PLL and Grid Synchronization - National Institute of...

PLL and Grid Synchronization

by

Subhash Joshi T G

Email: [email protected]

1 of 63

SUMMARY OF PRESENTATION

Unit vectors and its significance

Basics of transformations

Session 1: Unit Vectors For 3 Balanced/Unbalanced Grid Selection of reference variable for control

Selection of output parameter for PI-controller

Design of PI-controller constants

System bandwidth and method to select bandwidth

Unit vector under unbalanced grid condition

Test results

A Simple method for unit vector construction for balanced grid

Session 2: Unit Vectors For 1 Grid

Constructing two unit magnitude 900 displaced components

Mitigating the effect of grid frequency variation

Approximation method

Rigorous method

Test results 2 of 63

Unit Vectors

What is unit vector ?

Two unity magnitude fundamentalsinusoidal quantities, which are displacedby 900 from each other

One of the unit vector is in phase with gridvoltage

This should be free from harmonics

Phase angle error from grid voltage should be minimum

Significance of unit vector ?

STATCOM is an independent voltage source

Unit vector helps to synchronize STATCOM voltage and Gridvoltage

Also unit vector is used to extract the active and reactive powercomponent separately

It is also used to separate the individual harmonics in case ofactive power filters 3 of 63

Transformations

3 to - transformations

- to D-Q transformations

BY

BYR

XXX

XXXX

2

3

2

1

X

X

UU

UU

X

X

q

d

12

21

ρθ

β axis

α axis

Q axis

XβXres

Xα

D axis

U1 and U2 unity magnitude

components, where U2 lag U1 by 900

For grid connected system U1 is

aligned along R-phase grid voltage

α - axis

XB

XY

XR

θ

120o

-120o

β axis

4 of 63

SESSION 1

UNIT VECTORS FOR THREE PHASE BALANCED/UNBALANCED GRID

5 of 63

Unit vector for Balanced/Unbalanced

Grid condition

Let the grid voltages are,

Objective,

)120sin(

)120sin(

sin

tVV

tVV

tVV

ggB

ggY

ggR

tVVVV

tVVVVV

ggBY

ggBYR

cos2

3)(

2

3

sin2

3)(

2

1

3 -

VR

VY

VB

V

V

Apply three phase to two phase (-) transformation,

tU

tU

g

g

cos

sin

2

1

6 of 63


Grid condition

If U1 and U2 are known then V and V can be transformed to

D-Q axis

Assume U1 and U2 are known, then

When U1 gets synchronized with VR, then g=g’ and,

V

V

UU

UU

V

V

q

d

12

21

))sin((2

3

))cos((2

3

cos2

3

sin2

3

sincos

cossin

tV

tV

tV

tV

tt

tt

V

V

ggg

ggg

gg

gg

gg

gg

q

d

Let U1 is not synchronized to VR and its frequency is g’, then,

02

3g

q

d V

V

V

3

-VR

VY

VB

V

V

-

D-Q

U1 U2

Vd

Vq

7 of 63


Grid condition

If Vq = 0 is ensured, then U1 will be synchronized to VR

To ensure Vq = 0, feedback with simple PI-controller can be

used

PI-controller will be sufficient,

Since Vd and Vq are d.c. quantities and variation in e(t) with

time is minimal

3

-VR

VY

VB

V

V

-

D-Q

U1 U2

Vd

Vq+

-

0

s

kk i

p e(t)

8 of 63


Grid condition

What should be the output parameter of PI-controller ?

PI-controller performs well when the selected output parameter magnitude

swing is minimal

Our objective is to obtain U1 and U2, but they are sinusoidally varying

quantities and results in large swing in magnitude

The angle of U1 and U2 are also will be varying with time in large extend

Instead the frequency of U1 and U2 can be thought of selecting as output

parameter of PI-controller

tUtU cos ,sin 21

3

-VR

VY

VB

V

V

-

D-Q

U1 U2

Vd

Vq+

-

0

s

kk i

p e(t)

9 of 63


Grid condition

What should be the output parameter of PI-controller ?

If the grid frequency variation is minimal, then

PI-controller performance can be further improved by selecting the output

of PI-controller as ∆ (variation in grid frequency) instead of absolute grid

frequency,

3

-VR

VY

VB

V

V

-

D-Q

U1 U2

Vd

Vq+

-

0

s

kk i

p e(t) ∆

+

ref

s

1 t

Lookup

Table

U1

U2

-

ref is the nominal grid frequency

10 of 63


Grid condition

Design of PI-controller constants ?

