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Circular Shifts

Module by: Justin Romberg. E-mail the author

Summary: The module looks at circular shifting and how it is can be used as a tool to represent the shifting of a periodic sequence.

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The many properties of the DFT become really straightforward (very similar to the Fourier Series) once we have once concept down: Circular Shifts.

Circular shifts

We can picture periodic sequences as having discrete points on a circle as the domain

Figure 1
Figure 1 (fig1.png)

Shifting by mm, fn+m f n m , corresponds to rotating the cylinder mm notches ACW (counter clockwise). For m=-2 m -2 , we get a shift equal to that in the following illustration:

Figure 2: for m=-2 m -2
Figure 2 (fig2.png)
Figure 3
Figure 3 (fig3.png)

To cyclic shift we follow these steps:

1) Write fn f n on a cylinder, ACW

Figure 4: N=8 N 8
Figure 4 (fig4.png)

2) To cyclic shift by mm, spin cylinder m spots ACW fnf (( n + m )) N f n f (( n + m )) N

Figure 5: m=-3 m -3
Figure 5 (fig5.png)

Example 1

If fn=01234567 f n 0 1 2 3 4 5 6 7 , then f (( n - 3 )) N =34567012 f (( n - 3 )) N 3 4 5 6 7 0 1 2

It's called circular shifting, since we're moving around the circle. The usual shifting is called "linear shifting" (shifting along a line).

Notes on circular shifting

f (( n + N )) N =fn f (( n + N )) N f n Spinning NN spots is the same as spinning all the way around, or not spinning at all.

f (( n + N )) N =f (( n - ( N - m ) )) N f (( n + N )) N f (( n - ( N - m ) )) N Shifting ACW mm is equivalent to shifting CW Nm N m

Figure 6
Figure 6 (fig6.png)

f (( - n )) N f (( - n )) N The above expression, simply writes the values of fn f n clockwise.

Figure 7
(a) fn f n (b) f (( - n )) N f (( - n )) N
Figure 7(a) (fig7a.png)Figure 7(b) (fig7b.png)

Circular shifts and the DFT

Theorem 1: Circular Shifts and DFT

If fn DFT Fk f n DFT F k then f (( n - m )) N DFT e(i2πNkm)Fk f (( n - m )) N DFT 2 N k m F k (i.e. circular shift in time domain = phase shift in DFT)

Proof

fn=1Nk=0N1Fkei2πNkn f n 1 N k 0 N 1 F k 2 N k n
(1)
so phase shifting the DFT
fn=1Nk=0N1Fke(i2πNkn)ei2πNkn=1Nk=0N1Fkei2πNk(nm)=f (( n - m )) N f n 1 N k 0 N 1 F k 2 N k n 2 N k n 1 N k 0 N 1 F k 2 N k n m f (( n - m )) N
(2)

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