# Chirp Signal Transform and Its Properties

Authors: Mio Horai; Hideo Kobayashi; Takashi G. Nitta
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

The chirp signalexp
(
i
π
(
x
-
y
)
2
) is a typical example of CAZAC (constant amplitude zero autocorrelation) sequence. Using the chirp signals, the chirp z transform and the chirp-Fourier transform were defined in order to calculate the discrete Fourier transform. We define a transform directly from the chirp signals for an even or odd number N and the continuous version. We study the fundamental properties of the transform and how it can be applied to recursion problems and differential equations. Furthermore, when N is not prime and  N
=
M
L, we define a transform skipped L and develop the theory for it.


---

## Body

## 1. Introduction

Chirp signal has been widely used in technology, for example, the radar system [1], as the linear FM pulse signal, the spectrum analyser as the sweep signal, the communications as the chirp modulation signal [2], and CAZAC sequence. Since the CAZAC sequences have constant amplitude and zero autocorrelation properties, they are now widely used in the fields of channel estimation and time and frequency synchronizations for OFDM (orthogonal frequency division multiplexing) and CDMA (code division multiple access) techniques which are employed as the standard transmission techniques in the wireless communications systems [3–5]. For the discrete Fourier transform theory, attached the chirp signal and Fourier transform, two kinds of transforms were already defined, namely, chirp z transform and chirp-Fourier transform. The first transform is (i) Fourier transform, (ii) product with the chirp signal, the second one is (i) product with the chirp signal (ii) Fourier transform [6]. These transforms were investigated mainly for the calculation of discrete Fourier transform (DFT).In this paper, we define another transform directly treating the chirp signal, which is equal to (i) the product of the chirp signal, (ii) Fourier transform, and (iii) the product of the chirp signal, and which has a new meaning. Our motivation is totally different with the ones for the chirp z transform and the chirp-Fourier transform. We name it as chirp signal transform and consider the transform for an even or odd numberN, continuous case, and skipped version in the case of N
=
L
M. We calculate firstly some examples for the transform of discrete, continuous, and skipped versions. Secondly, we show the inverse transform for the transform elementary, not using usual Fourier transform theory. Thirdly we study the properties of the original chirp signal, which is fundamental for our theory. Finally, we apply the transform to the theory of recursion problems and ordinary differential and partial differential equations.Chirp z transform and chirp-Fourier transform are based on the “orthogonality” ofexp
⁡
⁡
(
i
2
π
x
y
), y
∈
R; on the other hand, our chirp signal transform is founded on the orthogonality of exp
⁡
⁡
(
i
π
(
x
-
y
)
2
), y
∈
R. These three kinds of transforms are changed to each other by the unitary transform exp
⁡
⁡
(
i
π
x
2
) and exp
⁡
⁡
(
i
π
y
2
); however, the represented meanings are completely different. The two transforms’ y is the meaning of the frequency and the remaining one’s y is the meaning of the position. The chirp signal exp
⁡
⁡
(
i
π
(
x
-
v
t
)
2
) is useful for the radar, because it can be generated by simple liner FM pulse which can increase the frequency bandwidth of pulse and accordingly improve the accuracy of range measurement. Moreover, the chirp signals (LFM or NLFM) have satisfied properties of ambiguity function for radar application. The chirp signal transform shows that any L
2-function is represented as a sum of such chirp signals with different central positions.Furthermore, letL and L
¯ be (
∂
/
∂
x
)
+
i
2
π
x and (
∂
/
∂
x
)
-
i
2
π
x. Since [
L
¯
,
L
]
=
4
π
i, the set {
L
,
L
¯
} generates the Lie algebra s
l
(
2
,
C
). Additionally, F
L
=
i
2
π
y
F and F
¯
L
¯
=
-
i
2
π
y
F
¯. It seems that the set {
L
,
L
¯
,
F
,
F
¯
} has a good property for representation theory. It is very similar to the construction for the solution of harmonic oscillator in quantum mechanics by L and L
¯. We expect to develop the algebraic theory for using the relationship of the differential operators L, L
¯ and the chirp transforms F, F
¯ in order to solve some differential equations.


