output voltage and are known as non-zero switching states and the remaining two



Similar documents
The Essence of Three-Phase PFC Rectifier Systems

Module 5. Three-phase AC Circuits. Version 2 EE IIT, Kharagpur

REAL-TIME LOAD BALANCING AND POWER FACTOR CORRECTION OF THREE-PHASE, FOUR-WIRE UNBALANCED SYSTEMS WITH DSTATCOM

A. Description: A simple queueing system is shown in Fig Customers arrive randomly at an average rate of

Released Assessment Questions, 2015 QUESTIONS

c b N/m 2 (0.120 m m 3 ), = J. W total = W a b + W b c 2.00

Summation Notation The sum of the first n terms of a sequence is represented by the summation notation i the index of summation

STUDY COURSE BACHELOR OF BUSINESS ADMINISTRATION (B.A.)

Swelling and Mechanical Properties of Hydrogels Composed of. Binary Blends of Inter-linked ph-responsive Microgel Particles

Repeated multiplication is represented using exponential notation, for example:

INVESTIGATION OF PARAMETERS OF ACCUMULATOR TRANSMISSION OF SELF- MOVING MACHINE

MATHEMATICS SYLLABUS SECONDARY 7th YEAR

An Intelligent E-commerce Recommender System Based on Web Mining

SINCLAIR COMMUNITY COLLEGE DAYTON, OHIO DEPARTMENT SYLLABUS FOR COURSE IN MAT INTERMEDIATE ALGEBRA I (3 CREDIT HOURS)

Chapter System of Equations

Measurement of Reactive Power in Three-Phase Electric Power Systems by Use of Walsh Functions

2 DIODE CLIPPING and CLAMPING CIRCUITS

We will begin this chapter with a quick refresher of what an exponent is.

α. Figure 1(iii) shows the inertia force and

Words Symbols Diagram. abcde. a + b + c + d + e

Ratio and Proportion

Design Consideration for Configuration, Topology & Control Schemes of DSTATCOM Implemented on Distribution Systems

Appendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:

Design of Hybrid Neural Network Model for Quality Evaluation of Object Oriented Software Modules

Practice Test 2. a. 12 kn b. 17 kn c. 13 kn d. 5.0 kn e. 49 kn

APPLIED THERMODYNAMICS TUTORIAL 2 GAS COMPRESSORS

Chapter. Contents: A Constructing decimal numbers

How To Write A Power Flow Model For A Power Plant

Fast Circuit Simulation Based on Parallel-Distributed LIM using Cloud Computing System

Application: Volume. 6.1 Overture. Cylinders

m n Use technology to discover the rules for forms such as a a, various integer values of m and n and a fixed integer value a.

Angles 2.1. Exercise Find the size of the lettered angles. Give reasons for your answers. a) b) c) Example

How To Balance Power In A Distribution System

n Using the formula we get a confidence interval of 80±1.64

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )

Labor Productivity and Comparative Advantage: The Ricardian Model of International Trade

Reasoning to Solve Equations and Inequalities

Time Series Analysis. Session III: Probability models for time series. Carlos Óscar Sánchez Sorzano, Ph.D. Madrid, July 19th 2006

Density Curve. Continuous Distributions. Continuous Distribution. Density Curve. Meaning of Area Under Curve. Meaning of Area Under Curve

End of term: TEST A. Year 4. Name Class Date. Complete the missing numbers in the sequences below.

Fundamentals of Cellular Networks

Calculus Cheat Sheet. except we make f ( x ) arbitrarily large and. Relationship between the limit and one-sided limits

Chapter 13 Volumetric analysis (acid base titrations)

If two triangles are perspective from a point, then they are also perspective from a line.

Applying Fuzzy Analytic Hierarchy Process to Evaluate and Select Product of Notebook Computers

MATHEMATICS FOR ENGINEERING BASIC ALGEBRA

3 Energy Non-Flow Energy Equation (NFEE) Internal Energy. MECH 225 Engineering Science 2

The remaining two sides of the right triangle are called the legs of the right triangle.

CHAPTER 31 CAPACITOR

Lecture 4: Cheeger s Inequality

MODELLING AND SIMULATION OF TWO-LEVEL SPACE VECTOR PWM INVERTER USING PHOTOVOLTAIC CELLS AS DC SOURCE

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100

WHAT HAPPENS WHEN YOU MIX COMPLEX NUMBERS WITH PRIME NUMBERS?

