Research Article Modeling of an EDLC with …downloads.hindawi.com/journals/mpe/2013/807034.pdfModeling of an EDLC with Fractional Transfer Functions Using Mittag-Leffler Equations

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 807034 7 pageshttpdxdoiorg1011552013807034

Research ArticleModeling of an EDLC with Fractional Transfer Functions UsingMittag-Leffler Equations

J J Quintana1 A Ramos2 and I Nuez1

1 Electronic and Automatic Engineering Department University of Las Palmas de GC 35017 Las Palmas Spain2 Process Engineering Department University of Las Palmas de GC 35017 Las Palmas Spain

Correspondence should be addressed to J J Quintana jjquintanadieaulpgces

Received 31 July 2013 Revised 30 October 2013 Accepted 31 October 2013

Academic Editor Bashir Ahmad

Copyright copy 2013 J J Quintana et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Electrochemical double-layer capacitors (EDLC) also known as supercapacitors or ultracapacitors are devices in which diffusionphenomena play an important role For this reason their modeling using integer-order differential equations does not yieldsatisfactory results The higher the temporal intervals are the more problems and errors there will be when using integer-orderdifferential equations In this paper a simple model of a real capacitor formed by an ideal capacitor and two parasitic resistorsone in series and the second in parallel is used The proposed model is based on the ideal capacitor adding a fractional behaviorto its capacity The transfer function obtained is simple but contains elements in fractional derivatives which makes its resolutionin the time domain difficult The temporal response has been obtained through the Mittag-Leffler equations being adapted to anyEDLC input signal Different charge and discharge signals have been tested on the EDLC allowing modeling of this device in thecharge rest and discharge stages The obtained parameters are few but identify with high precision the charge rest and dischargeprocesses in these devices

1 Introduction

In recent years the growing demand for new electrical energystoring systems has led to a remarkable development ofelectrochemical double-layer capacitors (EDLC) The EDLCare devices capable of storing energy and are characterized bytheir very rapid response during charge and discharge cycleswhich allows them to provide high power and to hold a highnumber of charge and discharge cycles

The complementary qualities of EDLC and batteries haveallowed the generation of numerous hybrid applicationsfor energy recovery or storage systems Energy storage inEDLC is not supported by chemical processes Moreoversupercapacitors have a high life cycle and do not needany maintenance From the above it can be concluded thatbecause of their characteristics these devices have raised greatinteresting expectations [1 2]

It is a usual procedure that the dynamics of real systemsare modeled by differential equations In most cases thedifferential equations are based on conventional derivatives

yielding sufficiently accuratemathematicalmodels Howeverthere are a variety of systems or phenomena in which math-ematical models based on ordinary differential equations donot provide satisfactory solutions One of these elements isthe EDLC It is this case when the application of modelsbased on fractional derivatives studied in the field of frac-tional calculus is very usefulThese equations are called frac-tional differential equations

Inmultiple fields of physics there are numerous examplesin which a differential equation is used as fractionalmodelingtool [3ndash5] This is very common when considering geometryconditions with fractal dimension and distributed parametersystems [6ndash8] In the field of electric power the currentthrough a capacitor is proportional to the noninteger-orderintegral for electric current [9 10] An electric networkcomposed of infinite RC elements can be modeled throughfractional differential equations [11 12]

In this paper a dynamic fractional model for EDLC isdeveloped based on a differential equation with fractional

2 Mathematical Problems in Engineering

derivatives This equation has been solved by means ofthe Laplace transform and in order to obtain the timedomain solution used the Mittag-Leffler function 119864

120572120573(119911)

The solution has been obtained for a unitary step input andfor a generic input This model has been applied to an EDLCfor its identification in long time periods obtaining verysatisfactory results

This paper is organized as follows Section 2 shows a briefreview of fractional calculus in Section 3 the fractionalmodelof the EDLC is deduced Section 4 expounds the experimentaldata and the discussion finally in Section 5 the conclusionsare presented

2 Brief Review of Fractional Calculus

In this section is exposed a brief mathematical backgroundof the fractional calculus [3] for understanding methodsresults and conclusions presented in this paper The fol-lowing are some of the most popular functions definitionsand properties of the fractional calculus

21 Riemann-Liouville Derivative Definition Riemann-Liou-ville definition expresses the fractional derivative as a timeconvolution integral

119886119863120572

119905119891 (119905) =

1

Γ (119899 minus 120572)

119889119899

119889119905119899int

119905

119886

119891 (120591)

(119905 minus 120591)120572minus119899+1

119889120591

(119899 minus 1 lt 120572 lt 119899)

(1)

where 119899 is an integer and 120572 is a real number and the frac-tional derivative order

22 CaputoDerivativeDefinition Caputo definition of a frac-tional derivative of a function is

119862

119886119863120572

119905119891 (119905) =

1

Γ (119899 minus 120572)int

119905

119886

119891(119899)

(120591)

(119905 minus 120591)120572+1minus119899

119889120591

(119899 minus 1 le 120572 lt 119899)

(2)


23 Grunwald-Letnikov Derivative Definition Grunwald-Letnikov definition is a numerical form of the fractionalderivative

119886119863120572

119905119891 (119905) = lim

ℎrarr0

1

ℎ120572

[(119905minus119886)ℎ]

sum

119895=0

(minus1)119895

(120572

119895)119891 (119905 minus 119895ℎ) (3)

24 Laplace Transform of the Fractional Derivative An inter-esting property of the fractional derivative operator is itsLaplace transform

119871 [0119863120572

119905119891 (119905)] = int

infin

0

119890minus119904119905

119886119863120572

119905119891 (119905) 119889119905

= 119904120572

119865 (119904) minus

119899minus1

sum

119896=0

119904119896

119886119863120572minus119896minus1

119905119891 (119905)

1003816100381610038161003816119905=0

(119899 minus 1 lt 120572 le 119899)

(4)

where if initial conditions are null it yields

119871 [0119863120572

119905119891 (119905)] = 119904

120572

119865 (119904) (5)

25 Mittag-Leffler Function The two-parameter function oftheMittag-Leffler type 119864

120572120573(119911) plays a very important role in

the fractional calculus and is defined as

119864120572120573

(119911) =

infin

sum

119896=0

119911119896

Γ (120572119896 + 120573)(120572 gt 0 120573 gt 0) (6)