## 2. Definitions, Notations, and Examples

In this section, we define chirp signal transform for discrete, continuous, and skipped versions.For an even (resp., odd)N and a constant number a with (
a
,
N
)
=
1, we define chirp signal transform F of the function over the set {
0,1
,
2
,
…
,
N
-
1
}. Consider
(1)
F
(
f
)
(
y
)
=
1
N
∑
x
=
0
N
-
1
exp
⁡
⁡
(
i
a
π
N
(
x
-
y
)
2
)
f
(
x
)
(
resp
.
,
1
N
∑
x
=
0
N
-
1
‍
exp
⁡
(
i
a
π
(
x
-
y
)
(
x
+
1
-
y
)
N
)
f
(
x
)
)
.
When N is odd especially, F
f
(
y
) can be also defined by (
1
/
N
)
∑
x
=
0
N
-
1
exp
⁡
⁡
(
(
i
2
π
(
x
-
y
)
(
x
+
1
-
y
)
)
/
N
)
f
(
x
).On the other hand, we also define chirp signal transformF for continuous version, for L
2
(
R
),
(2)
F
(
f
)
(
y
)
:
=
∫
-
∞
∞
exp
⁡
⁡
(
i
π
(
x
-
y
)
2
)
f
(
x
)
d
x
.
Since exp
⁡
⁡
(
i
π
(
x
-
y
)
2
)
=
exp
⁡
⁡
(
i
π
y
2
)
·
exp
⁡
⁡
(
-
i
2
π
x
y
)
exp
⁡
⁡
(
i
π
x
2
), the left hand side F
(
f
)
(
y
) is equal to
(3)
exp
⁡
⁡
(
i
π
y
2
)
(
∫
-
∞
∞
exp
⁡
⁡
(
-
i
2
π
x
y
)
(
exp
⁡
⁡
(
i
π
x
2
)
f
(
x
)
)
)
d
x
.
We denote exp
⁡
⁡
(
i
π
x
2
)
f
(
x
) to be (
U
f
)
(
x
) and the Fourier transform as F; then, F is also written as U
F
U. It is also the same for discrete cases. The transforms U
F and F
U are called as chirp z transform and chirp-Fourier transform [6].LetN be even and divided as L
·
M. Then, we write
(4)
e
L
,
N
(
x
)
=
{
exp
⁡
⁡
(
i
π
L
N
x
2
)
,
if
x
∣
L
0
,
if
x
∤
L
.
We define chirp signal transform F
L
,
N as (
F
L
,
N
f
)
(
y
)
=
(
1
/
M
)
∑
x
=
0
N
-
1
f
(
x
)
·
e
L
,
N
(
x
-
y
).IfN is odd, both c and d are even or odd with (
c
,
N
)
=
1, and N
=
L
M, then we define
(5)
e
L
,
N
(
x
)
=
{
exp
⁡
⁡
(
i
π
x
L
N
(
c
x
+
d
)
)
,
if
x
∣
L
0
,
if
x
∤
L
.In the following, we calculate some examples for the above chirp signal transformF.(i) Fora
<
b, let χ
[
a
,
b
] be the step function χ
[
a
,
b
]
(
x
)
=
1  (
x
∈
[
a
,
b
]
) or 0 (
x
∉
[
a
,
b
]
).Then, the chirp signal transform ofχ
[
a
,
b
] is the following:
(6)
(
F
χ
[
a
,
b
]
)
(
y
)
=
∫
-
∞
∞
exp
⁡
⁡
(
i
π
(
x
-
y
)
2
)
χ
[
a
,
b
]
(
x
)
d
x
=
∫
a
b
‍
exp
⁡
⁡
(
i
π
(
x
-
y
)
2
)
d
x
=
∫
a
b
∑
k
=
0
∞
i
π
(
x
-
y
)
2
k
k
!
d
x
=
∑
k
=
0
∞
(
∫
a
b
(
x
-
y
)
2
k
d
x
)
(
i
π
)
k
k
!
=
∑
k
=
0
∞
[
1
2
k
+
1
(
x
-
y
)
2
k
+
1
]
a
b
(
i
π
)
k
k
!
=
∑
k
=
0
∞
‍
(
i
π
)
k
k
!
(
2
k
+
1