CS99S Laboratory 2 Preparation Copyright W. J. Dally 2001 October 1, 2001

THE ARITHMETIC OF INTEGERS. - multiplication, exponentiation, division, addition, and subtraction

Outline. Numerical Analysis Boundary Value Problems & PDE. Exam. Boundary Value Problems. Boundary Value Problems. Solution to BVProblems

Laws of Exponents. net effect is to multiply with 2 a total of = 8 times

DlNBVRGH + Sickness Absence Monitoring Report. Executive of the Council. Purpose of report

MATH PLACEMENT REVIEW GUIDE

CHAPTER-10 WAVEFUNCTIONS, OBSERVABLES and OPERATORS

Authorized licensed use limited to: University of Illinois. Downloaded on July 27,2010 at 06:52:39 UTC from IEEE Xplore. Restrictions apply.

Quick Guide to Lisp Implementation

SOLVING EQUATIONS BY FACTORING

15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style

1 Computing the Standard Deviation of Sample Means

5 Boolean Decision Trees (February 11)

Calculating Principal Strains using a Rectangular Strain Gage Rosette

1 Fractions from an advanced point of view

Fuzzy Task Assignment Model of Web Services Supplier

Present and future value formulae for uneven cash flow Based on performance of a Business

Calculus Cheat Sheet. except we make f ( x ) arbitrarily large and. Relationship between the limit and one-sided limits

1. Definition, Basic concepts, Types 2. Addition and Subtraction of Matrices 3. Scalar Multiplication 4. Assignment and answer key 5.

Or more simply put, when adding or subtracting quantities, their uncertainties add.

Inter-domain Routing

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.

Graphs on Logarithmic and Semilogarithmic Paper

Two hours UNIVERSITY OF MANCHESTER SCHOOL OF COMPUTER SCIENCE. Date: Friday 16 th May Time: 14:00 16:00

SOLVING QUADRATIC EQUATIONS BY FACTORING

Trigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE The absolute value of the complex number z a bi is

Homework 3 Solutions

50 MATHCOUNTS LECTURES (10) RATIOS, RATES, AND PROPORTIONS

Now here is the important step

PREMIUMS CALCULATION FOR LIFE INSURANCE

EQUATIONS OF LINES AND PLANES

OUTLINE SYSTEM-ON-CHIP DESIGN. GETTING STARTED WITH VHDL August 31, 2015 GAJSKI S Y-CHART (1983) TOP-DOWN DESIGN (1)

Vectors Summary. Projection vector AC = ( Shortest distance from B to line A C D [OR = where m1. and m

CHAPTER 3 THE TIME VALUE OF MONEY

Space Vector Pulse Width Modulation Based Induction Motor with V/F Control

Section 7-4 Translation of Axes

Integration by Substitution

Regular Sets and Expressions

Transcription:

SPACE ECTOR MODULATION FOR THREE-LEG OLTAGE SOURCE INERTERS.1 THREE-LEG OLTAGE SOURCE INERTER The toology of three-leg voltge soure iverter is show i Fig..1. Beuse of the ostrit tht the iut lies must ever e shorted d the outut urret must lwys e otiuous voltge soure iverter ssume oly eight distit toologies. These toologies re show o Fig... Six out of these eight toologies rodue ozero outut voltge d re kow s o-zero swithig sttes d the remiig two toologies rodue zero outut voltge d re kow s zero swithig sttes. g L I C B A Fig..1. Toology of three-leg voltge soure iverter. 4

1() () 3() 4() 5() 6() 7() 8() Fig... Eight swithig stte toologies of voltge soure iverter.. OLTAGE SPACE ECTORS Se vetor modultio (SM) for three-leg SI is sed o the reresettio of the three hse qutities s vetors i two-dimesiol (α, β) le. This hs ee disussed i [1] d is illustrted here for the ske of omleteess. Cosiderig toology 1 of Fig.., whih is reeted i Fig..3 () we see tht the lie voltges,, d re give y 5

= g = 0.. (.1) = - g This e rereseted i the α, βle s show i Fig..3(), where voltges,, d re three lie voltge vetors disled 10 i se. The effetive voltge vetor geerted y this toology is rereseted s 1() i Fig..3(). Here the ottio refers to the three legs/hses,, eig either oeted to the ositive d ril () or to the egtive d ril (). Thus orresods to hse eig oeted to the ositive d ril d hses d eig oeted to the egtive d ril g Fig..3(). Toology 1-1() of voltge soure iverter. = g = 0 = - g - g - 1() g Fig..3(). Reresettio of toology 1 i the α, βle. 6

