This document discusses the four Fourier representations used to represent signals: Fourier series, discrete-time Fourier series, Fourier transform, and discrete-time Fourier transform. It explains why the frequencies of the sinusoids used in the representations are discrete or continuous depending on whether the signal is periodic or non-periodic. It also discusses why the integration/summation intervals and normalization factors differ between the four representations.

1. 1
Supplementary Note for Signals and Systems:
The Four Fourier Representations
Shang-Ho (Lawrence) Tsai
Department of Electrical Engineering
College of Electrical and Computer Engineering
National Chiao Tung University
Hsinchu Taiwan
E-mail: shanghot@mail.nctu.edu.tw
CONTENTS
I Goal of this supplemental note 2
II Represent a signal by combining weighted sinusoids 2
II-A When x is periodic and discrete . . . . . . . . . . . . . . . . . . . . . . . 4
II-B When x is periodic and continuous . . . . . . . . . . . . . . . . . . . . . . 5
II-C When x is nonperiodic and discrete . . . . . . . . . . . . . . . . . . . . . 5
II-D When x is nonperiodic and continuous . . . . . . . . . . . . . . . . . . . . 6
III Determine the weights (coefficients) of sinusoids 7
III-A When x is periodic and discrete . . . . . . . . . . . . . . . . . . . . . . . 9
III-B When x is nonperiodic and discrete . . . . . . . . . . . . . . . . . . . . . 10
III-C When x is periodic and continuous . . . . . . . . . . . . . . . . . . . . . . 11
III-D When x is nonperiodic and continuous . . . . . . . . . . . . . . . . . . . . 11
References 12
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2. 2
I. GOAL OF THIS SUPPLEMENTAL NOTE
This supplemental note is to help students who take “Signals and Systems” to clarify the
fundamental concept in Fourier representations. Moreover, it tries to help students to build an
intuition why the Fourier and the inverse representations are defined like that in the textbook.
After reading this note, students are expected to be able to answer the following questions:
• Why are the Fourier representations for discrete signals always periodic (DTFS and DTFT)?
• Why are the Fourier representations for periodic signals always discrete (FS and DTFS)?
• Why are the integration intervals for the Fourier representations sometimes from −∞ to ∞
and sometimes from −π to π (FT and DTFT)?
• Why are the integration intervals for the inverse Fourier representations sometimes from
−∞ to ∞ and sometimes from 0 to T (IFT and IFS)?
• Why are the summation elements for the Fourier representation sometimes from −∞ to ∞
and sometimes from 0 to N − 1 (FS and DTFS)?
• Why are the summation elements for the inverse Fourier representations sometimes from
−∞ to ∞ and sometimes from 0 to N − 1 (IDTFT and IDTFS)?
• Why are the normalization factors sometimes N, sometimes 2π, sometimes T, and some-
times 1 (For all four Fourier and inverse Fourier representations)?
• Why are the units for frequencies sometimes Hz, sometimes rad/sec and sometimes rad?
II. REPRESENT A SIGNAL BY COMBINING WEIGHTED SINUSOIDS
Since the existing problems of the Fourier representations are beyond our scope, we simply
assume the Fourier representations exist for the signals that we discuss herein. The basic concept
of the Fourier representation is to represent a signal x by combining (sum or integral) several
weighted sinusoids which are discrete or continuous in frequency.
Question 1. Suppose that x is periodic, are the frequencies of the sinusoids to represent x
discrete or continuous?
Solution: Let us consider that x is a discrete signal with fundamental period N first. In this
case, the sinusoids are discrete in time and are periodic. Let the periods of the sinusoids be Nk.
Since x is periodic, to represent x, all the periods of the sinusoids must satisfy
N = kNk, (1)
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3. 3
where k must be an integer equal or greater than 1. If (1) is not satisfied, the sinusoids do not
have a common period of N. As a result, there is no way that x can be completely represented
by the sinusoids. From (1), the relationship of the frequencies between x and the sinusoids is
given by
1/N = 1/(kNk). (2)
Define the fundamental (angle) frequency of x as
Ω0 = 2π/N (rads), (3)
and the (angle) frequencies of the sinusoids as
Ωk = 2π/Nk (rads). (4)
From (2), (3) and (4), we have
kΩ0 = Ωk. (5)
From (5) and using the fact that k is an integer, we know that the (angle) frequencies of the
sinusoids are discrete.