)
{
(
b
-
y
)
2
k
+
1
-
(
a
-
y
)
2
k
+
1
}
.(ii) We also calculate the chirp signal transform for the Gaussian distributionexp
⁡
⁡
(
-
π
x
2
). Consider
(7)
(
F
(
exp
⁡
(
-
π
x
2
)
)
)
(
y
)
=
∫
-
∞
∞
exp
⁡
⁡
(
-
π
x
2
)
exp
⁡
⁡
(
i
π
(
x
-
y
)
2
)
d
x
=
∫
-
∞
∞
exp
⁡
⁡
(
-
π
(
x
2
-
i
(
x
-
y
)
2
)
)
d
x
=
exp
⁡
⁡
(
-
π
1
-
i
2
y
2
)
×
(
∫
-
∞
∞
exp
⁡
(
-
π
(
1
-
i
)
(
x
+
i
1
-
i
y
)
2
)
d
x
)
.
Since exp
⁡
⁡
(
-
π
(
1
-
i
)
z
2
) is holomorphic and ∫
-
∞
∞
exp
⁡
⁡
(
-
π
(
1
-
i
)
(
x
+
(
i
/
(
1
-
i
)
)
y
)
2
)
d
x is constant up to y, we denote it as c
′.The above is equal toexp
⁡
⁡
(
-
π
(
(
1
-
i
)
/
2
)
y
2
)
(
∫
-
∞
∞
exp
⁡
⁡
(
-
π
(
1
-
i
)
x
2
)
d
x
). Therefore,
(8)
(
F
(
exp
⁡
⁡
(
-
π
x
2
)
)
)
(
y
)
=
c
′
exp
⁡
⁡
(
-
π
1
-
i
2
y
2
)
.WhenN is odd and N
=
L
M, we calculate the skipped chirp transform of f
(
x
)
=
exp
⁡
⁡
(
-
2
π
i
(
x
y
/
N
)
) at y
=
0 with respect to exp
⁡
(
i
π
(
x
(
x
+
d
)
/
L
N
)
), d
=
L
α, and x
=
L
m. Now, d is odd, and α is also odd. Consider
(9)
F
L
,
N
(
exp
⁡
⁡
(
-
2
π
i
x
k
N
)
)
(
0
)
=
∑
x
exp
⁡
⁡
(
i
π
L
m
(
L
m
+
L
α
)
L
N
)
exp
⁡
⁡
(
-
2
π
i
L
m
k
N
)
=
∑
exp
⁡
⁡
(
i
π
m
2
+
m
α
-
2
m
k
M
)
=
∑
m
=
0
M
-
1
exp
⁡
⁡
(
i
π
(
m
-
k
)
(
m
-
k
+
α
)
M
)
exp
⁡
⁡
(
i
π
k
(
α
-
k
)
M
)
=
C
exp
⁡
⁡
(
π
i
k
(
α
-
k
)
M
)
.Finally, we explain briefly that the discrete chirp signal transform becomes the continuous one, whenN increases to the infinity.LetH be an even positive integer, and 1
/
H is denoted by ϵ. Then, we define F
H for f in L
2
(
R
),
(10)
(
F
H
f
)
(
ϵ
m
)
=
1
H
∑
n
=
-
(
H
2
/
2
)
H
2
/
2
exp
⁡
⁡
(
i
π
(
n
-
m
)
2
H
2
)
f
(
ϵ
n
)
=
1
H
∑
n
=
-
(
H
2
/
2
)
H
2
/
2
exp
⁡
⁡
(
i
π
(
ϵ
n
-
ϵ
m
)
2
)
f
(
ϵ
n
)
.
Now,
(11)
(
F
f
)
(
y
)
=
∫
-
∞
∞
exp
⁡
⁡
(
i
π
(
x
-
y
)
2
)
f
(
x
)
d
x
=
lim
⁡
L
→
∞
⁡
∫
-
(
L
/
2
)
L
/
2
exp
⁡
⁡
(
i
π
(
x
-
y
)
2
)
f
(
x
)
d
x
=
lim
⁡
L
→
∞
⁡
lim
⁡
H
→
∞
⁡
1
H
∑
n
=
-
(
H
/
2
)
L
(
H
/
2
)
L
exp
⁡
⁡
(
i
π
(
ϵ
n
-
y
)
2
)
f
(
ϵ
n
)
.
If the final lim
⁡
L
→
∞
lim
⁡
H
→
∞ coincides to lim
⁡
H
→
∞ under L
=
H, the above F for the continuous case is the limit of the F
H for the discrete case. When H is odd, we can do the similar calculation. Furthermore, this correspondence can be represented as an infinitesimal lattice Fourier transform in nonstandard analysis [7–9].


