> − *M*−1 ∑ *m*=1

than the SLS based receiver. The SLS based receiver chooses only the dominant path.

AWGN Rayleigh Rayleigh, *L* = 2 Rayleigh, *L* = 4 Rayleigh, *L* = 8

AWGN Rayleigh Rayleigh, *L* = 2 Rayleigh, *L* = 4 Rayleigh, *L* = 8

*M* + 1 +

Figure 15 shows the gain that can be achieved using SLC. Figure 16 shows the gain that can be achieved using SLS. The SLC based receiver can collect more energy, but also more noise

*Eb*/*N*<sup>0</sup> in dB

*Eb*/*N*<sup>0</sup> in dB



 1 − *QD*  *a*0 *γ*, <sup>√</sup>*ρ*<sup>1</sup> *<sup>L</sup>*

MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture 19

(1 − Υ*D*(*γ*, *am*, *ρm*+1))

(1 − Υ*D*(*γ*, *am*, *ρm*))

*L*

*L* .

In order to calculate (39), we calculate the probability that *Y*max is in the interval (*ρm*, *ρm*+1]:

$$\begin{aligned} \mathbb{P}(\rho\_m < Y\_{\text{max}} \le \rho\_{m+1} | A = a\_m) \\ = \mathbb{P}\left(\bigcap\_{l=1}^L \{Y\_l < \rho\_{m+1} | A = a\_m\}\right) - \mathbb{P}\left(\bigcap\_{l=1}^L \{Y\_l < \rho\_m | A = a\_m\}\right). \end{aligned} \tag{40}$$

If all received energies *Yl* are independent, (40) reduces to:

$$\mathbb{P}\left(\bigcap\_{l=1}^{L}\{Y\_{l} < \rho\_{m} \,|\, A = a\_{m}\}\right) = \prod\_{l=1}^{L} \mathbb{P}\left(Y\_{l} < \rho\_{m} \,|\, A = a\_{m}\right). \tag{41}$$

Using the distribution function (14), the probability can be written as:

$$\mathbb{P}\left(Y\_{l} < \rho\_{\mathfrak{m}} \,|\, A = a\_{\mathfrak{m}}\right) = 1 - Q\_{\mathbb{D}}\left(a\_{\mathfrak{m}}\sqrt{\gamma\_{\prime}}\sqrt{\rho\_{\mathfrak{m}}}\right). \tag{42}$$

Combining (40), (41) and (42) and assume **P**(*A* = *am*) = 1/*M*, (39) can be written as:

$$P\_{\varepsilon}(\gamma, \mathbf{a}, \boldsymbol{\rho}, D, L) = 1 - \frac{1}{M} \sum\_{m=0}^{M-1} \left[ \prod\_{l=1}^{L} \left( 1 - Q\_{D} \left( a\_{m} \sqrt{\gamma\_{l}} \sqrt{\rho\_{m+1, l}} \right) \right) \right. $$

$$- \prod\_{l=1}^{L} \left( 1 - Q\_{D} \left( a\_{m} \sqrt{\gamma\_{l}} \sqrt{\rho\_{m, l}} \right) \right) \Big|\_{} \tag{43}$$

where *ρm*,*<sup>l</sup>* denotes the optimal interval thresholds which are based on the SNR *γl*. If all received energies *Yl* are independent and identical distributed, (41) reduces to:

$$\prod\_{l=1}^{L} \mathbb{P}\left(Y\_l < \rho\_m | A = a\_m\right) = \left(1 - Q\_D\left(a\_m \sqrt{\gamma} \sqrt{\rho\_m}\right)\right)^L. \tag{44}$$

In this case, (43) reduces to:

$$\begin{aligned} P\_{\boldsymbol{\varepsilon}}(\gamma, \mathbf{a}, \boldsymbol{\rho}, D, L) &= 1 - \frac{1}{M} \sum\_{m=0}^{M-1} \left[ (1 - Q\_D \left( a\_m \sqrt{\gamma\_\prime}, \sqrt{\rho\_{m+1}} \right))^L \right] \\ & \quad - \left( 1 - Q\_D \left( a\_m \sqrt{\gamma\_\prime}, \sqrt{\rho\_m} \right) \right)^L \right]. \end{aligned}$$

