CN102081114A - Instantaneous symmetrical component method-based current detection method for distribution static synchronous compensator (DSTATCOM) - Google Patents

Abstract

The invention discloses an improved instantaneous symmetrical component method-based current detection method for a distribution static synchronous compensator (DSTATCOM), which comprises the following steps of: 1, determining a symmetrical component method and a method for expressing an instantaneous value in a phasor time domain; 2, determining the instantaneous value; 3, implementing the improved instantaneous symmetrical component method; 4, establishing a model by adopting a MATLAB simulation tool to perform simulation analysis on a conclusion; 5, filtering, converting and processing data; and 6, subtracting obtained three phases of positive fundamental active currents iafp<+>, ibfp<+> and icfp<+> from a load current to obtain required comprehensive compensation command currents iac, ibc and icc comprising harmonic components, negative sequence components and reactive components. The method avoids a complex phase shift circuit, a phase locked loop circuit, and an address generator and a table lookup process of sine and cosine function table, and eliminates errors and faults caused thereby.

Description

A kind of DSTATCOM electric current detecting method based on instantaneous symmetrical component method

Technical field

The present invention relates to a kind of DSTATCOM electric current detecting method based on instantaneous symmetrical component method.

Background technology

Present symmetrical component method is based on the analytical approach of the three-phase asymmetrical system under the steady state conditions, it represents each preface component (positive sequence, negative phase-sequence and zero sequence) of the asymmetric variable of three-phase with the form of plural number or differential, need 90 ° phase shift could obtain each preface instantaneous components in the conversion process, but the instantaneous value of this moment is owing to comprised 90 ° phase shift, therefore instantaneous value can not be represented, real-time and the higher application scenario of accuracy requirement can not be satisfied in some three-phase asymmetrical systems each preface component detection.(Distrbution STATCOM DSTATCOM), quick and precisely detects the harmonic wave that meets in the electric current, idle and asymmetrical component, is that device is realized the effectively condition precedent of compensation for the power distribution network synchronous compensator plant.

Summary of the invention

The invention provides a kind of DSTATCOM electric current detecting method based on instantaneous symmetrical component method, it has avoided complicated phase-shift circuit, phase-locked loop circuit, the address generator of sin cos functions table and the process of tabling look-up, and has eliminated the sum of errors fault of bringing thus.

The present invention has adopted following technical scheme: a kind of DSTATCOM electric current detecting method based on instantaneous symmetrical component method, and it may further comprise the steps:

Step 1, the method for expressing of instantaneous value at first definite symmetrical component method and the phasor time domain: at first voltage phasor is adopted symmetrical component method to be decomposed into the phasor of three groups of symmetries, promptly positive sequence, negative phase-sequence and zero-sequence component are suc as formula (1)-Shi (3), the U in the formula _a, U _b, U _cBe three-phase voltage phasor, U _a ⁺, U _b ⁺, U _c ⁺, U _a ^-, U _b ^-, U _c ^-, U ₀, be respectively positive sequence, negative phase-sequence and zero-sequence component, α=e ^{J2 π/3},

$\left[\begin{array}{c}{U}_a^{+}\\ U_b^{+}\\ U_c^{+}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& \mathrm{\&alpha;}& {\mathrm{\&alpha;}}^2\\ {\mathrm{\&alpha;}}^2& 1& \mathrm{\&alpha;}\\ \mathrm{\&alpha;}& {\mathrm{\&alpha;}}^2& 1\end{array}\right]\left[\begin{array}{c}{U}_a\\ U_b\\ U_c\end{array}\right]---\left(1\right)$

$\left[\begin{array}{c}{U}_a^{-}\\ U_b^{-}\\ U_c^{-}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& {\mathrm{\&alpha;}}^2& \mathrm{\&alpha;}\\ \mathrm{\&alpha;}& 1& {\mathrm{\&alpha;}}^2\\ {\mathrm{\&alpha;}}^2& \mathrm{\&alpha;}& 1\end{array}\right]\left[\begin{array}{c}{U}_a\\ U_b\\ U_c\end{array}\right]---\left(2\right)$

U ₀＝(U _a+U _b+U _c)/3 (3)

The positive sequence phasor of three-phase voltage is represented with the form of real part and imaginary part, suc as formula (4),

$\left\{\begin{array}{c}{U}_a^{+}=\mathrm{Re}{U}_a^{+}+\mathrm{jIm}{U}_a^{+}\\ U_b^{+}=\mathrm{Re}{U}_b^{+}j+\mathrm{Im}{U}_b^{+}\\ U_c^{+}=\mathrm{Re}{U}_c^{+}+\mathrm{jIm}{U}_c^{+}\end{array}\right.---\left(4\right)$

When phasor is rotated in two-dimensional quadrature coordinate system α β with a certain frequencies omega, the instantaneous value of this phasor of projection position on this phasor arbitrary therein coordinate axis of any time, here choose the β axle, then the imaginary part of the voltage phasor in the formula (4) is represented the instantaneous value of each phasor;

With in formula (4) the substitution formula (1) and with its formal expansion with real part and imaginary part