Next, let us consider that x is a continuous signal with fundamental period T. In this case,
the sinusoids are also continuous in time and are periodic. By using similar argument as that
for discrete signal, we know that the periods of the sinusoids must also be multiples of T, i.e.
T = kTk, (6)
where Tk are the periods of the sinusoids and k must again be an integer equal or greater than
1. From (6), the relationship of frequencies between x and the sinusoids is given by
1/T = 1/(kTk) (Hz). (7)
Define the fundamental (angle) frequency of x as
ω0 = 2π/T (rads/sec), (8)
and the (angle) frequencies of the sinusoids as
ωk = 2π/Tk (rads/sec). (9)
From (7), (8) and (9), we have
kω0 = ωk. (10)
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4. 4
From (10) and using the fact that k is an integer, we know that the (angle) frequencies of the
sinusoids are discrete. From the discussion above, we can conclude the following property:
Property 1: The frequencies of the sinusoids to represent a period signal are discrete. We
usually say that a periodic signal have discrete frequency spectrum.
From Property 1, we know that to represent a periodic signal, the frequencies of the sinusoids
are discrete. Now we can write down the relationship between x and the sinusoids. Let us discuss
the relationship for discrete x and continuous x separately as follows:
A. When x is periodic and discrete
In this case, let us rewrite x as x[n] to reflect the fact that x is discrete in time. Since x[n]
is discrete in time, the sinusoids to represent x[n] should also be discrete in time. Since x[n] is
periodic, from Property 1 and (5), we let the sinusoids be ejkΩ0n
. Then, x[n] can be represented
by the combination of weighted sinusoids ejkΩ0n
, we have
x[n] =
k
X[k]ejkΩ0n
, (11)
where X[k] are the weights (or coefficients) at frequency kΩ0. To determine the range of k in
(11), we are asked how many different frequencies are sufficient to represent x[n]. To answer
this question, we notice that Ω0 = 2π/N. Hence, we have
ejkΩ0n
= ej 2π
N
kn
. (12)
From (12), we observe that ejkΩ0n
is periodic for k with period N, i.e.
ej(k+N)Ω0n
= ej 2π
N
(k+N)n
= ej 2π
N
kn
ej 2π
N
Nn
= ej 2π
N
kn
= ejkΩ0n
. (13)
Hence, from (13), we know that given a fundamental frequency Ω0 = 2π/N, we can generate
N different frequencies of sinusoids for all integer k, i.e.
ej0Ω0n
, ej1Ω0n
, ej2Ω0n
, · · · , ej(N−1)Ω0n
. (14)
Thus, it is sufficient to use N sinusoids in (17) to represent x[n]. Therefore, we can rewrite (11)
as
x[n] =
N−1
k=0
X[k]ejkΩ0n
, (15)
which leads to the definition of inverse Discrete-time Fourier Series (IDTFS).
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5. 5
B. When x is periodic and continuous
In this case, let us rewrite x as x(t) to reflect the fact that x is continuous in time. Since x(t)
is continuous in time, the sinusoids to represent x(t) should also be continuous in time. Since
x(t) is periodic, from Property 1 and (10), we let the sinusoids be ejkω0t
. Then, x(t) can be
represented by the combination of weighted sinusoids ejkω0t
, we have
x(t) =
k
X[k]ejkω0t
, (16)
where X[k] are the weights (or coefficients) at frequency kω0. Now let us determine the range
of k in (16). From (16), the values ejkω0t
= ejk 2π
T
t
for different k are different, since there is no
particular relationship between k and T. Hence, we do not have similar results as that in IDTFS
from Eqs. (11) to (15). Therefore, we know that given a fundamental frequency ω0 = 2π/T, we
can generate infinite different frequencies of sinusoids for all integer k, i.e.
