TI中文支持网MOVW DP, #_CNTL_2P2Z_Ref:n:
MOVL XAR0, @_CNTL_2P2Z_Ref:n: ; net pointer to Ref (XAR0)
MOVW DP,#_CNTL_2P2Z_Fdbk:n:
MOVL XAR1, @_CNTL_2P2Z_Fdbk:n: ; net pointer to Fdbk (XAR1)
MOVW DP,#_CNTL_2P2Z_DBUFF:n:
MOVL XAR4, #_CNTL_2P2Z_DBUFF:n: ; pointer to the DBUFF array (used internally by the module)
; calculate error (Ref – Fdbk)
MOVL ACC, *XAR0 ; ACC = Ref (Q24) = Q(24)
SUBL ACC, *XAR1 ; ACC = Ref(Q24) – Fdbk(Q24)= error(Q24)
LSL ACC, #6 ; Logical left shift by 6, Q{24}<<6 -> Q{30}
;store error in DBUFF MOVL *+XAR4[4], ACC ; e(n) = ACC = error Q{30} MOVW DP,#_CNTL_2P2Z_Out:n:
MOVL XAR2, @_CNTL_2P2Z_Out:n: ; net pointer to Out (XAR2)
MOVW DP,#_CNTL_2P2Z_Coef:n:
MOVL XAR3, @_CNTL_2P2Z_Coef:n: ; net pointer to Coef (XAR3)
ZAPA
MOV AR0,#8
MOVL XT, *+XAR4[AR0] ; XT = e(n-2)
QMPYL P, XT, *XAR3++ ; P = e(n-2) Q30 *B2{Q26} = I8Q24
MOVDL XT, *+XAR4[6] ; XT = e(n-1), e(n-2) = e(n-1)
QMPYAL P, XT, *XAR3++ ; P = e(n-1)Q30*B1{Q26} = Q24, ACC=e(n-2)*B2
MOVDL XT, *+XAR4[4] ; XT = e(n), e(n-1) = e(n)
QMPYAL P, XT, *XAR3++ ; P = e(n)Q30*B0{Q26}= Q24, ACC = e(n-2)*B2 + e(n-1)*B1
MOVL XT,*+XAR4[2] ; XT = u(n-2)
QMPYAL P, XT, *XAR3++ ; P = u(n-2)*A2, ACC = e(n-2)*B2 + e(n-1)*B1 + e(n)*B0
MOVDL XT,*+XAR4[0] ; XT = u(n-1), u(n-2) = u(n-1)
QMPYAL P, XT, *XAR3++ ; P = u(n-1)*A1, ACC = e(n-2)*B2 + e(n-1)*B1 + e(n)*B0 + u(n-2)*A2
ADDL ACC, @P ; ACC = e(n-2)*B2 + e(n-1)*B1 + e(n)*B0 + u(n-2)*A2 + u(n-1)*A1
只是将ACC赋值到了XR4[4]中， XR4[0]，XR4[2]，XR4[6]，XR4[8],的值是怎么来的，也就是DBUFF这个数组的值是怎么来的 u（n-1)和u（n-2）的值是如何计算得出的
汇编基础较弱，这里看不懂，希望能解答，谢谢了
