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# Data treatment - PowerPoint PPT Presentation

Data treatment . Collect 257 days, (2010/12~2011/1, 2011/3/15~2011/9) Outage raw data 475 times Case I: timer setting => 253 times Case II: insufficient radiation => 73 times Case III: system/ human error => 149 times Determine whether interpolation or not Data treatment algorithm.

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• Collect 257 days, (2010/12~2011/1, 2011/3/15~2011/9)

• Outage raw data 475 times

• Case I: timer setting => 253 times

• Case II: insufficient radiation => 73 times

• Case III: system/ human error => 149 times

• Determine whether interpolation or not

• Data treatment algorithm

Tstart: start time of outage, Tend: end time of outage, Tdur=Tstart - Tend

T={12:00, 13:00, 5:00, 7:00}, τ1= 10min, τ2=5min

Voff: the last voltage before outage Von: the first voltage after outage

S(AB) <-- △VoltageAB/ △tAB, S(BC) <-- △VoltageBC/ △tBC

========================================

Case1

IfTstart = T{i} ± τ 1 ANDTdur < τ2 then interpolate

Included;

Case2

if {Voff <11.6 AND Von>12.0 AND date < 2010/3/14}

OR {Voff <11.2 AND Von>11.5 AND date > 2010/3/14}

if 4.53<Rd1<14 AND 12.61<Rd2<23.71, then discard

else Included ;

Case 3

elseifTstart ∩ {t| 7pm<t<6pm} ≠ ΦOR Tend ∩ {t| 7pm<t<6pm} ≠ Φ, then discard

else interpolate

Included;

outage

V

B

C

A

D

t

• B(t) = B(t-1) + λ(t) – μ(t) ,if λ(t) – μ(t) ＜24

Bmax, drop(t) = λ(t) – μ(t)-24 ≧ 0

• B(t): energy in battery

• μ(t): energy consumption

• Bmax: energy in fully charged battery

B(t)

λ(t)

μ(t)

Derive [ λ(t) – μ(t)]

• Collect continued 2 days CASEII outage: 35

Derive Bmax

• Collect outage day due to insufficient radiation & work over last night : 18

• B(t-1) = actual lifetime – {λ(t) – μ(t)} < 15.78

Testing Bmax =15.78

• Collect outage day after continuous days which past nights

• 6 data sets (4/4, 4/17, 5/13, 5/24, 5/28, 9/21, )

• Collect CaseII outage and occur in night:

• Record Lifetime vs Battery voltage(V)

• Lifetime = 30.896V2 -727.51V + 4284

• B(t) = B(t-1) + λ(t) – μ(t) ; if B(t-1) + λ(t) – μ(t)<Bmax

Bmax, drop(t) = λ(t) – μ(t); if B(t-1) + λ(t) –

μ(t) ≧ Bmax

• Bmax = 25hr

One day radiation vsλ(t) – μ(t)

• Case I (253)=> interpolate 108 +81days

• Case II (73)=> discard 18 days

• Data ambiguous

• High amount radiation but short life time

• Case III (149)=> discard 8+9 days

• Empty:2 +3 (12/29 30, 8/5 8 9 )

• manual:3 (6/7, 7/16, 8/18)

• unknown:1 (5/11)

• no log:4 (12/10 14, 1/7, 3/14)

• Keep 224 out of 257 days for modeling

Tstart: start time of outage, Tend: end time of outage, Tdur=Tstart - Tend

T={12:00, 13:00, 5:00, 7:00}, τ1= 10min, τ2=5min

Voff: the last voltage before outage Von: the first voltage after outage

S(AB) <-- △VoltageAB/ △tAB, S(BC) <-- △VoltageBC/ △tBC

========================================

IfTstart = T{i} ± τ 1 ANDTdur < τ2 then interpolate

else if {Voff <11.6 AND Von>12.0 AND date < 2010/3/14} OR

{Voff <11.2 AND Von>11.5 AND date > 2010/3/14}

if 4.53<Rd1<14 AND 12.61<Rd2<23.71,

elseifTstart ∩ {t| 7pm<t<6pm} ≠ ΦOR Tend ∩ {t| 7pm<t<6pm} ≠ Φ, then discard

else interpolate

outage

V

B

C

A

D

t

• Occur 42 days in 165 days

• Reboot: 6

• Error: 3

• Empty: 13

• Manual: 3

• Unknown: 4(5/7, 9, 10, 11)

• No log: 12

• 8 days filtered in 42 days

• Empty:2(12/29 30),

• manual:1(6/7),

• unknown:1(5/11),

• no log:4(12/10 14, 1/7, 3/14)

• 10,14,29,30 Dec., 7 Jan., 14 Mar., 11 May, 7 Jun.

III類斷電

No

No

Yes

No

Yes

II類斷電

Yes

Yes

No

I類斷電

No

Yes

No

interpolation

interpolation

Yes

0< 有電壓值斜率差 <0.015

0.000315<III類斷電斜率差 <1.4365

smooth最大斜率差0.015

=>只要斜率差>0.015,挑出來,不能用

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BSpower consumption in one hr

BS 電壓與溫度相關性

• if (abs(mAB-mBC)>Threshold) || (abs(mBC-mCD)>Threshold)

filter out

Threshold = 0.019

• else if(13V<斷電前電壓) &&(duration > 19600 )

filter out

• else if(12V<斷電前電壓<12.5V) &&(duration > 10800 )

filter out

• else if (斷電前電壓<12V) &&(duration > 3600)

filterout

III類挑出不能用

• 12/14

• 12/29

• 一階微分用斜率來判斷

• |mAB-mBC|< δ

• |mBC-mCD|< δ

• 二階微分

• |F’’ABC- F’’BCD| < δ

D

C

B

A

C

B

D

A