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N x n c) y n =x n1 4 d) y n = 0, 0, 1, 0 ∆x n with ∆ denoting circular convolution Solution a) Since ej p 2 n x n =ej 2 p 4 n x n then DFT ej p 2 n x n =X k1 4 = 1 j,0,1j,1 b) In this case y n =Å1ÅÅÅ 2 ej 2 p 4 n x n ÅÅ1ÅÅ 2B h ʐM R X v ƃ { b g ^ A C e ̐ x ƍ i M ˗ i Q l i ́A 쓖 ̂ ̂ł B p ŃA X J v O X c i m } ƃ_ W o W p c ̃} W b N e v ł t ւ ŊȒP ɕύX ł ܂ jN 4 3 5 8 9 3 ; The Computation Domain Download Scientific Diagram ƒuƒƒbƒNŠ ‚èã‚° ƒƒ"ƒY "¯Œ^ ƒc[ ƒuƒƒbƒN ƒxƒŠ[ ƒVƒ‡[ƒg ƒAƒVƒ