Spectrally efficient IR-UWB pulse designs based on linear combinations of Gaussian Derivatives

Abstract

Pulse design is critical for impulse radio communications, as it determines the transmission efficiency with respect to regulation spectral limits. In this paper, we propose the design of several novel pulse shapes relying on combinations of Gaussian derivatives with the target of improving the spectral efficiency. A general model for maximizing the efficiency of dual and triplet couplings is presented, employing the interior point global optimization algorithm. Then, the spectrum peak frequency is derived in closed form for any combination of two Gaussians with consecutive orders of derivation. The parameters controlling the time properties of the generated waveforms have been adequately adjusted to reach the best compliance with the spectral mask. Novel pulses have been investigated providing a high efficiency using simple generation mechanisms, which can be practically implemented via analog circuits. An efficiency gain of more than 20% has been realized by our newly designed triplet combination over the conventional 5th order Gaussian derivative. Results demonstrated the advantage of the proposed pulse shapes in terms of the bit error rate performance for 2 Gbps OOK and 1 Gbps PPM in AWGN and multipath channels.

Appendices

Appendix A

Apply frequency derivative to the power spectral density in Eq. (24), then $S_{4,4}'(f)=$

$$\begin{aligned}&4 a^2 A_4^2 \Big [8 (2 \pi f)^7 2 \pi + (2 \pi f)^8 (-8 \pi ^2 f \sigma ^2) \Big ] e^{ -4 ( \pi f \sigma )^2 } {\cos ^2( \pi f \delta )} \nonumber \\&\quad - 4 a^2 A_4^2 \big (2\pi f \big )^8 e^{ -4 ( \pi f \sigma )^2 } 2 \cos ( \pi f \delta ) \sin ( \pi f \delta ) \pi \delta \end{aligned}$$

(44)

To find $f_{peak}$ state $S_{4,4}'(f)= 0$, given that for finite nonzero frequencies

$4 a^2 A_4^2 \big (2\pi f \big )^7 e^{ -4 ( \pi f \sigma )^2 } \ne 0$, then

$$\begin{aligned}&\Big [16 \pi - 2 \pi f (8 \pi ^2 f \sigma ^2) \Big ] {\cos ^2( \pi f \delta )} - 2 \big (2\pi f \big ) \nonumber \\&\quad \cos ( \pi f \delta ) \sin ( \pi f \delta ) \pi \delta = 0 \end{aligned}$$

(45)

Hence, either $\cos ( \pi f \delta )= 0$, or

$$\begin{aligned}&\Big [16 \pi - 2 \pi f (8 \pi ^2 f \sigma ^2) \Big ] {\cos ( \pi f \delta )} \nonumber \\&\quad - 2 \big (2\pi f \big ) \sin ( \pi f \delta ) \pi \delta = 0 \end{aligned}$$

(46)

It is assumed that the time delay $\delta$ should not exceed 10% of the pulse period (0.5 ns), in order to avoid a large truncation of the pulse energy, therefore, we assume that $\delta \le$ 50 ps. The power has to be concentrated in the UWB allocated band ([3.1-10.6] GHz), and the peak frequency is expected to lie at the mid interval of the desired band, as the 10 dB bandwidth has to be relatively large to cover a wide spectral margin under the regular mask. Hence, for practical values of $\delta$ and f, $\cos ( \pi f \delta )$ is different from 0, while for Eq. (46) it is sufficient the third order of Taylor series to approximate the sine and cosine functions:

$\sin (\pi f \delta )= (\pi f \delta )- (\pi f \delta )^3/6$

$\cos (\pi f \delta )= 1-(\pi f \delta )^2/2$

Following these approximations, Eq. (46) can be simplified to eventually reach a polynomial form:

$$\begin{aligned}&\Big [8 \pi - 2 \pi f (4 \pi ^2 f \sigma ^2) \Big ] \Big ( 1-(\pi f \delta )^2/2 \Big ) \nonumber \\&\quad - \big (2\pi f \big ) \Big ( (\pi f \delta )- (\pi f \delta )^3/6 \Big ) \pi \delta = 0 \end{aligned}$$

(47)

Expand and let $z=(\pi f)^2$, then

$$\begin{aligned} \delta ^2 \Big [12 \sigma ^2 + \delta ^2 \Big ] z^2 -6 \Big [ 3 \delta ^2 + 4 \sigma ^2 \Big ] z + 24 = 0 \end{aligned}$$

(48)

Which is a second degree equation, while

$$\begin{aligned} \Delta = 12 \Big [ 19 \delta ^4 - 24 \delta ^2 \sigma ^2 + 48 \sigma ^4 \Big ] \end{aligned}$$

(49)

Therefore,

$$\begin{aligned} z = \frac{ 6 \Big [ 3 \delta ^2 + 4 \sigma ^2 \Big ] \pm \sqrt{ 12 \Big [ 19 \delta ^4 - 24 \delta ^2 \sigma ^2 + 48 \sigma ^4 \Big ] }}{2 \delta ^2 \Big [12 \sigma ^2 + \delta ^2 \Big ] } \end{aligned}$$

(50)

Hence, to reach a peak frequency value within the [3.1-10.6] GHz band, the expression of $f_{peak}$ is as provided in Eq. (25).

