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Periodic Extension to DTFS

Module by: Roy Ha

Summary: This module looks at the periodic extensions of the DTFS coefficients and how the coefficients play a critical role in manipulating signals.

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Introduction

Now that we have an understanding of the discrete-time Fourier series (DTFS), we can consider the periodic extension of ck c k (the Discrete-time Fourier coefficients). Figure 1 shows a simple illustration of how we can represent a sequence as a periodic signal mapped over an infinite number of intervals.

Figure 1
(a) vectors
Figure 1(a) (fig1a.png)
(b) periodic sequences
Figure 1(b) (fig1b.png)

Exercise 1

Why does a periodic extension to the DTFS coefficients ck c k make sense?

Solution

Aliasing: b k =ⅇⅈ2πNkn b k 2 N k n

b k + N =ⅇⅈ2πNk+Nn=ⅇⅈ2πNknⅇⅈ2πn=ⅇⅈ2πNn= b k b k + N 2 N k N n 2 N k n 2 n 2 N n b k (1)
→ DTFS coefficients are also periodic with period NN.

Examples

Example 1: Discrete time square wave

Figure 2
Figure 2 (fig2.png)

Calculate the DTFS ck c k using:

ck=1N∑n=0N−1fnⅇ-ⅈ2πNkn c k 1 N n 0 N 1 f n 2 N k n (2)
Just like continuous time Fourier series, we can take the summation over any interval, so we have
c k =1N∑n=- N 1 N 1 ⅇ-ⅈ2πNkn c k 1 N n N 1 N 1 2 N k n (3)
Let m=n+ N 1 m n N 1 (so we can get a geometric series starting at 0)
c k =1N∑m=02 N 1 ⅇ-ⅈ2πNm− N 1 k=1Nⅇⅈ2πNk∑m=02 N 1 ⅇ-ⅈ2πNmk c k 1 N m 0 2 N 1 2 N m N 1 k 1 N 2 N k m 0 2 N 1 2 N m k (4)
Now, using the "partial summation formula"
∑n=0Man=1−aM+11−a n 0 M a n 1 a M 1 1 a (5)
c k =1Nⅇⅈ2πN N 1 k∑m=02 N 1 ⅇ-ⅈ2πNkm=1Nⅇⅈ2πN N 1 k1−ⅇ-ⅈ2πN2 N 1 +11−ⅇ-ⅈk2πN c k 1 N 2 N N 1 k m 0 2 N 1 2 N k m 1 N 2 N N 1 k 1 2 N 2 N 1 1 1 k 2 N (6)
Manipulate to make this look like a sinc function (distribute):
c k =1Nⅇ-ⅈk2π2Nⅇⅈk2πN N 1 +12−ⅇ-ⅈk2πN N 1 +12ⅇ-ⅈk2π2Nⅇⅈk2πN12−ⅇ-ⅈk2πN12=1Nsin2πk N 1 +12NsinπkN=digital sinc c k 1 N k 2 2 N k 2 N N 1 1 2 k 2 N N 1 1 2 k 2 2 N k 2 N 1 2 k 2 N 1 2 1 N 2 k N 1 1 2 N k N digital sinc (7)

note:

It's periodic! Figure 3, Figure 4, and Figure 5show our above function and coefficients for various values of N 1 N 1 .

Figure 3: N 1 =1 N 1 1
(a) Plot of fn f n . (b) Plot of ck c k .
Figure 3(a) (dtfs1.png)Figure 3(b) (dtfs1a.png)
Figure 4: N 1 =3 N 1 3
(a) Plot of fn f n . (b) Plot of ck c k .
Figure 4(a) (dtfs2.png)Figure 4(b) (dtfs2a.png)
Figure 5: N 1 =7 N 1 7
(a) Plot of fn f n . (b) Plot of ck c k .
Figure 5(a) (dtfs3.png)Figure 5(b) (dtfs3a.png)

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