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Find DFT of x(n)={1234} Using these results and not otherwise find DFT.
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written 5.6 years ago by | • modified 5.6 years ago |
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x1(n)={4123}x2(n)={2341}x3(n)={6464}
Solution:
Given:
x(n)={1,2,3,4}
By definition of DFT.
x(k)=∑N−1n=0x(n)wnkN K=0,1,⋯,N−1
∴x(k)=[11111−j−1j1−11−11j−1−j][1234]
x(k)=[10−2+2i−2−2−2j]
x1(n)={4,1,2,3}
x1(n)=x(n−1)
By using time shift
∴x1(k)=e−j2πlkNx(k)
Here L = 1
∴x1(k)=e−j2πR4x(k)
x1(k)=e−jπR/2x(k)
∴x1(0)=x(0)=10
x1(1)=e−jπ/2x(2)=−j(−2+2j)=2+2j
x2(2)=e−j2π/2x(2)=−1(−2)=2
x1(3)=e−j3π/2x(3)=j(−2−2j)=2−2j
∴x1(k)=[102+2j22−2j]
x2(n)={2341}
x2(n)=x(n+1)
By time shift property
X2(k)=ej2πk4x(r)
x2(b)=ejπk2x(R)
x2(0)=x(0)=10
x2(1)=ejπ/2x(1)=+j(−2+2j)=−2−2j
x2(2)=ejπx(2)=−1(−2)=2
x2(3)=ej3π/2x(3)=−j(−2−2j)=−2+2j
x3(n)={6,4,6,4}
x3(n)=x(n+1)+x(n−1)
By linearity and Time shift prop.
X3(k)=ej2πR4x(k)+e−−jπk4x(R)
=eπR4x(R)+e−jπk2x(k)
=[10+2+2j22−2j]+[10−2−2j2−2+2j]
x3(x)=[20040]
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