At time t=0, let grid and unit vectors (U1 and U2) frequency be g

At time t=0+, grid frequency changed from to and unit vector

frequency remains at g

3 to - transformation of grid voltage is,

gg

tVVtVV gggg cos

2

3 ,sin

2

3

- to D-Q transformation gives,

))sin((2

3

))cos((2

3

cos2

3

sin2

3

sincos

cossin

tV

tV

tV

tV

tt

tt

V

V

ggg

ggg

gg

gg

gg

gg

q

d

11 of 63


Grid condition


Assuming grid frequency variation is minimal, then

tt gggg )())sin((

Hence Vq is,

tVV gggq )(2

3

))sin((2

3

))cos((2

3

tV

tV

V

V

ggg

ggg

q

d

3

-VR(’g)

VY(’g)

VB(’g)

V(’g)

V(’g)-

D-Q

U1(g) U2(g)

Vd

Vq+

-

0

s

kk i

p e(t) ∆

+

ref

s

1

Lookup

Table

U1(g)

U2(g)

tg-

12 of 63


Grid condition



tt gggg )())sin((

Hence Vq is,

tVV gggq )(2

3

))sin((2

3

))cos((2

3

tV

tV

V

V

ggg

ggg

q

d

3

-VR(’g)

VY(’g)

VB(’g)

V(’g)

V(’g)-

D-Q

U1(g) U2(g)

Vd

Vq+

-

0

s

kk i

p e(t) ∆

+

ref

s

1

Lookup

Table

U1(g)

U2(g)

tg-

13 of 63


Grid condition



tt gggg )())sin((

Hence Vq is,

tVV gggq )(2

3

))sin((2

3

))cos((2

3

tV

tV

V

V

ggg

ggg

q

d

+-

tg

Vq+

-

0

s

kk i

p e(t) ∆

+

ref

s

1 tg

gV2

3

-

14 of 63


Grid condition


The block diagram can be further reduced by removing the quantities not

taking part in transients

ref is constant and no effect on transients

Once grid voltage is changed from to ,

can be assumed to be constant

Simplified block diagram is,

gg g

+-

tg

Vq+

-

0

s

kk i

p e(t) ∆

+

ref

s

1 tg

gV2

3

-+-Vqref=0

s

kk i

p s

1

gV2

3

15 of 63


Grid condition


Open loop transfer function of the system is given by,

+-Vqref=0

s

kk i

p s

1

gV2

3

ss

kskVsHsG

ip

g

1

2

3)()(

2

1

2

3)()(

i

pi

g

ks

kk

s

VsHsG

16 of 63


Grid condition


Find asymptotic bode plot of the system,

For zero:

Corner frequency,

Magnitude plot: +20dB/decade

Phase plot: +450/decade

For poles:

Corner frequency,

Magnitude plot: -40dB/decade

Phase plot: -1800

piz kk

ip k

2

1

2

3)()(

i

pi

g

ks

kk

s

VsHsG

17 of 63

pi kk

ik


Grid condition


Closed loop system will be stable,

If the open loop gain crosses 0dB (unity

gain) with -20dB/decade and ensuring

atleast -1350 Phase margin

If z > p, leads to a phase

close to -1800 makes system

prone to unstable

Hence to make system more

stable, keep z <= p

Since phase plot of zero has a

slope of +450/decade, for z =

p, the phase is -1350 leads to

phase margin of 450

2

1

2

3)()(

i

pi

g

ks

kk

s

VsHsG

18 of 63

ik


Grid condition


z = p implies,

One more component present

in the open loop transfer

function is,

i

p

i kk

k

ip kk

gV2

3

This will not affect the phase

plot, but gain cross over

frequency will be pushed further

away increasing the bandwidth

of the system

2

1

2

3)()(

i

pi

g

ks

kk

s

VsHsG

19 of 63

ik

-40dB/dec

20dB/dec

20dB/dec

Gain = 0dB

g

-1350


Grid condition

Find system bandwidth ?