The exponential function 119890119911 is a particular case of theMittag-

Leffler function

11986411

(119911) =

infin

sum

119896=0

119911119896

Γ (119896 + 1)= 119890119911

(7)

One of the main uses of the Mittag-Leffler function comesfrom the following Laplace transform

119871 [119905120572119896+120573minus1

119864(119896)

120572120573(plusmn119886119905120572

)] = int

infin

0

119890minus119904119905

119905120572119896+120573minus1

119864(119896)

120572120573(plusmn119886119905120572

) 119889119905

=119896119904120572minus120573

(119904120572 ∓ 119886)

119896+1

(8)

where 119896 indicates the order of the derivative of the Mittag-Leffler function

119864(119896)

120572120573(119911) =

119889119896

119889119911119896119864120572120573

(119911) (9)

Considering the specific case for 119896 = 0 yields

119871 [119905120573minus1

119864120572120573

(plusmn119886119905120572

)] =119904120572minus120573

119904120572 ∓ 119886 (10)

And by substituting theMittag-Leffler function the followingis obtained

119871[119905120573minus1

infin

sum

119896=0

(plusmn119886119905120572

)119896

Γ (120572119896 + 120573)] =

119904120572minus120573

119904120572 ∓ 119886 (11)

Which will be the relation used for calculating the solution inthe time domain

Mathematical Problems in Engineering 3

3 Dynamic Model of EDLC

Electric modeling of supercapacitors is an active line ofinvestigation with numerous contributions in the last fewyears [13ndash15] There are not yet totally satisfactory modelsapplicable to any operating mode of the EDLC allowing theEDLC behavior to be observed and simulated That is thereason why models based on heuristic techniques are mainlyused [14] Models based on the physics processes occurringin EDLC are also used [15] Most of the models obtainedmake use of a considerable number of variables either of ahigh number of passive elements or of fractional variablesmodels [16ndash18] Either way testing these devices during longtime periods has allowed observing their behavior and theirapplicability to energy storage systems Charge and dischargeare rapid processes occurring in a short time interval sincehigh currents are used

Due to diffusion processes resulting after the chargingphase The evolution of the processes occurring inside theEDLC generates a nonreversing thermal dissipation whichmakes the assessment of losses in the device possible Thisprocess would occur in both the EDLC charge and dischargephases

The model of EDLC has been considered as single input(current) single output (voltage) system The electric modelused comes from the model applied to ideal capacitors usinginteger-order differential equations Subsequently a modelbased on fractional order differential equations is proposed

31 Electric Model of a Conventional Capacitor The mathe-matical equation relating the current and the voltage at theterminals of a real capacitor can be deduced from an electriccircuit made of an ideal capacitor and two parasitic resistors[9 19]

Naming the series resistor as 1198771and the parallel one

as 1198772and operating the following equation is obtained

1198772sdot 119862

119889V (119905)119889119905

+ V (119905) = (1198771+ 1198772) sdot 119894 (119905) + 119877

1sdot 1198772sdot 119862

119889119894 (119905)

119889119905

(12)

with V(119905) being the voltage at the terminals of the EDLCand 119894(119905) its current The transfer function or impedance inthe Laplace domainwill be calculated as the quotient betweenvoltage and current

119866 (119904) =119881 (119904)

119868 (119904)= 1198771(

119904

119904 + (1 (1198772sdot 119862))

)

+1198771+ 1198772

1198772sdot 119862

(1

119904 + (1 (1198772sdot 119862))

)

(13)

Gathering the constants of the transfer function the followingis obtained

119866 (119904) =119881 (119904)

119868 (119904)= (1198961

119904

119904 + 119886+ 1198962

1

119904 + 119886) (14)

where

1198961= 1198771 119896

2=

1198771+ 1198772

1198772sdot 119862

119886 =1

1198772sdot 119862

(15)

(t)

i(t)

C EPR (R2)

ESR (R1)

Figure 1 Equivalent circuit of a real capacitor

The differential equation resulting from the transfer functioncan be solved for simple and complex signals using theexisting simulation programs

32 Electric Model of EDLC The model obtained for realcapacitor (12) fits well to the behavior of real capacitors in awide range of frequencies However if applied to an EDLCthe adjustment is unsatisfactory especially when long timeperiods are analyzed

Many papers [3 10 20] exposed the fractional behaviorof EDLC Thus by changing the ideal capacitor in the modelin Figure 1 for a fractional one the voltage and the currentacross such fractional capacitor are given by

119862 sdot 119863120572V (119905) = 119894 (119905) (16)

Applying this relation to the circuit of Figure 1 it gives adifferential equation of the form

0119863120572

119905V (119905) + 119886 sdot V (119905) = 119896

1sdot0119863120572

119905119894 (119905) + 119896

2sdot 119894 (119905) (17)

Applying Laplace transform (5) and assuming that the initialconditions are zero the following is obtained

119866 (119904) =119881 (119904)

119868 (119904)= (1198961

119904120572

119904120572 + 119886+ 1198962

1

119904120572 + 119886) (18)

This transfer function is similar to (14) differing only inthe fractional aspect of the capacitor In this case differ-ential equation (17) is of fractional order being definedby variable 120572 This equation can be solved for simple andcomplex signals input using Mittag-Leffler functions [3 21]

33 Solution to the Fractional Equation of EDLC In thissection the solution to (18) will be deduced in the timedomain for a variable input over time

331 Solution to a Unitary Step Input Multiplying transferfunction (18) by the unit step input 119894(119904) = 1119904 the responseof the system to this signal is obtained

V (119904) = 1198961

119904120572minus1

119904120572 + 119886+ 1198962

119904minus1

119904120572 + 119886

997904rArr V (119905) = 1198961119871minus1

[119904120572minus1

119904120572 + 119886] + 1198962119871minus1

[119904minus1

119904120572 + 119886]

(19)

Applying relation (11)

V (119905) = 1198961

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 1)+ 1198962sdot 119905120572

sdot

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 120572 + 1)

(20)

The response in the time domain to a unit step input isobtained


332 Solution to a Generic Input The response can beextended to any input For this purpose the superpositionof steps displaced by the sampling period of the input signalhas been used Thus for the first input signal the temporalresponse matches the response obtained by (20) So for thefirst time interval

V (119905) = 119894 (0) sdot (1198961

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962sdot 119905120572

sdot

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

0 le 119905 le 119879

(21)

