A Comparison of energy efficiency for UWB Modulations
Adil ELABBOUBI, Fouzia ELBAHHAR, Marc HEDDEBAUT, Yassine ELHILLALI
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A Comparison of energy efficiency for UWB Modulations Adil ELABBOUBI, Fouzia ELBAHHAR, Marc HEDDEBAUT, Yassine ELHILLALI 1 2 Outline Objectives UWB modulations descriptive The system model UWB modulations comparative
Adil ELABBOUBI, Fouzia ELBAHHAR, Marc HEDDEBAUT, Yassine ELHILLALI
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ο The growing demand of data from mobile
ο Researchers presented a concept called βGreen
ο The key techniques of Green Communication are:
ο
Cognitive Radio: improve the spectrum utilization efficiency and the network transmission performance.
ο
Network coding: remove the redundant routes optimize the throughput the effect of energy and bandwidth saving.
ο
Smart Grid: the combination of new communication techniques, hardware and software optimization to save energy.
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οThis work is a part of the third category of green
communications techniques.
οThe UWB system is chosen:
ο low energy consumption ο low complexity.
οUsing an Analytical model to compare the energy
consumption of modulation techniques.
οComparing the energy efficiency of some commonly
used UWB modulations in a multi-path environment.
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ο Transmitting a short pulse with a delay in time
ο The PPM signal:
πΉπ’ ππ‘
π(π’ β ππ
π β π ππ π β π[π ππ‘
]π)
+β π=ββ
ο The duration of a pulse: π
π and the
π
π.
ο π
π=πΎπ π (πΎ > 100), the symbol duration:
π‘ = ππ‘π π
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ο Transmitting a short pulse with different level
ο The PAM signal:
πΉπ’ ππ‘
π΅π[π ππ‘
]π(π’ β ππ π β π ππ π) +β π=ββ
,
πΉπ’ ππ‘ π΅π[π ππ‘ ] is one of the
possible amplitude, π΅π[π ππ‘
] = 2d π ππ‘
β 3 and πΉπ’ = πΉππ€ 3 ο Parameters are defined like the PPM case.
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ο Transmitting a short pulse with two different
ο The PSM signal: π‘ =
πΉπ’ ππ‘
ππ[π ππ‘
](π’ β ππ π β π ππ π) +β π=ββ
with ππ[π ππ‘
] is one of the
possible waveforms. ο Parameters are similar to the PPMβs ones.
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ο The time duration to transmit N bit: π = π
ππ +
ππ‘π + ππ’π
ο The energy needed to transmit N bit:
πΉ = π
πππ ππ + ππ‘πππ‘π + ππ’π ππ’π (π ππ β« ππ‘π β ππ‘π = 0)
ο The total energy consumed to transmit 1 bit : πΉπ = (ππ’+ππ)π
ππ+ππ’π π π’π
π
ππ’: the transmission power, π
π=πππ’+π ππ : the power of the
transceiver circuitry.