If the SNR is independent and identically *Rayleigh* distributed with *γ* = **E**(Γ*l*) for all *l*, the SEP in a fading channel with i.i.d. *Rayleigh* distributed channel gains can be written using the function Υ (24):

18 Will-be-set-by-IN-TECH

Thus, this receiver collects only a fraction of the total received energy. Using (9) and (38), the

In order to calculate (39), we calculate the probability that *Y*max is in the interval (*ρm*, *ρm*+1]:

 − **P**

 = *L* ∏ *l*=1

 *L*

*l*=1

√*γ*,

 *am* <sup>√</sup>*γl*,

√*γ*,

√*γ*,

√*γ*,

<sup>√</sup>*ρm*+1))*<sup>L</sup>*

<sup>√</sup>*ρm*))*<sup>L</sup>* .

**P**(*ρ<sup>m</sup>* < *Y*max ≤ *ρm*+1|*A* = *am*)**P**(*A* = *am*). (39)

{*Yl* < *ρm*|*A* = *am*}

√*ρm*). (42)

√*ρm*))*<sup>L</sup>* . (44)

, (43)

**P** (*Yl* < *ρm*| *A* = *am*). (41)

<sup>√</sup>*ρm*+1,*<sup>l</sup>*

<sup>√</sup>*ρm*,*<sup>l</sup>*

. (40)

SEP with SLS in the AWGN channel is:

= **P**

 *L*

**P** *L*

*l*=1

*l*=1

*Pe*(*γ*, **<sup>a</sup>**, <sup>ρ</sup>, *<sup>D</sup>*, *<sup>L</sup>*) = <sup>1</sup> <sup>−</sup> <sup>1</sup>

*L* ∏ *l*=1

*Pe*(*γ*, **<sup>a</sup>**, <sup>ρ</sup>, *<sup>D</sup>*, *<sup>L</sup>*) = <sup>1</sup> <sup>−</sup> <sup>1</sup>

In this case, (43) reduces to:

*Pe*(*γ*, **a**, ρ, *D*, *L*) = 1 − *Pc*(*γ*, **a**, ρ, *D*)

= 1 −

**P**(*ρ<sup>m</sup>* < *Y*max ≤ *ρm*+1|*A* = *am*)

If all received energies *Yl* are independent, (40) reduces to:

*M*−1 ∑ *m*=0

{*Yl* < *ρm*+1|*A* = *am*}

{*Yl* < *ρm*|*A* = *am*}

Using the distribution function (14), the probability can be written as:

**P** (*Yl* < *ρm*| *A* = *am*) = 1 − *QD* (*am*

*M*

received energies *Yl* are independent and identical distributed, (41) reduces to:

**P** (*Yl* < *ρm*|*A* = *am*) = (1 − *QD* (*am*

*M*

*M*−1 ∑ *m*=0 

If the SNR is independent and identically *Rayleigh* distributed with *γ* = **E**(Γ*l*) for all *l*, the SEP in a fading channel with i.i.d. *Rayleigh* distributed channel gains can be written using the

(1 − *QD* (*am*

− (1 − *QD* (*am*

Combining (40), (41) and (42) and assume **P**(*A* = *am*) = 1/*M*, (39) can be written as:

*M*−1 ∑ *m*=0

 *L* ∏ *l*=1 1 − *QD am* <sup>√</sup>*γl*,

− *L* ∏ *l*=1 1 − *QD*

where *ρm*,*<sup>l</sup>* denotes the optimal interval thresholds which are based on the SNR *γl*. If all

$$\begin{split} P\_{\varepsilon}(\overline{\gamma}, \mathbf{a}, \boldsymbol{\rho}, \boldsymbol{D}, \boldsymbol{L}) &= \frac{1}{M} \Big[ M + 1 + \left( 1 - Q\_{D} \left( a\_{0} \sqrt{\overline{\gamma}}, \sqrt{\rho\_{1}} \right) \right)^{L} \\ &\quad + \sum\_{m=1}^{M-2} \left( 1 - \mathbf{Y}\_{D}(\overline{\gamma}, a\_{m}, \rho\_{m+1}) \right)^{L} \\ &\quad - \sum\_{m=1}^{M-1} \left( 1 - \mathbf{Y}\_{D}(\overline{\gamma}, a\_{m}, \rho\_{m}) \right)^{L} \Big]. \end{split}$$

Figure 15 shows the gain that can be achieved using SLC. Figure 16 shows the gain that can be achieved using SLS. The SLC based receiver can collect more energy, but also more noise than the SLS based receiver. The SLS based receiver chooses only the dominant path.