As the input signal changes in the second period its responsecan be obtained by superposing to the previous response asecond step delayed in time a period 119879 resulting in

V (119905) = 119894 (0) sdot (1198961

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962sdot 119905120572

sdot

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

+ (119894 (119879) minus 119894 (0)) sdot (1198961

infin

sum

119895=0

(minus119886(119905 minus 119879)120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962(119905 minus 119879)

120572

infin

sum

119895=0

(minus119886(119905 minus 119879)120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

119879 le 119905 le 2119879

(22)

Generalizing (22) for a generic time interval 119873 the followingis obtained

V (119905) =

119873

sum

119896=0

(119894 (119896119879) minus 119894 ((119896 minus 1) 119879))

sdot [

[

1198961

infin

sum

119895=0

(minus119886 sdot (119905 minus 119896119879)120572

)119895

Γ (120572 sdot 119895 + 1)

+ 1198962(119905 minus 119896119879)

120572

sdot

infin

sum

119895=0

(minus119886 sdot (119905 minus 119896119879)120572

)119895

Γ (120572 sdot 119895 + 120572 + 1)

]

]

119873119879 le 119905 le (119873 + 1) 119879

(23)

where the initial conditions are zero

4 Experimental Results Identificationand Discussion

41 Instruments The experimental data have been obtainedin the laboratory making a test circuit and recording the

V

A

DC +minus

5Ω

DischargeChargeEDLC

Figure 2 Schematic diagram of the experimental circuit

electrical variables of the EDLC The analyzed EDLC man-ufactured by ELNA has a capacity of 47 F and a voltage of25 V

Figure 2 shows the electric circuit for the charge-dis-charge control A Crouzet Millenium III PLC has controlledthe charge and discharge switches Voltage and current datahave been recorded through a NI USB-6009 data logger byNational Instruments

42 Experimental Results Several tests have been performedto the EDLC in order to monitor voltage and current signalsBecause EDLC have memory effect just before each test theyhave been short-circuited for 24 hours to ensure zero initialconditions

The testswere performed as follows In a first phase lastinga few minutes was charged at a constant current between20mA and 09A until the voltage reached 24V in the secondphase the current is cut off and left to rest for 8 hours toallow the voltage stabilization and finally in the third phaseis discharged through a resistance of 5 ohm until it was fullydischarged

Although the sampling period of voltage and currenthas been recorded at 01 s sfor identification purposes ithas been considered a 5 s sampling time Figure 3 showsthe experimental data obtained The evolution of voltagein the EDLC throughout the process can be appreciated inFigure 3(a) while the current supplied at Figure 3(b) Theyare observed perfectly the three stages described above

43 Identification The coefficients for the proposed modelhave been defined according to transfer function (18) andimplemented using (23)

For identification the MATLAB simulation software wasused The function created has as input parameters 120572 119886 119896

1

and 1198962 Using (23) the voltage response is generated taking

all the current samples as input The voltage is comparedwith the experimental data through standard deviation (24)generating the output of the function Using this function inthe command fminsearch of MATLAB sought parametersare obtained For implementing (23) the terms 119895 of the summay be truncated without causing significant errors

The index chosen in this paper is the standard deviationdefined as

120590119889=

radicsum119873

119894=1(119910119894exp minus 119910

119894cal)2

119873 minus 1

(24)

where 119873 is the number of samples and subindexes whichrepresent the values obtained experimentally and calculatedaccording to the model


Volta

ge (V

)

Time (s)

2

15

1

05

0

minus05

times104

0 05 15 2 25 31

Charge (002A)-rest-discharge

(a)

0 05 15 2 25 3

Time (s)1

05

0

minus05

minus15

minus1Curr

ent (

A)

times104


(b)

Figure 3 Charge-rest-discharge at 20mA

0 1 2 3 4 5 6 7 817

19

21

23

25

27

Time (h)

Volta

ge (V

)

Charge (09A)-rest (best identification with

Calculated values

Experimental data

120572 = 1)

Figure 4 Fitting in the resting phase using integer derivatives

Table 1 Charge + rest identification parameters

Test 119868 (A) 120572 119886 1198961

1198962

120590119889

1 002A 0941 5974119864 minus 08 0239 0202 2079119864 minus 02

2 01 A 0952 8671119864 minus 08 0261 0187 1297119864 minus 02

3 05 A 0950 102119864 minus 07 0267 0170 1354119864 minus 02

4 05 A 0950 1020119864 minus 07 0250 0169 1338119864 minus 02

5 09 A 0953 9917119864 minus 08 0228 0164 1096119864 minus 02

The EDLC identification has been carried out calculatingthe parameters 120572 119886 119896

1 and 119896

2in (18) and has consisted

of two parts first from the beginning until just beforedischarged and second from the discharge onwards Thisis because the internal processes in the EDLC are different[22 23]

The data obtained for the charge and rest phases areshown in Table 1

Considering relation (15) between the parameters cal-culated with the resistors and the capacity in Figure 1 theparameters obtained for the charge are in Table 2

In order to compare the adjustment with traditionalmodel (14) in which the fractional index is an integer Table 3and Figure 4 show the best fit to the data of test number 5setting 120572 = 1

Table 2 Charge + rest electrical parameters

Test 119868 (A) 120572 119862 1198771

1198772

1 002A 0941 495 0239 3382119864 + 06

2 01 A 0952 535 0261 2155119864 + 06

3 05 A 0950 589 0267 1660119864 + 06

4 05 A 0950 591 0250 1660119864 + 06

5 09 A 0953 608 0228 1658119864 + 06

Table 3 Parameters calculated for test number 5 setting 120572 = 1

Test 120572 119862 1198771

1198772

1 1000 900 1262 2191119864 + 04

Table 4 Discharge identification parameters

Test 120572 119886 1198961

1198962

120590119889

1 0975 9600119864 minus 08 0235 0187 1391119864 minus 02

2 0981 9282119864 minus 08 0244 0185 1281119864 minus 02

3 0978 8427119864 minus 08 0267 0192 2068119864 minus 02

4 0976 1556119864 minus 07 0251 0202 2338119864 minus 02

5 0982 1032119864 minus 07 0240 0171 1285119864 minus 02

Table 5 Discharge electrical parameters

Test 120572 119862 1198771

1198772

1 0975 534 0235 1950119864 + 06

2 0981 540 0244 1996119864 + 06

3 0978 521 0267 2278119864 + 06

4 0976 495 0251 1297119864 + 06

5 0982 585 0240 1656119864 + 06

Operating in a similar way the parameters for thedischarge through the 5Ω resistor are in Table 4

