TWI423047B - Method for estimating parameters of induction machine by time-varied parameters - Google Patents
Method for estimating parameters of induction machine by time-varied parameters Download PDFInfo
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本發明係關於一種求解感應機參數之方法,特別是一種以時變參數疊代求解感應機參數的方法。The invention relates to a method for solving parameters of an induction machine, in particular to a method for solving parameters of an induction machine by time-varying parameter iteration.
感應機的定子為一繞組,其轉子在運轉時不需直流磁場電路。運用定子與轉子磁場之間的相對運動所感應的轉子電壓,來產生轉子電流;而藉由定子與轉子磁場相互作用,於感應機中產生感應轉矩。由於感應機的架構簡單與運轉容易,已成為最常使用的交流電動機。感應機控制的方法與系統的設計上皆需藉助於等效模型及其參數的幫助。相關模型及其參數取得的習知技術大致可分成離線估測與線上即時估測二大類:離線估測可經由標準的方法,如直流試驗、無載試驗及堵轉試驗來獲得;線上即時估測則可利用頻譜分析其頻率響應來估測其參數、利用參考模型的架構來估測,或利用數值方法的人工智慧的參數鑑定等方法。以上方法各具優缺點及其限制,而大多需要昂貴的量測設備且等待感應機轉速達到穩態之後才得以得到所有的參數。The stator of the induction machine is a winding, and its rotor does not require a DC magnetic field circuit during operation. The rotor voltage induced by the relative motion between the stator and the rotor magnetic field is used to generate the rotor current; and the stator and the rotor magnetic field interact to generate an induced torque in the induction machine. Due to its simple structure and easy operation, the induction machine has become the most commonly used AC motor. The method and system design of the induction machine control are all assisted by the equivalent model and its parameters. The known techniques for obtaining related models and their parameters can be roughly divided into two categories: offline estimation and online real-time estimation: offline estimation can be obtained through standard methods such as DC test, no-load test and stall test; online real-time estimation The measurement can use the frequency spectrum to analyze its frequency response to estimate its parameters, use the structure of the reference model to estimate, or use numerical methods to identify the parameters of artificial intelligence. Each of the above methods has its own advantages and disadvantages and limitations, and most of them require expensive measuring equipment and wait for the speed of the induction machine to reach a steady state before all parameters are obtained.
有鑑於此,本發明之主要目的在於提供一種求解感應機參數之方法,係以最小平方法,藉由梯度法疊代之最適值搜尋與計算,求得感應機的等效電路模型及等效機械模型之相關參數。此外,設定一組適當的初值,將有助於疊代過程的收斂與效率,本發明之另一目的將提供初值設定的一較佳方法。In view of this, the main object of the present invention is to provide a method for solving the parameters of the induction machine, which is to find the equivalent circuit model and equivalent of the induction machine by the least square method and the optimum value search and calculation by the gradient method. Related parameters of the mechanical model. Moreover, setting a suitable set of initial values will aid in the convergence and efficiency of the iterative process, and another object of the present invention is to provide a preferred method of initial value setting.
為達上述之目的,本發明所提出之求解感應機參數之方法,係包含下列步驟:首先,擷取感應機啟動時電壓、電流及轉速的時變信號;第二,計算該感應機在不同轉差率下的電阻及電抗;第三,設定一電路參數之目標函數;第四,設定該感應機的等效電路參數之初值,並藉由梯度法循環疊代與修正,當該目標函數的值收斂至一容許範圍值時,即可算出該感應機的等效電路模型之參數;第五:根據步驟一所得之啟動時的轉速信號及步驟四所得之等效電路參數,進行該感應機輸出轉矩之動態模擬,並計算其輸出轉矩;最後,根據步驟五所得之輸出轉矩與轉速,並設定一機械參數之目標函數,計算該感應電動機的等效機械模型之參數。如此可解出感應機的等效電路模型之參數:轉子電阻、定子電阻、轉子電抗、定子電抗及激磁電抗,及其等效機械模型之參數:慣量及阻尼。For the above purposes, the method for solving the parameters of the induction machine proposed by the present invention comprises the following steps: first, a time-varying signal of voltage, current and rotational speed when the induction machine is started; secondly, calculating the induction machine is different. Resistance and reactance under slip; third, setting the objective function of a circuit parameter; fourth, setting the initial value of the equivalent circuit parameter of the induction machine, and repeating the iteration and correction by the gradient method, when the target When the value of the function converges to a permissible range value, the parameters of the equivalent circuit model of the induction machine can be calculated; fifth: according to the rotational speed signal obtained at step 1 and the equivalent circuit parameters obtained in step four, The dynamic simulation of the output torque of the induction machine, and calculate its output torque; Finally, according to the output torque and speed obtained in step 5, and set the objective function of a mechanical parameter, calculate the parameters of the equivalent mechanical model of the induction motor. In this way, the parameters of the equivalent circuit model of the induction machine can be solved: rotor resistance, stator resistance, rotor reactance, stator reactance and excitation reactance, and the parameters of the equivalent mechanical model: inertia and damping.