$\left[\begin{array}{c}\mathrm{Re}{U}_a^{+}\\ \mathrm{Re}{U}_b^{+}\\ \mathrm{Re}{U}_c^{+}\end{array}\right]=\frac{1}{3}\left[\begin{array}{cccccc}1& 0& -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}& 1& 0& -\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}& \frac{\sqrt{3}}{2}& 1& 0\end{array}\right]\left[\begin{array}{c}\mathrm{Re}{U}_a\\ \mathrm{Im}{U}_a\\ \mathrm{Re}{U}_b\\ \mathrm{Im}{U}_b\\ \mathrm{Re}{U}_c\\ \mathrm{Im}{U}_c\end{array}\right]---\left(5\right)$

$\left[\begin{array}{c}\mathrm{Im}{U}_a^{+}\\ \mathrm{Im}{U}_b^{+}\\ \mathrm{Im}{U}_c^{+}\end{array}\right]=\frac{1}{3}\left[\begin{array}{cccccc}0& 1& \frac{\sqrt{3}}{2}& -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ -\frac{\sqrt{3}}{2}& -\frac{1}{2}& 0& 1& \frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}& 0& 1\end{array}\right]\left[\begin{array}{c}\mathrm{Re}{U}_a\\ \mathrm{Im}{U}_a\\ \mathrm{Re}{U}_b\\ \mathrm{Im}{U}_b\\ \mathrm{Re}{U}_c\\ \mathrm{Im}{U}_c\end{array}\right]---\left(6\right)$

By above two formulas as can be known, as long as detect the real part and the imaginary part instantaneous components of three-phase voltage phasor in real time, can obtain the real part of positive sequence voltage phasor and the instantaneous value of imaginary part, also can obtain simultaneously the instantaneous value in the positive sequence voltage phasor time domain, as the formula (6), in like manner can get instantaneous value in the time domain of negative phase-sequence and zero-sequence component;

Step 2, determine instantaneous value:

Adopt the characteristic of first-harmonic sinusoidal quantity to determine above-mentioned each transient component with two point sampling methods;

If the sampling period is T _s, t ₁Be a last sampling instant, t ₂Be current sampling instant, then t ₂-t ₂=T _s,

Corresponding voltage u instantaneous value is expressed as follows:

u ₁＝U _msin(α-ωT _s) (7)

u ₂＝U _msinα (8)

In the formula: U _mBe voltage peak, α is current voltage phase angle, and ω is an angular velocity;

Formula (7) is launched:

u ₁＝U _msin(α-ωT _s)

＝U _msinαcosωT _s-U _mcosαsinωT _s (9)

＝u ₂cosωT _s-U _mcosαsinωT _s

Can get by formula (9),

U _mcosα＝(u ₂cosωT _s-u ₁)/sinωT _s (10)

According to the contextual definition of aforementioned phasor and its instantaneous value, as can be known the value of formula (10), formula (8) equal the real part of voltage phasor and imaginary part doubly, that is:

$\mathrm{ReU}=U_m\cos \mathrm{\&alpha;}/\sqrt{2}---\left(11\right)$

$=(u_2\cos \mathrm{\&omega;}{T}_s-u_1)/\sqrt{2}\sin \mathrm{\&omega;}{T}_s$

$\mathrm{ImU}=U_m\sin \mathrm{\&alpha;}/\sqrt{2}=u_2/\sqrt{2}---\left(12\right)$

By formula (11) (12) as can be known, as sampling period T _sAfter determining, because ω T _sBe definite value, because T _sVery little, when the frequencies omega of electrical network first-harmonic fluctuates ω T in a certain small scope _sCan ignore, need continuous two sampled points can determine voltage phasor U;

The three-phase voltage phasor represented with the form of real part, imaginary part by formula (11) (12) respectively and put in order;

$\left[\begin{array}{c}\mathrm{Re}{U}_a\\ \mathrm{Im}{U}_a\\ \mathrm{Re}{U}_b\\ \mathrm{Im}{U}_b\\ \mathrm{Re}{U}_c\\ \mathrm{Im}{U}_c\end{array}\right]=\frac{1}{\sqrt{2}}\left[\begin{array}{cccccc}\frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s\\ 0& 0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(13\right)$

U in the formula _A1, u _A2, u _B1, u _B2, u _C1, u _C2Be respectively two continuous sampled points of three-phase voltage;

Step 3 is improved the realization of instantaneous symmetrical component method;

Formula (13) difference substitution formula (5) (6) and arrangement are obtained:

$\left[\begin{array}{c}\mathrm{Re}{U}_a^{+}\\ \mathrm{Re}{U}_b^{+}\\ \mathrm{Re}{U}_c^{+}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;$

$\left[\begin{array}{cccccc}\frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{\mathrm{<figure-callout\; id="1"\; label="T\; s"\; filenames="HSA00000363123700023.png"\; state="\{\{state\}\}">T</figure-callout>}}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{\mathrm{<figure-callout\; id="1"\; label="T\; s"\; filenames="HSA00000363123700023.png"\; state="\{\{state\}\}">T</figure-callout>}}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(14\right)$

$\left[\begin{array}{c}\mathrm{Im}{U}_a^{+}\\ \mathrm{Im}{U}_b^{+}\\ \mathrm{Im}{U}_c^{+}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;$

$\left[\begin{array}{cccccc}0& 1& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(15\right)$

According to formula (14) (15), utilize six data of the double collection of two point sampling methods can obtain current voltage phasor, the imaginary part of voltage phasor in the formula (15) be multiply by be

The instantaneous value of positive-sequence component can be asked for amplitude, phase angle and the sin cos functions thereof of voltage positive-sequence component easily according to formula (14) (15), is example mutually with α:

$U_a^{+}=\sqrt{{\left(\mathrm{Re}{U}_a^{+}\right)}^2+{\left(\mathrm{Im}{U}_a^{+}\right)}^2}---\left(16\right)$

${\mathrm{\&alpha;}}_{U_a^{+}}=\mathrm{arctg}(\mathrm{Im}{U}_a^{+}/\mathrm{Re}{U}_a^{+})---\left(17\right)$

$\sin {\mathrm{\&alpha;}}_{U_a^{+}}=\mathrm{Im}{U}_a^{+}/U_a^{+}---\left(18\right)$

$\cos {\mathrm{\&alpha;}}_{U_a^{+}}=\mathrm{Re}{U}_a^{+}/U_a^{+}---\left(19\right)$

In like manner, can obtain the transient expression formula of voltage negative phase-sequence air quantity and zero-sequence component real part and imaginary part;

$\left[\begin{array}{c}\mathrm{Re}{U}_a^{-}\\ \mathrm{Re}{U}_b^{-}\\ \mathrm{Re}{U}_c^{-}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;$

$\left[\begin{array}{cccccc}\frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{\mathrm{<figure-callout\; id="1"\; label="T\; s"\; filenames="HSA00000363123700023.png"\; state="\{\{state\}\}">T</figure-callout>}}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{\mathrm{<figure-callout\; id="1"\; label="T\; s"\; filenames="HSA00000363123700023.png"\; state="\{\{state\}\}">T</figure-callout>}}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(20\right)$

$\left[\begin{array}{c}\mathrm{Im}{U}_a^{-}\\ \mathrm{Im}{U}_b^{-}\\ \mathrm{Im}{U}_c^{-}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;$

$\left[\begin{array}{cccccc}0& 1& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(21\right)$

$\mathrm{Re}{U}_0=\frac{1}{3\sqrt{2}}(\mathrm{ctg\&omega;}{T}_s\&times;(u_{a2}+u_{b2}+u_{c2})---\left(22\right)$

$-(u_{a1}+u_{\mathrm{<figure-callout\; id="1"\; label="b"\; filenames="HSA00000363123700023.png"\; state="\{\{state\}\}">b</figure-callout>}1}+u_{c1})/\sin \mathrm{\&omega;}{T}_s)$

$\mathrm{Im}{U}_0=(u_{a2}+u_{b2}+u_{c2})/3\sqrt{2}---\left(23\right);$

Step 4 adopts the MATLAB emulation tool to set up model above-mentioned conclusion is carried out simulation analysis, and original state is set three-phase voltage symmetry, fundamental frequency f ₀=50HZ, sample frequency is F _s=12.8HZ, simulation time is set at 0.1s, when t=0.05s, disconnects the c phase voltage, the asymmetric transient state fault of aanalogvoltage three-phase, in the transient process in a sampling period, sequence voltage produces sudden change, when voltage produces sudden change, u in the formula (11) ₂Sudden change cause the increase of molecule difference, these big data are relatively and denominator small data sin (ω T _s) the rapid increase of the ReU that causes, finally cause the sudden change of each preface component of voltage, when data processing, data are carried out amplitude limiting processing and can eliminate sudden change;

Step 5, the filtering of data, conversion and processing: three-phase imbalance distortion voltage u _a, u _b, u _cThrough arrowband analog filter filtering harmonic wave composition, output fundamental frequency component u _Af, u _Bf, u _CfAfter the AD sample conversion, obtain the real part and the imaginary part of A phase voltage fundamental positive sequence through the simple transform operation of process according to formula (14), (15), can obtain in real time and the synchronous cosine and sine signal of three-phase system voltage fundamental positive-sequence component according to formula (16), (18), (19) at last, this kind sine and cosine generation circuit is avoided complicated unstable phase-locked loop circuit on the one hand, its cosine and sine signal directly calculates acquisition, on the other hand, adopt dsp processor to carry out data processing;

Step 6 is as ω t and voltage fundamental positive sequence voltage U _Af ⁺In the time of synchronously, load current i then _a, i _b, i _cIn change DC component in the dq0 coordinate system after through the synchronous coordinate conversion into the current component of three-phase voltage fundamental positive sequence same frequency

And harmonic component and negative sequence component change secondary and above AC compounent into, after the LPF low-pass filtering, the DC component of rotation transformed in the coordinate system of three phase static can obtain three-phase fundamental positive sequence current i _Af ⁺, i _Bf ⁺, i _Cf ⁺, detect reactive current, then zero setting i when the synchronous coordinate inversion as need _qGet final product, deduct gained three-phase fundamental positive sequence active current i with load current at last _Afp ⁺, i _Bfp ⁺, i _Cfp ⁺Can obtain required harmonic wave, negative phase-sequence and the idle comprehensive compensation instruction current i of comprising _Ac, i _Bc, i _Cc

The present invention has following beneficial effect: the present invention can avoid the process of tabling look-up of 90 ° phase-shift circuit, PPL phase-locked loop circuit and sin cos functions, eliminate error and the fault brought thus, can accurate detection go out and the synchronous fundamental active current component of system voltage fundamental positive sequence, thereby detect the instruction current that the DSTATCOM device compensates the required harmonic wave that comprises, negative phase-sequence and idle component.

