已知序列x(n)={1.2.5.4;n=0.1.2.3},求x(k)=DFT[x(n)]

Question

已知序列x(n)={1.2.5.4;n=0.1.2.3},求x(k)=DFT[x(n)]

Answer 1

Xk=sum xn*e^{-i*2*pi*k*n/N}
N=4
X1=x0*e^{-i*2*pi*1*0/4}+x1*e^{-i*2*pi*1*1/4}+x2*e^{-i*2*pi*1*2/4}+x3*e^{-i*2*pi*1*3/4}
=1+2*e^{-i*pi/2}+5*e^{-i*pi}+4*e^{-i*3pi/2}
=1+2(-i)+5(-1)+4(i)
=-4+2i
同理
X2=x0*e^{-i*2*pi*2*0/4}+x1*e^{-i*2*pi*2*1/4}+x2*e^{-i*2*pi*2*2/4}+x3*e^{-i*2*pi*2*3/4}
=1+2(-1)+5(1)+4(-1)
=0
X3=x0*e^{-i*2*pi*3*0/4}+x1*e^{-i*2*pi*3*1/4}+x2*e^{-i*2*pi*3*2/4}+x3*e^{-i*2*pi*3*3/4}
=1+2(i)+5(-1)+4(-i)
=-4-2i
X4=x0*e^{-i*2*pi*4*0/4}+x1*e^{-i*2*pi*4*1/4}+x2*e^{-i*2*pi*4*2/4}+x3*e^{-i*2*pi*4*3/4}
=1+2(1)+5(1)+4(1)
=12