**Figure 15.** BEP in the *Rayleigh* channel with SLC

**Figure 16.** BEP in the *Rayleigh* channel with SLS

### **3.5. Frequency selective fading channel**

Now we focus on the frequency selectivity of a channel and the effect on the SEP for an energy detection receiver. The received signal R is in the baseband equivalent model:

decomposition of **Σ***<sup>S</sup>* in (46) leads to

with *Gl* representing the elements of **G** and *G*<sup>∗</sup>

*<sup>l</sup>* = *GlG*<sup>∗</sup>

*fV*(*v*) =

*D* ∑ *i*=1

> *f*Γ� � *γ*� �

> > �−<sup>1</sup> ⎞ <sup>⎠</sup> *<sup>λ</sup>D*−<sup>2</sup> *j*

**E**(*V*) =

Introducing a change of variable in (48) yields:

Using the relative SNR *γ*� = *<sup>γ</sup>*

the following integral:

2*N*0*B*

*D* ∑ *j*=1 ⎛ <sup>⎝</sup>∏ *k*�=*j*

� *λ<sup>j</sup>* − *λ<sup>k</sup>*

*<sup>l</sup>* = |*Gl*|

*D* ∑ *j*=1

The average SNR *γ* is based on the expectation of the random variable *V*:

Because all random variables *Gi* are independent **E**(*V*) can be written as:

*λi***E**(|*Gi*|

⎛ <sup>⎝</sup>∏ *k*�=*j*

� *λ<sup>j</sup>* − *λ<sup>k</sup>*

*<sup>γ</sup>* <sup>=</sup> **<sup>E</sup>**(*V*)

*D* ∑ *i*=1

*λ<sup>i</sup>* with **E**(|*Gi*|

*f*Γ(*γ*) = 2*N*0*B fV* (2*N*0*Bγ*). (50)

�

<sup>2</sup>) = 2

*<sup>γ</sup>* , (50) can be expressed as:

= 2*N*0*Bγ fV*

The SEP in a fading channel can be calculated using (20). Combining (20) and (51) yields to

� ∞ 0 Q*D*

�

2*N*0*Bγγ*�

�� *a*2 *mγγ*� , <sup>√</sup>*ρ<sup>m</sup>* � *e* <sup>−</sup> <sup>2</sup>*N*0*<sup>B</sup> <sup>b</sup>λ<sup>j</sup> γγ*�

variable *V* is:

other. Therefore

19.147]:

The random variable *G*�

**S***H***S** = **S**�*H***U***H***ΛUS**�

with **Λ** being a diagonal matrix containing the eigenvalues *λ*1,..., *λ<sup>L</sup>* of **Σ***<sup>s</sup>* and the rows of

that **G** and **H** are equally distributed because the rows of **U** are orthonormal among each

**G** ∼ CN *<sup>L</sup>* (**0**, **Σ***s*).

That is a special case of the central *χ*<sup>2</sup> distribution which is equivalent to an exponential distribution. As a linear combination of independent and identically exponential distributed variates, *V* is general *Gamma* or general *Erlang* distributed with the PDF given as [21, eq.

*L* ∑ *l*=1

�−<sup>1</sup> ⎞ <sup>⎠</sup> *<sup>λ</sup>D*−<sup>2</sup> *<sup>j</sup> e* <sup>−</sup> *<sup>v</sup>*

*λlGlG*<sup>∗</sup>

MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture 21

, the random

*<sup>l</sup>* (47)

*<sup>λ</sup><sup>j</sup>* . (48)

. (51)

d*γ*� .

