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Periodic Extension to DTFS

Module by: Roy Ha. E-mail the author

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

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 =ei2πN(k+N)n=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=1Nn=0N1fne(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 =1Nn= 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 =1Nm=02 N 1 e(i2πN(m N 1 )k)=1Nei2πNkm=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=1aM+11a n 0 M a n 1 a M 1 1 a
(5)
c k =1Nei2πN N 1 km=02 N 1 e(i2πNk)m=1Nei2πN N 1 k1e(i2πN(2 N 1 +1))1e(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(ik2π2N)(eik2πN( N 1 +12)e(ik2πN( N 1 +12)))e(ik2π2N)(eik2πN12e(ik2πN12))=1Nsin2πk( N 1 +12)NsinπkN= 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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