signals and systems - eth z€¦ · discrete fourier series representation of a periodic signal...

Signals and Systems

Lecture 5: Discrete Fourier Series

Dr. Guillaume Ducard

Fall 2018

based on materials from: Prof. Dr. Raffaello D’Andrea

Institute for Dynamic Systems and Control

ETH Zurich, Switzerland

1 / 27

Outline

1 The Discrete Fourier SeriesDiscrete Fourier series representation of a periodic signalProperties of the discrete Fourier seriesDFS coefficients of real signals

2 Response to Complex Exponential SequencesComplex exponential as inputDFS coefficients of inputs and outputs

3 Relation between DFS and the DT Fourier TransformDefinitionExample

2 / 27

The Discrete Fourier SeriesResponse to Complex Exponential Sequences

Relation between DFS and the DT Fourier Transform

Discrete Fourier series representation of a periodic signalProperties of the discrete Fourier seriesDFS coefficients of real signals

Overview of Frequency Domain Analysis in Lectures 4 - 6

Tools for analysis of signals and systems in frequency domain:

The DT Fourier transform (FT): For general, infinitely long andabsolutely summable signals.⇒ Useful for theory and LTI system analysis.

The discrete Fourier series (DFS): For infinitely long butperiodic signals⇒ basis for the discrete Fourier transform.

The discrete Fourier transform (DFT): For general, finitelength signals.⇒ Used in practice with signals from experiments.

Underlying these three concepts is the decomposition of signals intosums of sinusoids (or complex exponentials).

3 / 27




Outline




4 / 27




Recall on periodic signals

A periodic signal displays a pattern that repeats itself, for exampleover time or space.

Recall

A periodic sequence x with period N is such that

x[n+N ] = x[n], ∀n

5 / 27




Overview of the Discrete Fourier Series

Analysis equation (Discrete Fourier Series)

DFS coefficients are obtained as:

X[k] =

N−1∑

n=0

x[n]e−jk 2πN

n.

Synthesis equation (Inverse Discrete Fourier Series)

x[n] =1

N

N−1∑

k=0

X[k]ejk2πN

n

6 / 27




Discrete Fourier series representation of a periodic signal

The Discrete Fourier Series (DFS) is an alternativerepresentation of a periodic sequence x with period N .

The periodic sequence x can be represented

as a sum of N complex exponentials with frequencies k 2πN, where

k = 0, 1, . . . , N−1:

x[n] =1

N

N−1∑

k=0

X[k]ejk2πN

n (1)

for all times n, where X[k] ∈ C is the kth DFS coefficient

corresponding to the complex exponential sequence ejk2πN

n.

7 / 27




Discrete Fourier series representation of a periodic signal

Remarks:

1 The frequency 2πN

is called the fundamental frequency and isthe lowest frequency component in the signal. (We will latershow that there is no loss of information in thisrepresentation.)

2 We only need N complex exponentials to represent a DTperiodic signal with period N .Indeed, there are only N distinct complex exponentials withfrequencies that are integer multiples of 2π

N:

ej(k+N) 2πN

n = ejk2πN

nejN2πN

n = ejk2πN

nej2πn = ejk2πN

n.

3 graphical representation (shown during class).

8 / 27




Discrete Fourier series coefficients

The DFS coefficients of Equation (1) are obtained from theperiodic signal x (the role of X and x are permuted with a minus sign in

the exponential) as:

X[k] =N−1∑

n=0

x[n]e−jk 2πN

n.

The DFS operator is denoted as Fs , where:

X = Fsx

and x = F−1s X.

The pairs are usually denoted as x[n] ←→ X[k].

9 / 27




Discrete Fourier series coefficients

Note that, like the underlying sequence x, X is periodic withperiod N :Proof:

X[k +N ] =

N−1∑

n=0

x[n]e−j(k+N) 2πN

n

=

N−1∑

n=0

x[n]e−jk 2πN

n e−j2πn︸︷︷︸

=1∀n

= X[k].

