CN103247996B - Compensation method for secondary current distortion caused by current transformer saturation - Google Patents

Abstract

The invention discloses a compensation method for secondary current distortion caused by current transformer saturation. The method comprises the steps as follows: 1), a secondary side distortion current signal II is sampled; 2), four continuous sampling points are taken in an undersaturated area of secondary side distortion current, and an aperiodic quantity amplitude at one of the sampling points is set to be B, so that the aperiodic sub-quantity IQ (N) of each sampling point is obtained; 3), the step 2) is repeated, n groups of four continuous sampling points are taken, and the aperiodic sub-quantity IQ (N) of each sampling point in a saturation area of the secondary side distortion current signal is calculated; 4), the amplitude Ip (N) of the periodic sub quantity of the undersaturated area of the secondary side distortion current is calculated; 5), the Ip (N) of the undersaturated area is subjected to the least square formula calculation; 6), the periodic sub-quantity amplitude of each sampling point in the saturation area is acquired; and 7), the periodic sub-quantity amplitude of each sampling point in the saturation area and the periodic sub-quantity amplitude of the saturation area of the secondary side distortion current are added together, so that compensation current I' (N) is obtained. The method has the advantages of small calculation quantity, high accuracy, real-time compensation and the like.

Description

The compensation method that CT saturation causes secondary current to distort

Technical field

The present invention relates to power train completely technical field of relay protection, particularly a kind of CT saturation compensation method of causing secondary current to distort.

Background technology

In order to ensure the operation of current system safety economy; monitoring and measuring must be carried out to the ruuning situation of circuit arrangement; current transformer is the transducer of the secondary device such as measuring instrument, the relaying protection acquisition electrical primary loop current information in electric power system; the small area analysis that the big current scaling transformation of primary system becomes secondary device directly to utilize by it; serve the effect that secondary device is protected; therefore be widely used in relay protection of power system, reliable, application is simple and the low cost advantage that to be current transformer main.

Due to the core characteristics of current transformer; Flux saturation can be reached after magnetic field energy reaches certain level; thus limit the progress of disease of electric current; cause CT saturation problem, electric power system high pressure and generation external area error etc. all can cause CT saturation, and the saturated secondary current that causes of current transformer distorts; make secondary current cannot true representation primary current; secondary device being had an immense impact on, as caused protective relaying device on-time delay, or bringing out misoperation phenomenon.

For reducing the adverse effect that CT saturation causes, Chinese scholars conducts in-depth research, and proposes detection method and the solution of various CT saturation.Wherein utilizing the distortion of software real-Time Compensation secondary current, make secondary current accurately reflect primary current, is the most simple and practical approach.

At present, the method that the secondary side distortion current caused for CT saturation compensates mainly contains following several: based on current transformer fundamental magnetization curve method, curve-fitting method and the method based on artificial neural net.

Fundamental magnetization curve method: the method by the transient magnetic flux calculated in real time is applied to magnetization curve, thus obtains exciting current, then is added on secondary side distortion current by the exciting current obtained, and obtains primary side current.But this method must depend on the magnetization curve of current transformer, and need further to study in the problem how determining initial magnetic flux.

Curve-fitting method: the method utilizes the non-saturated region electric current of fault starting stage to return the model parameter calculating fault current, and then compensates the secondary distortion current of saturation region.Method for parameter estimation does not affect by remanent magnetism, but compensation precision depends on the precision of adopted fault current model, and the impact being subject to degree of saturation is more serious.

Artificial neural network method: the method utilizes the ability of its programmable single-chip system Any Nonlinear Function, makes institute's establishing network be the inverse function of current transformer by training and study, and then completes the estimation to current transformer primary side current.But the scope of application problem of the acquisition of training sample and network need further research, and the difference of current transformer characteristic and load has limited to the extensive use of this method.

Summary of the invention

The object of the invention is to overcome the shortcoming of prior art and deficiency, there is provided that a kind of step is simple, amount of calculation is little and the compensation method that the CT saturation that precision is high causes secondary current to distort, this compensation method can in real time and compensate secondary side distortion current accurately.

Object of the present invention is achieved through the following technical solutions: the compensation method that CT saturation causes secondary current to distort, and comprises the following steps:

(1) to Current Transformer Secondary side distortion current signal I ₁sample, wherein the sampling period is T _s;

(2) in secondary side distortion current unsaturation district, 4 continuous print sampled point: K are got _f-3, K _f-2, K _f-1 and K _f, the aperiodic at one of them sampled point H place in these 4 sampled points is measured amplitude and is set as B, according to aperiodic component magnitude relation between continuous sampling point, draw the aperiodic component amplitude I of secondary side distortion current signal at each sampled point _q(N) be:

I _q(N)=B* (e ^-τ) ^n-H, N=1,2,3..., H=K _f-3, K _f-2, K _f-1 or K _f, K _f∈ [4, c];

N is each sampled point of secondary side distortion current, and wherein the individual sampled point of 1 to c belongs to the sampled point in unsaturation district; τ=T _s/ T _a, T _afor the damping time constant of attenuating dc component;

(3) repeat step (2), get n group 4 continuous print sampled points, by the secondary side distortion current signal I obtained that samples in step (2) ₁at K _f-3, K _f-2, K _f-1 and K _fthe amplitude of sample point carries out the calculating of following formula, draws with value:

$\left\{\begin{array}{c}{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}\right)}_f=\frac{I_1\left(K_f\right)-2I_1(K_f-1)\cos \mathrm{\ϵ}+I_1(K_f-2)}{2(1-\cos \mathrm{\ϵ})}\\ {\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}\right)}_f=\frac{I_1(K_f-1)-2I_1(K_f-2)\cos \mathrm{\ϵ}+I_1(K_f-3)}{2(1-\cos \mathrm{\ϵ})}\end{array}\right.,$ F=1,2...n, H=K _f-3, K _f-2, K _f-1 or K _f, 1≤n≤c-3, K _f∈ [4, c];

