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176 Chapter 5 THE DISCRETE FOURIER TRANSFORM
Solution
Note that x 1(n) isa3-point sequence, hence we will have to pad one zero to
make it a 4-point sequence before we perform the circular convolution. We will
compute this convolution in the time domain as well as in the frequency domain.
In the time domain we will use the mechanism of circular convolution, while in
the frequency domain we will use the DFTs.
• Time-domain approach: The 4-point circular convolution is given by
3∑
x 1(n) 4○ x 2(n) = x 1 (m) x 2 ((n − m)) 4
m=0
Thus we have to create a circularly folded and shifted sequence x 2((n−m)) N
for each value of n,multiply it sample by sample with x 1(m), add the samples
to obtain the circular convolution value for that n, and then repeat the
procedure for 0 ≤ n ≤ 3. Consider
x 1(m) ={1, 2, 2, 0} and x 2(m) ={1, 2, 3, 4}
for n =0
3∑
x 1(m) · x 2 ((0 − m)) 5
=
m=0
for n =1
3∑
x 1(m) · x 2 ((1 − m)) 5
=
m=0
for n =2
3∑
x 1(m) · x 2 ((2 − m)) 5
=
m=0
for n =3
3∑
x 1(m) · x 2 ((3 − m)) 5
=
m=0
=
=
=
=
3∑
[{1, 2, 2, 0}·{1, 4, 3, 2}]
m=0
3∑
{1, 8, 6, 0} =15
m=0
3∑
[{1, 2, 2, 0}·{2, 1, 4, 3}]
m=0
3∑
{2, 2, 8, 0} =12
m=0
3∑
[{1, 2, 2, 0}·{3, 2, 1, 4}]
m=0
3∑
{3, 4, 2, 0} =9
m=0
3∑
[{1, 2, 2, 0}·{4, 3, 2, 1}]
m=0
3∑
{4, 6, 4, 0} =14
m=0
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