Proeedig o similr lies the six o-zero voltge vetors (1-6) e show to ssume the ositios show i Fig..4. The tis of these vetors form regulr hexgo (dotted lie i Fig..4). We defie the re elosed y two djet vetors, withi the hexgo, s setor. Thus there re six setors umered 1-6 i Fig..4. 3 4 3 4 5 1 6 1 5 6 Fig..4. No-zero voltge vetors i the α, βle. Cosiderig the lst two toologies of Fig.. whih re reeted i Fig..5() for the ske of oveiee we see tht the outut lie voltges geerted y this toology re give y = 0 = 0 = 0.. (.) These re rereseted s vetors whih hve zero mgitude d hee re referred to s zero-swithig stte vetors or zero voltge vetors. They ssume the ositio t origi i the α, βle s show i Fig..5(). The vetors 1-8 re lled the swithig stte vetors (SSs). 7

7() 8() Fig..5(). Zero outut voltge toologies. = 0 = 0 = 0 7, 8 Fig..5(). Reresettio of the zero voltge vetors i the α, βle..3 SPACE ECTOR MODULATION The desired three hse voltges t the outut of the iverter ould e rereseted y equivlet vetor rottig i the outer lok wise diretio s show i Fig..6(). The mgitude of this vetor is relted to the mgitude of the outut voltge (Fig..6()) d the time this vetor tkes to omlete oe revolutio is the sme s the fudmetl time eriod of the outut voltge. 8

3 4 3 1 1 4 5 5 6 6 Fig..6(). Outut voltge vetor i the α, βle. 0 1 3 4 5 6 Fig..6(). Outut lie voltges i time domi. Let us osider the situtio whe the desired lie-to-lie outut voltge vetor is i setor 1 s show i Fig..7. This vetor ould e sythesized y the ulse-widthmodultio (PWM) of the two djet SS s 1() d (), the duty yle of eh eig d 1 d d, resetively, d the zero vetor ( 7() / 8() ) of duty yle d 0 [1]: d + d = = m e g jè 1 1.. (.3) d + d + d 1.. (.4) 1 0 = where, 0 m 0. 866, is the modultio idex. This would orresod to mximum lie-to-lie voltge of 1.0 g, whih is 15% more th ovetiol siusoidl PWM s show [1]. 9

() 7() 8() d d 1 θ 1 1() Fig..7. Sythesis of the required outut voltge vetor i setor 1. All SM shemes d most of the other PWM lgorithms [1,4], use (.3), (.4) for the outut voltge sythesis. The modultio lgorithms tht use o-djet SS s hve ee show to rodue higher THD d/or swithig losses d re ot lyzed here, lthough some of them, e.g. hysteresis, e very simle to imlemet d rovide fster trsiet resose. The duty yles d 1, d, d d 0, re uiquely determied from Fig..7, d (.3), (.4), the oly differee etwee PWM shemes tht use djet vetors is the hoie of the zero vetor(s) d the sequee i whih the vetors re lied withi the swithig yle. The degrees of freedom we hve i the hoie of give modultio lgorithm re: 1) The hoie of the zero vetor - whether we would like to use 7() or 8() or oth, ) Sequeig of the vetors 3) Slittig of the duty yles of the vetors without itroduig dditiol ommuttios. Four suh SM lgorithms re osidered i the ext setio, mely: 1) The right liged sequee ( SM1) 10

) The symmetri sequee (SM) 3) The ltertig zero vetor sequee ( SM3) 4) The highest urret ot swithed sequee (SM4). The modultio shemes re desried for the se whe the referee vetor is i setor 1: ll other ses re irulrly symmetri. These modultio shemes re lyzed d their reltive erforme with reset to swithig loss, THD d the ek-to-ek urret rile t the outut is ssessed. The lysis is erformed over the etire rge of modultio idex d for lod ower ftor gle vryig from 90 o to 90 o. All se vetor modultio shemes reseted here ssume digitl imlemettio d, hee, regulr smlig, i.e. ll duty yles re relulted t the egiig of the swithig yle, sed o the vlue of the referee voltge vetor t tht istt..4 MODULATION SCHEMES.4.1 RIGHT ALIGNED SEQUENCE (SM1) A simle wy to sythesize the outut voltge vetor is to tur-o ll the ottom (or to) swithes t the egiig of swithig yle d the to tur them off sequetilly so tht the zero vetor is slit etwee 7() d 8() eqully. This swithig sheme is show i Fig..8 for two smlig eriods. The sigls i the figure rereset the gtig sigls to the uer legs of the iverter. The sheme hs three swith tur-o s d three swith tur-off s withi swithig yle. The erforme of the left liged sequee, where the sequee of vetors is extly oosite to the right liged sequee, is exeted to e similr to the right liged sequee. 11