Wherein ε=T _sω, ω are the angular frequency of secondary side distortion current signal, I ₁(K _f), I ₁(K _f-1), I ₁(K _f-2) and I ₁(K _f-3) K is respectively _f, K _f-1, K _f-2 and K _fthe current amplitude of-3 sample point;

(4) by each in step (3) value addition is averaged and is obtained will be each value addition is averaged and is obtained then will with do division arithmetic and obtain e ^-τvalue, pass through with value obtain Be ^-τ:

$\left\{\begin{array}{c}{\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}\right)}_f}{n}\\ {\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}}\right)}_f}{n}\end{array}\right.,e^{-\mathrm{\τ}}=\frac{{\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}}{{\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}},f=\mathrm{1,2},...n,1\≤n\≤c-3,$

H=K _f-3, K _f-2, K _f-1 or K _f;

If H=K _f-3, then ${\mathrm{Be}}^{-\mathrm{\τ}}={\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]};$

If H=K _f-2, then ${\mathrm{Be}}^{-\mathrm{\τ}}={\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]};$

If H=K _f-1, then ${\mathrm{Be}}^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}\right)}^2}{{\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}};$

If H=K _f, then ${\mathrm{Be}}^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}\right)}^3}{{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}\right)}^2};$

(5) Be will obtained in step (3) ^-τand e ^-τsubstitute into I _q(N), in, the aperiodic component amplitude I of each sampled point in secondary side distortion current signal saturation region is then obtained _q(N):

I _q(N)=B* (e ^-τ) ^n-H=Be ^{-τ *}(e ^-τ) ^n-(H+1), H=K _f-3, K _f-2, K _f-1 or K _f, N=c+1, c+2...m;

Wherein the individual sampled point of c+1 to m is the sampled point in secondary side distortion current saturation region;

(6) the periodic component amplitude I of each sampled point in secondary side distortion current unsaturation district is obtained by the aperiodic component amplitude of each sampled point in secondary side distortion current signal unsaturation district _p(N):

I _p(N)=I ₁(N)-I _Q(N)，N∈[1,c]；

(7) by the periodic component amplitude I of each for secondary side distortion current sampled point _p(N) resolve into following trigonometric function and express formula:

I _p(N)=a ₁sin(ωNT _s)+a ₂cos(ωNT _s)；

Then the periodic component amplitude of each sampled point in secondary side distortion current unsaturation district obtained in step (6) is carried out least square method formula operation, obtain parameter a ₁and a ₂:

a=(X ^TX) ^-1X ^TI _P(N),N∈[1,c]；

Wherein a=[a ₁a ₂] ^Τ;

$X=\left[\begin{array}{cc}\sin \left(\mathrm{\ω}{T}_s\right)& \cos \left(\mathrm{\ω}{T}_s\right)\\ \·\·\·& \·\·\·\\ \sin \left(\mathrm{\ω}{\mathrm{NT}}_s\right)& \cos \left(\mathrm{\ω}{\mathrm{NT}}_s\right)\end{array}\right];$

(8) the parameter a will obtained in step (7) ₁and a ₂value be updated to following formula, obtain the periodic component amplitude of each sampled point in secondary side distortion current saturation region:

I _p(N)=a ₁sin(ωNT _s)+a ₂cos(ωNT _s)，N=c+1,c+2...m；

(9) the periodic component amplitude of the secondary side distortion current saturation region that the periodic component amplitude of each sampled point in secondary side distortion current saturation region step (8) obtained obtains with step (5) is carried out being added and is compensated the amplitude I ' (N) of current signal in each sampling:

I′(N)=I _Q(N)+I _p(N)，N=c+1,c+2...m。

Preferably, by K in described step (2) _fthe aperiodic component of-3 sample point is set as B, i.e. H=K _f-3, according to aperiodic component magnitude relation between continuous sampling point, draw the aperiodic component amplitude I of secondary side distortion current signal at each sampled point _q(N) be:

$I_Q\left(N\right)=B*{\left(e^{-\mathrm{\τ}}\right)}^{N-(K_f-3)}={\mathrm{Be}}^{-\mathrm{\τ}}*{\left(e^{-\mathrm{\τ}}\right)}^{N-(K_f-2)},N=\mathrm{1,2,3}...,K_f\∈[4,c];$

In described step (3) $\left\{\begin{array}{c}{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}\right)}_f={\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_f=\frac{I_1\left(K_f\right)-2I_1(K_f-1)\cos \mathrm{\ϵ}+I_1(K_f-2)}{2(1-\cos \mathrm{\ϵ})}\\ {\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}\right)}_f={\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_f=\frac{I_1(K_f-1)-2I_1(K_f-2)\cos \mathrm{\ϵ}+I_1(K_f-3)}{2(1-\cos )}\end{array}\right.,$ f=1,2...n，n≤c-3；

Obtain Be ^{-2 τ}and Be ^-τvalue is respectively:

$\left\{\begin{array}{c}{\mathrm{Be}}^{-\mathrm{\τ}}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_f}{n}\\ {\mathrm{Be}}^{-2\mathrm{\τ}}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_f}{n}\end{array}\right.,f=\mathrm{1,2}...n,n\≤c-3;$

By Be ^{-2 τ}with Be ^-τdo division arithmetic and obtain e ^-τvalue:

$e^{-\mathrm{\τ}}=\frac{{\mathrm{Be}}^{-2\mathrm{\τ}}}{{\mathrm{Be}}^{-\mathrm{\τ}}}.$

Further, by K in described step (2) _fthe periodic component of-3 sample point is set as according to the relation between neighbouring sample component dot cycle, draw the periodic component I of secondary side distortion current signal at each sampled point _p(N) be:

Further, it is characterized in that, described n gets 2, i.e. K in described step (3) _fvalue gets K respectively ₁and K ₂, by the secondary side distortion current signal I obtained that samples in step (2) ₁at K ₁, K ₁-1, K ₁-2 and K ₁the amplitude of-3 sample point and secondary side distortion current signal I ₁at K ₂, K ₂-1, K ₂-2 and K ₂the amplitude of-3 sample point carries out the calculating of following formula, draws (Be ^-τ) _f(Be ^{-2 τ}) _fvalue is:

$\left\{\begin{array}{c}{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_1=\frac{I_1(K_1-1)-2I_1(K_1-2)\cos \mathrm{\ϵ}+I_1(K_1-3)}{2(1-\cos \mathrm{\ϵ})}\\ {\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_1=\frac{I_1\left(K_1\right)-2I_1(K_1-1)\cos \mathrm{\ϵ}+I_1(K_1-2)}{2(1-\cos \mathrm{\ϵ})}\end{array}\right.,$

$\left\{\begin{array}{c}{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_2=\frac{I_1(K_2-1)-2I_1(K_2-2)\cos \mathrm{\ϵ}+I_1(K_2-3)}{2(1-\cos \mathrm{\ϵ})}\\ {\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_2=\frac{I_1\left(K_2\right)-2I_1(K_2-1)\cos \mathrm{\ϵ}+I_1(K_2-2)}{2(1-\cos \mathrm{\ϵ})}\end{array}\right..$

Further, described step (4) is middle by (Be ^-τ) ₁(Be ^-τ) ₂value addition is averaged and is obtained Be ^-τ; By (Be ^{-2 τ}) ₁(Be ^{-2 τ}) ₂value addition is averaged and is obtained Be ^{-2 τ}, then by Be ^{-2 τ}with Be ^-τdo division arithmetic and obtain e ^-τvalue:

$\left\{\begin{array}{c}{\mathrm{Be}}^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_2}{2}\\ {\mathrm{Be}}^{-2\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_2}{2}\end{array}\right.,e^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_2}{{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_2}.$

Further, it is characterized in that, described sampling period T _s=250 μ s.

The cardinal principle that the inventive method secondary side distortion current signal aperiodic component calculates is as follows:

(1), under electric power system normal operating condition, Current Transformer Secondary side current signal is:

I ₀=A ₀sin(ωt+θ)；

When electric power system is broken down, the network parameter of system is undergone mutation, fault current produces the change of fundamental voltage amplitude and phase place, but because system inductance has the characteristic suppressing current break, therefore often contain attenuating dc component in fault current, and the attenuating dc component in fault-current signal is the main cause causing CT saturation.

Fault-current signal detected, the secondary side current signal namely producing distortion is:

Wherein A ₀and A ₁be respectively the amplitude of current signal before and after fault, ω is system angle frequency, θ and be respectively before fault and the initial phase angle of fault after-current signal; B and T _abe respectively initial magnitude and the time constant of attenuating dc component.

Sample to above-mentioned secondary side distortion current, choose unsaturation district n group 4 continuous sampling points, wherein 4 sampled points are respectively K _f-3, K _f-2, K _f-1 and K _findividual sampled point.The aperiodic component at one of them sampled point H place of 4 sampled points is set as initial value B, and aperiodic component is set as:

According to the relation of aperiodic component amplitude between continuous sampling point, obtain the aperiodic component expression formula I of each sampled point _q(N) be:

I _q(N)=B* (e ^-τ) ^n-H=Be ^-τ* (e ^-τ) ^n-(H+1), N=1,2,3..., H=K _f-3, K _f-2, K _f-1 or K _f, K ∈ [4, c];

N is each sampled point of secondary side distortion current, and wherein the individual sampled point of 1 to c belongs to the sampled point in unsaturation district; Wherein ε=T _sω, τ=T _s/ T _a, T _sfor the sampling period, T _afor the damping time constant of attenuating dc component;

Can be drawn by above formula: as long as obtain I _q(N) Be in ^-τand e ^-τ, just can pass through I _q(N) the aperiodic component amplitude of each sampled point is obtained.

(2) Be under different H is calculated ^-τand e ^-τ:

(2-1) H=K is worked as _fwhen-3, the current amplitude I of four sampled points ₁(K _f), I ₁(K _f-1), I ₁(K _f-2) and I ₁(K _f-3) be respectively:

Simultaneous I ₁(K _f-1), I ₁(K _f-2) and I ₁(K _f-3) institute K is obtained _f(Be under value ^-τ) _fvalue:

${\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}}\right)}_f={\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_f=\frac{I_1(K_f-1)-2I_1(K_f-2)\cos \mathrm{\ϵ}+I_1(K_f-3)}{2(1-\cos \mathrm{\ϵ})},K_f\∈[4,c];$

Simultaneous I ₁(K _f), I ₁(K _f-1) and I ₁(K _f-2) got K is obtained _f(Be under value ^{-2 τ}) _fvalue:

${\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}}\right)}_f={\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_f=\frac{I_1(K_f-1)-2I_1(K_f-1)\cos \mathrm{\ϵ}+I_1(K_f-2)}{2(1-\cos \mathrm{\ϵ})},{f=\mathrm{1,2}...n,K}_f\∈[4,c].$

By each group of K _feach (Be under value ^-τ) _fvalue addition is averaged and is obtained Be ^-τ; Will each (Be ^{-2 τ}) _fvalue addition is averaged and is obtained Be ^{-2 τ}, then by Be ^{-2 τ}with Be ^-τdo division arithmetic and obtain e ^-τvalue;

