电子科技大学：《DSP算法实现技术与架构 VLSI Digital Signal Processing Systems Design and Implementation》课程教学资源（课件讲稿）Chapter 09 算法强度缩减 Algorithmic strength reduction in filters and transforms

9.2.1 Formulation of Parallel FIR using Polyphase Decomposition The input sequence can be decomposed into even-numbered part and odd-numbered part X(2)=0+x21+x222+X323+424+… =(x0+X2z2+X424+.)+z(x1+X322+Xz4+.) =X(z2)+zX(z2) similarly H(z)=H(z2)+z1H1(z2) 2021年2月 6

 The input sequence can be decomposed into even-numbered part and odd-numbered part  similarly 2021年2月 6 9.2.1 Formulation of Parallel FIR using Polyphase Decomposition ( ) ( ) ( ...) ( ...) ( ) ... 2 1 2 1 0 4 5 2 1 3 4 1 4 2 0 2 4 4 3 3 2 2 1 0 1 X z z X z x x z x z z x x z x z X z x x z x z x z x z                           ( ) ( ) ( ) 2 1 2 1 0 H z H z z H z   

/96 Y(z)=Y(z2)+z1Y(z2) =(X(z2)+z1X(22)H(z2)+z1H(z2) =X(z2)H(z2)+z(X(z2)H1(z2)+X(z2)H(z2) +z2X(22)H(z2) Y(z2)=X(z2)H(z2)+z2X1(z2)H1(z2) Y(z2)=X(z2)H1(z2)+X(z2)H(z2) 2021年2月 7

2021年2月 7 2 1 2 0 1 2 1 2 2 1 2 0 1 0 1 2 2 1 2 2 2 2 0 0 0 1 1 0 2 2 2 1 1 ( ) ( ) ( ) ( ( ) ( ))( ( ) ( )) ( ) ( ) ( ( ) ( ) ( ) ( )) ( ) ( ) Y z Y z z Y z X z z X z H z z H z X z H z z X z H z X z H z z X z H z               ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 0 2 1 2 1 2 0 2 1 2 1 2 1 2 2 0 2 0 2 0 Y z X z H z X z H z Y z X z H z z X z H z     

/966 Y-H.X [[医图 H, x(2k) HO y(2k) H1 x(2k+1) HO +y(2k+1) H1 z-2 Requires 2N multiplications and 2(N-1)additions. N/2*4 (N/2-1)*4+2 2021年2月 8

2021年2月 8  Requires 2N multiplications and 2(N-1) additions. N/2*4 (N/2-1)*4+2

/966 3-phase polyphase decomposition X(z)=X(z3)+z1X1(z3)+z2X2(z3) H(2)=H(2)+zH()+zH2() 2021年2月 9

 3-phase polyphase decomposition 2021年2月 9 ( ) ( ) ( ) ( ) 3 2 3 2 1 3 1 0 X z X z z X z z X z      ( ) ( ) ( ) ( ) 3 2 3 2 1 3 1 0 H z H z z H z z H z     

Y(2)=Y(z3)+z1Y(z3)+z2Y(z3) =(X(z3)+zX(z3)+z2X2(z3)H(z3)+zH1(z3)+z2H2(z3) =X(z3)H(z3)+z(X(z3)H1(z3)+X(z3)H(23) +z2(X1(z3)H1(z3)+X2(z3)H(z3)+X(z3)H2(z3) +z3(X(z3)H2(z3)+X(z3)H1(z3)+z4X2(z3)H2(z3) Y(z3)=Xo(z3)H(z3)+z3X(z3)H2(z3)+z3X(z3)H(z3) Y(z3)=X(z3)H(z3)+X(z3)H(z3)+zX2(z3)H2(z) Y,(z3)=X(z3)H2(z3)+X2(23)H(z3)+X(z3)H1(z3) 2021年2月 10

2021年2月 10 ( ( ) ( ) ( ) ( )) ( ) ( ) ( ( ) ( ) ( ) ( ) ( ) ( )) ( ) ( ) ( ( ) ( ) ( ) ( )) ( ( ) ( ) ( ))( ( ) ( ) ( )) ( ) ( ) ( ) ( ) 3 2 3 2 3 4 1 3 2 3 2 3 1 3 3 2 3 0 3 0 3 2 3 1 3 1 2 3 0 3 1 3 1 3 0 3 1 0 3 0 3 2 3 2 1 3 1 0 3 2 3 2 1 3 1 0 3 2 3 2 1 3 1 0 z X z H z X z H z z X z H z z X z H z X z H z X z H z X z H z z X z H z X z H z X z z X z z X z H z z H z z H z Y z Y z z Y z z Y z                            3 3 3 3 3 3 3 3 3 0 0 0 1 2 2 1 3 3 3 3 3 3 3 3 1 0 1 1 0 2 2 3 3 3 3 3 3 3 2 0 2 2 0 1 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Y z X z H z z X z H z z X z H z Y z X z H z X z H z z X z H z Y z X z H z X z H z X z H z            