Linear-Phase fir Transfer Functions Because of symmetry, we have h[o]h[8 h[l]=h7,h2]=h6],andh[3]=h[5 Thus we can write H(x)=h0](+3)+(x1+7) +和2](=2+6)+3(3+25)+414 z4{40(=4+4)+1(3+23) +和2(2+2-2)+h3(+21)+h4} Copyright C 2001, S K MitraCopyright © 2001, S. K. Mitra 6 Linear-Phase FIR Transfer Functions • Because of symmetry, we have h[0] = h[8], h[1] = h[7], h[2] = h[6], and h[3] = h[5] • Thus, we can write 8 1 7 H z h z h z z ( ) [0](1 ) [1]( ) − − − = + + + 2 6 3 5 4 2 3 4 − − − − − + h[ ](z + z ) + h[ ](z + z ) + h[ ]z { [ ]( ) [ ]( ) 4 4 4 3 3 0 1 − − − = z h z + z + h z + z [2]( ) [3]( ) [4]} 2 2 1 + h z + z + h z + z + h − −