Of which the following electric parameters are obtainedin Table 5

The following figures show graphically the fit of theproposed model with experimental data Figure 5 shows thatthe model has a good precision all over the time range testedFigures 6 and 8 display the model and the experimentaldata in the charging and discharging phases Figures 4 and 7show the fit when using transfer functions with integer and


01 2 3 4 5 6 7 8 90

05

1

15

2

25

3

Time (h)

Vol

tage

(V)


Figure 5 Experimental data of test number 5 versus proposedmodel

0 20 40 60 80 100 120 140 160 180 2000

05

1

15

2

25

3

Time (s)

Vol

tage

(V)

Charge at 09 A

Experimentaldata

Calculatedvalues

Figure 6 Detail of fitting in charging phase

fractional indices showing that the fit with the traditionalmodel gives unsatisfactory results In summary it can beobserved that the voltage deduced through the fractionalmodel adjusts very close to the real data even in long timeperiods

44 Discussion The self-discharge adjustment of EDLC aftercharging and after long time resting phases is a complex phe-nomenon which is not easily applicable to modeling throughinteger-order transfer functions In this work the EDLC hasbeenmodeledmaking use of the fractional derivative Takingthe electric circuit in Figure 1 as starting point and assumingthat the capacitor has a fractional index 120572 fractional transferfunction (18) has been deduced This function has beensolved in the time domain usingMittag-Leffler functions andimplemented in a simulation program through (23)

First of all considering Figure 5 it can be observed thatthe proposed transfer function adjusts well to the real datamodeling satisfactorily the self-discharge phenomenon

Unlike the traditional capacitor model which is definedby the value of two resistors and the capacity the proposedmodel adds a new parameter that is the fractional index 120572 ofthe fractional capacitor This new parameter models to agreater extent the self-discharging phenomenon

0 1 2 3 4 5 6 7 818

19

2

21

22

23

24

25

26

Time (h)

Vol

tage

(V)

Resting phase

Calculatedvalues

Experimentaldata

Figure 7 Fitting in the resting phase using the proposed model

8 8250

05

1

15

2

Time (h)

Vol

tage

(V)

Discharge

Calculatedvalues

Experimentaldata

Figure 8 Fitting in the discharging phase

Although constants in Tables 1 and 4 have been taken forthe identification of equations constants in Tables 2 and 5willprovide the data related to the value of the electric elements

In all cases the adjustment has been very precise with thetypical deviation being less than 003

The constancy of 1198771 1198772 and 119862 values for both charge

and discharge is especially noteworthyAt last the most significant difference in terms of charge

and discharge is given by the value of the fractional index 120572which is around 095 in the charge and around 098 in thedischarge

5 Conclusions

This work has consisted of EDLC modeling by implement-ing a simple fractional model allowing a very satisfac-tory response for long time periods Traditional modelsbased on transfer functions with integer coefficients yieldgood results in those cases when there is no self-dischargephenomenon Applying fractional mathematics implies anadditional parameter to the already necessary parameters inorder to define the basic model shown in Figure 1 which isthe fractional index 120572 According to the tests this index hasdifferent values for the charge and the discharge with those


values being apparently independent of the EDLC chargingcurrent

Thanks to the proposed transfer function and theMittag-Leffler functions it is possible to estimate the voltage at theEDLC terminals after a long time taking into account thecharging current and the way it has been charged

References

[1] A Burke ldquoUltracapacitors why how and where is the technol-ogyrdquo Journal of Power Sources vol 91 no 1 pp 37ndash50 2000

[2] B Conway Electrochemical Supercapacitors Scientific Funda-mentals and Technological Applications Kluwer Academic NewYork NY USA 1999

[3] I Podlubny Fractional Differential Equations vol 198 ofMath-ematics in Science and Engineering Academic Press San DiegoCalif USA 1999

[4] K B Oldham and J SpanierThe Fractional Calculus vol 1047Academic Press New York NY USA 1974

[5] K S Miller and B Ross An Introduction to the FractionalCalculus and Fractional Differential Equations John Wiley ampSons New York NY USA 1993

[6] A Oustaloup A Ballouk and B Bansard ldquoPartial differentialequations and non integer derivationrdquo in Proceedings of theInternational Conference on Systems Man and Cybernetics pp166ndash173 October 1993

[7] B Mbodje and G Montseny ldquoBoundary fractional derivativecontrol of the wave equationrdquo IEEE Transactions on AutomaticControl vol 40 no 2 pp 378ndash382 1995

[8] G Montseny J Audounet and B Mbodje ldquoOptimal model offractional integrators and application to systems with fadingmemoryrdquo in Proceedings of the International Conference onSystems Man and Cybernetics pp 65ndash70 October 1993

[9] S Westerlund and L Ekstam ldquoCapacitor theoryrdquo IEEE Trans-actions on Dielectrics and Electrical Insulation vol 1 no 5 pp826ndash839 1994

[10] N Bano and R A Hashmi ldquoElectrical studies of leaves overwide frequency rangerdquo IEEE Transactions on Dielectrics andElectrical Insulation vol 3 no 2 pp 229ndash232 1996

[11] G E Carlson and C A Halijak ldquoSimulation of the fractionalderivate operator and the fractional integral operator radic119904rdquo inProceedings of the Central State Simulation Council Meeting vol45 no 7 Kansas State University May 1961

[12] T C Haba M Martos G Ablart and P Bidan ldquoComposantselectroniques a impedance fractionnairerdquo in Proceedings of theESAIM conferece on Fractional Differential Systems ModelsMethods and Applications vol 5 pp 99ndash109 1998

[13] J J Quintana A Ramos and I Nuez ldquoIdentificaction of thefractional impedance of ultracapacitorsrdquo in Proceedings of theIFACWorkshop (FDArsquo06) Oporto Portugal 2006

[14] L Zubieta and R Bonert ldquoCharacterization of double-layercapacitors for power electronics applicationsrdquo IEEE Transac-tions on Industry Applications vol 36 no 1 pp 199ndash205 2000

[15] N Bertrand J Sabatier O Briat and J-M Vinassa ldquoEmbeddedfractional nonlinear supercapacitor model and its parametricestimation methodrdquo IEEE Transactions on Industrial Electron-ics vol 57 no 12 pp 3991ndash4000 2010