$\left\{\begin{array}{c}{\mathrm{Be}}^{-\mathrm{\τ}}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_f}{n}\\ {\mathrm{Be}}^{-2\mathrm{\τ}}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_f}{n}\end{array}\right.,e^{-\mathrm{\τ}}=\frac{{\mathrm{Be}}^{-2\mathrm{\τ}}}{{\mathrm{Be}}^{-\mathrm{\τ}}},f=\mathrm{1,2}...n,1\≤n\≤c-3;$

(2-2) H=K is worked as _fwhen-2, by K _fthe aperiodic component amplitude of-2 sample point is set as B, then I ₁(K _f), I ₁(K _f-1), I ₁(K _f-2) and I ₁(K _f-3) be respectively:

Simultaneous I ₁(K _f-1), I ₁(K _f-2) and I ₁(K _f-3) obtain (B) _fvalue,

${\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}}\right)}_f={\left(B\right)}_f=\frac{I_1(K_f-1)-2I_1(K_f-2)\cos \mathrm{\ϵ}+I_1(K_f-3)}{2(1-\cos \mathrm{\ϵ})},{f=\mathrm{1,2}...n;K}_f\∈[4,c].$

Simultaneous I ₁(K _f), I ₁(K _f-1) and I ₁(K _f-2) Be is obtained ^-τvalue:

${\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}}\right)}_f={\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_f=\frac{I_1\left(K_f\right)-2I_1(K_f-1)\cos \mathrm{\ϵ}+I_1(K_f-2)}{2(1-\cos \mathrm{\ϵ})},{f=\mathrm{1,2}...n;K}_f\∈[4,c].$

By each group of K _feach (B) under value _fvalue addition is averaged and is obtained B; Will each (Be ^-τ) _fvalue addition is averaged and is obtained Be ^-τ, then by Be ^-τdo division arithmetic with B and obtain e ^-τvalue;

$\left\{\begin{array}{c}B=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left(B\right)}_f}{n}\\ {\mathrm{Be}}^{-\mathrm{\τ}}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_f}{n}\end{array}\right.,e^{-\mathrm{\τ}}=\frac{{\mathrm{Be}}^{-\mathrm{\τ}}}{B},f=\mathrm{1,2}...n,1\≤n\≤c-3.$

(2-3) H=K is worked as _fwhen-1, by K _fthe aperiodic component amplitude of-1 sample point is set as B, then I ₁(K _f), I ₁(K _f-1), I ₁(K _f-2) and I ₁(K _f-3) be respectively:

Simultaneous I ₁(K _f-1), I ₁(K _f-2) and I ₁(K _f-3) (Be is obtained ^τ) _fvalue,

${\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}}\right)}_f={\left({\mathrm{Be}}^{\mathrm{\τ}}\right)}_f=\frac{I_1(K_f-1)-2I_1(K_f-2)\cos \mathrm{\ϵ}+I_1(K_f-3)}{2(1-\cos \mathrm{\ϵ})},{f=\mathrm{1,2}...n;K}_f\∈[4,c].$

Simultaneous I ₁(K _f), I ₁(K _f-1) and I ₁(K _f-2) obtain (B) _fvalue:

${\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}}\right)}_f={\left(B\right)}_f=\frac{I_1\left(K_f\right)-2I_1(K_f-1)\cos \mathrm{\ϵ}+I_1(K_f-2)}{2(1-\cos \mathrm{\ϵ})},{f=\mathrm{1,2}...n;K}_f\∈[4,c].$

By each group of K _feach (Be under value ^τ) _fvalue addition is averaged and is obtained Be ^τ; Will respectively (B) _fvalue addition is averaged and is obtained B, then by B and Be ^τdo division arithmetic and obtain e ^-τvalue;

$\left\{\begin{array}{c}B{e}^{\mathrm{\τ}}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{\mathrm{\τ}}\right)}_f}{n}\\ B=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left(B\right)}_f}{n}\end{array}\right.,e^{-\mathrm{\τ}}=\frac{B}{{\mathrm{Be}}^{\mathrm{\τ}}},f=\mathrm{1,2}...n,1\≤n\≤c-3;$

Then calculate ${\mathrm{Be}}^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{\left[(H-(K-3)\mathrm{\τ}-2\mathrm{\τ})\right]}\right)}^2}{{\mathrm{Be}}^{[(H-(K-3))\mathrm{\τ}-\mathrm{\τ}]}}=\frac{{\left(B\right)}^2}{{\mathrm{Be}}^{\mathrm{\τ}}}.$

(2-4) H=K is worked as _ftime, the aperiodic component amplitude by K sample point is set as B, then draw I according to above formula ₁(K _f), I ₁(K _f-1), I ₁(K _f-2) and I ₁(K _f-3) be respectively:

Simultaneous I ₁(K _f-1), I ₁(K _f-2) and I ₁(K _f-3) (Be is obtained ^{2 τ}) _fvalue,

${\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}}\right)}_f={\left({\mathrm{Be}}^{2\mathrm{\τ}}\right)}_f=\frac{I_1(K_f-1)-2I_1(K_f-2)\cos \mathrm{\ϵ}+I_1(K_f-3)}{2(1-\cos \mathrm{\ϵ})},{f=\mathrm{1,2}...n;K}_f\∈[4,c].$

Simultaneous I ₁(K _f), I ₁(K _f-1) and I ₁(K _f-2) (the Be obtained ^τ) _fvalue:

${\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}}\right)}_f={\left({\mathrm{Be}}^{\mathrm{\τ}}\right)}_f=\frac{I_1\left(K_f\right)-2I_1(K_f-1)\cos \mathrm{\ϵ}+I_1(K_f-2)}{2(1-\cos \mathrm{\ϵ})},{f=\mathrm{1,2}...n;K}_f\∈[4,c].$

By each group of K _feach (Be under value ^{2 τ}) _fvalue addition is averaged and is obtained Be ^{2 τ}; Will each (Be ^τ) _fvalue addition is averaged and is obtained Be ^τ, then by Be ^τwith Be ^{2 τ}do division arithmetic and obtain e ^-τvalue;