## 3. The Inverse Chirp Signal TransformF
-
1

In this section, we show the inverse transform of the chirp signal transform defined in the first section. LetN be an integer, and let K be the function space from {
0,1
,
…
,
N
-
1
} to the complex number field C. For f
,
g
∈
K, the L
2 inner product is defined by
(12)
(
f
,
g
)
=
∑
x
=
0
N
-
1
f
(
x
)
¯
g
(
x
)
.
If f
(
x
-
a
), 0
≤
a
≤
N
-
1 are orthogonal to each other, then an arbitrary function g
∈
K is represented as a linear combination of {
f
(
x
-
a
)
,
0
≤
a
≤
N
-
1
}. For example, in the case of even N, f
(
x
-
a
)
=
exp
⁡
⁡
(
i
a
π
(
(
x
-
a
)
2
/
N
)
) satisfies such property. Now,
(13)
(
F
¯
F
f
)
(
z
)
=
1
N
∑
y
=
0
N
-
1
(
∑
x
=
0
N
-
1
exp
⁡
⁡
(
-
i
a
π
(
y
-
z
)
2
N
)
×
exp
⁡
⁡
(
i
a
π
(
x
-
y
)
2
N
)
f
(
x
)
)
=
1
N
∑
x
=
0
N
-
1
(
∑
y
=
0
N
-
1
exp
⁡
⁡
(
i
a
π
(
x
+
z
-
2
y
)
(
x
-
z
)
N
)
)
f
(
x
)
.
Since ∑
y
=
0
N
-
1
exp
⁡
⁡
(
(
-
2
i
a
π
(
x
-
z
)
)
/
N
)
y is N for x
=
z, or 0 for x
≠
z, the above (
F
¯
F
f
)
(
z
) is equal to f
(
z
). Hence, F
¯
F
=
i
d, and F
-
1
=
F
¯.Similarly, whenN is odd, f
(
x
)
=
exp
⁡
⁡
(
i
a
π
(
x
(
x
+
c
)
/
N
)
) satisfies the same property; in fact, (
x
-
y
)
(
x
-
y
+
c
)
-
(
x
-
z
)
(
x
-
z
+
c
)
=
-
2
(
y
-
z
)
x
+
(
y
-
z
)
(
y
+
z
)
-
c
(
y
-
z
).Hence,(
F
-
1
f
)
(
y
)
=
(
1
/
N
)
∑
x
=
0
N
-
1
exp
⁡
⁡
(
-
(
i
a
π
(
x
-
y
)
(
x
-
y
-
c
)
/
N
)
)
f
(
x
).We remark that the suffix ofc is changed from positive to negative as F to F
-
1.Next, we prove thatF
¯
F
=
i
d for the continuous version. We try to prove it directly.Letg
t
(
x
) be exp
⁡
⁡
(
-
π
(
t
x
)
2
) for t
∈
R
-
{
0
}. Then, the Fourier transform of g
t is calculated as follows:
(14)
(
F
(
g
t
)
)
(
y
)
=
∫
-
∞
∞
exp
⁡
⁡
(
-
2
π
i
x
y
)
exp
⁡
⁡
(
-
π
(
t
x
)
2
)
d
x
=
(
∫
-
∞
∞
exp
⁡
⁡
(
-
π
(
t
x
+
i
y
t
)
2
)
d
x
)
×
exp
⁡
⁡
(
-
π
(
y
t
)
2
)
,
and we put t
x as x
′
(15)
=

(

∫
-
∞
∞
‍
exp
⁡
(
-
π
(
x
′
+
i
y
t
)
2
)
d
x
′

)
1
|
t
|
exp
⁡
(
-
π
(
y
t
)
2
)