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ο For transmission: πππ’ = π
ππ + π πππ + π ππππ’ π
πππ = Ξ±ππ’ (Ξ± = ΞΎ Ξ· β 1)
ο For PAM and PPM:
ππ = ππππ΅ + ππππ¦ + ππππ’ + π ππ + π ππππ + π π΅πΈπ·
ο For PSM:
ππ = ππππ΅ + 2(ππππ¦ + ππππ’) + π ππ + π ππππ + π π΅πΈπ·
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π =
β π¦ ππ¦ +β
ο In case of non-severe fading: π
π β€ 2πβ1π! ππππ exp
2 β
ο In case of severe fading: π
π β πΞ»1+(1βπ)Ξ»2 πππ
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ο π π‘ = ππ‘π π, so π π‘ = πΎ πΆ. π ππ = ππ π‘ thus π ππ = ππΎ πΆ (ππ‘ = 1) ο In case of non-severe fading for n=1:
π
π = 1 πππ exp
(
π2 2 β π) so SNR= 1 ππ exp
(
π2 2 β π)
SNR=
πΉπ‘ π0 = πΉπ’ π»ππ0 = ππ’π
π‘
π»ππ0 then ππ’π π‘ = π0 1 ππ exp
(
π2 2 β π)π»π (π»π = πππππ»1)
ππ’π
ππ = π0 1 ππ exp
(
π2 2 β π)π»πN ο Total energy consumption:
1 ππ exp
π2 2 β π)π»π+ ππ,πππβππππ π
ππ
π
1 ππ exp
π2 2 β π)π»π+ ππ,πππβππππ π
ππ
π
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ο In case of severe fading: π
π β πΞ»1+(1βπ)Ξ»2 πππ
so πππ = πΞ»1+(1βπ)Ξ»2
ππ
SNR= πΉπ‘
π0 = πΉπ’ π»ππ0 = ππ’π
π‘
π»ππ0 then ππ’π π‘ = π0 πΞ»1+(1βπ)Ξ»2 ππ
π»π ππ’π
ππ = π0 πΞ»1+(1βπ)Ξ»2 ππ
π»ππ
ο Total energy consumption:
πΞ»1+(1βπ)Ξ»2 π
π
π»π+
ππ,πππβππππ π
ππ
π
πΞ»1+(1βπ)Ξ»2 π
π
π»π+
ππ,πππβππππ π
ππ
π
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π
π = π
2πππ. π¦ π
β π¦ ππ¦ +β
ο In case of non-severe fading: π π β€ 2πβ1π! (2πππ)π exp
(π2 π2
2 β ππ)
β π β π (Lognormal approximation)
ο In case of severe fading: π π β πΞ»1+(1βπ)Ξ»2 2πππ
(coxian approximation)
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ο we have π π‘ = ππ‘π π, so π π‘ = πΎ πΆ. π ππ = ππ π‘ thus π ππ = ππΎ πΆ ο In case of non-severe fading for n=1:
π
π = 1 2πππ exp
(
π2 2 β π) Alors SNR= 1 2ππ exp
(
π2 2 β π)
SNR=
πΉπ‘ π0 = πΉπ’ π»ππ0 = ππ’π
π‘
π»ππ0 donc ππ’π π‘= π0 1 2ππ exp π2 2 β π π»π
ππ’π
ππ = π0 1 2ππ exp
(
π2 2 β π)π»πN ο Total energy consumption:
πΉπ = (1 + Ξ±)π0
1 2ππ exp
(π2
2 β π)π»π+ ππ,ππ΅πβππππ π
ππ
π
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ο In case of severe fading: π
π β πΞ»1+(1βπ)Ξ»2 2πππ
so πππ = πΞ»1+(1βπ)Ξ»2
2ππ
SNR= πΉπ‘
π0 = πΉπ’ π»ππ0 = ππ’π
π‘
π»ππ0 then ππ’π π‘ = π0 πΞ»1+(1βπ)Ξ»2 2ππ
π»π
ππ = π0 πΞ»1+(1βπ)Ξ»2 2ππ
ο Total energy consumption:
2ππ
π»π+
ππ,ππ΅πβππππ π
ππ
π
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π
π = 4 π»πΌπ
Ξ· = 0.6 π = 3.5 π0 = β170dBm/Hz πΆ = 500 ππΌπ π = 106 π
ππ = 25.2 ππ
ππππ΅ = 7.68 ππ ππππ¦ = 15 ππ ππππ’ = 2.5 ππ π
π΅πΈπ· = 7.6 ππ
πππ»π΅ = 12 ππ ππΉπΈ = 10.8 ππ π
ππππ’ = π ππππ’π
= 2.5 ππ ππ = 40 ππΆ π»1 = 28 ππΆ π
π = 10β3
πΎ=500 π = β0.0039 π = 0.6883 Ξ»1 = 4.9 Ξ»2 = 65.44 p=1.0617 d=10
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