$\left\{\begin{array}{c}{\mathrm{Be}}^{2\mathrm{\τ}}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{2\mathrm{\τ}}\right)}_f}{n}\\ {\mathrm{Be}}^{\mathrm{\τ}}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{\mathrm{\τ}}\right)}_f}{n}\end{array}\right.,e^{-\mathrm{\τ}}=\frac{{\mathrm{Be}}^{\mathrm{\τ}}}{{\mathrm{Be}}^{2\mathrm{\τ}}},f=\mathrm{1,2}...n,1\≤n\≤c-3;$

Then calculate ${\mathrm{Be}}^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{[(H-(K-3))\mathrm{\τ}-2\mathrm{\τ}]}\right)}^3}{{\left({\mathrm{Be}}^{[(H-(K-3))\mathrm{\τ}-\mathrm{\τ}]}\right)}^2}=\frac{{\left({\mathrm{Be}}^{\mathrm{\τ}}\right)}^3}{{\left({\mathrm{Be}}^{2\mathrm{\τ}}\right)}^2}.$

The present invention, by getting many groups 4 continuous print sampled points, obtains different K _funder value with value, then gets their mean value, to obtain e more accurately ^-τand Be ^-τvalue.

The present invention has following advantage and effect relative to prior art:

(1) the inventive method first utilizes the aperiodic component of one or more groups 4 sampled points of unsaturation district to obtain the aperiodic component of each sampled point in unsaturation district and saturation region, then the periodic component amplitude of each sampled point in unsaturation district is obtained by the aperiodic component amplitude of each sampled point of unsaturation district, again the amplitude of each for secondary side distortion current unsaturation district sampled point periodic component is carried out least square method computing, obtain the periodic component amplitude of each sampled point in saturation region, finally the aperiodic component amplitude of each for saturation region sampled point and periodic component amplitude are added, obtain final offset current.Without the need to carrying out Taylor expansion computing to aperiodic component in the inventive method, can be realized by least square method computing, therefore little, the step of the inventive method amount of calculation simply and improve the precision of calculating to a certain extent.

(2) sampled point that the inventive method utilizes only relates to unsaturation district; therefore, it is possible to the secondary distortion current produced when satisfying to current transformer carries out compensating in real time and accurately; time delay is very little, can effectively avoid protective relaying device to produce the phenomenon of time delay or malfunction.

(3) the inventive method does not rely on magnetization curve, the iron core remanent magnetism summation current transformer parameter of CT, thus can not limit to its range of application.

Accompanying drawing explanation

Fig. 1 is current system fault model schematic diagram of the present invention.

Fig. 2 to be remanent magnetism be in 0% situation secondary side saturation current signal, unsaturation current signal and the saturation current signal through overcompensation.

Fig. 3 to be remanent magnetism be in 80% situation secondary side saturation current signal, unsaturation current signal and the saturation current signal through overcompensation.

Fig. 4 to be remanent magnetism be in-80% situation secondary side saturation current signal, unsaturation current signal and the saturation current signal through overcompensation.

Embodiment

Below in conjunction with embodiment and accompanying drawing, the present invention is described in further detail, but embodiments of the present invention are not limited thereto.

Embodiment

Be illustrated in figure 1 the electric power system that the present embodiment uses, wherein S ₁and S ₂be one group of equivalent AC current source, E _s1=230 ∠ 0 ° Kv; E _s2=230 ∠ 30 ° Kv; Impedance Z _s1=Z _s2=1.600+j8.664 Ω; Transmission line total length is 200Km, every kilometer of positive sequence impedance Z _l1with zero sequence impedance Z _l0be respectively: Z _l1=0.013+j0.293 Ω, Z _l0=0.386+j0.293 Ω; This work of electric power system frequency is f=50Hz.The sample frequency of sampling to Current Transformer Secondary side distortion current signal is 4000Hz, i.e. each cycle 80 sampled points.

The current transformer that the present embodiment adopts is Jiles-Atherton model, and wherein primary side and secondary side no-load voltage ratio are 200:20; In the t=0.04s moment, there is the fault of single-phase short circuit ground connection in this power system transmission line and bus distance 100Km place.

The compensation method that the present embodiment causes secondary current to distort to this electric power system CT saturation, comprises the following steps:

(1) sampling apparatus is with T _sthe sampling period of=250 μ s is to Current Transformer Secondary side distortion current signal I ₁sample.

(2) in secondary side distortion current unsaturation district, 4 continuous print sampled point K are got _f-3, K _f-2, K _f-1 and K _f, by K _fthe aperiodic of-3 sample point measures amplitude and is set as B, simultaneously by K _fthe periodic component amplitude of-3 sample point is set as according to the relation between the relation of aperiodic component amplitude between continuous sampling point and neighbouring sample component amplitude dot cycle, draw the aperiodic component I of secondary side distortion current signal at each sampled point _qand periodic component I (N) _p(N) be respectively:

$I_Q\left(N\right)=B*{\left(e^{-\mathrm{\τ}}\right)}^{N-(K_f-3)},N=\mathrm{1,2,3}...,K_f\∈[4,c];$

Wherein N represents N number of sampled point of secondary side distortion current, and wherein the individual sampled point of 1 to c belongs to the sampled point in unsaturation district.

Wherein ε=T _sω, ω are the angular frequency of secondary side distortion current signal, ω=2 π f=100 π.τ=T _s/ T _a, T _afor the damping time constant of attenuating dc component; A ₁and be respectively K _fthe current amplitude of-3 sampled points and initial angle.

