Question

4. Which of the following digital filters, expressed in terms of either their impulse response h[n] or their frequency response H (ejw), can be implemented in practice. Briefly justify the basis of your answer in each case. (a) (2 points) h[n] = 3-nu[n] (b) (2 points) H(ejw) = 2 + 3e-jw (c) (2 points) h[n] = 8[n + 1] – 8[n – 1] (d) (2 points) H(ejw) = e-j5w sin( 4w) sin()

Answer 1

4n) hem) = 3"um) = () um) HICSW) = a hin son naroo . men) - Si no lo nao #tew), anno (Hyun even 0 5-00 neo t - - - - #keb) = (a wote wa) alle ewt [ ( swer J . n=D L b> how) = 2 + 30 w . NOTE: DTFT t - A fen) al. MenFOTET xcesw) aln-no) LOTET Who xpow] is &in-no) es euro, . By using above properties of atat we have to find heno. 1

: How)= a +30 J I. O TF1 I hen) = 2.500 +3.5(0-1)). OTET A Son Of An [here A=2] A. Sln-no) or to the who, A.-3, noa! hen) = Sin+1) - Sin-1) DO TFT HO) SWE), _ Jul? c) OLET them). Gw w Slo-no) Orta Swno Slntno) QTFT eywno Lososovo 40 1 - | Heem) a aj slow sino - 80 - 160 n. 2j. dy Hewo) -Skolyw) sino) HLW) = 150 sin ( (5+ 2)W) : Seno) NOTE: ...Offs. Sin (N; ++)w) . Sinkin sin(w) ANI till. het is proove.

xcelw, - debe hace won n=-Ni + N = m. PN - WM-N)) ww, za kavu, oh - F. e m=o 090 xl qiw) = awna za cewe jong med zewni s 1-10 w2N1717: F. 207 z o ? 1-2=17 ( no la Li- words as west k) y est , **) pro burity? 28/2 a 1/2 que un gran - sin ( + 13/W) SR( ) xow) = sin (Nit?) w) ) sin(%) ;. rin) 2. ! " : q a - 5 -4 4 5 >n hin) = 211-5) hin) - n=0 23 9 10 on

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