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Discrete-Time Fourier Series
Lecture 11
Dr. Görkem Saygılı
Department of Biomedical Engineering
Ankara University
Signals & Systems, 2018-2019 Fall
Content of This Lecture:
In this lecture, we will learn how to represent discrete time peri-
odic signals with complex exponentials.
Later, we will learn how to extend this to aperiodic signals (Dis-
crete Time Fourier Transform).
Response of LTI to Complex Exponentials:
If we can represent a signal as a linear combination of basic signals
then the response of an LTI system to this signal will be the linear
combination of responses to each of these basic signals.
For LTI systems these basic signals are complex exponentials (for
continuous time: est, for discrete time: zn). Because:
I Complex exponentials exit an LTI system with only a change
(if any) in their amplitude.
I Broad classes of signals can be represented with linear
combination of complex exponentials.
Eigenfunctions of LTI Systems:
By definition, an eigenfunction of a system is a signal such that
the response of the system to that signal is just the signal itself
multiplied with a constant.
Complex exponentials are eigenfunctions of LTI systems.
They enter an LTI system and leave the system as they are but
just multiplied by a constant.
Periodic Signals:
Remember, if signal x[n] is periodic with period N0, then x[n] =
x[n + N0].
I Fundamental period: N0 where N = kN0 and k ∈ Z (sec).
I Fundamental frequency: Ω0 = 2πf0 (radians/sec).
I Fundamental period: N0 = 2π
Ω0
Harmonically-Related Complex Exponentials:
σk[n] = ejk22π
N
n
, k ∈ Z (1)
where N is the period.
There are only N distinct ejk 2π
N
n
.
ejk 2π
N
n
= ej(k+lN) 2π
N
n
where l ∈ Z
Representing A Periodic Signal with Complex Exponentials:
We can represent a discrete time periodic signal x[n] with harmon-
ically related complex exponentials as:
x[n] =
k0+N−1
X
k=k0
akejk 2π
N
n
(2)
This representation is called Fourier Series representation of x[n]
which is a continuous time periodic signal with a fundamental pe-
riod of N0.
Important Remarks:
x[n] =
k0+N−1
X
k=k0
akejk 2π
N
n
(3)
I In this representation, ak’s are called Fourier series
coefficients (or spectral coefficients)
I Almost all periodic discrete time signals that are common in
engineering can be written in this form.
Finding Fourier Series Coefficients (ak’s):
We can calculate ak’s by evaluating the following summation:
ak =
1
N
N−1
X
n=0
x[n]e−jk 2π
N
n
(4)
The boundary encapsulates any period of the signal.
Properties of Fourier Series Coefficients - 1:
If a DT periodic signal satisfies:
x[n] = x∗
[n] (5)
Hence x(t) is a real CT periodic signal. Then,
ak = a∗
−k (6)
Properties of Fourier Series Coefficients - 2:
If x[n] is real and even then ak is also real and even:
x[n] = x[n] ⇒ ak = a−k (7)
If x[n] is real and odd then ak is purely imaginary and odd:
x[n] = −x[n] ⇒ ak = −a−k (8)
Linear Combination of Two CT Periodic Signals:
x[n] =
k0+N−1
X
k=k0
akejk 2π
N
n
, (9)
y[n] =
k0+N−1
X
k=k0
bkejk 2π
N
n
, (10)
Ax[n] + By[n] =
k0+N−1
X
k=k0
(Aak + Bbk)ejk 2π
N
n
(11)

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bme301_lec_11.pdf

  • 1. Discrete-Time Fourier Series Lecture 11 Dr. Görkem Saygılı Department of Biomedical Engineering Ankara University Signals & Systems, 2018-2019 Fall
  • 2. Content of This Lecture: In this lecture, we will learn how to represent discrete time peri- odic signals with complex exponentials. Later, we will learn how to extend this to aperiodic signals (Dis- crete Time Fourier Transform).
  • 3. Response of LTI to Complex Exponentials: If we can represent a signal as a linear combination of basic signals then the response of an LTI system to this signal will be the linear combination of responses to each of these basic signals. For LTI systems these basic signals are complex exponentials (for continuous time: est, for discrete time: zn). Because: I Complex exponentials exit an LTI system with only a change (if any) in their amplitude. I Broad classes of signals can be represented with linear combination of complex exponentials.
  • 4. Eigenfunctions of LTI Systems: By definition, an eigenfunction of a system is a signal such that the response of the system to that signal is just the signal itself multiplied with a constant. Complex exponentials are eigenfunctions of LTI systems. They enter an LTI system and leave the system as they are but just multiplied by a constant.
  • 5. Periodic Signals: Remember, if signal x[n] is periodic with period N0, then x[n] = x[n + N0]. I Fundamental period: N0 where N = kN0 and k ∈ Z (sec). I Fundamental frequency: Ω0 = 2πf0 (radians/sec). I Fundamental period: N0 = 2π Ω0
  • 6. Harmonically-Related Complex Exponentials: σk[n] = ejk22π N n , k ∈ Z (1) where N is the period. There are only N distinct ejk 2π N n . ejk 2π N n = ej(k+lN) 2π N n where l ∈ Z
  • 7. Representing A Periodic Signal with Complex Exponentials: We can represent a discrete time periodic signal x[n] with harmon- ically related complex exponentials as: x[n] = k0+N−1 X k=k0 akejk 2π N n (2) This representation is called Fourier Series representation of x[n] which is a continuous time periodic signal with a fundamental pe- riod of N0.
  • 8. Important Remarks: x[n] = k0+N−1 X k=k0 akejk 2π N n (3) I In this representation, ak’s are called Fourier series coefficients (or spectral coefficients) I Almost all periodic discrete time signals that are common in engineering can be written in this form.
  • 9. Finding Fourier Series Coefficients (ak’s): We can calculate ak’s by evaluating the following summation: ak = 1 N N−1 X n=0 x[n]e−jk 2π N n (4) The boundary encapsulates any period of the signal.
  • 10. Properties of Fourier Series Coefficients - 1: If a DT periodic signal satisfies: x[n] = x∗ [n] (5) Hence x(t) is a real CT periodic signal. Then, ak = a∗ −k (6)
  • 11. Properties of Fourier Series Coefficients - 2: If x[n] is real and even then ak is also real and even: x[n] = x[n] ⇒ ak = a−k (7) If x[n] is real and odd then ak is purely imaginary and odd: x[n] = −x[n] ⇒ ak = −a−k (8)
  • 12. Linear Combination of Two CT Periodic Signals: x[n] = k0+N−1 X k=k0 akejk 2π N n , (9) y[n] = k0+N−1 X k=k0 bkejk 2π N n , (10) Ax[n] + By[n] = k0+N−1 X k=k0 (Aak + Bbk)ejk 2π N n (11)