(3) repeat step (2), get 2 groups of 4 continuous print sampled points, namely get 2 different K _fvalue, is respectively K ₁and K ₂, when f gets 1 and 2 respectively, by the secondary side distortion current signal I obtained that samples in step (2) ₁at K ₁, K ₁-1, K ₁-2 and K ₁the amplitude of-3 sample point and secondary side distortion current signal I ₁at K ₂, K ₂-1, K ₂-2 and K ₂the amplitude of-3 sample point carries out the calculating of following formula, draws K ₁and K ₂(Be under value ^-τ) _f(Be ^{-2 τ}) _fvalue:

$\left\{\begin{array}{c}{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_1=\frac{I_1(K_1-1)-2I_1(K_1-2)\cos \mathrm{\ϵ}+I_1(K_1-3)}{2(1-\cos \mathrm{\ϵ})}\\ {\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_1=\frac{I_1\left(K_1\right)-2I_1(K_1-1)\cos \mathrm{\ϵ}+I_1(K_1-2)}{2(1-\cos \mathrm{\ϵ})}\end{array}\right.,K_1\∈[4,c];$

$\left\{\begin{array}{c}{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_2=\frac{I_1(K_2-1)-2I_1(K_2-2)\cos \mathrm{\ϵ}+I_1(K_2-3)}{2(1-\cos \mathrm{\ϵ})}\\ {\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_2=\frac{I_1\left(K_2\right)-2I_1(K_2-1)\cos \mathrm{\ϵ}+I_1(K_2-2)}{2(1-\cos \mathrm{\ϵ})}\end{array}\right.,K_2\∈[4,c].$

(4) by (Be in step (3) ^-τ) ₁(Be ^-τ) ₂value addition is averaged and is obtained Be ^-τ; By (Be ^{-2 τ}) ₁(Be ^{-2 τ}) ₂value addition is averaged and is obtained Be ^{-2 τ}, then by Be ^{-2 τ}with Be ^-τdo division arithmetic and obtain e ^-τvalue;

$\left\{\begin{array}{c}{\mathrm{Be}}^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_2}{2}\\ {\mathrm{Be}}^{-2\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_2}{2}\end{array}\right.,e^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_2}{{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_2}.$

(5) Be will obtained in step (3) ^-τand e ^-τsubstitute into the I of step (2) _q(N), in, the aperiodic component amplitude I of each sampled point in secondary side distortion current signal saturation region is then obtained _q(N):

$I_Q\left(N\right)=B*{\left(e^{-\mathrm{\τ}}\right)}^{N-(K_f-3)}={\mathrm{Be}}^{-\mathrm{\τ}}*{\left(e^{-\mathrm{\τ}}\right)}^{N-(K_f-2)},N=c+1,c+2...m,K_f\∈[4,c];$

Wherein the individual sampled point of c+1 to m is the sampled point in secondary side distortion current saturation region.

(6) by the aperiodic component of each sampled point in the secondary side distortion current signal unsaturation district that obtains in step (5), the amplitude I of each sampled point periodic component in secondary side distortion current unsaturation district is calculated _p(N):

I _p(N)=I ₁(N)-I _Q(N)，N∈[1,c]。

(7) the periodic component amplitude I of each sampled point of secondary side distortion current _p(N) trigonometric function resolved into is:

The periodic component amplitude of each sampled point in secondary side distortion current unsaturation district is carried out following least square method formula operation, obtains parameter a ₁and a ₂:

a=(X ^TX) ^-1X ^TI _P(N),N∈[1,c]；

Wherein a=[a ₁a ₂] ^Τ;

$X=\left[\begin{array}{cc}\sin \left(\mathrm{\ω}{T}_s\right)& \cos \left(\mathrm{\ω}{T}_s\right)\\ \·\·\·& \·\·\·\\ \sin \left(\mathrm{\ω}{\mathrm{NT}}_s\right)& \cos \left(\mathrm{\ω}{\mathrm{NT}}_s\right)\end{array}\right].$

(8) the parameter a will obtained in step (7) ₁and a ₂value be updated to following formula, obtain the periodic component amplitude of each sampled point in secondary side distortion current saturation region,

I _p(N)=a ₁sin(ωNT _s)+a ₂cos(ωNT _s)，N=c+1,c+2...m。

(9) periodic component of each sampled point in secondary side distortion current saturation region that periodic component value and the step (5) of each sampled point in secondary side distortion current saturation region step (8) obtained obtain is carried out being added and is compensated electric current I ' (N):

I′(N)=I _Q(N)+I _p(N)，N=c+1,c+2...m。

Being respectively iron core remanent magnetism is as shown in Figure 2, Figure 3 and Figure 4 in 0%, 80% and-80% three kind of situation, adopts the current signal that the present embodiment method obtains.The distortion current signal that what wherein dotted line represented record when being CT saturation, namely there occurs the secondary side current signal of distortion; What solid line represented is the normal current signal that current transformer primary side corresponds to secondary side; What circle represented is utilize the present embodiment method to be compensated current signal I ' (N).

Error between the normal current signal that compensating current signal and current transformer primary side correspond to secondary side is called compensating error, calculates the compensating error in Fig. 2 to Fig. 4 according to following compensating error computing formula:

$\mathrm{\β}\%=\frac{\sqrt{\frac{1}{M}{\mathrm{\Σ}}_{N=1}^M{(i_1\left(N\right)-I^{\′}\left(N\right))}^2}}{\max \left(i_1\right)-\min \left(i_1\right)}\×100,N=\mathrm{1,2,3}....$

Wherein M=80 refers to the sampled point number in one-period T=1/f=0.02s, and I ' (N) is compensating current signal, i ₁then for current transformer primary side corresponds to the normal current signal of secondary side.

Show that the compensating error of Fig. 2, Fig. 3 and Fig. 4 is respectively 0.33%, 0.39% and 0.18% by above formula.Visible, the method for the present embodiment can realize the compensation of degree of precision, and not by the impact of iron core remanent magnetism.