,
as
∫
-
∞
∞
‍
exp
⁡
(
-
π
(
x
′
+
i
y
t
)
2
)
d
x
′
=
1
,
=
1
|
t
|
exp
⁡
(
-
π
(
y
t
)
2
)
.
Now, lim
⁡
t
→
∞
g
t
(
x
)
=
1 and lim
⁡
t
→
0
(
1
/
|
t
|
)
exp
⁡
(
-
π
(
y
/
t
)
2
) is equal to ∞ for y
=
0 and 0 for y
≠
0. We denote (
1
/
|
t
|
)
exp
⁡
⁡
(
-
π
(
y
/
t
)
2
) to be δ
t
(
y
); then, (
F
(
g
t
)
)
(
y
) is equal to δ
t
(
y
).Theorem 1.
F
¯
F
f
=
f for f
∈
L
2
(
R
).Proof.
Consider(16)
F
¯
(
F
(
f
)
)
(
z
)
=
∫
-
∞
∞
exp
⁡
⁡
(
-
i
π
(
y
-
z
)
2
)
(
F
(
f
)
(
y
)
)
d
y
=
lim
⁡
t
→
∞
⁡
∫
-
∞
∞
g
t
(
y
)
exp
⁡
⁡
(
-
i
π
(
y
-
z
)
2
)
(
F
(
f
)
(
y
)
)
d
y
=
lim
⁡
t
→
∞
⁡
∫
-
∞
∞
g
t
(
y
)
exp
⁡
⁡
(
-
i
π
(
y
-
z
)
2
)
×
(
∫
-
∞
∞
exp
⁡
⁡
(
i
π
(
x
-
y
)
2
)
f
(
x
)
d
x
)
d
y
;by Fubini’s theorem,
(17)
=
lim
⁡
t
→
∞
⁡
∫
-
∞
∞
‍
f
(
x
)
(
∫
-
∞
∞
‍
g
t
(
y
)
exp
⁡
(
-
i
π
(
y
-
z
)
2
)
×
exp
⁡
(
i
π
(
x
-
y
)
2
)
d
y
∫
-
∞
∞
‍
)
d
x
=
lim
⁡
t
→
∞
⁡
∫
-
∞
∞
‍
f
(
x
)
(
∫
-
∞
∞
g
t
(
y
)
exp
⁡
⁡
(
-
i
2
π
(
x
-
z
)
y
)
d
y
)
×
exp
⁡
(
i
π
(
x
2
-
z
2
)
)
d
x
=
lim
⁡
t
→
∞
⁡
∫
-
∞
∞
f
(
x
)
δ
t
(
x
-
z
)
exp
⁡
⁡
(
i
π
(
x
2
-
z
2
)
)
d
x
=
lim
⁡
t
→
∞
⁡
∫
-
∞
∞
‍
(
f
(
x
)
·
exp
⁡
(
i
π
(
x
2
-
z
2
)
)
)
δ
t
(
x
-
z
)
d
x
=
f
(
z
)
·
exp
⁡
⁡
(
0
)
=
f
(
z
)
.
Finally, we show the inverse transform of the chirp signal transform for skipped version. WhenN is even, we obtain the following.Theorem 2.
Functionse
L
,
N
(
x
-
a
), 0
≤
a
≤
N
-
1 are orthogonal to each other.Proof.
Fora
,
b, if a
-
b
∤
L; then, (
e
L
,
N
(
x
-
a
)
,
e
L
,
N
(
x
-
b
)
)
=
0 by the definition. If a
-
b
∣
L, then a
-
b is written as L
·
α. Consider
(18)
∑
x
=
0
N
-
1
e
L
,
N
¯
(
x
-
a
)
e
L
,
N
(
x
-
b
)
=
∑
x
=
0
N
-
1
e
L
,
N
¯
(
x
)
e
L
,
N
(
x
-
(
b
-
a
)
)
=
∑
x
=
0
N
-
1
e
L
,
N
¯
(
x
)
e
L
,
N
(
x
-
L
α
)
=
∑
m
=
0
(
N
/
L
)
-
1
e
L
,
N
¯
(
L
m
)
e
L
,
N
(
L
m
-
L
α
)
=
∑
m
=
0
(
N
/
L
)
-
1
exp
⁡
⁡
(
-
i
π
(
L
m
)
2
L
N
)
exp
⁡
⁡
(
i
π
(
L
m
-
L
α
)
2
L
N
)
=
∑
m
=
0
(
N
/
L
)
-
1
exp
⁡
⁡
(
i
π
L
2
L
N
(
α
2
-
2
α
m
)
)
=
exp
⁡
⁡
(
i
π
L
N
α
2
)
∑
m
=
0
(
N
/
L
)
-
1
exp
⁡
⁡
(
-
i
2
π
N
L
·
α
m
)
=
{
0
,
if
L
·
α
∤
N
exp
⁡
⁡
(
i
π
L
N
α
2
)
·
M
,
if
L
·
α
∣
N
.
Hence, ifα
=
0, then (
e
L
,
N
,
e
L
,
N
)
=
M; otherwise, (
e
L
,
N
(
x
-
a
)
,
e
L
,
N
(
x
-
b
)
)
=
0.We assume thatN is odd. We obtain the following.Theorem 3.
Functionse
L
,
N
(
x
-
a
), 0
≤
a
≤
N
-
1 are orthogonal to each other.Proof.
We denoted to be d
=
L
β. For a
,
b, if a
-
b
∤
L, then (
e
L
,
N
(
x
-
a
)
,
e
L
,
N
(
x
-
b
)
)
=
0 by the definition. If a
-
b
∣
L, then a
-
b is written as L
·
α. Consider
(19)
∑
x
=
0
N
-
1
e
L
,
N
¯
(
x
-
a
)
e
L
,
N
(
x
-
b
)
=
∑
x
=
0
N
-
1
e
L
,
N
¯
(
x
)
e
L
,
N
(
x
-
(
b
-
a
)
)
=
∑
x
=
0
N
-
1
e
L
,
N
¯
(
x
)
e
L
,
N
(
x
-
L
α
)
=
∑
m
=
0
(
N
/
L
)
-
1
e
L
,
N
¯
(
L
m
)
e
L
,
N
(
L
m
-
L
α
)
=
∑
m
=
0
(
N
/
L
)
-
1
exp
⁡
⁡
(
-
i
π
L
m
(
c
L
m
+
L
β
)
L
N
)
×
exp
⁡
⁡
(
i
π
(
L
m
-
L
α
)
(
c
(
L
m
-
L
α
)
+
L
β
)
L
N
)
=
exp
⁡
⁡
(
-
π
i
L
L
N
α
β
)
∑
m
=
0
(
N
/
L
)
-
1
exp
⁡
⁡
(
-
2
π
i
N
c
L
α
m
)
=
0
since (
c
,
N
)
=
1.
Hence, ifα
=
0, then (
e
L
,
N
,
e
L
,
N
)
=
M; otherwise, (
e
L
,
N
(
x
-
a
)
,
e
L
,
N
(
x
-
b
)
)
=
0.The above theorem leads to the same property as the even case. Hence, for both cases, we obtain(20)
(
F
L
,
N
-
1
f
)
(
y
)
=
1
M
∑
x
=
0
N
-
1
e
¯
L
,
N
(
y
-
x
)
f
(
x
)
.


