Linear-Phase fr Transfer Functions Because of symmetry, we have ho]=h 8 h1]=h[7],h[2]=h6],andh[3]=h[5 Thus we can write H()=h0(+23)+1(x+z7) +2(2+6)+h33+25)+4-4 =z4{h0(=4+24)+h](=3+x3) +h2](z2+2-2)+h3](z+2-)+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 − −