ConclusionsWhen working with the DFS, it is therefore common practice toonly consider one period of the sequence X[k], that is: only theN DFS coefficients as

X[k] for k = 0, 1, . . . , N−1.10 / 27




Proof that the DFS operator in invertible - Part I

The following identity highlights the orthogonality of complex exponentials:

1

N

N−1∑

n=0

ej(r−k) 2π

Nn =

1 for r − k = mN, m ∈ Z

0 otherwise.

Proof:Case 1 : r − k = mN . In this case, we have

ejmN 2π

Nn = e

j2πmn = 1 for all m,n

∴1

N

N−1∑

n=0

ej(r−k) 2π

Nn =

1

N

N−1∑

n=0

1 =1

NN= 1.

Case 2 : r − k 6= mN . Define l := r − k. We then have

1

N

N−1∑

n=0

ejl 2π

Nn =

1

N

1− ejl2πN

N

1− ejl2πN

,

the above equation is a geometric series and ejl2πN 6= 1 since l 6= mN . For

l 6= mN , we therefore have 1N

1−ej2πl

1−ejl 2π

N

= 0.11 / 27




Proof that the DFS operator in invertible - Part II

In order to prove that F−1s is the inverse transform of Fs , we need to show:

1 FsF−1s = I

2 and F−1s Fs = I , where I is the identity operator.

We now show that FsF−1s = I :

FsF−1s X[k] =

N−1∑

n=0

(

1

N

N−1∑

r=0

X[r]ejr2πN

n

)

e−jk 2π

Nn

=

N−1∑

r=0

X[r]

1

N

N−1∑

n=0

ej(r−k) 2π

Nn

︸︷︷︸

.

From above equation, the term in underbrace is equal to 1 for r = k mod N

and 0 otherwise. Thus:

FsF−1s X[k] =

N−1∑

r=0

X[r]

(

1

N

N−1∑

n=0

ej(r−k) 2π

Nn

)

= X[k mod N ] = X[k].

12 / 27




In a similar way, it can also be shown that F−1s Fs = I.

Remark:As both x and X are periodic with period N , we can sum over anyN consecutive values (denoted as 〈N〉):

x[n] =1

N

∑

k=〈N〉

X[k]ejk2πN

n and X[k] =∑

n=〈N〉

x[n]e−jk 2πN

n.

13 / 27




Outline




14 / 27




Properties of the discrete Fourier series

Linearity

a1x1[n]+ a2x2[n] ←→ a1X1[k]+ a2X2[k]

Parseval’s theorem (recall: period is N)

N−1∑

n=0

|x[n]|2 =1

N

N−1∑

k=0

|X[k]|2

15 / 27




Proof of Parseval’s theorem

1

N

N−1∑

k=0

|X[k]|2 =?

=1

N

N−1∑

k=0

X∗[k]X[k] (where ∗ denotes the complex conjugate)

=1

N

N−1∑

k=0

(N−1∑

n=0

x∗[n]ejk

2πN

n

)

X[k]

=1

N

N−1∑

k=0

N−1∑

n=0

x∗[n]X[k]ejk

2πN

n

=N−1∑

n=0

x∗[n]

1

N

N−1∑

k=0

X[k]ejk2πN

n

︸︷︷︸

x[n]

=

N−1∑

n=0

x∗[n]x[n] =

N−1∑

n=0

|x[n]|2

16 / 27




Outline




17 / 27




DFS coefficients of real signals: x[n] ∈ R for all n1 So far: x[n] ∈ C for all n.

2 Now we consider: x[n] ∈ R for all n; (most often case in practice)

Recall: DFS coefficients of a periodic signal x : X[k] =N−1∑

n=0

x[n]e−jk 2πN

n.

Letting k = N − γ, where γ is an integer, leads to

X[N − γ] =N−1∑

n=0

x[n]e−j(N−γ) 2πN

n

=

N−1∑

n=0

x[n] e−j2πn

︸︷︷︸=1∀n

ejγ 2π

Nn =

N−1∑

n=0

x[n]ejγ2πN

n =

N−1∑

n=0

x∗[n]ejγ

2πN

n = X∗[γ],

We used that for a real signal x, x[n] = x∗[n].