為使對本發明內容有進一步了解,下文中茲配合相關實施例及圖示作詳細說明。請參考圖1,繪示本發明之方法流程圖,主要包含下列步驟:擷取感應機啟動時一次測電壓v(n)、一次測電流I(n)及轉速wr
(n)的時間序列信號,感應機等效電路可由阻抗對轉速的變化求得(步驟110)。由於感應機暫態時間常數很短,以致於由電感所引起的暫態將在
起動初期快速降到可忽略的範圍,於是電壓電流的特性將由穩態阻抗所主導,則感應機之穩態等效電路模型請參考圖2,其中R s
為定子電阻、R r
為轉子電阻、X m
為激磁電抗、X s
定子等效電抗、且X r
為轉子等效電抗;s
為轉差率,其由感應機啟動起之時間序列定義為;可得計算感應機之一次側電阻與電抗在不同轉差率下分別為(步驟120):
本發明係採用最小平方法以求取相關的電路參數,其最適值之搜尋係以梯度法進行。梯度法的主要原理是以目標函數的梯度為其搜尋的方向,以處理多變數非線性函數的最小化搜尋,藉由一連串的疊代過程,使設定的結果能趨近實際值。梯度係指目標函數的一次微分,對於感應機等效電路參數的求解,可設定其目標函數為(步驟130)
E R 與E X 為具有極小值的函數,而在極小值時此二函數對每個參數的梯度均為0。當在極小值附近可得一不為0的梯度,則該梯度可設為最佳解之搜尋方向。本發明係以R s 、R r 作為搜尋E R 的最小值,以X s 、X r 、X m 搜尋E X 的最小值,可建立輔助方程式 E R and E X are functions with a minimum value, and at a minimum value, the gradient of each of the two functions is zero for each parameter. When a gradient other than 0 is obtained near the minimum value, the gradient can be set to the search direction of the optimal solution. In the present invention, R s , R r are used as the minimum value of searching E R , and X s , X r , X m are used to search for the minimum value of E X , and an auxiliary equation can be established.
M =(R r /s )2 +(X m +X r )2 M = ( R r / s ) 2 + ( X m + X r ) 2
則電阻與電抗的誤差量分別為Then the error between the resistance and the reactance is
則搜尋E R 及E X 最小值的梯度可表示為Then the gradient searching for the minimum values of E R and E X can be expressed as
藉由上述五方程式修正阻抗參數,以求取目標函數的最小值,且每一個參數的下一個狀態為The impedance parameter is modified by the above five equations to obtain the minimum value of the objective function, and the next state of each parameter is
其中η Rs 、η Rr 、η Xs 、η Xr 、及η Xm 分別為對應各個電路參數加速因子;並根據上述五方程式修正各個參數值,使得目標函數逐漸收斂。Where η Rs , η Rr , η Xs , η Xr , and η Xm are respectively corresponding to each circuit parameter acceleration factor; and the respective parameter values are corrected according to the above five equations, so that the objective function gradually converges.
