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Given a sequence x(n) for 0≤n≤3, where x(0)=1,x(1)=3,x(2)=3, and x(3)=4. Evaluate its DFT X(k).
1 Answer
written 2.3 years ago by |
Solution:
Since N=4,W4=e−jπ/2, then using:
X(k)=∑3n=0x(n)Wkn4=∑3n=0x(n)e−jπkn2
Thus, for k=0
X(0)=3∑n=0x(n)e−j0=x(0)e−j0+x(1)e−j0+x(2)e−j0+x(3)e−j0=x(0)+x(1)+x(2)+x(3)=1+2+3+4=10 for k=1XX(1)=3∑n=0x(n)e−jπn2=x(0)e−j0+x(1)e−jπ2+x(2)e−jπ+x(3)e−j3π2=x(0)−jx(1)−x(2)+jx(3)=1−j2−3+j4=−2+j2 for k=2X(2)=3∑n=0x(n)e−jπn=x(0)e−j0+x(1)e−jπ+x(2)e−j2π+x(3)e−j3π=x(0)−x(1)+x(2)−x(3)=1−2+3−4=−2 and for k=3X(3)=3∑n=0x(n)e−j3πn2=x(0)e−j0+x(1)e−j3π2+x(2)e−j3π+x(3)e−jjπ2=x(0)+jx(1)−x(2)−jx(3)=1+j2−3−j4=−2−j2