State Estimation Introduction 2.0 Exact Pseudo-measurments

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1 State Estimation. Introduction In these notes, we eplore two very practical and related issues in regards to state estimation: - use of pseudo-measurements - network observability. Eact Pseudo-measurments It is important to keep in mind that the objective of state estimation is to obtain a computer model that accurately represents the current conditions in the power system. So if we can think of ways to improve the model using something other than actual measurements, we should feel free to do that. Psuedo-measurements are not measurements but are used in the state-estimation algorithm as if they were. If we can know with certainty that a particular pseudo-measurement is accurate, we should use it as it will increase the accuracy of our state estimate.

2 he most common eact pseudo-measurement is the bus injection at a substation that has no generation and serves no load. Figure below illustrates. Bus p Fig. In Fig., bus p has no generation or load. We therefore know the real and reactive power injection of this bus with precision; it is. nd so we can add two more measurements to the measurements that we actually have: zi hi i zi hi i

3 where: Measurements h P h i i z i P p, inj z i Q p, inj 4 G cos B sin n p, inj Vp Vk pk p k pk p k k n 5 G sin B cos Qp, inj Vp Vk pk p k pk p k k 6 We recognize the summations of eqs. 5 and 6 as the power flow equations for real and reactive power injection, respectively. he terms η i and η i+ are zero-mean Gaussian distributed errors for the pseudo-measurements. We can account for the fact that these pseudomeasurements are eact by letting σ i and σ i+ be very small and therefore the corresponding weights, /σ i, and /σ i+, to be very large he weighted least-square estimation algorithm is then carried out as usual.

4 . Observability Recall our very first eample at the beginning of the first set of state estimation notes. It is below. In the circuit given of Fig., current injections I, I, and voltage E are unknown. Let R =R =R =. Ω. he measurements are: meter : i, =. mpere meter : i, =-. mpere meter : i, =.8 mpere meter V: e=. volt he problem is to determine the state of the circuit, which in this case is nodal voltages v, v, and the voltage e across the voltage source. R + V - Node Node + I I R R e - Node Fig. 4

5 5 o determine the state of the circuit v, v, and e, we wrote each one of the measurements in terms of the states. We then epressed these four equations in matri form:..8.. e v v 7 Let s denote terms in eq. 7 as,, and b, so: b 8 We solved eq. 8 using: b b G b I 9 First, the gain matri is given as G he inverse of the gain matri is then found from Matlab as 4 G

6 6 he pseudo-inverse is then 4 4 I G hen we obtained the least squares estimate of the states from the 4 measurements as b I Question: What would happen to this problem if we lost a measurement? Let s say that we lost, the measurement on the current flowing from bus to bus. Let s see what happens. o solve it, we just remove the first equation corresponding to, in eq. 7, the first row of the -matri and the top element in b-vector...8. e v v 4

7 7 ctually, here the matri is and therefore we can solve eactly as:..8. e v v 5 But let s go ahead and use eq. 9 to see what happens. First, the gain matri is given as G 6 he inverse of the gain matri is then G 7 he pseudo-inverse is then I G 8 he least squares estimate of the states from the measurements is then

8 I b Compared to the solution of eq., our estimate can be assumed to be less accurate since it is based on fewer measurements, but at least we still did obtain a solution. Question: What if we lost two measurements? Let s say that we lost, the measurement on the current flowing from bus to bus, and, the measurement on the current flowing from bus to bus. Let s see what happens. o solve it, we just remove the first two equations corresponding to, in eq. 7, the first row of the -matri and the top element in b- vector. v.8 v. e 8

9 he matri is once again non-square, so we must use our least-squares procedure. First, the gain matri is given as G 6 he inverse of the gain matri is, however, singular it s determinant is zero or equivalently, it has a zero eigenvalue. s a result, it can not be inverted. In this case, our process must stop since we need G - to evaluate, as indicated in eq. 9, repeated below for convenience. I b G b b What is the problem here? 9 he basic problem is that we do not have enough measurements. In this case, the system is said to be unobservable. his means that despite the availability of some measurements, it is not possible to provide an estimate of the states with those available measurements. 9

10 Eample: Consider the system in Fig.. his is the same system considered in our previous set of state estimation notes and in your homework assignment. here are real power measurements taken at P, P, and P. But we assume that the measurements P and P fail so that we only have the measurement P =.6 pu. s in previous eamples, assume all voltages are. per unit, all measurement devices have σ=., and that the bus angle is reference. he state vector is =[θ θ ] as before. Bus Bus P P P Bus Fig.

