SOLUTIONS TO ASSIGNMENT NO.1 3. The given nonreursive signl proessing struture is shown s X 1 1 2 3 4 5 Y 1 2 3 4 5 X 2 There re two ritil pths, one from X 1 to Y nd the other from X 2 to Y. The itertion long suh pth is given y T smple = T m +5 T We n redue this time period y dividing the whole iruit into two prts s elow nd using trnsposition on eh prt. Trnsposing prt 1 yields X 1 5 4 3 2 1 Y Thus equivlent rodst struture is given y X 1 5 4 3 2 1 Y 5 4 3 2 1 X 2 The itertion period for this rodst iruit is T smple = T m + T
Numer of smples proessed per seond (throughput) is given y 1/ (T m + T ). 7. () The 6 th -order FIR filter is given y y= x + x(n-4) +x(n-6) This iruit is represented in lok digrm y x 4 2 There re two ritil pths from x to y.the itertion long ritil pth is given y T smple = T m +2 T To limit the lok period y one multiply-dd time we use trnsposition on the iruit ove nd get the new iruit s x y 4 2 y The new time period of eh output smple for the new trnsposed iruit is given y T smple = T m + T 7(). Blok rhiteture for the iruit 6 th -order FIR filter of lok size of 3 using prllel proessing is given y equtions y(3k) = x(3k) +x(3k-4)+x(3k-6) y(3k+1) = x(3k+1) +x(3k-3)+x(3k-5) y(3k+2) = x(3k+2) +x(3k-2)+x(3k-4)
x(3k+2) x(3k+1) x(3k) 2 y(3k) 2 y(3k+1) 2 y(3k+1) Given tht T m = 3T Rerrnging the pipelined prllel filter struture suh tht the lok period is one-fourth of multiply-dd time is given y
x(3k+2) x(3k+1) x(3k) 2 y(3k-9) y(3k-8) y(3k-7) The ove struture shows the pipelined prllel filter struture, in whih the multiplier hs een roken into three prts m1, m2 nd m3. Eh prt hs the sme omputtion time s ddition omputtion. The pipelining lthes re pled long the feed forwrd utsets shown y dotted lines.
8(). The given reursive filter is x = x(n-2) + u The iruit ove is given y 2 u Breking up the multiply-dd opertion into 2 omponents is done y using two Multiply Add Components (MAC) whih is given y the figure x u MAC Output = u + Z In our iruit MAC is represented y Z u MAC x x(n-2) 2 x = x(n-2) + u Redistriuting the dely elements in the loop is given y u MAC1 MAC2 x x(n-1) u
8(). The given eqution is y = y(n-2) + v First slow down the iruit y repling the 2-dely with 4-dely nd then interleve the two omputtions. There ppers n idle yle in every two yles. These idle yles re used to operte v...v 2,u 2,v 1,u 1 MAC x 1,y 1,x 2,y 2,x 3..., 4 Time 1 2 3 4 5 6 7 8 9. Input u1 v1 u2 v2 u3 v3 u4 v4 v5... Output x1 y1 x2 y2 x3 y3 x4 y4 x5.. Pipelining the multiply-dd opertion y 4 stges we get u MAC1 x MAC2 MAC3 MAC4 9. Given tht Threshold voltge V t =.4 V Initil Voltge supply V o = 5 V Let the level of pipelining e M Let β is the voltge redution ftor i.e. supply voltge n e redued to βv o for the pipelined system. From (1) nd (1) we get β 2 <=1/5 => β = 5 =.447
Suppose tht the pipelined system nd the originl system hve the sme smple rte, we hve M= β (V o - V t ) 2 (βv o - V t ) 2 Using the vlues of V o, V t & β nd solving for M we get 1. Given tht M=.447(5-.4) 2 (.447 * 5 -.4) 2 = 2.8 = 3 Therefore, the system should e pipelined t 3 level. Sustitute M=3 into (3.9) nd solve for β =.427. The supply voltge for pipelined system is βv o =.427* 5 V = 2.14 Volts. As required, filter () nd filter () hve equl lok period, therefore: C hrge() V = C hrge() V k(v - V t ) 2 k(v - V t ) 2 From the filter struture we know T ritil =9T for filter (), T ritil =4T For filter(), C hrge() = Propgtion dely of iruit A(T ritil() ) = V * (V -V t ) 2 C hrge() Propgtion dely of iruit B (T ritil() ) V * (V -V t ) 2 Sustitute the vlues of V = 4 V nd V t =.5 V, we hve: 36(V ) 2-85V + 9 = V 1 =2.25 Volt =.11 Volt-------disrded V 2 Compre to filter(),the rtio of power sved y filter () is 1- (V ) 2 = 1-2.25 2 = 68.34% (V ) 4 2