## 4. Properties of the Chirp Signal

In this section, we show some properties of the chirp signalexp
⁡
⁡
(
i
π
(
x
-
y
)
2
) which is fundamental for our chirp signal transform.Since(
∂
/
∂
x
)
(
exp
⁡
(
i
π
(
x
-
y
)
2
)
)
=
2
π
i
(
x
-
y
)
exp
⁡
(
i
π
(
x
-
y
)
2
), (
(
∂
/
∂
x
)
-
i
2
π
x
)
(
exp
⁡
⁡
(
i
π
(
x
-
y
)
2
)
)
=
(
-
i
2
π
y
)
exp
⁡
⁡
(
i
π
(
x
-
y
)
2
).Now, we write the differential operators(
∂
/
∂
x
)
+
i
2
π
x, (
∂
/
∂
x
)
-
i
2
π
x as L, L
¯. The differential operators L and L
¯ satisfy the following properties:
(21)
L
¯
L
=
d
2
d
x
2
+
(
4
π
x
2
+
2
π
i
)
,
L
L
¯
=
d
2
d
x
2
+
(
4
π
x
2
-
2
π
i
)
,
[
L
¯
,
L
]
=
L
¯
L
-
L
L
¯
=
4
π
i
.Then, the chirp signalexp
⁡
⁡
(
i
π
(
x
-
y
)
2
) is the eigen function of L
¯ with the eigen value -
i
2
π
y. Let Φ satisfy the equation L
Φ
=
0. Then, L
¯
L
(
L
¯
Φ
)
=
L
¯
(
L
L
¯
Φ
)
=
L
¯
(
4
π
i
+
L
¯
L
)
Φ
=
4
π
i
L
¯
Φ. Similarly, L
¯
L
(
L
¯
n
Φ
)
=
n
·
4
π
i
L
¯
n
Φ; that is, L
¯
n
Φ is the eigen function for L
¯
L with eigen value n
·
4
π
i.LetN be an odd integer, and let c, d be integers. Then, we define f
N
,
c
,
d
(
x
) as exp
⁡
(
-
(
i
π
x
(
c
x
+
d
)
/
N
)
). We have the following.Theorem 4.
The functionf
N
,
c
,
d
(
x
) is periodic iff both c and d are odd or even.Proof.
Consider(22)
x
(
c
x
+
d
)
-
(
x
+
N
)
(
c
(
x
+
N
)
+
d
)
=
-
2
c
x
N
+
N
(
c
N
+
d
)
≡
0
(
mod
⁡
2
N
)
iffc
N
+
d is even. Since N is odd and both c and d are odd or even, c
N
+
d is always even.Theorem 5.
f
N
,
c
,
d
(
x
-
a
), 0
≤
a
≤
N
-
1 are orthogonal to each other, iff a
∤
N.Proof.
Consider(23)
∑
x
=
0
N
-
1
f
¯
(
x
)
f
(
x
+
1
)
=
∑
x
=
0
N
-
1
exp
⁡
⁡
(
-
i
π
x
(
a
x
+
b
)
N
)
×
exp
⁡
⁡
(
i
π
(
x
+
1
)
(
a
(
x
+
1
)
+
b
)
N
)
=
exp
⁡
⁡
(
π
i
(
a
+
b
)
N
)
∑
n
=
0
N
-
1
exp
⁡
⁡
(
2
π
i
N
a
n
)
,
exp
⁡
⁡
(
2
π
i
a
/
N
)
≠
1
iff
a
∤
N
.
Secondly, we calculate the Fourier transform for the chirp signal. Consider(24)
1
N
∑
x
=
0
N
-
1
exp
⁡
⁡
(
-
i
π
x
2
N
)
exp
⁡
(
-
i
2
π
x
y
N
)
=
1
N
∑
x
=
0
N
-
1
exp
⁡
(
-
i
π
N
(
x
2
+
2
x
y
)
)
=
1
N
∑
x
=
0
N
-
1
exp
⁡
(
-
i
π
N
(
(
x
+
y
)
2
-
y
2
)
)
=
(
1
N
∑
x
=
0
N
-
1
exp
⁡
(
-
i
π
N
(
x
+
y
)
2
)
)
exp
⁡
(
i
π
y
2
N
)
;Since  exp
⁡
(
-
i
(
π
/
N
)
x
2
) is periodic,
(25)
=
(
1
N
∑
x
=
0
N
-
1
exp
⁡
(
-
i
π
N
x
2
)
)
exp
⁡
(
i
π
y
2
N
)
;we denote (
1
/
N
)
∑
x
=
0
N
-
1
‍
exp
⁡
(
-
i
(
π
/
N
)
x
2
) by C
(26)
=
C
exp
⁡
(
i
π
y
2
N
)
.
Now, since Fourier transform preserves the norm, |
C
|
=
1.
Hence, the Fourier transform forexp
⁡
⁡
(
-
i
(
π
x
2
/
N
)
) is equal to C
exp
⁡
(
i
(
π
y
2
/
N
)
).
Now, we assumeN is even, L divides N, and N
=
L
M. Consider
(27)
f
N
,
L
(
x
)
=
{
exp
⁡
(
π
i
x
2
L
N
)
,
x
=
L
m
(
x
∣
L
)
0
,
x
≠
L
m
(
x
∤
L
)
.
We calculate the usual Fourier transform of f
N
,
L
(
x
). Consider
(28)
(
F
f
N
,
L
)
(
k
)
=
1
N
∑
x
=
0
N
-
1
f
N
,
L
(
x
)
exp
⁡
(
-
2
π
i
x
k
N
)
=
1
N
∑
m
=
0
M
-
1
exp
⁡
(
π
i
(
L
m
)
2
L
N
)
exp
⁡
(
-
2
π
i
L
m
k
N
)
=
1
N
∑
m
=
0
M
-
1
exp
⁡
(
π
i
m
2
M
)
exp
⁡
(
-
2
π
i
m
k
M
)
=
1
N
∑
m
=
0
M
-
1
exp
⁡
(
π
i
(
m
-
k
)
2
M
)
exp
⁡
(
-
π
i
k
2
M
)
=
(
1
N
∑
m
=
0
M
-
1
exp
⁡
(
π
i
m
2
M
)
)
exp
⁡
(
-
π
i
k
2
M
)
.
Next, we calculate the Fourier transform off
N
,
L
(
x
-
a
). Consider
(29)
F
f
N
,
L
(
x
-
a
)
(
y
)
=
1
N
∑
x
f
(
x
-
a
)
exp
⁡
(
-
2
π
i
x
y
N
)
,
x
′
:
=
x
-
a
,
=
1
N
∑
x
′
f
(
x
′
)
exp
⁡
(
-
2
π
i
x
′
+
a
N
y
)
=
1
N
∑
x
′
f
(
x
′
)
exp
⁡
(
-
2
π
i
x
′
y
N
)
exp
⁡
(
-
2
π
i
a
y
N
)
.
Letf
(
x
,
t
)
=
exp
⁡
(
i
π
(
x
-
v
t
)
2
). Then, (
∂
/
∂
x
)
f
=
2
i
π
(
x
-
v
t
)
f and (
∂
/
∂
t
)
f
=
-
2
i
π
(
x
-
v
t
)
v
f. Hence, (
∂
/
∂
t
)
f
=
-
v
(
∂
/
∂
x
)
f
=
-
2
i
π
(
x
-
v
t
)
v
f. It is a chirp signal with central position v
t, which is a wave of x and t.