*<sup>l</sup>* denoting the complex conjugate of *Gl*. Note

<sup>4</sup>*N*0*B*. (49)

<sup>2</sup>) = 2.

<sup>2</sup> is central *χ*<sup>2</sup> distributed with two degrees of freedom.

unitary matrix **U** being the corresponding eigenvectors. Substituting **G** = **US**�

*V* = **S***H***S** = **G***H***ΛG** =

$$\mathcal{R} = \underbrace{xH}\_{\mathbf{S}} + W\_{\mathbf{M}}$$

where H = (*H*1,..., *HL*) *<sup>T</sup>* denotes the random channel impulse response, circular symmetric complex white Gaussian noise W ∼ CN *<sup>L</sup>* � **0**, *σ*<sup>2</sup> *n***I** � and x denotes the convolution matrix, containing shifted versions of the transmitted signal:

$$\mathbf{x} = \begin{pmatrix} s\_1 & 0 & \dots & 0 \\ s\_2 & s\_1 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ s\_L & s\_{L-1} & \dots & s\_1 \end{pmatrix}.$$

The effect of intersymbol interference has not been taken into account by choosing a gap between symbols larger than the channel delay spread. Using the quadratic form of S the SNR Γ at the receiver can be written as:

$$
\Gamma = \frac{E\_R}{N\_0} = \frac{\mathbf{S}^H \mathbf{S}}{4N\_0 B}.
$$

### *3.5.1. NLOS fading channel*

In an environment with no line of sight between transmitter and receiver, the expectation of the random samples of the channel impulse response H is zero:

$$H \sim \mathcal{CN}\_D(\mathbf{0}, \Sigma) \quad \text{with } \Sigma(i, j) = h\_i h\_j e^{-\frac{|i - j|}{\delta}}.$$

where *β* > 0 describes the correlation of the elements in the covariance matrix **Σ**. If the samples of the channel impulse response are uncorrelated, **Σ** is a diagonal matrix. The expectation the noise free received signal is also zero:

$$\mathbf{S} = x H \sim \mathcal{CN}\_D \left( \mathbf{0} / \Sigma\_s \right) \quad \text{with } \Sigma\_s = x \Sigma x^H.$$

Let **Σ** 1 2 *<sup>s</sup>* denote a matrix that fulfils **Σ** 1 2 *s* **Σ** 1 2 *<sup>s</sup>* = **Σ***s*, then a whitened vector **S**� ∼ CN *<sup>L</sup>* (**0**,**I**) is defined as

$$
\Sigma\_s^{-\frac{1}{2}} \mathbf{S} = \mathbf{S}'.\tag{45}
$$

Note that the existence of **Σ** 1 2 *<sup>s</sup>* is guaranteed for any positive definite matrix **Σ***s*. To calculate the distribution of instantaneous SNR Γ, we need to analyse the distribution of **S***H***S** first. Using (45) yields

$$\mathbf{S}^{H}\mathbf{S} = \mathbf{S}^{\prime H}\boldsymbol{\Sigma}\_{s}\mathbf{S}^{\prime}.\tag{46}$$

Using eigenvalue decomposition and special properties of the central *χ*2-distribution, a closed form expression for the PDF of **r***H***r** can be found [21]. Performing an eigenvalue decomposition of **Σ***<sup>S</sup>* in (46) leads to

20 Will-be-set-by-IN-TECH

Now we focus on the frequency selectivity of a channel and the effect on the SEP for an energy

+W,

*<sup>T</sup>* denotes the random channel impulse response, circular symmetric

⎞

⎟⎟⎟⎠ .

<sup>−</sup> <sup>|</sup>*i*−*j*<sup>|</sup> *<sup>β</sup>* ,

*<sup>s</sup>* = **Σ***s*, then a whitened vector **S**� ∼ CN *<sup>L</sup>* (**0**,**I**) is

*<sup>s</sup>* is guaranteed for any positive definite matrix **Σ***s*. To calculate the

. (45)

. (46)

� and x denotes the convolution matrix,

<sup>R</sup> <sup>=</sup> ���� xH S

> � **0**, *σ*<sup>2</sup> *n***I**

*s*<sup>1</sup> 0 ... 0 *s*<sup>2</sup> *s*<sup>1</sup> ... 0

*sL sL*−<sup>1</sup> ... *s*<sup>1</sup>

<sup>=</sup> <sup>S</sup>*H*<sup>S</sup> <sup>4</sup>*N*0*B*.