· · · , e−j2ω0n
, e−j1ω0n
, ej0ω0n
, ej1ω0n
, ej2ω0n
, · · · . (17)
Thus, we need infinite frequencies of sinusoids to represent x(t). Therefore, we can rewrite (16)
as
x(t) =
∞
k=−∞
X[k]ejkω0t
, (18)
which is the definition of inverse Fourier Series (IFS).
What happens if x is nonperiodic? Under such situation, Property 1 no longer holds. Hence,
the frequencies of the sinusoids are no longer multiples of the fundamental frequencies. Thus,
the frequencies of the sinusoids are continuous if x is nonperiodic. We usually say that the
frequency spectrum of a nonperiodic signal is continuous. In this case, to represent x, we need
to use integration instead of summation to combine the sinusoids. Again, x can be discrete or
continuous. Let us discuss these two cases separately as follows:
C. When x is nonperiodic and discrete
In this case, let us rewrite x as x[n] to reflect the fact that x is discrete in time. Since x[n]
is discrete in time, the sinusoids to represent x[n] should also be discrete in time. In addition,
now x[n] is nonperiodic, hence the frequencies of the sinusoids are continuous. From discussion
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6. 6
above, it is reasonable to let the sinusoids be ejΩn
, where the unit of Ω is rads. Thus, x[n] can
be represented by the combination of weighted sinusoids given by
x[n] =
Ω
A(Ω)ejΩn
dΩ, (19)
where A(Ω) is the weight for sinusoid at frequency Ω. To determine the range of Ω in (19), we
notice that since n is an integer, we know ejΩn
is periodic for Ω with period 2π, i.e.
ej(Ω+2π)n
= ejΩn
ej2πn
= ejΩn
.
Hence, it is sufficient to use the sinusoids with frequencies in the range where −π < Ω ≤ π to
represent x[n]. Hence, we can rewrite (19) as
x[n] =
π
−π
A(Ω)ejΩn
dΩ. (20)
By defining new weights as
X(ejΩ
) = 2πA(Ω), (21)
we can rewrite (20) as
x[n] =
1
2π
π
−π
X(ejΩ
)ejΩn
dΩ, (22)
which is the definition of inverse Discrete-time Fourier Transform (IDTFT). Please note that the
reason to write the new weights as X(ejΩ
) instead of X(Ω) is to reflect the fact that X(ejΩ
)
is periodic with period 2π. We will explain later why X(ejΩ
) is period. In addition, we will
explain later why we need to define the new weights.
D. When x is nonperiodic and continuous
In this case, let us rewrite x as x(t) to reflect the fact that x is continuous in time. Since
x(t) is continuous in time, the sinusoids to represent x(t) should also be continuous in time. In
addition, now x[n] is nonperiodic, hence the frequencies of the sinusoids are continuous. From
discussion above, it is reasonable to let the sinusoids be ejωt
, where the unit of ω is rad/sec.
Thus, x(t) can be represented by the combination of weighted sinusoids given by
x(t) =
ω
A(ω)ejωt
dω, (23)
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7. 7
where A(ω) is the weight for sinusoid at frequency ω. Please note since t is not an integer,
different values for ω will lead to different frequencies of sinusoids ejωt
. Hence, we need to use
the sinusoids with all possible ω to represent x(t). Hence, we rewrite (23) as
x(t) =
∞
−∞
A(ω)ejωt
dω, (24)
By defining new weights as
X(jω) = 2πA(ω), (25)
we can rewrite (24) as
x(t) =
1
2π
∞
−∞
X(jω)ejωt
dω, (26)
which is the definition of inverse Fourier Transform (IFT). We will explain later why we need
to define the new weights.
III. DETERMINE THE WEIGHTS (COEFFICIENTS) OF SINUSOIDS
In the previous section, we already introduced four different Fourier representations to repre-
sent signals by combining weighted sinusoids. Except the weights, we actually have everything
to represent signals using sinusoids. In this section, we would like to determine the weights. The
detailed derivation of how to determine the weights is beyond our scope. Instead, we give simple
concept to help students gain insight into how to determine the weights. First, let us introduce
the concept of basis.
Definition: A basis B of a vector space V is a linearly independent subset of V that spans (or
generates) V.