設定較佳的初值將提高梯度法疊代收斂的效率,因此本發明提供一計算感應機等效電路參數初值的方法如下(步驟140):請參考圖3,為感應機起動時所可能發生的阻抗的對應其轉差率,其中轉子轉速常常無法達到同步轉速,而本發明即針對此事實而求取近似的初值。當轉差率s =1時,阻抗值為R (1)及X (1);當轉差率s =0.5時,電阻值R (0.5);當接近同步轉速時,可得二個接近0的轉差率s 1及s 2,其所對應的電抗值分別為X (s 1)及X (s 2)。在起動之初,s 介於0.5至1之間,因X m 遠大於X s 及R r ,且電阻呈線性改變,可得一近似關係R (s )≒R s +R r /s ,因此可得電阻的初值為R r 0 =0.5(R (0.5)-R (1))且R s 0 =R (1)-R r 0 ;此時電抗變化很小,且其值約為X s 與X r 之和,故設定X s 與X r 相同的初值,即X s 0 =X r 0 =0.5X (1);感應機的激磁電抗可在無載轉速獲得,當感應機到達同步轉速時,電抗值為X (0)=X s +X m ,但在實際上,使感應機達到同步轉速將增加量測的複雜性,本發明採用外插方式,在不增加量測的複雜度下達到精確的計算成果。當感應機接近同步轉速時電抗呈線性增加,因此可以利用接近同步轉速的數個取樣值來估算同步轉速時的電抗,根據外插法可得同步轉速時的激磁電抗值為Setting a preferred initial value will improve the efficiency of the cascade convergence of the gradient method. Therefore, the present invention provides a method for calculating the initial value of the equivalent circuit parameter of the induction machine as follows (step 140): Please refer to FIG. 3, which is possible when the induction machine is started. The resulting impedance corresponds to its slip rate, where the rotor speed often fails to reach the synchronous speed, and the present invention seeks an approximate initial value for this fact. When the slip ratio s =1, the impedance values are R (1) and X (1); when the slip ratio s = 0.5, the resistance value R (0.5); when approaching the synchronous speed, two close to 0 are obtained. The slip ratios s 1 and s 2 have corresponding reactance values of X ( s 1) and X ( s 2), respectively. At the beginning of the start, s is between 0.5 and 1. Since X m is much larger than X s and R r and the resistance changes linearly, an approximate relationship R ( s ) ≒ R s + R r / s is obtained . The initial value of the available resistance is R r 0 =0.5( R (0.5)- R (1)) and R s 0 = R (1)- R r 0 ; the reactance change is small at this time, and its value is about X. The sum of s and X r , so set the initial value of X s and X r , ie X s 0 = X r 0 =0.5 X (1); the magnetizing reactance of the induction machine can be obtained at the no-load speed, when the induction machine arrives At synchronous speed, the reactance value is X (0) = X s + X m , but in practice, bringing the induction machine to synchronous speed will increase the complexity of the measurement. The present invention uses the extrapolation method without increasing the measurement. Accurate calculation results are achieved under complexity. When the induction machine approaches the synchronous speed, the reactance increases linearly. Therefore, it is possible to estimate the reactance at the synchronous speed by using several sampling values close to the synchronous speed. The exciting reactance value at the synchronous speed can be obtained according to the extrapolation method.
經過上述步驟即可算得所有感應機電路參數的初始值:R s 0 、R r 0 、X s 0 、X r 0 、X m 0 。Through the above steps, the initial values of all the sensor circuit parameters can be calculated: R s 0 , R r 0 , X s 0 , X r 0 , X m 0 .