[16] R Martın J J Quintana A Ramos and I de la Nuez ldquoMod-eling of electrochemical double layer capacitors by means offractional impedancerdquo Journal of Computational and Nonlinear

Dynamics vol 3 no 2 Article ID 021303 pp 21303ndash214002008

[17] V Musolino L Piegari and E Tironi ldquoNew full-frequency-range supercapacitormodel with easy identification procedurerdquoIEEE Transactions on Industrial Electronics vol 60 no 1 pp112ndash120 2013

[18] Y Wang T Hartley C Lorenzo and J AdamsModeling Ultra-capacitors as Fractional-Order Systems New Trends in Nan-otechnology and Fractional Calculus Applications 2010

[19] R L Spyker ldquoClassical equivalent circuit parameters for adouble-layer capacitorrdquo IEEE Transactions on Aerospace andElectronic Systems vol 36 no 3 pp 829ndash836 2000

[20] A Dzielinski G Sarwas and D Sierociuk ldquoTime domain vali-dation of ultracapacitor fractional order modelrdquo in Proceedingsof the 49th IEEE Conference on Decision and Control (CDC rsquo10)pp 3730ndash3735 December 2010

[21] T J Freeborn B Maundy and A S Elwakil ldquoMeasurementof supercapacitor fractional-order model parameters fromvoltage-excited step responserdquo IEEE Journal on Emerging andSelected Topics in Circuits and Systems vol 3 no 3 pp 416ndash4242013

[22] R de Levie ldquoOn porous electrodes in electrolyte solutions ICapacitance effectsrdquo Electrochimica Acta vol 8 no 10 pp 751ndash780 1963

[23] M Kaus J Kowal and D U Sauer ldquoModelling the effects ofcharge redistribution during self-discharge of supercapacitorsrdquoElectrochimica Acta vol 55 no 25 pp 7516ndash7523 2010

Submit your manuscripts athttpwwwhindawicom

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Algebra

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


derivatives This equation has been solved by means ofthe Laplace transform and in order to obtain the timedomain solution used the Mittag-Leffler function 119864

120572120573(119911)

The solution has been obtained for a unitary step input andfor a generic input This model has been applied to an EDLCfor its identification in long time periods obtaining verysatisfactory results

This paper is organized as follows Section 2 shows a briefreview of fractional calculus in Section 3 the fractionalmodelof the EDLC is deduced Section 4 expounds the experimentaldata and the discussion finally in Section 5 the conclusionsare presented

2 Brief Review of Fractional Calculus

In this section is exposed a brief mathematical backgroundof the fractional calculus [3] for understanding methodsresults and conclusions presented in this paper The fol-lowing are some of the most popular functions definitionsand properties of the fractional calculus

21 Riemann-Liouville Derivative Definition Riemann-Liou-ville definition expresses the fractional derivative as a timeconvolution integral

119886119863120572

119905119891 (119905) =

1

Γ (119899 minus 120572)

119889119899

119889119905119899int

119905

119886

119891 (120591)

(119905 minus 120591)120572minus119899+1

119889120591

(119899 minus 1 lt 120572 lt 119899)

(1)


22 CaputoDerivativeDefinition Caputo definition of a frac-tional derivative of a function is

119862

119886119863120572

119905119891 (119905) =

1

Γ (119899 minus 120572)int

119905

119886

119891(119899)

(120591)

(119905 minus 120591)120572+1minus119899

119889120591

(119899 minus 1 le 120572 lt 119899)

(2)


23 Grunwald-Letnikov Derivative Definition Grunwald-Letnikov definition is a numerical form of the fractionalderivative

119886119863120572

119905119891 (119905) = lim

ℎrarr0

1

ℎ120572

[(119905minus119886)ℎ]

sum

119895=0

(minus1)119895

(120572

119895)119891 (119905 minus 119895ℎ) (3)

24 Laplace Transform of the Fractional Derivative An inter-esting property of the fractional derivative operator is itsLaplace transform

119871 [0119863120572

119905119891 (119905)] = int

infin

0

119890minus119904119905

119886119863120572

119905119891 (119905) 119889119905

= 119904120572

119865 (119904) minus

119899minus1

sum

119896=0

119904119896

119886119863120572minus119896minus1

119905119891 (119905)

1003816100381610038161003816119905=0

(119899 minus 1 lt 120572 le 119899)

(4)

where if initial conditions are null it yields

119871 [0119863120572

119905119891 (119905)] = 119904

120572

119865 (119904) (5)

25 Mittag-Leffler Function The two-parameter function oftheMittag-Leffler type 119864

120572120573(119911) plays a very important role in

the fractional calculus and is defined as

119864120572120573

(119911) =

infin

sum

119896=0

119911119896

Γ (120572119896 + 120573)(120572 gt 0 120573 gt 0) (6)

The exponential function 119890119911 is a particular case of theMittag-

Leffler function

11986411

(119911) =

infin

sum

119896=0

119911119896

Γ (119896 + 1)= 119890119911

(7)

One of the main uses of the Mittag-Leffler function comesfrom the following Laplace transform

119871 [119905120572119896+120573minus1

119864(119896)

120572120573(plusmn119886119905120572

)] = int

infin

0

119890minus119904119905

119905120572119896+120573minus1

119864(119896)

120572120573(plusmn119886119905120572

) 119889119905

=119896119904120572minus120573

(119904120572 ∓ 119886)

119896+1

(8)

where 119896 indicates the order of the derivative of the Mittag-Leffler function

119864(119896)

120572120573(119911) =

119889119896

119889119911119896119864120572120573

(119911) (9)

Considering the specific case for 119896 = 0 yields

119871 [119905120573minus1

119864120572120573

(plusmn119886119905120572

)] =119904120572minus120573

119904120572 ∓ 119886 (10)

And by substituting theMittag-Leffler function the followingis obtained

119871[119905120573minus1

infin

sum

119896=0

(plusmn119886119905120572

)119896

Γ (120572119896 + 120573)] =

119904120572minus120573

119904120572 ∓ 119886 (11)

Which will be the relation used for calculating the solution inthe time domain









1198772sdot 119862

119889V (119905)119889119905

+ V (119905) = (1198771+ 1198772) sdot 119894 (119905) + 119877

1sdot 1198772sdot 119862

119889119894 (119905)

119889119905

(12)


119866 (119904) =119881 (119904)

