Question

For a signal x(n)=sin(2*pi*n/5) defined for n=0to7, evaluate the Fast Fourier Transform using signal flow graph. (Use decimation in Time Algorithm).                                               

Answer 1

8 (0) 60. 365 -XCI) - 7,622 -10. 312 न x (2) -२.५१-j2. 125 • ६६) - rcl) - (3) 40. 278 - 51.03 न (५) ८ 0.815 - x() - 0.278 + j1.03 Rs) - 2cs) १८) K17 (6) ८ -२.५१ +12. 125 जोर । 2 का 1621 jo. 312

(423 4s 3 (บวน 100% - ouds o- () Ups - (1) (r) - * (L) xs - (2) 1 (1) 2 Sn () พ. (s) - shา (f) = 0 2. (6) - ciๆ (47) ) - Sin () e ose acos tj sint - 4 - 4 4. 4, C, 4] cts] 2 10) + 1Cy) - 2 co)* 4(y) 2. Sin( ): -0.9 2 - 2 ty) “ 1 ) 2 1(0) - (9) - Sfๆ ()0.5 7 2 CA) + x (6) , sitty) + Srn(3) : 1 2 ( - Le) , Sๆ ฯ? -Sh (29 - -0.36 - ๔ct) + ats) 2 st1 cr) + 0 2 0:ษ LCI) - its) 2 sh ()-0 2 0-9 4.3) 4 zcy) - Sn() + Sn( ) 2 - 1 zc9) - xc?) • sn (6) - Silly) - 6

bis aut al wees 0.59 be = ax + Quy way a 0.95 + 30.36 - ai - acum s-2049 bu 2 a, -Ryu = 0.95 -30.36 bs = qs + ay wag om -0.225 hog a got a won 0.95 by FQ; - w - 2.125 be a q nao wame=0.95 1.622 -30. 31) 5 0.278 — jf03 XCO) = b; + 6_480 0.3655 ☆ CV) 7 62 + bo w a Angos or45r3054 MC2) a bat boy wā ar 2049-12125 *(3) aby of bo wg -+62 30. 312 621 0.312 x(4) = b; ab wo a 0.815 *(5) - 6,- bele a 6+4+5*jte $ 0.278 +3:03 XC6) aba-bawa 249 +2.125 XC7) ab. -6.W? 162 4; 0,312