Description of drawings

Fig. 1 is that DSTATCOM offset current of the present invention detects theory diagram

Fig. 2 is the oscillogram of the present invention's three system voltages

Fig. 3 is a three-phase positive sequence voltage oscillogram of the present invention

Fig. 4 is a three-phase negative/positive voltage oscillogram of the present invention

Fig. 5 is a residual voltage oscillogram of the present invention

Fig. 6 is a B phase positive sequence voltage transient state process of the present invention

Embodiment

The invention discloses a kind of DSTATCOM electric current detecting method based on instantaneous symmetrical component method, it may further comprise the steps:

Step 1, the method for expressing of instantaneous value at first definite symmetrical component method and the phasor time domain:

At first voltage phasor is adopted symmetrical component method to be decomposed into the phasor of three groups of symmetries, promptly positive sequence, negative phase-sequence and zero-sequence component are suc as formula (1)-Shi (3), the U in the formula _a, U _b, U _cBe three-phase voltage phasor, U _a ⁺, U _b ⁺, U _c ⁺, U _a ^-, U _b ^-, U _c ^-, U ₀, be respectively positive sequence, negative phase-sequence and zero-sequence component, α=e ^{J2 π/3},

$\left[\begin{array}{c}{U}_a^{+}\\ U_b^{+}\\ U_c^{+}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& \mathrm{\&alpha;}& {\mathrm{\&alpha;}}^2\\ {\mathrm{\&alpha;}}^2& 1& \mathrm{\&alpha;}\\ \mathrm{\&alpha;}& {\mathrm{\&alpha;}}^2& 1\end{array}\right]\left[\begin{array}{c}{U}_a\\ U_b\\ U_c\end{array}\right]---\left(1\right)$

$\left[\begin{array}{c}{U}_a^{-}\\ U_b^{-}\\ U_c^{-}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& {\mathrm{\&alpha;}}^2& \mathrm{\&alpha;}\\ \mathrm{\&alpha;}& 1& {\mathrm{\&alpha;}}^2\\ {\mathrm{\&alpha;}}^2& \mathrm{\&alpha;}& 1\end{array}\right]\left[\begin{array}{c}{U}_a\\ U_b\\ U_c\end{array}\right]---\left(2\right)$

U ₀＝(U _a+U _b+U _c)/3 (3)

The positive sequence phasor of three-phase voltage is represented with the form of real part and imaginary part, suc as formula (4),

$\left\{\begin{array}{c}{U}_a^{+}=\mathrm{Re}{U}_a^{+}+\mathrm{jIm}{U}_a^{+}\\ U_b^{+}=\mathrm{Re}{U}_b^{+}j+\mathrm{Im}{U}_b^{+}\\ U_c^{+}=\mathrm{Re}{U}_c^{+}+\mathrm{jIm}{U}_c^{+}\end{array}\right.---\left(4\right)$

When phasor is rotated in two-dimensional quadrature coordinate system α β with a certain frequencies omega, the instantaneous value of this phasor of projection position on this phasor arbitrary therein coordinate axis of any time, here choose the β axle, then the imaginary part of the voltage phasor in the formula (4) is represented the instantaneous value of each phasor;

With in formula (4) the substitution formula (1) and with its formal expansion with real part and imaginary part

$\left[\begin{array}{c}\mathrm{Re}{U}_a^{+}\\ \mathrm{Re}{U}_b^{+}\\ \mathrm{Re}{U}_c^{+}\end{array}\right]=\frac{1}{3}\left[\begin{array}{cccccc}1& 0& -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}& 1& 0& -\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}& \frac{\sqrt{3}}{2}& 1& 0\end{array}\right]\left[\begin{array}{c}\mathrm{Re}{U}_a\\ \mathrm{Im}{U}_a\\ \mathrm{Re}{U}_b\\ \mathrm{Im}{U}_b\\ \mathrm{Re}{U}_c\\ \mathrm{Im}{U}_c\end{array}\right]---\left(5\right)$

$\left[\begin{array}{c}\mathrm{Im}{U}_a^{+}\\ \mathrm{Im}{U}_b^{+}\\ \mathrm{Im}{U}_c^{+}\end{array}\right]=\frac{1}{3}\left[\begin{array}{cccccc}0& 1& \frac{\sqrt{3}}{2}& -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ -\frac{\sqrt{3}}{2}& -\frac{1}{2}& 0& 1& \frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}& 0& 1\end{array}\right]\left[\begin{array}{c}\mathrm{Re}{U}_a\\ \mathrm{Im}{U}_a\\ \mathrm{Re}{U}_b\\ \mathrm{Im}{U}_b\\ \mathrm{Re}{U}_c\\ \mathrm{Im}{U}_c\end{array}\right]---\left(6\right)$

By above two formulas as can be known, as long as detect the real part and the imaginary part instantaneous components of three-phase voltage phasor in real time, can obtain the real part of positive sequence voltage phasor and the instantaneous value of imaginary part, also can obtain simultaneously the instantaneous value in the positive sequence voltage phasor time domain, as the formula (6), in like manner can get instantaneous value in the time domain of negative phase-sequence and zero-sequence component;

Step 2, determine instantaneous value:

Adopt the characteristic of first-harmonic sinusoidal quantity to determine above-mentioned each transient component with two point sampling methods;

If the sampling period is T _s, t ₁Be a last sampling instant, t ₂Be current sampling instant, then t ₂-t ₂=T _s,

Corresponding voltage u instantaneous value is expressed as follows:

u ₁＝U _msin(α-ωT _s) (7)

u ₂＝U _msinα (8)