Conclusions:

For a real signal x, we have that X[N − k] = X∗[k].

Take γ = 0 → X[N ] = X∗[0], γ = N → X[0] = X∗[N ].By periodicity X[0] = X[N ]. Thus X[0] = X∗[0].

For a real signal, X[0] is therefore always real.18 / 27




DFS coefficients of real signals

Example

Consider the periodic signal :

x = . . . ,1

2↑

, 2,1

2, 1, . . . ,

Questions:

1 What is the period of such signal ?

2 Compute the Discrete Fourier Series coefficients.

19 / 27



Complex exponential as inputDFS coefficients of inputs and outputs

Outline




20 / 27




Consider a periodic input sequence u[n] with period N .When this input is applied to the LTI system G for all time n, the resultingoutput sequence y[n] = Gu[n] is also periodic (period N), and thus has aDFS representation.Rewriting the sequences using their DFS representations, it follows that:

1

N

N−1∑

k=0

Y [k]ejk2πN

n

= G

1

N

N−1∑

k=0

U [k]ejk2πN

n

.

Using linearity, this can be rewritten as:

∀n,1

N

N−1∑

k=0

Y [k] ejk2πN

n =1

N

N−1∑

k=0

G(

U [k]ejk2πN

n)

=1

N

N−1∑

k=0

H(z = ejk 2π

N ) U [k] ejk2πN

n,

where H(z) is the transfer function of system G.

21 / 27




Outline




22 / 27




1

N

N−1∑

k=0

Y [k] ejk2πN

n =1

N

N−1∑

k=0

H(z = ejk 2π

N ) U [k] ejk2πN

n.

Comparing the coefficients of the complex exponential terms leads to, for allk ∈ Z, lead to the following relationship between the DFS coefficients:

Y [k] = H(ejk2πN )U [k].

This result shows that:

the DFS coefficients of the output sequence’s : Y [k] are related to the DFScoefficients of the input U [k] by the system’s transfer functionH(z), sampled

at z = ejk2πN .

Remark:Note that this is equivalent to sampling the discrete-time Fourier transformH(Ω) of the system’s impulse response h[n] at discrete frequencies Ω = k 2π

N

for k ∈ Z.

23 / 27



DefinitionExample

Outline




24 / 27



DefinitionExample


Consider the Fourier series representation of a periodic signal x with period N :

x[n] =1

N

N−1∑

k=0

X[k]ejk2πN

n

for all times n. In Lecture 4, we saw how to extend the Fourier transform todeal with sequences that are not absolutely summable. Using that result (andrecalling that a rigorous treatment would need an understanding of the theoryof distributions), we can find the Fourier transform of x as follows:

X(Ω) =

∞∑

n=−∞

1

N

N−1∑

k=0

X[k]ejk2πN

ne−jΩn

=1

N

N−1∑

k=0

X[k]∞∑

n=−∞

ejn(k 2π

N−Ω)=

2π

N

N−1∑

k=0

X[k]δ(Ω− k2π

N).

1 Note that one would have to rigorously show that the order of summation can be swapped.

2 The DT Fourier transform of a periodic signal is therefore a finite sum of scaled and shifted Dirac delta

functions.

3 Note that the delta functions are located at the finite frequencies of the DFS, and scaled by the DFS

coefficients X[k].25 / 27



DefinitionExample

Outline




26 / 27



DefinitionExample

Example : Consider the absolutely summable sequence x1[n], the sequencex2[n] which is periodic with N = 6, and the magnitudes of their DT Fouriertransforms.

We see that X2(Ω) is composed of Dirac functions located at the discretefrequencies of the DFS representation of x2[n]:

x2[n] =16

∑5k=0 X2[k]e

jk 2π6

n

27 / 27

signals and systems - eth z€¦ · discrete fourier series representation of a periodic signal...

Documents