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Linear Convolution Using the DFT 181
⎡
=
⎢
⎣
=
∞∑
r=−∞ m=0
[ ∞
∑
r=−∞
N∑
1−1
x 1 (m)x 2 (n − m − rN)
⎥
⎦ R N (n)
} {{ }
x 3(n−rN)
]
x 3 (n − rN)
⎤
R N (n) using (5.43)
This analysis shows that, in general, the circular convolution is an aliased
version of the linear convolution. We observed this fact in Example 5.15.
Now since x 3 (n) isanN =(N 1 + N 2 − 1)-point sequence, we have
x 4 (n) =x 3 (n); 0 ≤ n ≤ (N − 1)
which means that there is no aliasing in the time domain.
Conclusion: If we make both x 1 (n) and x 2 (n) N = N 1 + N 2 − 1point
sequences by padding an appropriate number of zeros, then the circular
convolution is identical to the linear convolution.
□ EXAMPLE 5.16 Let x 1(n) and x 2(n) bethe following two 4-point sequences.
x 1(n) ={1, 2, 2, 1} , x 2(n) ={1, −1, −1, 1}
a. Determine their linear convolution x 3(n).
b. Compute the circular convolution x 4(n) sothat it is equal to x 3(n).
Solution
We will use MATLAB to do this problem.
a. MATLAB Script:
>> x1 = [1,2,2,1]; x2 = [1,-1,-1,1]; x3 = conv(x1,x2)
x3 = 1 1 -1 -2 -1 1 1
Hence the linear convolution x 3(n) isa7-point sequence given by
x 3(n) ={1, 1, −1, −2, −1, 1, 1}
b.We will have to use N ≥ 7. Choosing N =7,wehave
>> x4 = circonvt(x1,x2,7)
x4 = 1 1 -1 -2 -1 1 1
Hence
x 4 = {1, 1, −1, −2, −1, 1, 1} = x 3(n)
□
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