In the formula: U _mBe voltage peak, α is current voltage phase angle, and ω is an angular velocity;

Formula (7) is launched:

u ₁＝U _msin(α-ωT _s)

＝U _msinαcosωT _s-U _mcosαsinωT _s

(9)

＝u ₂cosωT _s-U _mcosαsinωT _s

Can get by formula (9),

U _mcosα＝(u ₂cosωT _s-u ₁)/sinωT _s (10)

According to the contextual definition of aforementioned phasor and its instantaneous value, the value of formula (10), formula (8) equals the voltage phase as can be known

The real part of amount and imaginary part doubly, that is:

$\mathrm{ReU}=U_m\cos \mathrm{\&alpha;}/\sqrt{2}$

$=(u_2\cos \mathrm{\&omega;}{T}_s-u_1)/\sqrt{2}\sin \mathrm{\&omega;}{T}_s---\left(11\right)$

$\mathrm{ImU}=U_m\sin \mathrm{\&alpha;}/\sqrt{2}=u_2/\sqrt{2}---\left(12\right)$

By formula (11) (12) as can be known, as sampling period T _sAfter determining, because ω T _sBe definite value, because T _sVery little, when the frequencies omega of electrical network first-harmonic fluctuates ω T in a certain small scope _sCan ignore, need continuous two sampled points can determine voltage phasor U;

The three-phase voltage phasor represented with the form of real part, imaginary part by formula (11) (12) respectively and put in order;

$\left[\begin{array}{c}\mathrm{Re}{U}_a\\ \mathrm{Im}{U}_a\\ \mathrm{Re}{U}_b\\ \mathrm{Im}{U}_b\\ \mathrm{Re}{U}_c\\ \mathrm{Im}{U}_c\end{array}\right]=\frac{1}{\sqrt{2}}\left[\begin{array}{cccccc}\frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s\\ 0& 0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(13\right)$

U in the formula _A1, u _A2, u _B1, u _B2, u _C1, u _C2Be respectively two continuous sampled points of three-phase voltage;

Step 3 is improved the realization of instantaneous symmetrical component method;

Formula (13) difference substitution formula (5) (6) and arrangement are obtained:

$\left[\begin{array}{c}\mathrm{Re}{U}_a^{+}\\ \mathrm{Re}{U}_b^{+}\\ \mathrm{Re}{U}_c^{+}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;$

$\left[\begin{array}{cccccc}\frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(14\right)$

$\left[\begin{array}{c}\mathrm{Im}{U}_a^{+}\\ \mathrm{Im}{U}_b^{+}\\ \mathrm{Im}{U}_c^{+}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;$

$\left[\begin{array}{cccccc}0& 1& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(15\right)$

According to formula (14) (15), utilize six data of the double collection of two point sampling methods can obtain current voltage phasor, the imaginary part of voltage phasor in the formula (15) be multiply by be

The instantaneous value of positive-sequence component can be asked for amplitude, phase angle and the sin cos functions thereof of voltage positive-sequence component easily according to formula (14) (15), is example mutually with α:

$U_a^{+}=\sqrt{{\left(\mathrm{Re}{U}_a^{+}\right)}^2+{\left(\mathrm{Im}{U}_a^{+}\right)}^2}---\left(16\right)$

${\mathrm{\&alpha;}}_{U_a^{+}}=\mathrm{arctg}(\mathrm{Im}{U}_a^{+}/\mathrm{Re}{U}_a^{+})---\left(17\right)$

$\sin {\mathrm{\&alpha;}}_{U_a^{+}}=\mathrm{Im}{U}_a^{+}/U_a^{+}---\left(18\right)$

$\cos {\mathrm{\&alpha;}}_{U_a^{+}}=\mathrm{Re}{U}_a^{+}/U_a^{+}---\left(19\right)$

In like manner, can obtain the transient expression formula of voltage negative phase-sequence air quantity and zero-sequence component real part and imaginary part;

$\left[\begin{array}{c}\mathrm{Re}{U}_a^{-}\\ \mathrm{Re}{U}_b^{-}\\ \mathrm{Re}{U}_c^{-}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;$

$\left[\begin{array}{cccccc}\frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(20\right)$

$\left[\begin{array}{c}\mathrm{Im}{U}_a^{-}\\ \mathrm{Im}{U}_b^{-}\\ \mathrm{Im}{U}_c^{-}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;---\left(20\right)$

$\left[\begin{array}{cccccc}0& 1& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]$

$\mathrm{Re}{U}_0=\frac{1}{3\sqrt{2}}(\mathrm{ctg\&omega;}{T}_s\&times;(u_{a2}+u_{b2}+u_{c2})---\left(22\right)$

$-(u_{a1}+{\mathrm{<figure-callout\; id="1"\; label="u\; b"\; filenames="HSA00000363123700023.png"\; state="\{\{state\}\}">u</figure-callout>}}_b+u_{c1})/\sin \mathrm{\&omega;}{T}_s)$