The present embodiment also can by the K in four sampled points _f-2, K _f-1 or the periodic component amplitude of Kf be set as B, simultaneously by K _f-2, K _f-1 or K _fthe periodic component of sample point is set as or then according to the relation between the relation of aperiodic component amplitude between continuous sampling point and neighbouring sample component dot cycle, calculate aperiodic component amplitude and the periodic component amplitude of other three sampled points, thus obtain the aperiodic component amplitude of all sampled points of secondary side distortion current signal.

Above-described embodiment is the present invention's preferably execution mode; but embodiments of the present invention are not restricted to the described embodiments; change, the modification done under other any does not deviate from Spirit Essence of the present invention and principle, substitute, combine, simplify; all should be the substitute mode of equivalence, be included within protection scope of the present invention.

Claims (6)

1. the CT saturation compensation method that causes secondary current to distort, is characterized in that, comprise the following steps:

(1) to Current Transformer Secondary side distortion current signal I ₁sample, wherein the sampling period is T _s;

(2) in secondary side distortion current unsaturation district, 4 continuous print sampled point: K are got _f-3, K _f-2, K _f-1 and K _f, the aperiodic at one of them sampled point H place in these 4 sampled points is measured amplitude and is set as B, according to aperiodic component magnitude relation between continuous sampling point, draw the aperiodic component amplitude I of secondary side distortion current signal at each sampled point _q(N) be:

I _q(N)=B* (e ^-τ) ^n-H, N=1,2,3..., H=K _f-3, K _f-2, K _f-1 or K _f, f=1,2...n, 1≤n≤c-3, K _f∈ [4, c];

N is each sampled point of secondary side distortion current, and wherein the individual sampled point of 1 to c belongs to the sampled point in unsaturation district; τ=T _s/ T _a, T _afor the damping time constant of attenuating dc component;

(3) repeat step (2), get n group 4 continuous print sampled points, by the secondary side distortion current signal I obtained that samples in step (2) ₁at K _f-3, K _f-2, K _f-1 and K _fthe amplitude of sample point carries out the calculating of following formula, draws with value:

$\left\{\begin{array}{c}{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}\right)}_f=\frac{I_1\left(K_f\right)-{2I}_1(K_f-1)\cos \mathrm{\ϵ}+I_1(K_f-2)}{2(1-\cos \mathrm{\ϵ})}\\ {\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}\right)}_f=\frac{I_1(K_f-1)-{2I}_1(K_f-2)\cos \mathrm{\ϵ}+I_1(K_f-3)}{2(1-\cos \mathrm{\ϵ})}\end{array}\right.,$ F=1,2...n, H=K _f-3, K _f-2, K _f-1 or K _f, 1≤n≤c-3, K _f∈ [4, c];

Wherein ε=T _sω, ω are the angular frequency of secondary side distortion current signal, I ₁(K _f), I ₁(K _f-1), I ₁(K _f-2) and I ₁(K _f-3) K is respectively _f, K _f-1, K _f-2 and K _fthe current amplitude of-3 sample point;

(4) by each in step (3) value addition is averaged and is obtained will be each value addition is averaged and is obtained then will with do division arithmetic and obtain e ^-τvalue, pass through with value obtain Be ^-τ:

$\left\{\begin{array}{c}{\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}\right)}_f}{n}\\ {\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}\right)}_f}{n}\end{array}\right.,e^{-\mathrm{\τ}}=\frac{{\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}}{{\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}},$ F=1,2...n, 1≤n≤c-3, H=K _f-3, K _f-2, K _f-1 or K _f;

If H=K _f-3, then ${\mathrm{Be}}^{-\mathrm{\τ}}={\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]};$

If H=K _f-2, then ${\mathrm{Be}}^{-\mathrm{\τ}}={\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]};$

If H=K _f-1, then ${\mathrm{Be}}^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}\right)}^2}{{\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}};$

If H=K _f, then ${\mathrm{Be}}^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{\left[(H-(K_f-3))\mathrm{\τ}2\mathrm{\τ}\right]}\right)}^3}{{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}}\right)}^2};$

(5) Be will obtained in step (3) ^-τand e ^-τsubstitute into I _q(N), in, the aperiodic component amplitude I of each sampled point in secondary side distortion current signal saturation region is then obtained _q(N):

I _q(N)=B* (e ^-τ) ^n-H=Be ^-τ* (e ^-τ) ^n-(H+1), H=K _f-3, K _f-2, K _f-1 or K _f, N=c+1, c+2...m;

Wherein the individual sampled point of c+1 to m is the sampled point in secondary side distortion current saturation region;

(6) the periodic component amplitude I of each sampled point in secondary side distortion current unsaturation district is obtained by the aperiodic component amplitude of each sampled point in secondary side distortion current signal unsaturation district _p(N):

I _p(N)＝I ₁(N)-I _Q(N)，N∈[1,c]；

Wherein I ₁(N) that represent is Current Transformer Secondary side distortion current signal I ₁in the value at correspondence each sampled point N place;

(7) by the periodic component amplitude I of each for secondary side distortion current sampled point _p(N) resolve into following trigonometric function and express formula:

I _p(N)＝a ₁sin(ωNT _s)+a ₂cos(ωNT _s)；

Then the periodic component amplitude of each sampled point in secondary side distortion current unsaturation district obtained in step (6) is carried out least square method formula operation, obtain parameter a ₁and a ₂:

a＝(X ^TX) ^-1X ^TI _P(N),N∈[1,c]；

Wherein a=[a ₁a ₂] ^Τ;

$X=\left[\begin{array}{cc}\sin \left({\mathrm{\ωT}}_s\right)& \cos \left({\mathrm{\ωT}}_s\right)\\ ...& ...\\ \sin \left({\mathrm{\ωNT}}_s\right)& \cos \left({\mathrm{\ωNT}}_s\right)\end{array}\right];$