本發明藉由動態模擬計算感應機的轉矩。感應機等效電路的參數可由前述的步驟獲得,而在定子參考架構下,三相感應機的動態模型可表為The present invention calculates the torque of the induction machine by dynamic simulation. The parameters of the equivalent circuit of the induction machine can be obtained by the aforementioned steps, and under the stator reference architecture, the dynamic model of the three-phase induction machine can be expressed as
v qs =(R s +L s p )i qs +L m pi qr v qs =( R s + L s p ) i qs + L m pi qr
v ds =(R s +L s p )i ds +L m pi dr v ds =( R s + L s p ) i ds + L m pi dr
v qr =L m pi qs -ω r L m i ds +(R r +L r p )i qr -ω r L r i dr v qr = L m pi qs -ω r L m i ds +( R r + L r p ) i qr -ω r L r i dr
v dr =ω r L m i qs +L m pi ds +ω r L r i qr +(R r +L r p )i dr v dr =ω r L m i qs + L m pi ds +ω r L r i qr +( R r + L r p ) i dr
其中i qs ,i ds 為定子電流;i qr ,i dr 為轉子電流;v qs ,v ds 為定子電壓;v qr ,v dr 為轉子電壓;p 為微分因子();L r 、L m 、L s 分別為轉子電感(Xr /ws )、激磁電感(Xm /ws )、及定子電感(Xs /ws )。因此可進一步得到輸出轉矩為(步驟150)Where i qs , i ds is the stator current; i qr , i dr is the rotor current; v qs , v ds is the stator voltage; v qr , v dr is the rotor voltage; p is the differential factor ( L r , L m , L s are the rotor inductance (X r /w s ), the magnetizing inductance (X m /w s ), and the stator inductance (X s /w s ), respectively. Therefore, the output torque can be further obtained (step 150).
T =3PL m (i dr i qs -i qr i ds ) T = 3 PL m ( i dr i qs - i qr i ds )
其中P 為電動機的極數。Where P is the number of poles of the motor.
慣量J 與黏滯阻尼B 係為感應機等效機械模型之重要參數,共同決定了其輸出轉矩與轉速之間的關係;亦即當輸出轉矩與轉速為已知的情況下,則可進一步逆推得慣量與黏滯阻尼。假設感應機產生的扭力只造成感應機的速度改變,並沒有帶動其他機械負載,則相對應的轉速符合下列微分方程式The inertia J and the viscous damping B are important parameters of the equivalent mechanical model of the induction machine, which jointly determine the relationship between the output torque and the rotational speed; that is, when the output torque and the rotational speed are known, Further inversely push the inertia and viscous damping. Assuming that the torque generated by the induction machine only causes the speed of the induction machine to change, and does not drive other mechanical loads, the corresponding speed corresponds to the following differential equation.
以離散資料表示:Expressed as discrete data:
J (ω r (n )-ω r (n -1))+B ω r (n )=T (n ),n=1,2,... J (ω r ( n )-ω r ( n -1)) + B ω r ( n )= T ( n ), n=1, 2,...
當考慮系統為線性時,其中慣量及黏滯阻尼是不變的,要獲得最恰當的參數可設定目標函數為When considering that the system is linear, where the inertia and viscous damping are constant, to obtain the most appropriate parameters, the objective function can be set to
當目標函數最小時,則有最適當的慣量J 與黏滯阻尼B ,亦即對上一方程式慣量J 與黏滯阻尼B 的梯度皆為0,因此慣量J 與黏滯阻尼B 可由下式獲得(步驟160)When the objective function is the smallest, there is the most appropriate inertia J and viscous damping B , that is, the gradient of the upper inertia J and the viscous damping B is 0, so the inertia J and the viscous damping B can be obtained by (Step 160)
經由上述方法與步驟,可解出感應機的等效電路模型之參數:轉子電阻、定子電阻、轉子電抗、定子電抗及激磁電抗,及其等效機械模型之參數:慣量及阻尼。Through the above methods and steps, the parameters of the equivalent circuit model of the induction machine can be solved: rotor resistance, stator resistance, rotor reactance, stator reactance and excitation reactance, and parameters of the equivalent mechanical model: inertia and damping.
為了讓本發明之上述及其他目的、特徵、優點能更明顯易懂,下文將特舉本發明較佳實施例,並配合所附圖式,詳細說明如下。The above and other objects, features and advantages of the present invention will become more <RTIgt;
假設有一三相、4極、0.5馬力、60Hz的感應機,請參考圖4,為其在起動階段的電壓、電流對於轉速的實測曲線數據。在起動之初,電壓產生些許的壓降,隨著轉速上升電壓逐漸回復到正常值;此時由於轉差的關係,轉子的應電勢尚未建立完成,造成較大的電流。當轉速接近額定轉速時,電流則降到一個相當小的值。起動階段阻抗相位約變化在30°至70°之間。在剛起動及達穩定後,電抗都較電阻為大。在接近額定轉速時,電阻較電抗為大。Assuming a three-phase, 4-pole, 0.5-horsepower, 60 Hz induction machine, please refer to Figure 4 for the measured curve data of voltage and current for the starting speed. At the beginning of the start-up, the voltage produces a slight voltage drop. As the speed rises, the voltage gradually returns to the normal value. At this time, due to the slip relationship, the potential of the rotor has not been established, resulting in a large current. When the speed is close to the rated speed, the current drops to a fairly small value. The phase of the impedance during the start-up phase varies between 30° and 70°. After the start and stability, the reactance is larger than the resistance. When the rated speed is approaching, the resistance is larger than the reactance.