d 0 / d 1 d d 0 / T s d 0 / d 1 d d 0 / T s Fig..8. Phse gtig sigls i SM1..4. Symmetri Sequee (SM) This sheme hs ee show i revious works [4] to hve the lowest THD. This is euse of the symmetry i the swithig wveform s e see i Fig..9. The umer of ommuttios i oe smlig eriod is six. Sie this sheme hs the sme umer of swithigs s SM1, with three swith tur-os d three swith tur-offs, their swithig losses re exeted to e similr. d 0 4 d 1 d d 0 T s d d 1 d 0 4 d 0 4 d 1 d d 0 T s d d 1 d 0 4 Fig..9. Phse gtig sigls i SM. 1

.4.3 Altertig Zero etor Sequee (SM3) I this sheme, kow s DI sequee i literture [6], the zero vetors 7() d 8() re used ltertively i djet yles so tht the effetive swithig frequey is hlved, s show i Fig..10. d 1 d d 0 T s d d 1 d 0 T s Fig..10. Phse gtig sigls i SM3. However, the smlig eriod is still T s, sme s i the other shemes. The swithig losses for this sheme re exeted to e idelly 50% s omred to those of the revious two shemes d THD sigifitly higher due to the existee of the hrmois t hlf of the smlig frequey..4.4 Highest Curret Not-Swithed Sequee (SM4) This sheme, kow s DD sequee i literture [6], is sed o the ft tht the swithig losses re roximtely roortiol to the mgitude of the urret eig swithed d hee it would e dvtgeous to void swithig the iverter leg rryig the highest istteous urret. This is ossile i most ses, euse ll djet SS s differ i the stte of swithes i oly oe leg. Hee, y usig oly oe zero vetor, 7() or 8() withi give setor oe of the legs does ot hve to e swithed t ll, s show i Fig..11 (). 13

T s T s d 1 d d 0 d 1 d d 0 Fig..11(). Phse gtig sigls i SM4. However, sie the hoie of the o-zero SSs is sed o the desired outut voltge vetor d the hse d mgitude of the urret re determied y the lod, it is ot lwys ossile to void swithig the hse rryig the highest urret. I suh se the hse rryig the seod highest urret is ot swithed d the swithig losses re still redued. For exmle, the hoie of zero vetors i setor 1 is determied usig the flowhrt i Fig..11(). Y I > I d 0 7() N d 0 8() Fig..11(). Choie of zero vetor i setor 1..5 ANALYSIS I this setio the THD, swithig losses d ek-to-ek urret rile of ll the shemes re lyzed over the etire rge of modultio idex, d over vryig lod ower ftor gles. 14

.5.1 Totl Hrmoi Distortio The totl hrmoi distortio (THD) of eriodi voltge whih e rereseted y the Fourier series = e = 1 jωt is defied s THD = =.. (.5) 1 d its weighted totl hrmoi distortio is defied s WTHD = =.. (.6) 1 The lysis of THD is doe sed o ovel lgorithm where the Fourier oeffiiets of ll the ulses of give lie/hse voltge i oe fudmetl time eriod re summed u. The lysis is vlid for ll itegrl f s /f o, where f s is the swithig frequey d f o is the fudmetl frequey t the outut of the iverter. Fig..1 shows the lot of tyil hse voltge i oe fudmetl time eriod. The THD of this wveform is lulted y first deomosig the wveform of Fig..1 ito series is ulses whih resemle Fig..13. Phse Phse oltge oltge T s T o Time Time Fig..1. Tyil Phse voltge i oe fudmetl time eriod. 15