(8) the parameter a will obtained in step (7) ₁and a ₂value be updated to following formula, obtain the periodic component amplitude of each sampled point in secondary side distortion current saturation region:

I _p(N)＝a ₁sin(ωNT _s)+a ₂cos(ωNT _s)，N＝c+1,c+2...m；

(9) the periodic component amplitude of the secondary side distortion current saturation region that the periodic component amplitude of each sampled point in secondary side distortion current saturation region step (8) obtained obtains with step (5) is carried out being added and is compensated the amplitude I ' (N) of current signal in each sampling:

I′(N)＝I _Q(N)+I _p(N)，N＝c+1,c+2...m。

2. the CT saturation according to claim 1 compensation method that causes secondary current to distort, is characterized in that, by K in described step (2) _fthe aperiodic component of-3 sample point is set as B, i.e. H=K _f-3, according to aperiodic component magnitude relation between continuous sampling point, draw the aperiodic component amplitude I of secondary side distortion current signal at each sampled point _q(N) be:

$I_Q\left(N\right)=B*{\left(e^{-\mathrm{\τ}}\right)}^{N-(K_f-3)}={\mathrm{Be}}^{-\mathrm{\τ}}*{\left(e^{-\mathrm{\τ}}\right)}^{N-(K_f-2)},$ N＝1,2,3...，K _f∈[4,c]；

In described step (3) $\left\{\begin{array}{c}{\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-2\mathrm{\τ}]}\right)}_f={\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_f=\frac{I_1\left(K_f\right)-{2I}_1(K_f-1)\cos \mathrm{\ϵ}+I_1(K_f-2)}{2(1-\cos \mathrm{\ϵ})}\\ {\left({\mathrm{Be}}^{[(H-(K_f-3))\mathrm{\τ}-\mathrm{\τ}]}\right)}_f={\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_f=\frac{I_1(K_f-1)-{2I}_1(K_f-2)\cos \mathrm{\ϵ}+I_1(K_f-3)}{2(1-\cos \mathrm{\ϵ})}\end{array}\right.,$ f＝1,2...n，n≤c-3；

Obtain Be ^{-2 τ}and Be ^-τvalue is respectively:

$\left\{\begin{array}{c}{\mathrm{Be}}^{-\mathrm{\τ}}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_f}{n}\\ {\mathrm{Be}}^{-2\mathrm{\τ}}=\frac{\underset{f=1}{\overset{n}{\mathrm{\Σ}}}{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_f}{n}\end{array}\right.,f=\mathrm{1,2}...n,n\≤c-3;$

By Be ^{-2 τ}with Be ^-τdo division arithmetic and obtain e ^-τvalue:

$e^{-\mathrm{\τ}}=\frac{{\mathrm{Be}}^{-2\mathrm{\τ}}}{{\mathrm{Be}}^{-\mathrm{\τ}}}.$

3. the CT saturation according to claim 2 compensation method that causes secondary current to distort, is characterized in that, by K in described step (2) _fthe periodic component of-3 sample point is set as according to the relation between neighbouring sample component dot cycle, draw the periodic component I of secondary side distortion current signal at each sampled point _p(N) be:

Wherein A ₁and be respectively K _fthe current amplitude of-3 sampled points and initial angle.

4. the CT saturation according to claim 2 compensation method that causes secondary current to distort, it is characterized in that, described n gets 2, i.e. K in described step (3) _fvalue gets K respectively ₁and K ₂, by the secondary side distortion current signal I obtained that samples in step (2) ₁at K ₁, K ₁-1, K ₁-2 and K ₁the amplitude of-3 sample point and secondary side distortion current signal I ₁at K ₂, K ₂-1, K ₂-2 and K ₂the amplitude of-3 sample point carries out the calculating of following formula, draws (the Be under two K values ^-τ) _f(Be ^{-2 τ}) _fvalue is:

$\left\{\begin{array}{c}{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_1=\frac{I_1(K_1-1)-{2I}_1(K_1-2)\cos \mathrm{\ϵ}+I_1(K_1-3)}{2(1-\cos \mathrm{\ϵ})}\\ {\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_1=\frac{I_1\left(K_1\right)-{2I}_1(K_1-1)\cos \mathrm{\ϵ}+I_1(K_1-2)}{2(1-\cos \mathrm{\ϵ})}\end{array}\right.,$

$\left\{\begin{array}{c}{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_2=\frac{I_1(K_2-1)-{2I}_1(K_2-2)\cos \mathrm{\ϵ}+I_1(K_2-3)}{2(1-\cos \mathrm{\ϵ})}\\ {\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_2=\frac{I_1\left(K_2\right)-{2I}_1(K_2-1)\cos \mathrm{\ϵ}+I_1(K_2-2)}{2(1-\cos \mathrm{\ϵ})}\end{array}\right..$

5. the CT saturation according to claim 4 compensation method that causes secondary current to distort, is characterized in that, by (Be in described step (4) ^-τ) ₁(Be ^-τ) ₂value addition is averaged and is obtained Be ^-τ; By (Be ^{-2 τ}) ₁(Be ^{-2 τ}) ₂value addition is averaged and is obtained Be ^{-2 τ}, then by Be ^{-2 τ}with Be ^-τdo division arithmetic and obtain e ^-τvalue:

$\left\{\begin{array}{c}{\mathrm{Be}}^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_2}{2}\\ {\mathrm{Be}}^{-2\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_2}{2}\end{array}\right.,e^{-\mathrm{\τ}}=\frac{{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-2\mathrm{\τ}}\right)}_2}{{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_1+{\left({\mathrm{Be}}^{-\mathrm{\τ}}\right)}_2}.$

6. the CT saturation according to any one of claim 1 to 5 compensation method that causes secondary current to distort, is characterized in that, described sampling period T _s=250 μ s.