11 a Determine the vector of measurement epressions h, the derivative epressions h H, and the weighting matri R. b Develop H R H, h z R H b and comment on our ability to solve Δ=b for Δ. Solution: a ` sin 5 sin cos b V V g V V g V P h Since =[θ θ ], the derivative epressions are 5cos 5cos h h h H and with σ =.,. R. b First get. 5cos 5cos 5cos 5cos 5cos 5cos 5cos 5cos H R H Now get b.

12 ` ` ` sin 5cos.cos sin 5cos.cos 5sin.6 5cos 5cos h z R H b herefore the equation we need to solve is ` ` sin 5 cos.cos sin 5 cos.cos 5cos 5cos 5cos 5cos or ` ` sin 5cos.cos sin 5cos.cos 5cos 5cos 5cos 5cos he above equation solves according to ` ` sin 5cos.cos sin 5cos.cos 5cos 5cos 5cos 5cos o perform the necessary matri inversion, we require the matri determinant, which is: 5cos 5cos 5cos 5cos 5cos 5cos det 4 4

13 he matri is singular and we cannot solve the equation. In general, we say that the power system is observable if H R H is non-singular. Linear algebra tet books [, pg. 46], [] define matri rank: he maimum number of linearly independent rows in an m n matri is the row rank of, and the maimum number of linearly independent columns in an m n matri is the column rank of. If is m n, then: row rank m & col rank n. For any matri : row rank=col rank rank. Observability analysis from H: ssume for a power system state estimation problem that m>n more measurements than states. he power system is observable if matri H R H is nonsingular, and this occurs if H has rank n, where n is the number of columns of H.

14 Most state estimators will perform an observability analysis. If the network is unobservable, it may be the case that some pockets or islands of the network are still observable. hus we also need to be able to identify observable parts of the system from unobservable parts of the system. Doing so will will enable us to determine from which part of the network we need to obtain additional measurements. he topic of observability analysis is wellcovered in [, ch 7]. 4. pproimate pseudo-measurements key step in state estimation is to test for observability. If the network is not observable, i.e., if we do not have enough independent measurements, then we will not be able to obtain a network model. 4

15 When the state estimator detects that the network is unobservable, it can make use of approimate pseudo-measurements. Eamples of such approimate pseudo-measurements include: Information obtained from plant operators over phone or . Information obtained from previous measurements. Information obtained from a load flow calculation. In using approimate pseudo-measurements, it is generally good practice to pair it with a relatively large variance in the weighting matri, given that it is in fact approimate. 5. Quality of the state estimate lthough the state estimator provides the best estimate of the states given all available measurements, there remains the rather important question of: How good is the state estimate? 5

16 Clearly, if the state estimate is very poor, we will not want to use it even if it is the best. here are two issues related to state estimate quality. Bad data: Some data might be completely erroneous, polluted with some kind of gross error beyond the normal meter error. For eample, a transducer may be wired incorrectly or is malfunctioning and gives a negative number or a zero or a number that is times what it should be. Meter error: We have already discussed that meter error will always be present. We should check to ensure that there is not one or more meters in a crucial location having magnitude large enough to cause the entire state estimate to have unacceptable quality. Your tet provides an overview of these topics in Section.6. Detection and Identification of Bad Measurements, which depends on Hypothesis esting. 6

17 [] W. Brogan, Modern Control heory, Prentice-Hall, 98. [] G. Strang, Linear lgebra and its applications, Harcourt-Brace, 988. []. Monticelli, State estimation in electric power systems, a generalized approach, Kluwer,