## 5. Application for Recursion Formulas and Ordinary Differential Equations

In this section, we develop some applications of the chirp signal transform.(i) For a functionf from {
0,1
,
…
,
N
-
1
} to C, we calculate the chirp signal transform of f
(
x
+
1
) with even N, a
=
1. Consider
(30)
(
F
f
(
x
+
1
)
)
(
y
)
=
1
N
∑
x
=
0
N
-
1
‍
exp
⁡
(
i
π
(
x
-
y
)
2
N
)
f
(
x
+
1
)
=
exp
⁡
(
i
π
2
y
+
1
N
)
×
{
1
N
∑
x
′
=
1
N
exp
⁡
(
i
π
(
x
′
-
y
)
2
N
)
∑
x
′
=
1
N
exp
⁡
(
i
π
(
x
′
-
y
)
2
N
)
×
(
exp
⁡
(
-
i
2
π
x
′
N
)
f
(
x
′
)
)
}
=
exp
⁡
(
i
π
2
y
+
1
N
)
{
F
(
exp
⁡
⁡
(
-
i
2
π
x
N
)
f
(
x
)
)
}
(
y
)
.
We apply this formula to the following equation for a constant C and a given function g
(
x
):
(31)
f
(
x
+
1
)
+
C
exp
⁡
(
-
i
2
π
x
N
)
f
(
x
)
=
g
(
x
)
.
We transform this equation by the chirp signal transform using formula (30),
(32)
(
exp
⁡
(
i
π
2
y
+
1
N
)
+
C
)
F
(
exp
⁡
(
-
i
2
π
x
N
)
f
(
x
)
)
(
y
)
=
F
(
g
(
x
)
)
(
y
)
.
We assume C is not equal to -
exp
⁡
⁡
(
i
π
(
(
2
y
+
1
)
/
N
)
) for any y
∈
{
0,1
,
…
,
N
-
1
}. For example, the absolute value of C is not equal to 1. Consider
(33)
F
(
exp
⁡
(
-
i
2
π
x
N
)
f
(
x
)
)
(
y
)
=
1
exp
⁡
(
i
π
(
(
2
y
+
1
)
/
N
)
)
+
C
F
(
g
(
x
)
)
(
y
)
;
hence, f
(
x
)
=
exp
⁡
⁡
(
i
(
2
π
x
/
N
)
)
·
F
¯
(
(
1
/
(
exp
⁡
⁡
(
i
π
(
(
2
y
+
1
)
/
N
)
)
+
C
)
)
F
(
g
)
(
y
)
)
(
x
).Similarly, we calculate the chirp signal transform ofexp
⁡
⁡
(
i
(
2
π
(
x
+
k
)
k
/
N
)
)
f
(
x
+
k
). Consider
(34)
F
(
exp
⁡
(
i
2
π
(
x
+
k
)
k
N
)
f
(
x
+
k
)
)
(
y
)
=
1
N
∑
x
=
0
N
-
1
exp
⁡
(
i
π
(
x
-
y
)
2
N
)
×
exp
⁡
(
i
2
π
(
x
+
k
)
k
N
f
(
x
+
k
)
)
,
(
(
x
-
y
)
2
+
2
(
x
+
k
)
k
=
(
x
+
k
-
y
)
2
+
2
k
y
+
k
2
)
=
exp
⁡
(
i
π
(
2
k
y
+
k
2
)
N
)
×
(
1
N
∑
x
=
0
N
-
1
exp
⁡
(
i
π
(
x
+
k
-
y
)
2
N
)
f
(
x
+
k
)
)
=
exp
⁡
(
i
π
(
2
y
+
k
)
k
N
)
(
F
f
)
(
y
)
.
Let (