Defined as the frequency at

which the closed-loop

magnitude is equal to -3 dB

For phase of -1200, the open

loop bandwidth and closed loop

bandwidth are found to be

closer

Since gain at is 0dB, the

gain cross over frequency of

open loop system is,

ig k

ik

ip kk

20 of 63

ik

-40dB/dec

20dB/dec

20dB/dec

Gain = 0dB

g

-1350


Grid condition


The modified crossover

frequency considering 3/2Vg is,

Gain at is,

Gain at new crossover

frequency ( ) is 0dB

Slope during this time is

-20dB/decade

gV

2

3log20

ig k

g

20

loglog

2

3log200

ig

g

k

V

ip kk

21 of 63

ik

-40dB/dec

20dB/dec

20dB/dec

Gain = 0dB

g

-1350


Grid condition


The modified crossover

frequency considering 3/2Vg is,

Closed loop bandwidth (appx.)

is,

Once the bandwidth is fixed

proportional constant (kp) can

be found

igg kV2

3

pgkVBW2

3

ip kk

22 of 63


Grid condition

How to fix the bandwidth ?

Bandwidth is decided by the harmonics present in Vd and Vq components

as well as the response time requirement in transient conditions

For a balanced three wire system the minimum harmonics expected on

Vd and Vq are 300Hz (transformation of 5th and 7th harmonics)

Here bandwidth of 30Hz to 60Hz will be sufficient for proper

attenuation of harmonics in grid voltages

g

pV

BWk

23

ip kk

2

pi kk

In Summary:

Based on the application and transient requirement fix the bandwidth

If grid voltage peak is known, proportional constant (kp) can be found

Once kp is known integral constant ki can be computed

23 of 63


Grid condition

Unit vector under unbalanced grid condition ?

Under unbalanced grid conditions, the grid voltage contains fundamental

positive sequence as well as fundamental negative sequence components

Let us construct the unit vector such that it is synchronized with

fundamental positive sequence component

As earlier fundamental positive sequence present in the grid voltage when

transformed to D-Q reference frame reflected as d.c. component and

fundamental negative sequence present in the grid voltage reflected as

100Hz component

Since we required only d.c. component in Vq, 100Hz component can be

easily removed by using simple low pass filter of corner frequency around

10Hz

24 of 63


Grid condition

Unit vector under unbalanced grid condition ?

C is the corner frequency of low pass filter

Low pass filter will not affect the information of fundamental positive

sequence, as it is a d.c. quantity

3

-VR

VY

VB

V

V

-

D-Q

U1 U2

Vd

Vq+

-

0

s

kk i

p e(t) ∆

+

ref

s

1

Lookup

Table

U1

U2

t-

Cs

s

25 of 63


Grid condition

Test results

Transient performance analysis carried out by simulation using PSIM

simulation package

0

-200

-400

200

400

VR VY VB

0.06 0.08 0.1 0.12 0.14

Time (s)

0

-0.5

-1

0.5

1

cosrow sinrow

50% step voltage sag

appeared in VR and VY

Unit vectors

3 Grid

Voltage restored26 of 63


Grid condition

Test results

Transient performance analysis carried out by simulation using PSIM

simulation package

0

-200

-400

200

400

VR VY VB

0

-0.5

-1

0.5

1

cosRow sinRow

0.36 0.38 0.4 0.42 0.44 0.46

Time (s)

0

100

200

300

400

theta

Step change in frequency (from 50Hz to 30Hz)

appeared in VR, VY and VB

Unit vectors

3 Grid

t

27 of 63


Grid condition

Test results

Unit vector implemented for unbalanced grid condition in TI’s DSP

TMS320F2812

VR and VB are nominal, VY=0

Grid voltagesUnit vectorGrid voltage

Unit vector

Balanced grid voltage

Unit vector t

Unit vector and angle t 28 of 63

A Simple method for

Unit vector in Balanced Grid condition

A simple method exists based on trigonometry to compute

unit vector for balanced grid condition

Transform three phase grid voltage to - co-ordinate

tVtVtVtV gggg cos2

3 ,sin

2

3

3

-VR

VY

VB

V

V

Cs

s

Cs

s

V’

V’

The output of Low Pass Filter (LPF) is given by,

sVs

ssVsV

s

ssV

cc

,

29 of 63

A Simple method for


Output of LPF in time domain is,

and are always 900 displacedirrespective of grid frequency and cornerfrequency of LPF