The the Fourier omoets for eh of these ulses re lulted d summed u to get the effetive Fourier omoets of the etire wveform i oe fudmetl time eriod. This roedure is illustrted here for SM1. Cosiderig ulse show i Fig..13 with fudmetl eriod T o = 1/f o (Fig..1) its Fourier oeffiiets t frequey f o re give y m π = (si( ( dm Ts + T π T o m Tm π )) si( T o )).. (.7) m Tm π π = ( os( ) os( ( dm Ts + Tm ))) π T T o o.. (.8) oltge m T m d m Ts T s Time Fig..13. Outut voltge ulse. where d m is the ulse duty yle, T m is the ulse dely time, m is the ulse mgitude, d T s = 1, is the smlig time eriod. f s Deomositio of the hse voltge swithig wveform for SM1 (Fig..13) to oti series of ulses s i Fig..1 is two ste roess. At first the wveform is deomosed to oti steed ulses s show i Fig..14(), the this steed ulse wveform is further deomosed to oti ulses, similr to Fig..13, s show i Fig..14() d Fig..14(). 16

Cosider the lod to e wye oeted d led. The the hse voltges, with reset to eutrl oit (N) hose suh tht + + =0 i y setor for y lod e otied from = - 3, = - 3, = - 3 The hse voltge () i setor1 durig differet SSs (8,1,,7) for SM1 is show i Tle.1 Tle.1. Phse voltge i setor 1: SM1. SS s 8() 1() () 7() duty yle d 0 / d 1 d d 0 / 0 g /3 g /3 0 Fig..14 shows the deomositio of tyil hse voltge i the k th swithig itervl from the egiig of the hse voltge eriod, where f s 0 k - 1.. (.9) f o g /3 g /3 g /3 g /3 0 0 0 d 0 / d 1 d d 0 / d 0 / d 1 d 0 /+d d 0 /+d 1 d d 0 / (k-1)t s kt s (k-1)t s kt s (k-1)t s kt s () () () first ulse Fig..14. Phse voltge deomositio i oe smlig eriod. Comrig eh of the omoet ulses with Fig..13 oe fid tht for the 17

dm = d 1, Tm = k - g m =, 3 d + T s, 1 0.. (.10) d for the seod ulse d m m = d, T g = 3 m d0 = k - 1 + + d1 T, s.. (.11) The duty yles d 1, d d d 0 re lulted usig (.3) d (.4). The Fourier omoets of these ulses re otied from (.7) d (.8). Thus the Fourier oeffiiets of the hse voltge e foud usig: = fs -1 f o k = 0 k + fs -1 f o k = 0 k.. (.1) The Fourier oeffiiets of the lie urret e otied y I =.. (.13) lie _ urret z where z is the imede of give hse. Altertively, WTHD of lie voltge ould rereset THD of lie urret [,3]. Fig..15 shows the THD of the simulted lie urret d lie voltge d the WTHD of lie voltge for ll the shemes over the etire rge of modultio idex, for the iverter rmeters g = 400, f o = 60 Hz, f s = 160, R(lod) = 3, L(lod) = 1mH. Aedix B lists the MATLAB ode, used to geerte the redited results i Fig..15. 18

0 0.866 Modultio idex(m) () THD THD THD THD 0 Modultio idex(m) () 0.866 0 Modultio idex(m) () 0.866 Fig..15. oltge d urret distortio s futio of modultio idex. ) THD of lie urret 19

) WTHD of lie urret ) THD of lie voltge These results gree with wht is foud i literture. It is iterestig to ote tht the THD (lie urret d voltge) for ll the shemes dereses with irese i modultio idex. This is euse of the irese i the fudmetl omoet of the voltge/urret with irese i modultio idex; the other higher order hrmois eig reltively ostt. It lso e see tht SM (symmeti) hs the lest THD. This e ssoited with the symmetry i the swithig wveform. By itroduig symmetry (SM), the umer of hse voltge ulses i give swithig time-eriod is douled s omred to the other shemes. This would mke the overter look s if it were oertig t twie the swithig frequey. Hee this results i redued THD d redued ek-to-ek rile i the lod urret..5.. Swithig Losses The swithig losses re ssumed to e roortiol to the rodut of the voltge ross the swith d the urret through the swith t the istt of swithig. Sie the voltge ross the swith is the us voltge, it is osidered to e ostt. Thus the losses re roortiol to the urret durig swithig. As first roximtio, the swithig rile is egleted d the losses re estimted sed o the umer of ommuttios required for eh swithig sheme d the urret t the istt of swithig. Losses for sheme SM4 re deedet o the lod ower ftor d their loss erforme hs ee otimized usig the flow hrt 0