The effect of intersymbol interference has not been taken into account by choosing a gap between symbols larger than the channel delay spread. Using the quadratic form of S the

In an environment with no line of sight between transmitter and receiver, the expectation of

where *β* > 0 describes the correlation of the elements in the covariance matrix **Σ**. If the samples of the channel impulse response are uncorrelated, **Σ** is a diagonal matrix. The

<sup>S</sup> <sup>=</sup> xH ∼ CN *<sup>D</sup>* (0, **<sup>Σ</sup>***s*) with **<sup>Σ</sup>***<sup>s</sup>* <sup>=</sup> <sup>x</sup>**Σ**x*H*.

distribution of instantaneous SNR Γ, we need to analyse the distribution of **S***H***S** first. Using

**S***H***S** = **S**�*H***Σ***s***S**�

Using eigenvalue decomposition and special properties of the central *χ*2-distribution, a closed form expression for the PDF of **r***H***r** can be found [21]. Performing an eigenvalue

> **<sup>Σ</sup>**<sup>−</sup> <sup>1</sup> <sup>2</sup> *<sup>s</sup>* **S** = **S**�

H ∼ CN *<sup>D</sup>* (0, **Σ**) with **Σ**(*i*, *j*) = *hihje*

detection receiver. The received signal R is in the baseband equivalent model:

**x** =

the random samples of the channel impulse response H is zero:

expectation the noise free received signal is also zero:

1 2

*<sup>s</sup>* denote a matrix that fulfils **Σ**

⎛

⎜⎜⎜⎝

. . . . . . ... . . .

<sup>Γ</sup> <sup>=</sup> *ER N*0

**3.5. Frequency selective fading channel**

complex white Gaussian noise W ∼ CN *<sup>L</sup>*

SNR Γ at the receiver can be written as:

containing shifted versions of the transmitted signal:

where H = (*H*1,..., *HL*)

*3.5.1. NLOS fading channel*

Let **Σ** 1 2

defined as

(45) yields

Note that the existence of **Σ**

$$\mathbf{S}^{H}\mathbf{S} = \mathbf{S}'^{H}\mathbf{U}^{H}\mathbf{A}\mathbf{U}\mathbf{S}'$$

with **Λ** being a diagonal matrix containing the eigenvalues *λ*1,..., *λ<sup>L</sup>* of **Σ***<sup>s</sup>* and the rows of unitary matrix **U** being the corresponding eigenvectors. Substituting **G** = **US**� , the random variable *V* is:

$$V = \mathbf{S}^H \mathbf{S} = \mathbf{G}^H \mathbf{A} \mathbf{G} = \sum\_{l=1}^L \lambda\_l \mathbf{G}\_l \mathbf{G}\_l^\* \tag{47}$$

with *Gl* representing the elements of **G** and *G*<sup>∗</sup> *<sup>l</sup>* denoting the complex conjugate of *Gl*. Note that **G** and **H** are equally distributed because the rows of **U** are orthonormal among each other. Therefore

$$\mathbf{G} \sim \mathcal{CN}\_L \left( \mathbf{0}, \Sigma\_{\mathbf{s}} \right).$$

The random variable *G*� *<sup>l</sup>* = *GlG*<sup>∗</sup> *<sup>l</sup>* = |*Gl*| <sup>2</sup> is central *χ*<sup>2</sup> distributed with two degrees of freedom. That is a special case of the central *χ*<sup>2</sup> distribution which is equivalent to an exponential distribution. As a linear combination of independent and identically exponential distributed variates, *V* is general *Gamma* or general *Erlang* distributed with the PDF given as [21, eq. 19.147]:

$$f\_V(v) = \sum\_{j=1}^D \left( \prod\_{k \neq j} \left( \lambda\_j - \lambda\_k \right)^{-1} \right) \lambda\_j^{D-2} e^{-\frac{v}{\lambda\_j}}.\tag{48}$$