At the zero crossing of , will bein peak and vice versa

Unit magnitude of and can beobtained by dividing the term by its ownmagnitude,

t

VtVt

VtV g

cg

cg

g

cg

cgcos2

3

,sin23

2222

c

g

tanwhere,

)(tV )(tV

)(tV )(tV

)(tV)(tV

Unit vectors

30 of 63

A Simple method for


Block diagram of unit vector construction

3

-VR

VY

VB

V

V

Cs

s

Cs

s

V’(t)

V’(t)

ZCD

ZCD

S&H

S&H

ABS

Trigger

X1 / Y1

X1

Y1

X2

Y2

Trigger

ABS

X2 / Y2 ttU gcos)(2

ttU gsin1

31 of 63

A Simple method for


Let us perform the following operations,

Phase angle () is minimum when c=250

For c=250, is zero when grid frequency is 50Hz

t

V

tVtVtF

g

gc

cg

sin2

3

)()()(

22

1

t

V

tVtVtF

g

gc

cg

cos2

3

)()()(

22

2

gc

gc

tanwhere,

32 of 63

A Simple method for


Complete block diagram of unit vector construction

where, c=250

3

-VR

VY

VB

V

V

Cs

s

Cs

s

V’

V’

ZCD

ZCD

S&H

S&H

ABS

Trigger

X1 / Y1

X1

Y1

X2

Y2

Trigger

ABS

X2 / Y2 ttU gcos)(2

ttU gsin1

-+

++

This method will introduce a small phase angle error when grid

frequency varies

Each harmonics is reduced by when compared to

fundamental21

2

h33 of 63

A Simple method for


Test results

Grid voltage

Unit vector

34 of 63

SESSION 2

UNIT VECTORS FOR SINGLE PHASE GRID

35 of 63

Overview of the presentation




Rigorous method

36 of 63





Rigorous method

37 of 63

Let R-phase grid voltage be, VR = Vg sin(St)

Let the above voltage is passed through a LPF of cornerfrequency, c

Using Laplace analysis

Innovative method for constructingunit vector

VR FR+

-

sc

t

VS

CS

Cgsin

22FR(t)|steady

C

S

tanand

222222

1

SS

S

S

C

CCS

SCg

s

s

ss

V

FR(s)

tce

sc

scgV

22FR(t)|transient=

The steady state output, FR(t) is given by

Transient term of FR(t) is given by,

38 of 63

constructing unit vector (contd.)

Let the FR(t) is again pass through aLPF of same corner frequency, c

Using Laplace analysis

VRFR

+

-

sc

FR’+

-

sc

22

22

222

22

222

2 12

112

1

S

SC

S

C

C

CS

C

C

CS

SCgRss

s

ssVsF

t

VtF S

CS

Cg

R cos22

2

The steady state output, FR’(t) is given by

and

Transient term of FR’(t) is given by,

tcetscc

sc

scgV

2222

22

2

FR’(t)|transient=

22

22

sinCS

SC

22

2cos

CS

SC

,

t

VS

CS

Cgsin

22FR(t)|steady

VR = Vg sin(St)

39 of 63


Subtract FR’(t) from FR(t), let theresult be FR’’(t)

The steady state output, FR’’(t) is given by

and

Transient term of FR’’(t) is given by,

VRFR

+

-

sc

FR’+

-

sc

+

-

FR’’

t

VtF S

CS

SCg

R sin22

FR’’(t)|transient=

tcetc

sc

cs

sc

scgV

22

22

22

22

22

sinCS

SC

22

2cos

CS

SC

,

VR = Vg sin(St)

t

VtF S

CS

Cg

R cos22

2

40 of 63


Comparing steady state term of FR’(t) and FR’’(t)

FR’(t) and FR’’(t) are always 900

displaced irrespective of gridfrequency and corner frequency ofLPF

At the zero crossing of FR’’(t), FR’(t)will be in peak and vice versa

Unit magnitude of FR’(t) and FR’’(t)can be obtained by dividing the termby its own magnitude

VR

FR

FR’+

-+

-+

-

sc

sc

FR’’

t

VtF S

CS

Cg

R cos22

2

t

VtF S

CS

SCg

R sin22

41 of 63


ZCD

ZCD

S&H

S&H

ABS

Trigger

F1= -cos(St+)

F2 =sin(St+)

X / YX

Y

X

Y

VR

FR

+

-

sc FR

’

+

-

sc

+

-

FR’’

Trigger

ABS

X / Y

Inference from the above result

F2(t) is phase shifted from grid voltageby an angle

VR = Vg sin(St)

42 of 63


Inference from the above result (contd.)