請參考圖5,為不同轉差率下的阻抗。感應機由靜止至額定轉速時的電阻與電抗變化,證實了暫態項對感應機起動階段的影響極小,穩態項足以提供一個簡單明暸且足夠精確的結果。利用本發明所提供設定感應機的參數初值之方程式,可得其等效電路參數初值分別為R r 0 =0.5(R (0.5)-R (1))=8.68Ω,R s 0 =R (1)-R r =23.37Ω,X s 0 =X r 0 =0.5X (1)=13.52Ω,及Please refer to Figure 5 for the impedance at different slip rates. The resistance and reactance changes of the induction machine from standstill to rated speed confirm that the transient term has minimal effect on the start-up phase of the induction machine, and the steady-state term is sufficient to provide a simple and accurate enough accuracy. By using the equation for setting the initial value of the parameter of the induction machine provided by the invention, the initial value of the equivalent circuit parameter is obtained as R r 0 = 0.5 ( R (0.5) - R (1)) = 8.68 Ω, R s 0 = R (1) - R r = 23.37 Ω, X s 0 = X r 0 = 0.5 X (1) = 13.52 Ω, and
以此初值代入梯度法之疊代計算感應機的電阻及電抗,其結果請參考表1,顯示梯度法收斂的結果:經過4000次的疊代程序,可得收斂的結果,且可發現初值設定已達相當近似實際值的結果,其中雜訊及非線性因素使得目標函數無法達到0。The initial value is substituted into the gradient method to calculate the resistance and reactance of the induction machine. The results are shown in Table 1. The result of the convergence of the gradient method is shown: after 4000 iterations, the convergence result can be obtained, and the initial result can be found. The value setting has reached a result that is fairly close to the actual value, where the noise and nonlinear factors make the objective function unable to reach zero.
本實施例以上述所得之參數結果進行系統動態行為的模擬,並與實際值比較。模擬結果請參考圖6,顯示二者電流的比較,且電流的標準差為0.339,可發現兩者相當接近;由此亦可發現,暫態將會在一週內衰減到極小的值,因此圖6中電流會呈現穩態的結果。證實穩態項主導電流的變化。而由該感應機之輸出轉矩與轉速,可得其轉動慣量J 及黏滯阻尼B 各為0.38(g。m2 )及0.61(mN。m/(rad/sec))。圖7顯示了轉速的比較。轉速的標準差為8.388,為一相當低的值。證實本方法所得的參數相當符合實際情況。In this embodiment, the dynamic behavior of the system is simulated by the parameter results obtained above, and compared with actual values. For the simulation results, please refer to Figure 6, which shows the comparison of the currents, and the standard deviation of the current is 0.339. It can be found that the two are quite close; it can also be found that the transient will decay to a minimum value within one week, so the graph The current in 6 will show a steady state result. It is confirmed that the steady state term dominates the change in current. From the output torque and the rotational speed of the induction machine, the moment of inertia J and the viscous damping B are each 0.38 (g.m 2 ) and 0.61 (mN.m/(rad/sec)). Figure 7 shows a comparison of the rotational speeds. The standard deviation of the rotational speed is 8.388, which is a relatively low value. It is confirmed that the parameters obtained by the method are quite in line with the actual situation.
雖然本發明以前述之較佳實施例揭露如上,然其並非用以限定本發明,任何熟習相像技藝者,在不脫離本發明之精神和範圍內,所作更動與潤飾之等效替換,仍為本發明之專利保護範圍內。While the present invention has been described above in terms of the preferred embodiments thereof, it is not intended to limit the invention, and the equivalent of the modification and retouching of the present invention is still within the spirit and scope of the present invention. Within the scope of patent protection of the present invention.