Theorem: If B (b1, b2, · · · ) is an orthogonal basis (B is a basis and the vectors in B are
orthogonal) and span the vector space V, the unique representation of any vector v in V can be
expressed as
v =
k
b†
kv
ck
bk, (27)
where b†
k is complex-conjugate of bk.
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8. 8
From (27), we can regard ck = b†
kv as the projection of v on bk. Since v is a linear
combination of bk, we can also regard this projection ck as a “weight” of bk to represent
v. In this case, we may explain (27) as that vector v is the sum of weighted bk.
Example: The most famous orthogonal basis used in our real life is the Cartesian coordinate
system. The basis B in the three-dimension Cartesian coordinate system is define as
b1 = (1 0 0)t
, b2 = (0 1 0)t
, and b3 = (0 0 1)t
.
According to the theorem, a vector v = (2 3 4)t
can be written as
v = b†
1vb1 + b†
2vb2 + b†
3vb3 = 2b1 + 3b2 + 4b3,
where 2 is the weight for b1, 3 is the weight for b2 and 4 is the weight for b3, to represent v.
Similar to the vector space, for continuous functions, we may regard the basis B as that it
contains several vectors (may be finite or infinite) and each vector has infinite elements. Since the
functions are continuous, we can use similar representation in (27) and represent a continuous
function v(t) as
v(t) =
k τ
b∗
k(τ)v(τ)dτ
ck
bk(t), (28)
where b∗
k(τ) is conjugate of bk(τ). From (28), we can regard ck = τ
b∗
k(τ)v(τ)dτ as the projection
of v(t) on bk(t). Since v(t) is obtained by combining bk(t), we can also regard this projection
ck as a “weight” of bk(t) to represent v(t). In this case, we may explain (28) as that function
v(t) is the sum of weighted bk(t).
When the signal x(t) or x[n] is nonperiodic, the weight is no longer discrete. In this case, we
will replace ck by cf , where f is a continuous function of f. Hence, the expression in (28) is
rewritten as
v(t) =
f τ
b∗
f (τ)v(τ)dτ
cf
bf (t)df, (29)
With the concept of (27), (28) and (29), we can explain how to obtain the weights for the
four Fourier representations.
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9. 9
A. When x is periodic and discrete
Referring to (15), our goal is to determine the weights X[k]. Comparing (15) and (27), we
can regard ejkΩ0n
as an orthogonal basis to represent x[n]. Hence, referring to (27), the weights
of ejkΩ0n
are obtained by projecting x[n] on ejkΩ0n
, i.e.
ck =
n
x[n]e−jkΩ0n
. (30)
Since x[n] is periodic with period N, the sinusoids ejkΩ0n
also have a common period of N, the
projection only involves a period of N elements for x[n] and ejkΩ0n
. We can rewrite (30) as
ck =
N−1
n=0
x[n]e−jkΩ0n
. (31)
To make the pair of DTFS and IDTFS lead to an identity operator, i.e.
x[n] = IDTFS {DTFS {x[n]}} ,
we define new weights according to (31) as
X[k] =
1
N
N−1
n=0
x[n]e−jkΩ0n
, (32)
which leads to the weight definition of DTFS, i.e DTFS. It can be easily verified that the IDTFS
and DTFS pair in (15) and (32) indeed leads to an identity operator.
Question 2. Is X[k] in (32) periodic or nonperiodic?
Solution: Since n is an integer, e−jkΩ0n
= e−j 2π
N
kn
is periodic for k with period N, i.e.
e−j(k+N)Ω0n
= e−j 2π
N
(k+N)n
= e−j 2π
N
kn
e−j 2π
N
Nn
= e−j 2π
N
kn
= e−jkΩ0n
,
we know that X[k] in (32) is also periodic with period N, i.e.
