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172 Chapter 5 THE DISCRETE FOURIER TRANSFORM
Therefore we first convert x(n) into its periodic extension ˜x(n), and
then shift it by m samples to obtain
˜x(n − m) =x ((n − m)) N
(5.35)
This is called a periodic shift of ˜x(n). The periodic shift is then converted
into an N-point sequence. The resulting sequence
˜x(n − m)R N (n) =x ((n − m)) N
R N (n) (5.36)
is called the circular shift of x(n). Once again to visualize this, imagine
that the sequence x(n)iswrapped around a circle. Now rotate the circle
by k samples and unwrap the sequence from 0 ≤ n ≤ N − 1. Its DFT
is given by
DFT [x ((n − m)) N
R N (n)] = W km
N X(k) (5.37)
□ EXAMPLE 5.11 Let x(n) =10(0.8) n , 0 ≤ n ≤ 10 be an 11-point sequence.
a. Sketch x((n + 4)) 11R 11(n), that is, a circular shift by 4 samples toward the
left.
b. Sketch x((n − 3)) 15R 15(n), that is, a circular shift by 3 samples toward the
right, where x(n) isassumed to be a 15-point sequence.
Solution
We will use a step-by-step graphical approach to illustrate the circular shifting
operation. This approach shows the periodic extension ˜x(n) =x((n)) N of x(n),
followed by a linear shift in ˜x(n) toobtain ˜x(n − m) =x((n − m)) N , and finally
truncating ˜x(n − m) toobtain the circular shift.
a. Figure 5.16 shows four sequences. The top-left shows x(n), the bottom-left
shows ˜x(n), the top-right shows ˜x(n+4), and finally the bottom-right shows
x((n + 4)) 11R 11(n). Note carefully that as samples move out of the [0,N− 1]
window in one direction, they reappear from the opposite direction. This is
the meaning of the circular shift, and it is different from the linear shift.
b. In this case the sequence x(n) istreated as a 15-point sequence by padding
4 zeros. Now the circular shift will be different than when N = 11. This
is shown in Figure 5.17. In fact the circular shift x ((n − 3)) 15
looks like a
linear shift x(n − 3).
□
To implement a circular shift, we do not have to go through the
periodic shift as shown in Example 5.11. It can be implemented directly
in two ways. In the first approach, the modulo-N operation can be used
on the argument (n − m) inthe time domain. This is shown below in the
cirshftt function.
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