If corner frequency of LPF (c) is set

equal to grid frequency (s), i.e. c = s :

phase shift = 0

F2(t) is in phase with grid voltage

and F1(t) is lagging the grid voltage by 900

VR = Vg sin(St)

Fix the corner frequency of LPF (c) is equal to s = 250rad/sec, where grid frequency fs = 50Hz

Let Grid frequency varies a maximum of 10% (45Hz to55Hz)

Phase shift, = 60 for -10% of grid frequency variation

Loose the proper synchronization

F1(t) = -cos(St+)

F2(t) = sin(St+)

22

22

sinCS

SC

22

2cos

CS

SC

43 of 63


Phase error with variation in grid frequency

-7

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

7

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

% ge of grid frequency variation

ph

ase

erro

r

1ph without comp

6.0

44 of 63





Rigorous method

45 of 63



Give a variation for the grid frequency s in the peakof FR’’(t)

Substitute c= s and solving will giveVR = Vg sin(St)

t

VtF S

CS

SCg

R sin22

Peak of FR’’(t), F’’R(peak), is more or less independent of

grid frequency variation

F’’R(peak) Vg/2 for c = s

2

211

1

2''

ss

sgV

sc

ssRF

1 tocompare negligible is 2

21 if

2 ''

s

gV

sc

ssRF

46 of 63


Mitigating the effect of grid frequency variation (contd.)

VR = Vg sin(St)

Vgcos = 2F’’R(peak)

Solve the following mathematical relation

F1(t) = -cos(St+)

F2(t) = sin(St+) tVVtVt SggSgS sinsincoscossin

tVVtVt SggSgS cossinsincoscos

The following trignometric relation can beused to eliminate the phase angle

t

VtF S

CS

SCg

R sin22

22

22

sinCS

SC

22

2cos

CS

SC

22

22

22

2

)(

22

CS

SCgg

CS

Cg

gpeakR VVV

VF

F’’R(peak) Vg/2 for c = s

22

''

sc

scgV

peakRF

t

VtF S

CS

Cg

R cos22

2

sin2 )( ggpeakR VVF

)()( 22sin peakRpeakRg FFV

22

2'

sc

cgV

peakRF

47 of 63


Mitigating the effect of grid frequency variation (contd.)VR = Vg sin(St)

Substituting


tVVtVt SggSgS sinsincoscossin

Rewriting the relation

t

VtF S

CS

SCg

R sin22

t

VtF S

CS

Cg

R cos22

2


)()()(1 22cos)2(sinsin peakRpeakRSpeakRSSg FFtFtUtV

22

''

sc

scgV

peakRF

22

2'

sc

cgV

peakRF

tspeakR

FtR

FtR

FU cos)(

2)('2)(''21

t

VU S

CS

SCSCgsin

222

22

1

S

SC

tanand

48 of 63



Substituting




t

VtF S

CS

SCg

R sin22

t

VtF S

CS

Cg

R cos22

2


22

''

sc

scgV

peakRF

22

2'

sc

cgV

peakRF

)()()(2 22sin)2(coscos peakRpeakRSpeakRSSg FFtFtUtV

tFtFtFU SpeakRSpeakRR sin2cos2)(2 )()(2

t

VU S

CS

SCSCgcos

222

22

2

S

SC

tanand

49 of 63


Comparing steady state term of U1(t) and U2(t)

Inference from the above result

U1(t) and U2(t) are 900 displaced

At the zero crossing of U1(t), U2(t) will be in peak and viceversa

Unit magnitude of U1(t) and U2(t) can be obtained bydividing the term by its own magnitude

U1(t) is phase shifted from grid voltage by angle (-)

t

VU S

CS

SCSCgsin

222

22

1

t

VU S

CS

SCSCgcos

222

22

2

VR = Vg sin(St)


50 of 63


Inference from the above result (contd.)