$\mathrm{Im}{U}_0=(u_{a2}+u_{b2}+u_{c2})/3\sqrt{2}---\left(23\right);$

Step 4 adopts the MATLAB emulation tool to set up model above-mentioned conclusion is carried out simulation analysis, and original state is set three-phase voltage symmetry, fundamental frequency f ₀=50HZ, sample frequency is F _s=12.8HZ, simulation time is set at 0.1s, when t=0.05s, disconnects the c phase voltage, the asymmetric transient state fault of aanalogvoltage three-phase, in Fig. 2, during t＜0.05s, three-phase voltage waveform symmetry, the u of this moment _c=0, because the two point sampling methods of employing, the transient state transit time that only needs a sampling period, each preface component of voltage can reach steady state (SS), three-phase voltage positive-sequence component waveform and three-phase voltage waveform are in full accord in Fig. 3, and three-phase negative/positive voltage and residual voltage are zero, in Fig. 4 and Fig. 5, at t=0.05s constantly, disconnect the C phase voltage, in Fig. 3, Fig. 4, Fig. 5, under the steady state (SS) suddenly, voltage negative sequence component and zero-sequence component are not 0 owing to three-phase imbalance, and the filtration time here is t=1/T _S=78.125e ^-6 _s, filter process amplifies as Fig. 6, and in the transient process in a sampling period, sequence voltage produces sudden change, when voltage produces sudden change, u in the formula (11) ₂Sudden change cause the increase of molecule difference, these big data are relatively and denominator small data sin (ω T _s) the rapid increase of the ReU that causes, finally cause the sudden change of each preface component of voltage, when data processing, data are carried out amplitude limiting processing and can eliminate sudden change;

Step 5, in Fig. 1, the filtering of data, conversion and processing: three-phase imbalance distortion voltage u _a, u _b, u _cThrough arrowband analog filter filtering harmonic wave composition, output fundamental frequency component u _Af, u _Bf, u _CfAfter the AD sample conversion, obtain the real part and the imaginary part of A phase voltage fundamental positive sequence through the simple transform operation of process according to formula (14), (15), can obtain in real time and the synchronous cosine and sine signal of three-phase system voltage fundamental positive-sequence component according to formula (16), (18), (19) at last, this kind sine and cosine generation circuit is avoided complicated unstable phase-locked loop circuit on the one hand, its cosine and sine signal directly calculates acquisition, on the other hand, adopt dsp processor to carry out data processing;

Step 6 is as ω t and voltage fundamental positive sequence voltage U _Af ⁺In the time of synchronously, load current i then _a, i _b, i _cIn change DC component in the dq0 coordinate system after through the synchronous coordinate conversion into the current component of three-phase voltage fundamental positive sequence same frequency

And harmonic component and negative sequence component change secondary and above AC compounent into, after the LPF low-pass filtering, the DC component of rotation transformed in the coordinate system of three phase static can obtain three-phase fundamental positive sequence current i _Af ⁺, i _Bf ⁺, i _Cf ⁺, detect reactive current, then zero setting i when the synchronous coordinate inversion as need _qGet final product, deduct gained three-phase fundamental positive sequence active current i with load current at last _Afp ⁺, i _Bfp ⁺, i _Cfp ⁺Can obtain required harmonic wave, negative phase-sequence and the idle comprehensive compensation instruction current i of comprising _Ac, i _Bc, i _Cc

Claims (1)

1. DSTATCOM electric current detecting method based on the instantaneous symmetrical component method of modified, it may further comprise the steps:

Step 1, the method for expressing of instantaneous value at first definite symmetrical component method and the phasor time domain:

At first voltage phasor is adopted symmetrical component method to be decomposed into the phasor of three groups of symmetries, promptly positive sequence, negative phase-sequence and zero-sequence component are suc as formula (1)-Shi (3), the U in the formula _a, U _b, U _cBe three-phase voltage phasor, U _a ⁺, U _b ⁺, U _c ⁺, U _a ^-, U _b ^-, U _c ^-, U ₀, be respectively positive sequence, negative phase-sequence and zero-sequence component, α=e ^{J2 π/3},

$\left[\begin{array}{c}{U}_a^{+}\\ U_b^{+}\\ U_c^{+}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& \mathrm{\&alpha;}& {\mathrm{\&alpha;}}^2\\ {\mathrm{\&alpha;}}^2& 1& \mathrm{\&alpha;}\\ \mathrm{\&alpha;}& {\mathrm{\&alpha;}}^2& 1\end{array}\right]\left[\begin{array}{c}{U}_a\\ U_b\\ U_c\end{array}\right]---\left(1\right)$

$\left[\begin{array}{c}{U}_a^{-}\\ U_b^{-}\\ U_c^{-}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& {\mathrm{\&alpha;}}^2& \mathrm{\&alpha;}\\ \mathrm{\&alpha;}& 1& {\mathrm{\&alpha;}}^2\\ {\mathrm{\&alpha;}}^2& \mathrm{\&alpha;}& 1\end{array}\right]\left[\begin{array}{c}{U}_a\\ U_b\\ U_c\end{array}\right]---\left(2\right)$

U ₀＝(U _a+U _b+U _c)/3 (3)

The positive sequence phasor of three-phase voltage is represented with the form of real part and imaginary part, suc as formula (4),

$\left\{\begin{array}{c}{U}_a^{+}=\mathrm{Re}{U}_a^{+}+\mathrm{jIm}{U}_a^{+}\\ U_b^{+}=\mathrm{Re}{U}_b^{+}j+\mathrm{Im}{U}_b^{+}\\ U_c^{+}=\mathrm{Re}{U}_c^{+}+\mathrm{jIm}{U}_c^{+}\end{array}\right.---\left(4\right)$

