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From: AmiHK on 28 Jan 2010 07:58 Hi All, I have been recently placed on a new project related to Power Systems.I am new in this field and not getting a proper place to start with.So please give me your advise. The project is an Active Shunt Type Power Filter used to take out harmonics from the power line. We have an existing product selling in the market,but with an old controller design, TI DSP(TMS320C40)<=====>Dual Ram<=====>TMS320F240(TI MCU). So we have thought of migrating on a new controller platform. The existing source code we have is in assembly,and the documentation written is not up to the mark(I mean the derivations..etc) All that i can see in the docs are final equations with missing explanation.Below is the section I am not able to understand, mainly the equation (1),(2).....(5) their derivations. ---------------start here ------------------ Recursive real-time harmonics and reactive power detection algorithm routine flow chart shown in Figure 4-20. F240 controlled by the A / D converter samples the input data to the form of signed integers stored in the designated memory unit. C32 first converts integer to floating-point number, and then executes the recursive real-time detection of harmonic and reactive power compensation algorithm instructions, the calculation is as described below According to symmetrical component method, in three-phase three-wire system, the instantaneous load current ia, ib, ic can be decomposed into positive sequence component group and the negative sequence components Group. To use the index, 1 positive sequence, 2 negative sequence,k harmonic frequency,I Indicated that peak current, ¢ first phase angle, N sampling frequency cycle points,n Counts for the sampling points n=0,1,2,…N-1. So, after sampling the load current can be expressed as ix(n) = ∑k=1to ∞ [I1k sin(2∏nk/N + ￠1k -2l∏/3) + I2k sin(2∏nk/N + ￠2k +2l∏/3)] In which l = 0,1,2 for x=a, x=b, x=c Substituting in the following formula ixypk(n) = {ixypk(n-1) – 2*[ ix(n-N)- ix(n)] * sin(2∏nk/N - 2l∏/3)}/N ixyqk(n) = {ixyqk(n-1) – 2*[ ix(n-N)- ix(n)] * cos(2∏nk/N - 2l∏/3)}/N -------------------------(1) For the fundamental wave can be calculated Matrix [A11(n) B11(n) , A21(n) B21(n)] = 1/3 matrix [iayp1(n)+ ibyp1(n)+ icyp1(n) iayq1(n)+ ibyq1(n)+ icyq1(n) , 2iayp1(n)- ibyp1(n)- icyp1(n) 2iayq1(n)- ibyq1(n)- icyq1(n)] ------------------(2) B11 A11,Represent the fundamental positive-sequence reactive current components and fundamental positive-sequence active component of the amplitude. Suppose ix11(n) = A11(n)* sin(2∏n/N + 2l∏/3) + B11(n)* cos(2∏n/N + 2l∏/3) --------(3) ix21(n) = A21(n)* sin(2∏n/N + 2l∏/3) + B21(n)* cos(2∏n/N + 2l∏/3) --------(4) ix1(n) =ixyp1(n)* sin(2∏n/N + 2l∏/3) + ixyq1(n)* cos(2∏n/N + 2l∏/3) --------(5) If only the active power filter to compensate harmonic and fundamental negative sequence, instead of reactive power compensation, then use equation (3). Calculated to be fundamental positive sequence current signals ix11, Then the load current signal minus the ix11 command signal can be obtained icx*(n ) = ix(n) - ix11.If the demand compensation after the network side of a symmetrical three-phase fundamental current, with power factor 1,with equation (3) makes A11 = 0, Which can be fundamental positive sequence of the active component ipx11, With the load current signal minus ipx11,will get the command signal icx*(n ) = ix(n) - ipx11 If the active power filter to compensate harmonics and not just reactive and negative sequence compensation,use equation (5), Calculated fundamental current signal ix1. Then the load current signal minus the ix1,will get the command signal. For the harmonic,don't need to distinguish positive sequence component & negative sequence component. If the active power filter is used to compensate for nominated sub-harmonic,then by using equation (1),with ixyqk(n), ixypk(n) be calculated as ixk(n) =ixyqk(n)* cos(2∏kn/N ) + ixypk(n)* sin(2∏kn/N ) And is a sub harmonic current signal. ------------- end here ----------------------------------- Thanks in advance |