119868 (119904)= 1198771(

119904

119904 + (1 (1198772sdot 119862))

)

+1198771+ 1198772

1198772sdot 119862

(1

119904 + (1 (1198772sdot 119862))

)

(13)


119866 (119904) =119881 (119904)

119868 (119904)= (1198961

119904

119904 + 119886+ 1198962

1

119904 + 119886) (14)

where

1198961= 1198771 119896

2=

1198771+ 1198772

1198772sdot 119862

119886 =1

1198772sdot 119862

(15)

(t)

i(t)

C EPR (R2)

ESR (R1)





119862 sdot 119863120572V (119905) = 119894 (119905) (16)


0119863120572

119905V (119905) + 119886 sdot V (119905) = 119896

1sdot0119863120572

119905119894 (119905) + 119896

2sdot 119894 (119905) (17)


119866 (119904) =119881 (119904)

119868 (119904)= (1198961

119904120572

119904120572 + 119886+ 1198962

1

119904120572 + 119886) (18)




V (119904) = 1198961

119904120572minus1

119904120572 + 119886+ 1198962

119904minus1

119904120572 + 119886

997904rArr V (119905) = 1198961119871minus1

[119904120572minus1

119904120572 + 119886] + 1198962119871minus1

[119904minus1

119904120572 + 119886]

(19)


V (119905) = 1198961

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 1)+ 1198962sdot 119905120572

sdot

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 120572 + 1)

(20)




V (119905) = 119894 (0) sdot (1198961

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962sdot 119905120572

sdot

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

0 le 119905 le 119879

(21)


V (119905) = 119894 (0) sdot (1198961

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962sdot 119905120572

sdot

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

+ (119894 (119879) minus 119894 (0)) sdot (1198961

infin

sum

119895=0

(minus119886(119905 minus 119879)120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962(119905 minus 119879)

120572

infin

sum

119895=0

(minus119886(119905 minus 119879)120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

119879 le 119905 le 2119879

(22)


V (119905) =

119873

sum

119896=0

(119894 (119896119879) minus 119894 ((119896 minus 1) 119879))

sdot [

[

1198961

infin

sum

119895=0

(minus119886 sdot (119905 minus 119896119879)120572

)119895

Γ (120572 sdot 119895 + 1)

+ 1198962(119905 minus 119896119879)

120572

sdot

infin

sum

119895=0

(minus119886 sdot (119905 minus 119896119879)120572

)119895

Γ (120572 sdot 119895 + 120572 + 1)

]

]

119873119879 le 119905 le (119873 + 1) 119879

(23)




V

A

DC +minus

5Ω

DischargeChargeEDLC









1




120590119889=

radicsum119873

119894=1(119910119894exp minus 119910

119894cal)2

119873 minus 1

(24)



Volta

ge (V

)

Time (s)

2

15

1

05

0

minus05

times104

0 05 15 2 25 31


(a)

0 05 15 2 25 3

Time (s)1

05

0

minus05

minus15

minus1Curr

ent (

A)

times104


(b)


0 1 2 3 4 5 6 7 817

19

21

23

25

27

Time (h)

Volta

ge (V

)


Calculated values

Experimental data

120572 = 1)



Test 119868 (A) 120572 119886 1198961

1198962

120590119889

1 002A 0941 5974119864 minus 08 0239 0202 2079119864 minus 02

2 01 A 0952 8671119864 minus 08 0261 0187 1297119864 minus 02

3 05 A 0950 102119864 minus 07 0267 0170 1354119864 minus 02

4 05 A 0950 1020119864 minus 07 0250 0169 1338119864 minus 02

5 09 A 0953 9917119864 minus 08 0228 0164 1096119864 minus 02


1 and 119896







Test 119868 (A) 120572 119862 1198771

1198772

1 002A 0941 495 0239 3382119864 + 06

2 01 A 0952 535 0261 2155119864 + 06

3 05 A 0950 589 0267 1660119864 + 06

4 05 A 0950 591 0250 1660119864 + 06

5 09 A 0953 608 0228 1658119864 + 06


Test 120572 119862 1198771

1198772

1 1000 900 1262 2191119864 + 04


Test 120572 119886 1198961

1198962

120590119889

1 0975 9600119864 minus 08 0235 0187 1391119864 minus 02

2 0981 9282119864 minus 08 0244 0185 1281119864 minus 02

3 0978 8427119864 minus 08 0267 0192 2068119864 minus 02

4 0976 1556119864 minus 07 0251 0202 2338119864 minus 02

5 0982 1032119864 minus 07 0240 0171 1285119864 minus 02


Test 120572 119862 1198771

1198772

1 0975 534 0235 1950119864 + 06

2 0981 540 0244 1996119864 + 06

3 0978 521 0267 2278119864 + 06

4 0976 495 0251 1297119864 + 06

5 0982 585 0240 1656119864 + 06





01 2 3 4 5 6 7 8 90

05

1

15

2

25

3

Time (h)

Vol

tage

(V)



0 20 40 60 80 100 120 140 160 180 2000

05

1

15

2

25

3

Time (s)

Vol

tage

(V)

Charge at 09 A

Experimentaldata

Calculatedvalues






0 1 2 3 4 5 6 7 818

19

2

21

22

23

24

25

26

Time (h)

Vol

tage

(V)

Resting phase

Calculatedvalues

Experimentaldata


8 8250

05

1

15

2

Time (h)

Vol

tage

(V)

Discharge

Calculatedvalues

Experimentaldata







5 Conclusions





References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra

















1198772sdot 119862

119889V (119905)119889119905

+ V (119905) = (1198771+ 1198772) sdot 119894 (119905) + 119877

1sdot 1198772sdot 119862

119889119894 (119905)

119889119905

(12)


119866 (119904) =119881 (119904)

119868 (119904)= 1198771(

119904

119904 + (1 (1198772sdot 119862))

)

+1198771+ 1198772

1198772sdot 119862

(1

119904 + (1 (1198772sdot 119862))

)

(13)


119866 (119904) =119881 (119904)

119868 (119904)= (1198961

119904

119904 + 119886+ 1198962

1

119904 + 119886) (14)

where

1198961= 1198771 119896

2=

1198771+ 1198772

1198772sdot 119862

119886 =1

1198772sdot 119862

(15)

(t)

i(t)

C EPR (R2)

ESR (R1)