X[k + N] = X[k]. (33)
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10. 10
B. When x is nonperiodic and discrete
Referring to (22), our goal is to obtain the weight X(ejΩ
). Comparing (22) and (27), we can
regard ejΩn
as a basis to represent x[n]. Hence, referring to (27), the weights of ejΩn
are obtained
by projecting x[n] on ejΩn
, i.e.
c(Ω) =
n
x[n]e−jΩn
. (34)
Note that since the frequency spectrum of a nonperiodic signal is continuous, it is more appro-
priate to use c(Ω) instead of ck to represent the weights in (34). Since x[n] is nonperiodic, the
projection in (34) involves infinite elements for x[n] and ejΩn
. Hence we can rewrite (34) as
c(Ω) =
∞
n=−∞
x[n]e−jΩn
. (35)
To make the pair of DTFT and IDTFT lead to an identity operator, i.e.
x[n] = IDTFT {DTFT {x[n]}} ,
we already multiplied 1/2π in the definition of IDTFT in (22). Hence, there is no need to add
normalization factor for the weights in (35). It can be easily verified that the IDTFT and DTFT
pair in (22) and (35) indeed leads to an identity operator.
Question 3. Is c(Ω) in (35) periodic or nonperiodic?
Solution: Since n is an integer, e−jΩn
= e−j(Ω+2π)n
is periodic for Ω with period 2π, i.e.
e−j(Ω+2π)n
= e−jΩn
e−j2πn
= e−jΩn
,
we know that c(Ω) is periodic with period 2π, i.e.
c(Ω + 2π) = c(Ω). (36)
To reflect the fact that the weights of DTFT are periodic, we define a new weight X(ejΩ
) =
c(Ω). Then, we have
X(ejΩ
) =
∞
n=−∞
x[n]e−jΩn
, (37)
which leads to the weight definition of DTFT, i.e DTFT.
From Question 2 and Question 3, we can conclude the following important property.
Property 2: The weights of Fourier representations for discrete signals are periodic. We
usually say that the frequency spectrum of a discrete signal is periodic.
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11. 11
C. When x is periodic and continuous
Referring to (18), our goal is to obtain the weights X[k]. Comparing (18) and (28), we can
regard ejkω0t
as an orthogonal basis to represent x(t). Hence, referring to (28), the weights of
ejkω0t
are obtained by projecting x(t) on ejkω0t
, i.e.
ck =
t
x(t)e−jkω0t
dt. (38)
Since x(t) is periodic with period T and the sinusoids ejkω0t
also have a common period of T,
the projection only involves a period of T for x(t) and ejkω0t
. Hence we can rewrite (38) as
ck =
T
0
x(t)e−jkω0t
dt. (39)
To make the pair of FS and IFS lead to an identity operator, i.e.
x(t) = IFS {FS {x(t)}} ,
we define new weights according to (39) as
X[k] =
1
T
T
0
x(t)e−jkω0t
dt, (40)
which leads to the weight definition of FS, i.e FS. It can be easily verified that the IFS and FS
pair in (18) and (40) indeed leads to an identity operator.
D. When x is nonperiodic and continuous
Referring to (26), our goal is to obtain the weight X(jω). Comparing (26) and (29), we can
regard ejωt
as a basis to represent x(t). Hence, referring to (29), the weights of ejωt
are obtained
by projecting x(t) on ejωt
, i.e.
c(ω) =
t
x(t)e−jωt
dt. (41)
Note that since the frequency spectrum of a nonperiodic signal is continuous, it is more appro-
priate to use c(ω) instead of ck to represent the weights in (41). Since x(t) is nonperiodic, the
projection in (41) involves infinite period for x(t) and ejωt
. Hence we can rewrite (41) as
c(ω) =
∞
−∞
x(t)e−jωt
dt. (42)
To make the pair of FT and IFT lead to an identity operator, i.e.
x(t) = IFT {FT {x(t)}} ,
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12. 12
we already multiplied 1/2π in the definition of IFT in (26). Hence, there is no need to add
normalization factor for the weights in (42). Letting X(jω) = c(ω), we have
X(jω) =
∞
−∞
x(t)e−jωt
dt, (43)
which leads to the weight definition of FT, i.e FT. It can be easily verified that the IFT and FT
pair in (26) and (43) indeed leads to an identity operator.
REFERENCES
[1] S. Haykin and B. V. Veen, Signals and Systems, 2nd Edition, John Wiley & Sons, Inc., 2003.
[2] G. Strang, Linear Algebra and Its Applications, 3nd Edition, Brooks Cole, Inc., 1988.
April 13, 2009 DRAFT