步驟110 擷取感應機啟動時一次測電壓、電流及轉速的時變信號:v(n)、i(n)、wr (n)。Step 110: Time-varying signals for measuring voltage, current, and speed at the start of the induction machine: v(n), i(n), w r (n).
步驟120 計算該感應機在不同轉差率下的電阻及電抗:R(s)、X(s)。Step 120 calculates the resistance and reactance of the induction machine at different slip ratios: R(s), X(s).
步驟130 設定電路參數之目標函數。Step 130 sets the objective function of the circuit parameters.
步驟140 設定該感應機等效電路參數之初值,藉由梯度法循環疊代算出該等效電路模型之參數:R s 、R r 、X s 、X r 、X m 。Step 140: setting an initial value of the equivalent circuit parameter of the induction machine, and calculating parameters of the equivalent circuit model by using a gradient method cyclic iteration: R s , R r , X s , X r , X m .
步驟150 根據步驟110所得之啟動時變信號,進行感應機之動態模擬並計算其輸出轉矩:T(n)。Step 150: According to the start time change signal obtained in step 110, perform dynamic simulation of the induction machine and calculate its output torque: T(n).
步驟160 設定一機械參數之目標函數,計算該感應機的等效機械模型之參數。Step 160 sets an objective function of a mechanical parameter and calculates a parameter of the equivalent mechanical model of the induction machine.
圖1為本發明之求解感應機參數的方法流程圖。1 is a flow chart of a method for solving parameters of an induction machine according to the present invention.
圖2為感應機之穩態等效電路。Figure 2 shows the steady-state equivalent circuit of the induction machine.
圖3為感應機起動時所可能發生的阻抗對應其轉差率之曲線圖,以作為初值設定的參考。Figure 3 is a graph of the impedance that may occur when the induction machine is started, corresponding to its slip rate, as a reference for the initial value setting.
圖4為本實施例之感應機在起動階段的電壓、電流對於轉速的實測數據曲線。4 is a measured data curve of the voltage and current of the induction machine in the starting phase with respect to the rotational speed of the embodiment.
圖5為本實施例之感應機在不同轉差率下的阻抗比較。FIG. 5 is a comparison of impedances of the induction machine of the present embodiment at different slip rates.
圖6為本實施例之電流的實測量值與本法模擬求解之比較。Fig. 6 is a comparison of the measured value of the current of the present embodiment with the simulation method of the present method.
圖7為本實施例之轉速的實測量值與本法模擬求解之比較。Fig. 7 is a comparison of the actual measured value of the rotational speed of the present embodiment with the simulation solution of the present method.
步驟110 擷取感應機啟動時一次測電壓、電流及轉速的時變信號:v(n)、i(n)、wr (n)。Step 110: Time-varying signals for measuring voltage, current, and speed at the start of the induction machine: v(n), i(n), w r (n).
步驟120 計算該感應機在不同轉差率下的電阻及電抗:R(s)、X(s)。Step 120 calculates the resistance and reactance of the induction machine at different slip ratios: R(s), X(s).
步驟130 設定電路參數之目標函數。Step 130 sets the objective function of the circuit parameters.
步驟140 設定該感應機等效電路參數之初值,藉由梯度法循環疊代算出該等效電路模型之參數:R s 、R r 、X s 、X r 、X m 。Step 140: setting an initial value of the equivalent circuit parameter of the induction machine, and calculating parameters of the equivalent circuit model by using a gradient method cyclic iteration: R s , R r , X s , X r , X m .
步驟150 根據步驟110所得之啟動時變信號,進行感應機之動態模擬並計算其輸出轉矩:T(n)。Step 150: According to the start time change signal obtained in step 110, perform dynamic simulation of the induction machine and calculate its output torque: T(n).
步驟160 設定一機械參數之目標函數,計算該感應機的等效機械模型之參數。Step 160 sets an objective function of a mechanical parameter and calculates a parameter of the equivalent mechanical model of the induction machine.
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