119862 sdot 119863120572V (119905) = 119894 (119905) (16)


0119863120572

119905V (119905) + 119886 sdot V (119905) = 119896

1sdot0119863120572

119905119894 (119905) + 119896

2sdot 119894 (119905) (17)


119866 (119904) =119881 (119904)

119868 (119904)= (1198961

119904120572

119904120572 + 119886+ 1198962

1

119904120572 + 119886) (18)




V (119904) = 1198961

119904120572minus1

119904120572 + 119886+ 1198962

119904minus1

119904120572 + 119886

997904rArr V (119905) = 1198961119871minus1

[119904120572minus1

119904120572 + 119886] + 1198962119871minus1

[119904minus1

119904120572 + 119886]

(19)


V (119905) = 1198961

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 1)+ 1198962sdot 119905120572

sdot

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 120572 + 1)

(20)




V (119905) = 119894 (0) sdot (1198961

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962sdot 119905120572

sdot

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

0 le 119905 le 119879

(21)


V (119905) = 119894 (0) sdot (1198961

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962sdot 119905120572

sdot

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

+ (119894 (119879) minus 119894 (0)) sdot (1198961

infin

sum

119895=0

(minus119886(119905 minus 119879)120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962(119905 minus 119879)

120572

infin

sum

119895=0

(minus119886(119905 minus 119879)120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

119879 le 119905 le 2119879

(22)


V (119905) =

119873

sum

119896=0

(119894 (119896119879) minus 119894 ((119896 minus 1) 119879))

sdot [

[

1198961

infin

sum

119895=0

(minus119886 sdot (119905 minus 119896119879)120572

)119895

Γ (120572 sdot 119895 + 1)

+ 1198962(119905 minus 119896119879)

120572

sdot

infin

sum

119895=0

(minus119886 sdot (119905 minus 119896119879)120572

)119895

Γ (120572 sdot 119895 + 120572 + 1)

]

]

119873119879 le 119905 le (119873 + 1) 119879

(23)




V

A

DC +minus

5Ω

DischargeChargeEDLC









1




120590119889=

radicsum119873

119894=1(119910119894exp minus 119910

119894cal)2

119873 minus 1

(24)



Volta

ge (V

)

Time (s)

2

15

1

05

0

minus05

times104

0 05 15 2 25 31


(a)

0 05 15 2 25 3

Time (s)1

05

0

minus05

minus15

minus1Curr

ent (

A)

times104


(b)


0 1 2 3 4 5 6 7 817

19

21

23

25

27

Time (h)

Volta

ge (V

)


Calculated values

Experimental data

120572 = 1)



Test 119868 (A) 120572 119886 1198961

1198962

120590119889

1 002A 0941 5974119864 minus 08 0239 0202 2079119864 minus 02

2 01 A 0952 8671119864 minus 08 0261 0187 1297119864 minus 02

3 05 A 0950 102119864 minus 07 0267 0170 1354119864 minus 02

4 05 A 0950 1020119864 minus 07 0250 0169 1338119864 minus 02

5 09 A 0953 9917119864 minus 08 0228 0164 1096119864 minus 02


1 and 119896







Test 119868 (A) 120572 119862 1198771

1198772

1 002A 0941 495 0239 3382119864 + 06

2 01 A 0952 535 0261 2155119864 + 06

3 05 A 0950 589 0267 1660119864 + 06

4 05 A 0950 591 0250 1660119864 + 06

5 09 A 0953 608 0228 1658119864 + 06


Test 120572 119862 1198771

1198772

1 1000 900 1262 2191119864 + 04


Test 120572 119886 1198961

1198962

120590119889

1 0975 9600119864 minus 08 0235 0187 1391119864 minus 02

2 0981 9282119864 minus 08 0244 0185 1281119864 minus 02

3 0978 8427119864 minus 08 0267 0192 2068119864 minus 02

4 0976 1556119864 minus 07 0251 0202 2338119864 minus 02

5 0982 1032119864 minus 07 0240 0171 1285119864 minus 02


Test 120572 119862 1198771

1198772

1 0975 534 0235 1950119864 + 06

2 0981 540 0244 1996119864 + 06

3 0978 521 0267 2278119864 + 06

4 0976 495 0251 1297119864 + 06

5 0982 585 0240 1656119864 + 06





01 2 3 4 5 6 7 8 90

05

1

15

2

25

3

Time (h)

Vol

tage

(V)



0 20 40 60 80 100 120 140 160 180 2000

05

1

15

2

25

3

Time (s)

Vol

tage

(V)

Charge at 09 A

Experimentaldata

Calculatedvalues






0 1 2 3 4 5 6 7 818

19

2

21

22

23

24

25

26

Time (h)

Vol

tage

(V)

Resting phase

Calculatedvalues

Experimentaldata


8 8250

05

1

15

2

Time (h)

Vol

tage

(V)

Discharge

Calculatedvalues

Experimentaldata







5 Conclusions





References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











V (119905) = 119894 (0) sdot (1198961

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962sdot 119905120572

sdot

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

0 le 119905 le 119879

(21)


V (119905) = 119894 (0) sdot (1198961

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962sdot 119905120572

sdot

infin

sum

119895=0

(minus119886 sdot 119905120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

+ (119894 (119879) minus 119894 (0)) sdot (1198961

infin

sum

119895=0

(minus119886(119905 minus 119879)120572

)119895

Γ (120572 sdot 119895 + 1)

+1198962(119905 minus 119879)

120572

infin

sum

119895=0

(minus119886(119905 minus 119879)120572

)119895

Γ (120572 sdot 119895 + 120572 + 1))

119879 le 119905 le 2119879

(22)


V (119905) =

119873

sum

119896=0

(119894 (119896119879) minus 119894 ((119896 minus 1) 119879))

sdot [

[

1198961

infin

sum

119895=0

(minus119886 sdot (119905 minus 119896119879)120572

)119895

Γ (120572 sdot 119895 + 1)

+ 1198962(119905 minus 119896119879)

120572

sdot

infin

sum

119895=0

(minus119886 sdot (119905 minus 119896119879)120572

)119895

Γ (120572 sdot 119895 + 120572 + 1)

]

]

119873119879 le 119905 le (119873 + 1) 119879

(23)




V

A

DC +minus

5Ω

DischargeChargeEDLC









1




120590119889=

radicsum119873

119894=1(119910119894exp minus 119910

119894cal)2

119873 minus 1

(24)