When phasor is rotated in two-dimensional quadrature coordinate system α β with a certain frequencies omega, the instantaneous value of this phasor of projection position on this phasor arbitrary therein coordinate axis of any time, here choose the β axle, then the imaginary part of the voltage phasor in the formula (4) is represented the instantaneous value of each phasor;

With in formula (4) the substitution formula (1) and with its formal expansion with real part and imaginary part

$\left[\begin{array}{c}\mathrm{Re}{U}_a^{+}\\ \mathrm{Re}{U}_b^{+}\\ \mathrm{Re}{U}_c^{+}\end{array}\right]=\frac{1}{3}\left[\begin{array}{cccccc}1& 0& -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}& 1& 0& -\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}& \frac{\sqrt{3}}{2}& 1& 0\end{array}\right]\left[\begin{array}{c}\mathrm{Re}{U}_a\\ \mathrm{Im}{U}_a\\ \mathrm{Re}{U}_b\\ \mathrm{Im}{U}_b\\ \mathrm{Re}{U}_c\\ \mathrm{Im}{U}_c\end{array}\right]---\left(5\right)$

$\left[\begin{array}{c}\mathrm{Im}{U}_a^{+}\\ \mathrm{Im}{U}_b^{+}\\ \mathrm{Im}{U}_c^{+}\end{array}\right]=\frac{1}{3}\left[\begin{array}{cccccc}0& 1& \frac{\sqrt{3}}{2}& -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ -\frac{\sqrt{3}}{2}& -\frac{1}{2}& 0& 1& \frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}& -\frac{\sqrt{3}}{2}& -\frac{1}{2}& 0& 1\end{array}\right]\left[\begin{array}{c}\mathrm{Re}{U}_a\\ \mathrm{Im}{U}_a\\ \mathrm{Re}{U}_b\\ \mathrm{Im}{U}_b\\ \mathrm{Re}{U}_c\\ \mathrm{Im}{U}_c\end{array}\right]---\left(6\right)$

By above two formulas as can be known, as long as detect the real part and the imaginary part instantaneous components of three-phase voltage phasor in real time, can obtain the real part of positive sequence voltage phasor and the instantaneous value of imaginary part, also can obtain simultaneously the instantaneous value in the positive sequence voltage phasor time domain, as the formula (6), in like manner can get instantaneous value in the time domain of negative phase-sequence and zero-sequence component;

Step 2, determine instantaneous value:

Adopt the characteristic of first-harmonic sinusoidal quantity to determine above-mentioned each transient component with two point sampling methods;

If the sampling period is T _s, t ₁Be a last sampling instant, t ₂Be current sampling instant, then t ₂-t ₂=T _s,

Corresponding voltage u instantaneous value is expressed as follows:

u ₁＝U _msin(α-ωT _s) (7)

u ₂＝U _msinα (8)

In the formula: U _mBe voltage peak, α is current voltage phase angle, and ω is an angular velocity;

Formula (7) is launched:

u ₁＝U _msin(α-ωT _s)

＝U _msinαcosωT _s-U _mcosαsinωT _s (9)

＝u ₂cosωT _s-U _mcosαsinωT _s

Can get by formula (9),

U _mcosα＝(u ₂cosωT _s-u ₁)/sinωT _s (10)

According to the contextual definition of aforementioned phasor and its instantaneous value, as can be known the value of formula (10), formula (8) equal the real part of voltage phasor and imaginary part doubly, that is:

$\mathrm{ReU}=U_m\cos \mathrm{\&alpha;}/\sqrt{2}---\left(11\right)$

$=(u_2\cos \mathrm{\&omega;}{T}_s-u_1)/\sqrt{2}\sin \mathrm{\&omega;}{T}_s$

$\mathrm{ImU}=U_m\sin \mathrm{\&alpha;}/\sqrt{2}=u_2/\sqrt{2}---\left(12\right)$

By formula (11) (12) as can be known, as sampling period T _sAfter determining, because ω T _sBe definite value, because T _sVery little, when the frequencies omega of electrical network first-harmonic fluctuates ω T in a certain small scope _sCan ignore, need continuous two sampled points can determine voltage phasor U;

The three-phase voltage phasor represented with the form of real part, imaginary part by formula (11) (12) respectively and put in order;

$\left[\begin{array}{c}\mathrm{Re}{U}_a\\ \mathrm{Im}{U}_a\\ \mathrm{Re}{U}_b\\ \mathrm{Im}{U}_b\\ \mathrm{Re}{U}_c\\ \mathrm{Im}{U}_c\end{array}\right]=\frac{1}{\sqrt{2}}\left[\begin{array}{cccccc}\frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s\\ 0& 0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(13\right)$

U in the formula _A1, u _A2, u _B1, u _B2, u _C1, u _C2Be respectively two continuous sampled points of three-phase voltage;

Step 3 is improved the realization of instantaneous symmetrical component method;

Formula (13) difference substitution formula (5) (6) and arrangement are obtained:

$\left[\begin{array}{c}\mathrm{Re}{U}_a^{+}\\ \mathrm{Re}{U}_b^{+}\\ \mathrm{Re}{U}_c^{+}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;$

$\left[\begin{array}{cccccc}\frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(14\right)$

$\left[\begin{array}{c}\mathrm{Im}{U}_a^{+}\\ \mathrm{Im}{U}_b^{+}\\ \mathrm{Im}{U}_c^{+}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;---\left(15\right)$

$\left[\begin{array}{cccccc}0& 1& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]$