The average SNR *γ* is based on the expectation of the random variable *V*:

$$
\overline{\gamma} = \frac{\mathbb{E}(V)}{4N\_0 B}.\tag{49}
$$

Because all random variables *Gi* are independent **E**(*V*) can be written as:

$$\mathbb{E}(V) = \sum\_{i=1}^{D} \lambda\_i \mathbb{E}(|G\_i|^2) = 2 \sum\_{i=1}^{D} \lambda\_i \quad \text{with } \mathbb{E}(|G\_i|^2) = 2.$$

Introducing a change of variable in (48) yields:

$$f\_{\Gamma}(\gamma) = 2\mathrm{N}\_{0}\mathcal{B} \, f\_{V} \left( 2\mathrm{N}\_{0}\mathcal{B}\gamma \right). \tag{50}$$

Using the relative SNR *γ*� = *<sup>γ</sup> <sup>γ</sup>* , (50) can be expressed as:

$$f\_{\Gamma'}\left(\gamma'\right) = 2N\_0 B \overline{\gamma} \, f\_V\left(2N\_0 B \overline{\gamma} \gamma'\right). \tag{51}$$

The SEP in a fading channel can be calculated using (20). Combining (20) and (51) yields to the following integral:

$$2N\_0 B \sum\_{j=1}^D \left( \prod\_{k \neq j} \left( \lambda\_j - \lambda\_k \right)^{-1} \right) \lambda\_j^{D-2} \int\_0^\infty \mathcal{Q}\_D \left( \sqrt{a\_m^2 \overline{\gamma} \gamma'}, \sqrt{\rho\_m} \right) e^{-\frac{2N\_0 B}{\hbar \lambda\_j} \overline{\gamma} \gamma'} \, \mathrm{d} \gamma' \, \mathrm{d} \overline{\gamma}$$

#### 22 Will-be-set-by-IN-TECH 22 Ultra-Wideband Radio Technologies for Communications, Localization and Sensor Applications MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture <sup>23</sup>

In the case of partial CSI, the integral can be solved using [40, eq. 12] with adequate substitutions:

*3.5.2. LOS fading channel*

*Erlang* distribution for *V* in section 3.5.1.

with the mean **M** = (*M*1,..., *MD*) with *Mi* =

realise an efficient and low complex interference mitigation.

**4.1. Interference robustness of energy detection**

the MARCUM-Q-function in (20).

normal distribution

UWB system is shown.

not guaranteed.

In a channel model with LOS, **H** is expected to have a non-zero mean vector **M**. Therefore *V* is equal in distribution to a linear combination of noncentral *χ*<sup>2</sup> distributed variates. Unfortunately, there is no such convenient correspondence for that distribution as the general

MIRA – Physical Layer Optimisation for the Multiband Impulse Radio UWB Architecture 23

Even though the PDF of a linear combination of noncentral *χ*<sup>2</sup> distributed variates is given in [48], there is no known closed-form expression for the integral of the product of this PDF and

The average SEP may still be computed using a discrete approximation of the PDF of *V* (histogram). This histogram can be generated with arbitrary precision by sampling the PDF given in [48] or by performing a Monte-Carlo-simulation based on the distribution of **H**. For the Monte-Carlo-simulations the samples of the LOS channel impulse response **H** follow a

**H** ∼ CN *<sup>L</sup>* (**M**, **Σh**)

As the unlicensed MIR-UWB system has no exclusive frequency range there is an increased interference potential from present and from future radio systems operating in the same frequency range. Hence, the performance of the MIR-UWB system can be reduced and a reliable communication cannot be guaranteed at any time. For this reason it is imperative to

Section 4 is structured as follows: In section 4.1 an analysis of the interference robustness of an energy detection receiver is presented. This allows the identification of suitable MIR-UWB system parameters which can be configured preferably robust against interferences before initial operation of the MIR-UWB system. The following section 4.2 deals with coexistence-based approaches which are focused on an efficient and adaptive interference mitigation with low complexity. As the mitigation of narrowband interference (NBI) is a crucial issue of the MIR-UWB section 4.3 analyses the non-linear Teager-Kaiser (TK) operation. Thereby, the potential to mitigate NBI and to integrate the TK operation into the existing MIR