Volta

ge (V

)

Time (s)

2

15

1

05

0

minus05

times104

0 05 15 2 25 31


(a)

0 05 15 2 25 3

Time (s)1

05

0

minus05

minus15

minus1Curr

ent (

A)

times104


(b)


0 1 2 3 4 5 6 7 817

19

21

23

25

27

Time (h)

Volta

ge (V

)


Calculated values

Experimental data

120572 = 1)



Test 119868 (A) 120572 119886 1198961

1198962

120590119889

1 002A 0941 5974119864 minus 08 0239 0202 2079119864 minus 02

2 01 A 0952 8671119864 minus 08 0261 0187 1297119864 minus 02

3 05 A 0950 102119864 minus 07 0267 0170 1354119864 minus 02

4 05 A 0950 1020119864 minus 07 0250 0169 1338119864 minus 02

5 09 A 0953 9917119864 minus 08 0228 0164 1096119864 minus 02


1 and 119896







Test 119868 (A) 120572 119862 1198771

1198772

1 002A 0941 495 0239 3382119864 + 06

2 01 A 0952 535 0261 2155119864 + 06

3 05 A 0950 589 0267 1660119864 + 06

4 05 A 0950 591 0250 1660119864 + 06

5 09 A 0953 608 0228 1658119864 + 06


Test 120572 119862 1198771

1198772

1 1000 900 1262 2191119864 + 04


Test 120572 119886 1198961

1198962

120590119889

1 0975 9600119864 minus 08 0235 0187 1391119864 minus 02

2 0981 9282119864 minus 08 0244 0185 1281119864 minus 02

3 0978 8427119864 minus 08 0267 0192 2068119864 minus 02

4 0976 1556119864 minus 07 0251 0202 2338119864 minus 02

5 0982 1032119864 minus 07 0240 0171 1285119864 minus 02


Test 120572 119862 1198771

1198772

1 0975 534 0235 1950119864 + 06

2 0981 540 0244 1996119864 + 06

3 0978 521 0267 2278119864 + 06

4 0976 495 0251 1297119864 + 06

5 0982 585 0240 1656119864 + 06





01 2 3 4 5 6 7 8 90

05

1

15

2

25

3

Time (h)

Vol

tage

(V)



0 20 40 60 80 100 120 140 160 180 2000

05

1

15

2

25

3

Time (s)

Vol

tage

(V)

Charge at 09 A

Experimentaldata

Calculatedvalues






0 1 2 3 4 5 6 7 818

19

2

21

22

23

24

25

26

Time (h)

Vol

tage

(V)

Resting phase

Calculatedvalues

Experimentaldata


8 8250

05

1

15

2

Time (h)

Vol

tage

(V)

Discharge

Calculatedvalues

Experimentaldata







5 Conclusions





References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Volta

ge (V

)

Time (s)

2

15

1

05

0

minus05

times104

0 05 15 2 25 31


(a)

0 05 15 2 25 3

Time (s)1

05

0

minus05

minus15

minus1Curr

ent (

A)

times104


(b)


0 1 2 3 4 5 6 7 817

19

21

23

25

27

Time (h)

Volta

ge (V

)


Calculated values

Experimental data

120572 = 1)



Test 119868 (A) 120572 119886 1198961

1198962

120590119889

1 002A 0941 5974119864 minus 08 0239 0202 2079119864 minus 02

2 01 A 0952 8671119864 minus 08 0261 0187 1297119864 minus 02

3 05 A 0950 102119864 minus 07 0267 0170 1354119864 minus 02

4 05 A 0950 1020119864 minus 07 0250 0169 1338119864 minus 02

5 09 A 0953 9917119864 minus 08 0228 0164 1096119864 minus 02


1 and 119896







Test 119868 (A) 120572 119862 1198771

1198772

1 002A 0941 495 0239 3382119864 + 06

2 01 A 0952 535 0261 2155119864 + 06

3 05 A 0950 589 0267 1660119864 + 06

4 05 A 0950 591 0250 1660119864 + 06

5 09 A 0953 608 0228 1658119864 + 06


Test 120572 119862 1198771

1198772

1 1000 900 1262 2191119864 + 04


Test 120572 119886 1198961

1198962

120590119889

1 0975 9600119864 minus 08 0235 0187 1391119864 minus 02

2 0981 9282119864 minus 08 0244 0185 1281119864 minus 02

3 0978 8427119864 minus 08 0267 0192 2068119864 minus 02

4 0976 1556119864 minus 07 0251 0202 2338119864 minus 02

5 0982 1032119864 minus 07 0240 0171 1285119864 minus 02


Test 120572 119862 1198771

1198772

1 0975 534 0235 1950119864 + 06

2 0981 540 0244 1996119864 + 06

3 0978 521 0267 2278119864 + 06

4 0976 495 0251 1297119864 + 06

5 0982 585 0240 1656119864 + 06





01 2 3 4 5 6 7 8 90

05

1

15

2

25

3

Time (h)

Vol

tage

(V)



0 20 40 60 80 100 120 140 160 180 2000

05

1

15

2

25

3

Time (s)

Vol

tage

(V)

Charge at 09 A

Experimentaldata

Calculatedvalues






0 1 2 3 4 5 6 7 818

19

2

21

22

23

24

25

26

Time (h)

Vol

tage

(V)

Resting phase

Calculatedvalues

Experimentaldata


8 8250

05

1

15

2

Time (h)

Vol

tage

(V)

Discharge

Calculatedvalues

Experimentaldata







5 Conclusions





References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










01 2 3 4 5 6 7 8 90

05

1

15

2

25

3

Time (h)

Vol

tage

(V)



0 20 40 60 80 100 120 140 160 180 2000

05

1

15

2

25

3

Time (s)

Vol

tage

(V)

Charge at 09 A

Experimentaldata

Calculatedvalues






0 1 2 3 4 5 6 7 818

19

2

21

22

23

24

25

26

Time (h)

Vol

tage

(V)

Resting phase

Calculatedvalues

Experimentaldata


8 8250

05

1

15

2

Time (h)

Vol

tage

(V)

Discharge

Calculatedvalues

Experimentaldata







5 Conclusions





References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Documents

Research Article Modeling of an EDLC with …downloads.hindawi.com/journals/mpe/2013/807034.pdfModeling of an EDLC with Fractional Transfer Functions Using Mittag-Leffler Equations