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Circular Shifts

Module by: Justin Romberg

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.

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
fig1.png
Figure 1
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:
fig2.png
Figure 2: for m=-2 m -2
fig3.png
Figure 3
To cyclic shift we follow these steps:
1) Write fn f n on a cylinder, ACW
fig4.png
Figure 4: N=8 N 8
2) To cyclic shift by mm, spin cylinder m spots ACW fn→f (( n + m )) N → f n f (( n + m )) N
fig5.png
Figure 5: m=-3 m -3
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 N-m N m
fig6.png
Figure 6
f (( - n )) N f (( - n )) N The above expression, simply writes the values of fn f n clockwise.
fig7a.pngfig7b.png
Subfigure 7.1: fn f n
Subfigure 7.2: f (( - n )) N f (( - n )) N
Figure 7

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-ⅈ2πNkmFk f (( n - m )) N ↔ DFT 2 N k m F k (i.e. circular shift in time domain = phase shift in DFT)
Proof
fn=1N∑k=0N-1Fkⅇⅈ2πNkn f n 1 N k 0 N 1 F k 2 N k n (1)
so phase shifting the DFT
fn=1N∑k=0N-1Fkⅇ-ⅈ2πNknⅇⅈ2πNkn=1N∑k=0N-1Fkⅇⅈ2πNkn-m=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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