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Find IDFT of the sequence X (K) = (5,0,1-j,0,1,0,1+j,0)
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written 2.3 years ago by |
Solution:
We have,
x(n)=1N∑N−1k=0X(k)ej2πkn/Nn=0,1,….,N−1
For N=8
x(n)=18∑N−2k=0X(k)ejπn/4n=0,1,….7
For n=0;x(0)=∑7k=0X(k)=18[5+0+1−j+0+1+j+0]=1
For n=1;x(1)=18∑7k=0X(k)ejπk/4=18[5+−(1−j)(j)+1(−1)+(1+j)(−j)]=6/8=0.75
For n=2;x(2)=18∑7k=0X(k)ejπk/2=18[5+(1−j)(−1)+1(1)+(1+j)(−1)]=4/8=0.5
For n=3;x(3)=18∑7k=0X(k)ej3πk/4=18[5+(1−j)(−j)+1(−1)+(1+j)(j)]=2/8=0.25
For n=4;x(4)=18∑7k=0X(k)ej5πk/4=18[5+(1−j)(1)+1(1)+(1+j)(1)]=8/8=1
For n=5;x(5)=18∑7k=0X(k)ej5πk/4=18[5+(1−j)(j)+(1)(−1)+(1+j)(−j)]=6/8=0.75
For n=6;x(6)=18∑7k=0X(k)ej3πk/2=18[5+(1−j)(−1)+1(1)+(1+j)(−j)]=4/8=0.5
For n=7;x(7)=18∑7k=0ej7πk/4=18[5+(1−j)(−j)+1(1)+(1+j)(j)]=2/8=0.25
x(n)={1,0.75,0.5,0.25,1,0.75,0.5,0.25}
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