Appendix B

The frequency derivative of $S_{5,6}(f)$ provided in (31) is

$$\begin{aligned} S_{5,6}'(f)= & {} \big (2\pi f\big )^{10} e^{ -4 ( \pi f \sigma )^2 } {A_5}^2 \times \nonumber \\&\Big [-4 a_5 a_6 \pi \sqrt{\frac{2}{11}}\, \sigma \big ( \sin (2\pi f \delta ) + 2 \pi f \delta \cos (2\pi f \delta ) \big ) \nonumber \\&+ 8{\pi }^2 f {a_6}^2 \frac{2}{11} \sigma ^2 \Big ]\nonumber \\&+{A_5}^2\Big [20\pi (2\pi f\big )^{9} e^{ -4 ( \pi f \sigma )^2 } + \big (2\pi f\big )^{10} e^{ -4 ( \pi f \sigma )^2 }\nonumber \\&\big (-8{\pi }^2 f {\sigma }^2 \big ) \Big ] \nonumber \\&\times \Big [ {a_5}^2 - 4 a_5 \sin (2\pi f \delta ) a_6 \sqrt{\frac{2}{11}}\, \sigma \pi f \nonumber \\&+ {a_6}^2 \frac{2}{11} \sigma ^2 (2\pi f)^2 \Big ] \end{aligned}$$

(51)

At the peak frequency, $S_{5,6}'(f)=0$, and knowing that $A_5 \ne 0$, then for finite nonzero frequencies:

$$\begin{aligned}&-4 a_5 a_6 \pi \sqrt{\frac{2}{11}}\, \sigma \Big ( \sin (2\pi f \delta ) 2 \pi f + 2 f^2 \pi \cos (2\pi f \delta ) 2 \pi \delta \Big ) \nonumber \\&\quad + 16{\pi }^3 f^2 {a_6}^2 \frac{2}{11} \sigma ^2 + \Big (20\pi - 16{\pi }^3 f^2 {\sigma }^2 \Big ) \times \nonumber \\&\quad \Big ( {a_5}^2 - 4 a_5 \sin (2\pi f \delta ) a_6 \sqrt{\frac{2}{11}}\, \sigma \pi f + {a_6}^2 \frac{2}{11} \sigma ^2 (2\pi f)^2 \Big )=0 \end{aligned}$$

(52)

What determines the pulse design is the relative amplitude of candidate waveforms, hence $a_5$ can be fixed at 1 for simplicity, while the resultant pulse dynamics will depend on the value of $a_6$ with respect to $a_5=1$. Therefore

$$\begin{aligned}&- a_6 \sqrt{\frac{2}{11}}\, \sigma \Big ( \sin (2\pi f \delta ) 2 \pi f + 4 f^2 {\pi }^2 \cos (2\pi f \delta ) \delta \Big )\nonumber \\&\quad + 4{\pi }^2 f^2 {a_6}^2 \frac{2}{11} \sigma ^2 + \Big (5 - 4{\pi }^2 f^2 {\sigma }^2 \Big ) \times \nonumber \\&\quad \Big ( 1 - 4 \sin (2\pi f \delta ) a_6 \sqrt{\frac{2}{11}}\, \sigma \pi f + {a_6}^2 \frac{2}{11} \sigma ^2 4{\pi }^2 f^2 \Big )=0 \end{aligned}$$

(53)

While in “Appendix A”, the third order Taylor series expansion of the sinusoids yielded an acceptable level of accuracy, a fourth order expansion is needed in this “Appendix” for an improved accuracy:

$\sin (2\pi f \delta )= (2\pi f \delta )- (2\pi f \delta )^3/6$

$\cos (2\pi f \delta )= 1-(2\pi f \delta )^2/2 + (2\pi f \delta )^4/24$

Following these approximations, Eq. (53) can be simplified to eventually reach a polynomial form:

$$\begin{aligned}&- 2 a_6 \sqrt{\frac{2}{11}}\, \sigma \delta (2\pi f)^2 + 4 a_6 \sqrt{\frac{2}{11}}\, \sigma (2\pi f)^4 {\delta }^3/6 \nonumber \\&\quad - a_6 \sqrt{\frac{2}{11}}\, \sigma (2\pi f)^6 {\delta }^5 /24 \nonumber \\&\quad + (2{\pi } f)^2 {a_6}^2 \frac{2}{11} \sigma ^2 + 5 + 5 \Big (-2 a_6 \sqrt{\frac{2}{11}} \sigma \delta \nonumber \\&\quad + {a_6}^2 \frac{2}{11} {\sigma }^2 \Big )(2 \pi f)^2 \nonumber \\&\quad + 5 \times 2 a_6 \sqrt{\frac{2}{11}} \sigma {\delta }^3 (2\pi f)^4 /6 - (2{\pi } f)^2 {\sigma }^2 \nonumber \\&\quad - {\sigma }^2 \Big (-2 a_6 \sqrt{\frac{2}{11}} \sigma \delta + {a_6}^2 \frac{2}{11} {\sigma }^2 \Big )(2 \pi f)^4\nonumber \\&\quad - 2 {\sigma }^2 a_6 \sqrt{\frac{2}{11}} \sigma {\delta }^3 (2\pi f)^6 /6 =0 \end{aligned}$$

(54)

Let $U=(2\pi f)^2$, then arrange to obtain:

$$\begin{aligned}&\Big [ - 12 a_6 \sqrt{\frac{2}{11}}\, \sigma \delta + 6 {a_6}^2 \frac{2}{11} \sigma ^2 - {\sigma }^2 \Big ] U \nonumber \\&\quad + \sqrt{\frac{2}{11}} a_6 \sigma \Big [ 14 {\delta }^3 /6 + 2 {\sigma }^2 \delta - {a_6} \sqrt{\frac{2}{11}} {\sigma }^3 \Big ] U^2 \nonumber \\&\quad + a_6 \sqrt{\frac{2}{11}} {\delta }^3 \sigma \Big [ - {\delta }^2 /24 - {\sigma }^2 /3 \Big ] U^3 + 5 = 0 \end{aligned}$$

(55)