2012 IEEE Power and Energy Society General Meeting 2012
DOI: 10.1109/pesgm.2012.6343981
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Evaluation of residue based power oscillation damping control of inter-area oscillations for static power sources

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Cited by 7 publications
(5 citation statements)
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“…For comparison with the DDC-WADC, conventional WADC (C-WADC) is designed based on residue method [24]. The modal analysis reveals three loosely damped modes (indexed as Mode 1, 2 and 3) with damping ratio as 0.0842, 0.021, 0.012 and frequency as 0.706 Hz, 0.980 Hz and 1.089Hz respectively.…”
Section: A Case: Variable Operating Conditionsmentioning
confidence: 99%
“…For comparison with the DDC-WADC, conventional WADC (C-WADC) is designed based on residue method [24]. The modal analysis reveals three loosely damped modes (indexed as Mode 1, 2 and 3) with damping ratio as 0.0842, 0.021, 0.012 and frequency as 0.706 Hz, 0.980 Hz and 1.089Hz respectively.…”
Section: A Case: Variable Operating Conditionsmentioning
confidence: 99%
“…Figure 2(b) shows the control strategies of the DFIGs. The energy of the wind is converted into the mechanical energy of the DFIGs by the WT, whose captured power (P t ) may be expressed as, ΔP t = P wloss ΔC P ( , t ) (15) where P wloss is the lost power of the wind, C P is the coefficient of the power utilization which is affected by the rotor speeds of the WT (ω t) and the pitch angle (β). To maintain the optimal rotor speeds of the WT, β is controlled by the PAC, which also regulates P t and the torque of the WT (T t ).…”
Section: Modelling Of Dfigmentioning
confidence: 99%
“…The parameters and computation times corresponding to the PODs are given in Table 6. The ranges of the POD parameters are given as follow: K POD [5,15], T d1 [0.4s, 0.6s]. T d2 [0.05s, 0.20s].…”
Section: Validation To Control Effect Of Pod With Time-domain Analysismentioning
confidence: 99%
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“…APPENDIX A PROOF OF LEMMA III.1The difference equation can be written from(17) as ∆y(k + 1) = f e (y(k), .., y(k − n y ))− f e (y(k − 1), .., y(k − n y − 1))(24) By adding and subtracting f e (y(k − 1), .., y(k − n y ))) ∆y(k + 1) = f e (y(k), .., y(k − n y )) − f e (y(k − 1),y(k − 1).., y(k − n y ))) + f e (y(k − 1), y(k − 1), .., y(k − n y ))) − f e (y(k − 1), .., y(k − n y − 1)) (25)From assumption A1 we have the f e differentiable. Further using differential mean value theorem (∂f e * /∂y(k) = f e (y(k), .)…”
mentioning
confidence: 99%