A basic issue of the MIR-UWB's energy detection receiver is its high sensitivity with respect to interferences passing the analogue front-end. A significant reduction of the instantaneous Signal-to-Interference-and Noise Ratio (SINR) can occur so that a reliable communication is

For this reason it is required to investigate the interference robustness of an OOK and BPPM specific energy detection [7, 9]. The analysis bases on an analytical investigation of the interference robustness of an energy detector within an arbitrary but fixed subband. Thereby, dependencies between system- and interference specific parameters can be identified which

**4. Interference investigations for non-coherent multiband UWB**

*<sup>κ</sup>*

*<sup>κ</sup>*+<sup>1</sup> <sup>∀</sup> *<sup>i</sup>* and the covariance matrix **<sup>Σ</sup><sup>h</sup>** <sup>=</sup> <sup>1</sup>

*<sup>κ</sup>*+<sup>1</sup> **I**.

$$\Theta\_{D}(\overline{\gamma}\_{i}N\_{0},\lambda\_{i}B,a\_{m},\rho\_{m}) = \frac{1}{\overline{\gamma}}\sum\_{j=1}^{D} \left(\prod\_{k\neq j} \left(\lambda\_{j} - \lambda\_{k}\right)^{-1}\right) \lambda\_{j}^{D-2} \mathbf{e}^{-\frac{\rho\_{m}}{2}} \left\{ \left(1 + \frac{2N\_{0}B}{a\_{m}^{2}\lambda\_{j}}\right)^{D-1} \right.$$

$$\cdot \left[\exp\left(\frac{\rho\_{m}}{2}\frac{1}{1 + \frac{2N\_{0}B}{a\_{m}^{2}\lambda\_{j}}}\right) - \sum\_{n=0}^{D-2} \frac{1}{n!} \left(\frac{\rho\_{m}}{2}\frac{1}{1 + \frac{2N\_{0}B}{a\_{n}^{2}\lambda\_{j}}}\right)^{n}\right] + \sum\_{n=0}^{D-2} \frac{1}{n!} \left(\frac{\rho\_{m}}{2}\right)^{n}\right\}. \tag{52}$$

Using the function Θ*<sup>D</sup>* (52), the average SEP in a frequency selective fading channel can be expressed in closed form [3]:

$$\begin{split} \overline{P}\_{\epsilon}(\overline{\gamma}\_{\prime}N\_{0},B,a,\rho,\lambda,M,D) &= \frac{1}{M} \Bigg[ M - 1 + \mathcal{Q}\_{D}\left(0, \sqrt{\rho\_{1}}\right) \\ &+ \sum\_{m=1}^{M-1} \Theta\_{D}(\overline{\gamma}\_{\prime}N\_{0},\lambda,B,a\_{m},\rho\_{m}) \\ &- \sum\_{m=1}^{M-2} \Theta\_{D}(\overline{\gamma}\_{\prime}N\_{0},\lambda,B,a\_{m},\rho\_{m+1}) \Bigg]. \end{split}$$

This equation is only valid if two constraints are met:


If there exist two or more identical eigenvalues, one might rearrange (47) and decrease *D* used in the subsequent calculations to meet the second constraint. Figure 17 shows the BEP in the frequency selective fading channel with i.i.d. and correlated channel gains.

**Figure 17.** BEP for OOK in the frequency selective fading channel

### *3.5.2. LOS fading channel*

22 Will-be-set-by-IN-TECH

In the case of partial CSI, the integral can be solved using [40, eq. 12] with adequate

�−<sup>1</sup> ⎞ <sup>⎠</sup> *<sup>λ</sup>D*−<sup>2</sup> *<sup>j</sup>* <sup>e</sup><sup>−</sup> *<sup>ρ</sup><sup>m</sup>* 2 ⎧ ⎨ ⎩

> 1 1 + <sup>2</sup>*N*0*<sup>B</sup> a*2 *<sup>m</sup>λ<sup>j</sup>*

*<sup>M</sup>* <sup>−</sup> <sup>1</sup> <sup>+</sup> <sup>Q</sup>*<sup>D</sup>* (0, <sup>√</sup>*ρ*1)