According to formula (14) (15), utilize six data of the double collection of two point sampling methods can obtain current voltage phasor, the imaginary part of voltage phasor in the formula (15) be multiply by be

The instantaneous value of positive-sequence component can be asked for amplitude, phase angle and the sin cos functions thereof of voltage positive-sequence component easily according to formula (14) (15), is example mutually with α:

$U_a^{+}=\sqrt{{\left(\mathrm{Re}{U}_a^{+}\right)}^2+{\left(\mathrm{Im}{U}_a^{+}\right)}^2}---\left(16\right)$

${\mathrm{\&alpha;}}_{U_a^{+}}=\mathrm{arctg}(\mathrm{Im}{U}_a^{+}/\mathrm{Re}{U}_a^{+})---\left(17\right)$

$\sin {\mathrm{\&alpha;}}_{U_a^{+}}=\mathrm{Im}{U}_a^{+}/U_a^{+}---\left(18\right)$

$\cos {\mathrm{\&alpha;}}_{U_a^{+}}=\mathrm{Re}{U}_a^{+}/U_a^{+}---\left(19\right)$

In like manner, can obtain the transient expression formula of voltage negative phase-sequence air quantity and zero-sequence component real part and imaginary part;

$\left[\begin{array}{c}\mathrm{Re}{U}_a^{-}\\ \mathrm{Re}{U}_b^{-}\\ \mathrm{Re}{U}_c^{-}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;$

$\left[\begin{array}{cccccc}\frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}\\ \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s+\sqrt{3}}{2}& \frac{1}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\mathrm{ctg\&omega;}{T}_s-\sqrt{3}}{2}& \frac{-1}{\sin \mathrm{\&omega;}{T}_s}& \mathrm{ctg\&omega;}{T}_s\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(20\right)$

$\left[\begin{array}{c}\mathrm{Im}{U}_a^{-}\\ \mathrm{Im}{U}_b^{-}\\ \mathrm{Im}{U}_c^{-}\end{array}\right]=\frac{1}{3\sqrt{2}}\&times;$

$\left[\begin{array}{cccccc}0& 1& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1& \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}\\ \frac{\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{-\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& \frac{-\sqrt{3}}{2\sin \mathrm{\&omega;}{T}_s}& \frac{\sqrt{3}\mathrm{ctg\&omega;}{T}_s-1}{2}& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{a1}\\ u_{a2}\\ u_{b1}\\ u_{b2}\\ u_{c1}\\ u_{c2}\end{array}\right]---\left(21\right)$

$\mathrm{Re}{U}_0=\frac{1}{3\sqrt{2}}(\mathrm{ctg\&omega;}{T}_s\&times;(u_{a2}+u_{b2}+u_{c2})$

$-(u_{a1}+u_{b1}+u_{c1})/\sin \mathrm{\&omega;}{T}_s)---\left(22\right)$

$\mathrm{Im}{U}_0=(u_{a2}+u_{b2}+u_{c2})/3\sqrt{2}---\left(23\right);$

Step 4 adopts the MATLAB emulation tool to set up model above-mentioned conclusion is carried out simulation analysis, and original state is set three-phase voltage symmetry, fundamental frequency f ₀=50HZ, sample frequency is F _s=12.8HZ, simulation time is set at 0.1s, when t=0.05s, disconnects the c phase voltage, the asymmetric transient state fault of aanalogvoltage three-phase, in the transient process in a sampling period, sequence voltage produces sudden change, when voltage produces sudden change, u in the formula (11) ₂Sudden change cause the increase of molecule difference, these big data are relatively and denominator small data sin (ω T _s) the rapid increase of the ReU that causes, finally cause the sudden change of each preface component of voltage, when data processing, data are carried out amplitude limiting processing and can eliminate sudden change;

Step 5, the filtering of data, conversion and processing: three-phase imbalance distortion voltage u _a, u _b, u _cThrough arrowband analog filter filtering harmonic wave composition, output fundamental frequency component u _Af, u _Bf, u _CfAfter the AD sample conversion, obtain the real part and the imaginary part of A phase voltage fundamental positive sequence through the simple transform operation of process according to formula (14), (15), can obtain in real time and the synchronous cosine and sine signal of three-phase system voltage fundamental positive-sequence component according to formula (16), (18), (19) at last, this kind sine and cosine generation circuit is avoided complicated unstable phase-locked loop circuit on the one hand, its cosine and sine signal directly calculates acquisition, on the other hand, adopt dsp processor to carry out data processing;

Step 6 is as ω t and voltage fundamental positive sequence voltage U _Af ⁺In the time of synchronously, load current i then _a, i _b, i _cIn change DC component in the dq0 coordinate system after through the synchronous coordinate conversion into the current component of three-phase voltage fundamental positive sequence same frequency

And harmonic component and negative sequence component change secondary and above AC compounent into, after the LPF low-pass filtering, the DC component of rotation transformed in the coordinate system of three phase static can obtain three-phase fundamental positive sequence current i _Af ⁺, i _Bf ⁺, i _Cf ⁺, detect reactive current, then zero setting i when the synchronous coordinate inversion as need _qGet final product, deduct gained three-phase fundamental positive sequence active current i with load current at last _Afp ⁺, i _Bfp ⁺, i _Cfp ⁺Can obtain required harmonic wave, negative phase-sequence and the idle comprehensive compensation instruction current i of comprising _Ac, i _Bc, i _Cc

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