Fix the corner frequency of LPF (c) is equalto s = 250 rad/sec, where grid frequency fs= 50Hz

Let Grid frequency varies a maximum of10% (45Hz to 55Hz)

Give a variation and substitute c= s inthe phase angle (-)

VR = Vg sin(St)

22

22

sinCS

SC

22

2cos

CS

SC


tantan1

tantantan

S

S

SC

SS

1

2

1

tan

2

S

SC

tan

1 tocompare negligible is3

2

1 and

22 if

ss

51 of 63



Phase shift, (-) = 0.320 for -10% grid frequency variation


-3

-2

-1

0

1

2

3

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10


ph

ase

erro

r

1ph with comp (appx)

0.32

52 of 63


ZCD

ZCD

S&H

S&H

ABS

Trigger

X / Y

XY

XY

VR

FR

+

-

sc FR

’(t)+

-

+

-

FR’’(t)

Trigger

ABS X / Y

sc

FR’’(peak)

F1(t) = -cos(St+)

F2(t) = sin(St+)

tspeakR

FtR

FtR

FU cos)(

2)('2)(''21


+

+-

+

+-

FR’(peak) F1(t)

F2(t)

ZCD

ZCD

S&H

S&H

ABS

Trigger

X / Y

XY

XY

Trigger

ABS X / Y

U1(t)

U1(t)

U2(t)

U2(t)

cos

sin53 of 63





Rigorous method

54 of 63



The approximation made in the earlierderivation is:

The above method is Approximation method

Another method of finding out Vgsin(st) is:

22

22

sinCS

SC

22

''

sc

scgV

peakRF

sin2 )( ggpeakR VVF

22

2'

sc

cgV

peakRF

1 tocompare negligible is

2

21 if

2''

s

gV

ss

scRF

SC

CS

Cg

peakRpeakR

VFF

22 SC

CS

Cg

peakRpeakR

VFF

22

22

2

22

222

22

CS

Cg

SC

CS

Cg

peakR

peakRpeakRpeakRpeakR

V

V

F

FFFF

&

sin

22

22

g

CS

SCg VV

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Substituting


tVVtVt SggSgS sinsincoscossin


t

VtF S

CS

SCg

R sin22

t

VtF S

CS

Cg

R cos22

2

22

''

sc

scgV

peakRF

22

2'

sc

cgV

peakRF

tspeakR

FtR

FtR

FU cos)(

2)('2)(''21

t

F

FFtFtVU S

peakR

peakRpeakR

RSg cos2sin)(

2

)(

2

)(

1

t

F

FFtFtVU S

peakR

peakRpeakR

SpeakRSg sincos2cos)(

2

)(

2

)(

)(2


)(

2)(

2)(

sin

peakRF

peakRF

peakRF

gV


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Phase shift, (-) = 00 for any grid frequency


-3

-2

-1

0

1

2

3

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10


ph

ase

erro

r

1ph with comp (rigorous)

0

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ZCD

ZCD

S&H

S&H

ABS

Trigger

X / YX

Y1

X

Y2

VR

FR

+

-

sc FR

’(t)+

-

+

-

FR’’(t)

Trigger

ABS

X / Y

sc

FR’’(peak)

F1(t) = -cos(St+)

F2(t) = sin(St+)

+

+

+

-

FR’(peak)

ZCD

ZCD

S&H

S&H

ABS

Trigger

X / Y

XY

XY

Trigger

ABS X / Y

U1(t)

U1(t)

U2(t)

U2(t)

cos

sin

+-

Y12

Y22 X

X

Y

F1(t)

F2(t)2

2

t

F

FFtFtVU S

peakR

peakRpeakR

RSg cos2sin)(

2

)(

2

)(

1

t

F

FFtFtVU S

peakR

peakRpeakR

SpeakRSg sincos2cos)(

2

)(

2

)(

)(2

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Test results

Grid frequency (simulated using function generator)is varied at t=t1 from 50Hz to 45Hz


Grid voltage and unit vector without

compensation for grid frequency variation

Grid voltage and unit vector with


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Test results

Steady state waveform at 25Hz and 75Hz withapproximation method


Frequency = 25Hz Frequency = 75Hz

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Test results

Step change of frequency from 25Hz and 75Hz aswell as from 75Hz to 25Hz


Grid voltage and unit vector without


Grid voltage and unit vector with

compensation for grid frequency variation61 of 63

Conclusions

Single grid voltage is considered for the construction of unitvector

Two unity magnitude fundamental sinusoidal quantities,which are displaced by 900 from each other

One of the unit vector is in phase with grid voltageirrespective of the grid frequency variation

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Thank you

63 of 63

PLL and Grid Synchronization - National Institute of...

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