⎞ ⎠

Θ*D*(*γ*, *N*0,λ, *B*, *am*, *ρm*)

Θ*D*(*γ*, *N*0,λ, *B*, *am*, *ρm*+1)

*<sup>n</sup>*⎤ ⎦ +

� 1 +

> *D*−2 ∑ *n*=0

2*N*0*B a*2 *<sup>m</sup>λ<sup>j</sup>*

1 *n*! � *ρm* 2 �*n* ⎫ ⎬ ⎭

> � .

�*D*−<sup>1</sup>

. (52)

substitutions:

<sup>Θ</sup>*D*(*γ*, *<sup>N</sup>*0, <sup>λ</sup>, *<sup>B</sup>*, *am*, *<sup>ρ</sup>m*) = <sup>1</sup>

⎛ ⎝ *ρm* 2

· ⎡ ⎣exp

expressed in closed form [3]:

*γ*

1 1 + <sup>2</sup>*N*0*<sup>B</sup> a*2 *<sup>m</sup>λ<sup>j</sup>*

*Pe*(*γ*, *<sup>N</sup>*0, *<sup>B</sup>*, <sup>a</sup>, <sup>ρ</sup>,λ, *<sup>M</sup>*, *<sup>D</sup>*) = <sup>1</sup>

This equation is only valid if two constraints are met:

2. all eigenvalues *λ<sup>j</sup>* are pairwise disjunct.

BEP

10-6

**Figure 17.** BEP for OOK in the frequency selective fading channel

10-4

10-2

100

*D* ∑ *j*=1 ⎛ <sup>⎝</sup>∏ *k*�=*j*

> ⎞ ⎠ −

� *λ<sup>j</sup>* − *λ<sup>k</sup>*

*D*−2 ∑ *n*=0

1 *n*!

Using the function Θ*<sup>D</sup>* (52), the average SEP in a frequency selective fading channel can be

+ *M*−1 ∑ *m*=1

− *M*−2 ∑ *m*=1

If there exist two or more identical eigenvalues, one might rearrange (47) and decrease *D* used in the subsequent calculations to meet the second constraint. Figure 17 shows the BEP in the

*Eb*/*N*<sup>0</sup> in dB

0 5 10 15 20 25 <sup>10</sup>-8

*M* �

1. The receiver may only use *ρ* to determine the decision threshold (partial CSI) and

frequency selective fading channel with i.i.d. and correlated channel gains.

*D* = 2, *β* = 0 *D* = 4, *β* = 0 *D* = 8, *β* = 0 *D* = 4, *β* = 5 *D* = 4, *β* = 10 ⎛ ⎝ *ρm* 2

In a channel model with LOS, **H** is expected to have a non-zero mean vector **M**. Therefore *V* is equal in distribution to a linear combination of noncentral *χ*<sup>2</sup> distributed variates. Unfortunately, there is no such convenient correspondence for that distribution as the general *Erlang* distribution for *V* in section 3.5.1.

Even though the PDF of a linear combination of noncentral *χ*<sup>2</sup> distributed variates is given in [48], there is no known closed-form expression for the integral of the product of this PDF and the MARCUM-Q-function in (20).

The average SEP may still be computed using a discrete approximation of the PDF of *V* (histogram). This histogram can be generated with arbitrary precision by sampling the PDF given in [48] or by performing a Monte-Carlo-simulation based on the distribution of **H**. For the Monte-Carlo-simulations the samples of the LOS channel impulse response **H** follow a normal distribution

$$\mathbf{H} \sim \mathcal{CN}\_L(\mathbf{M}, \Sigma\_\mathbf{h})$$

with the mean **M** = (*M*1,..., *MD*) with *Mi* = *<sup>κ</sup> <sup>κ</sup>*+<sup>1</sup> <sup>∀</sup> *<sup>i</sup>* and the covariance matrix **<sup>Σ</sup><sup>h</sup>** <sup>=</sup> <sup>1</sup> *<sup>κ</sup>*+<sup>1</sup> **I**.