reseted i Fig..11() for the etire rge of lod ower ftor gle. Fig..16, shows the loss erforme hrteristis of ll the shemes. From Fig..16 we see tht SM4 hs 50 % redutio i losses t high lod ower ftors (>0.866) d the svigs i the losses redue to 37% t low lod ower ftors. Reltive Reltive swithig losses swithig loss Lod Lod ower ftor gle Fig..16. Reltive swithig losses s futio of lod ower ftor gle..5.3. Pek-to-Pek Curret Rile The ek-to-ek vlue of the urret rile, t mximum vlue of lod urret, is imortt i the desig of idutors - to e used s filters. Some simle formule hve ee derived elow for the urte estimtio of the rile for ll the shemes. Mximum outut urret rile ours whe the volt-seod ross the outut idutors is the lrgest. Cosider hse i Fig..1 d ssume tht outut voltge A (Fig..1) vries slowly with reset to the swithig frequey. The for ssive lod, the mximum volt-seod ross the idutor i hse ours whe the referee vetor i Fig..7 is ollier with the swithig vetor 1(). At this istt the duty yle d = 0 d duty yle d 1 = m, d the resultt hse voltge will e s show i Fig..17. 1

Phse voltge m v d 1 = m d 0 = 1- m d 1 T s T s d 0 T s Time Fig..17. Phse voltge ulse t its ek low frequey vlue. Due to the ssumtio tht the outut voltge A (Fig..1) is ostt, the voltge ross the idutor is Sie d L A ( ) v = - -.. (.14) = 3 d v 1 g ; ( ) m m =.. (.15) ( ) v I = L d T 0 s.. (.16) Thus the ek-to-ek urret rile is give y I = g 3L ( 1- m) m T s.. (.17) Equtio (17) is true for SM1, SM3, d SM4. For SM the duty yle d 0 is slit i hlf d hee I g = 3L 1 ( -m) m T s.. (.18)

Fig..18 shows the lot of the reltive ek-to-ek rile urret, sed o the ove lysis d sed o the simultio results usig SABER, for the etire rge of modultio idex. Reltive ek-to-ek rile 0 Modultio idex (m) 0.866 Fig..18. Reltive ek-to-ek urret rile..6 Simultio d Exerimetl Results The iruit i Fig..1 ws simulted usig SABER. Aedix A gives the listig of the se vetor modultio temlte develoed i MAST for SM1. The iruit i Fig..1 ws simulted usig idel swithes, RCD suers, d idel diodes, usig the followig rmeters: g = 16, I lod = 6.5 A, f o = 108 Hz, f s = 3888 Hz, m = 0.6. The outut urret wveforms d their setr re show i Fig..19. Fig..0 shows the wveforms of the outut urret d setrum of outut urret otied from exerimetl iverter ruig uder similr oditios. The figures show firly good greemet etwee simultio d exerimetl results. 3

SM1 SM SM3 SM4 Fig..19. Simulted lie urrets d setrum of lie urrets ( f s /f o = 36). 4

Fig..0. Exerimetl lie urrets d setrum of lie urrets (f s /f o = 36). 5

.7 PERFORMANCE SUMMARY Tle. summrizes the erforme of four se vetor modultio shemes. It e lerly see tht sheme with high THD hs low losses d vie vers d these hrteristis re lod deedet. This ould e trslted s trde-off to e mde etwee the size of the het sik d size of the filters. At low swithig frequeies SM ould e used, sie losses re ot very ritil. At high swithig frequeies SM4 is referred, eseilly t high lod ower ftors due to the 50% redutio i swithig losses. SM3 ould e used t low lod ower ftors. The right liged sequee (SM1) (or the left-liged sequee) does ot seem to hve y rtiulr dvtge if the overter is hrd swithed. However, if softswithig is itrodued the this sheme is rtiulrly useful euse here ll the three legs re eig swithed t the sme time. It hs lso ee oserved tht the erforme of SM4 e imroved y itroduig symmetry. This result s i redued THD d i redued ek-to-ek rile i the lod urret; with the swithig losses remiig the sme.. Tle.. Reltive erforme of vrious modultio shemes (three-leg). Modultio Shemes SM1 SM SM3 SM4 No of ommuttios i T s 6 6 3 4 Reltive Losses 1 1 0.5 0.5-0.63 ** Domit hrmoi f s f s f s / f s THD t low mod idex Lest THD t high mod. idex Lest Highest Reltive ek-to-ek rile t I mx 1 0.5 1 1 Numer of swithig sttes i T s 4 7 3 3 ** Deedig o the lod ower ftor * Imortt for digitl modultor imlemettio 6