Question

Given x(n) = 2^n and N = 8 find X(k) using DIT fft algorithm?

Answer 1

Answer:

enter image description here. Output of stage – 1. S 1(0)=x(0)+x(4)=1+5=6. S1(1)=x(0)-x(4)=1-5=-4. S1( 2)=x(2)+x(6)=3+7=10. S1(3)=x(2)-x(6)=3-7=-4

Explanation:

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Answer 2

Answer:

X(K)=36,−4+9.65j,−4+4j,−4+1.65j,−4,−4−1.65j,−4−4j,−4−9.65j

Explanation:

Output of stage 1

S1(0)=x(0)+x(4)=1+5=6

S1(1)=x(0)-x(4)=1-5=-4

S1(2)=x(2)+x(6)=3+7=10

S1(3)=x(2)-x(6)=3-7=-4

S1(4)=x(1)+x(5)=2+6=8

S1(5)=x(1)-x(5)=2-6=-4

S1(6)=x(3)+x(7)=4+8=12

S1(7)=x(3)-x(7)=4-8=-4

Output of stage 2

S2(0)=S1(0)+W08S1(2)=6+(1)(10)=16

S2(1)=S1(1)+W28S1(3)=−4+(−j)(−4)=−4+4j

S2(2)=S1(0)−W08S1(2)=6−(1)(10)=−4

S2(3)=S1(1)−W28S1(3)=−4−(−j)(−4)=−4−4j

S2(4)=S1(4)+W08S1(6)=8+(1)(12)=20

S2(5)=S1(5)+W28S1(7)=−4+(−j)(−4)=−4+4j

S2(6)=S1(4)−W08S1(6)=8−(1)(12)=−4

S2(7)=S1(5)−W28S1(7)=−4−(−j)(−4)=−4−4j

Final output

X(0)=S2(0)+W08S2(4)=16+(1)(20)

X(1)=S2(1)+W18S2(5)=−4+4j+(0.707−j0.707)(−4+4j)

X(2)=S2(2)+W28S2(6)=−4+(−j)(−4)

X(3)=S2(3)+W38S2(7)=−4−4j+(−0.707−j0.707)(−4−4j)

X(4)=S2(0)−W68S2(4)=16−(1)(20)

X(5)=S2(1)−W18S2(5)=−4+4j−(0.707−j0.707)(−4+4j)

X(6)=S2(2)−W28S2(6)=−4(−j)(−4)

X(7)=S2(3)−W38S2(7)=−4−4j−(−0.707−j0.707)(−44j)

X(K)=36,−4+9.65j,−4+4j,−4+1.65j,−4,−4−1.65j,−4−4j,−4−9.65j

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