T
n
f
)
(
x
) be defined by f
(
x
+
n
). Then, T
n
exp
⁡
⁡
(
i
π
(
(
x
-
y
)
2
/
N
)
) is equal to exp
⁡
⁡
(
i
π
(
(
x
+
n
-
y
)
2
/
N
)
), that is, exp
⁡
⁡
(
i
π
(
(
x
-
(
y
-
n
)
)
2
/
N
)
).We consider the following equation:(35)
∑
k
=
0
m
‍
C
k
exp
⁡
(
i
2
π
(
x
+
k
)
k
N
)
T
k
f
(
x
)
=
g
(
x
)
.
We put the chirp signal transform of it. Consider
(36)
F
(
∑
k
=
0
m
C
k
exp
⁡
(
i
2
π
(
x
+
k
)
k
N
)
T
k
f
(
x
)
)
=
F
(
g
(
x
)
)
.
By equality (34),
(37)
(
∑
k
=
0
m
C
k
exp
⁡
(
i
π
(
2
y
+
k
)
k
N
)
)
(
F
f
)
(
y
)
=
F
(
g
)
(
y
)
.
If ∑
k
=
0
m
‍
C
k
exp
⁡
⁡
(
i
(
(
π
(
2
y
+
k
)
k
)
/
N
)
)
≠
0, for y
∈
{
0,1
,
…
,
N
-
1
}, then
(38)
(
F
f
)
(
y
)
=
1
∑
k
=
0
m
C
k
exp
⁡
⁡
(
i
(
(
π
(
2
y
+
k
)
k
)
/
N
)
)
F
(
g
)
(
y
)
.Hence,(39)
f
=
F
-
1
(
1
∑
k
=
0
m
‍
C
k
exp
⁡
⁡
(
i
(
(
π
(
2
y
+
k
)
k
)
/
N
)
)
F
(
g
)
(
y
)
)
.(ii) We put the assumption thatlim
⁡
x
→
±
∞
f
(
x
)
=
0. Consider
(40)
(
F
f
′
(
x
)
)
(
y
)
=
∫
-
∞
∞
exp
⁡
(
i
π
(
x
-
y
)
2
)
f
′
(
x
)
d
x
=
[
f
(
x
)
exp
⁡
(
i
π
(
x
-
y
)
2
)
]
-
∞
∞
-
i
2
π
∫
-
∞
∞
‍
(
x
-
y
)
f
(
x
)
exp
⁡
(
i
π
(
x
-
y
)
2
)
d
x
;by the assumption,
(41)
=
-
i
2
π
∫
-
∞
∞
‍
(
x
f
(
x
)
)
exp
⁡
(
i
π
(
x
-
y
)
2
)
d
x
+
i
2
π
y
∫
-
∞
∞
‍
f
(
x
)
exp
⁡
(
i
π
(
x
-
y
)
2
)
d
x
=
-
i
2
π
F
(
x
f
(
x
)
)
(
y
)
+
i
2
π
y
F
(
f
(
x
)
)
(
y
)
.
Hence, F
(
f
′
(
x
)
)
(
y
)
=
-
i
2
π
F
(
x
f
(
x
)
)
(
y
)
+
i
2
π
y
F
(
f
(
x
)
)
(
y
) and
(42)
F
(
(
d
d
x
+
i
2
π
x
)
f
)
(
y
)
=
i
2
π
y
F
(
f
)
(
y
)
.
We write the differential operator (
d
/
d
x
)
+
i
2
π
x as L; then, F
(
L
f
)
(
y
)
=
i
2
π
y
F
(
f
)
(
y
). We consider the following differential equation, for coefficients C
k
(
0
≤
k
≤
n
) and a given function g
(
x
):
(43)
∑
k
=
0
n
C
k
L
k
f
=
g
.We apply the chirp signal transform:(44)
F
(
∑
k
=
0
n
C
k
L
k
f
)
=
F
(
g
)
,
∑
k
=
0
n
C
k
F
(
L
k
f
)
=
F
(
g
)
,
(
∑
k
=
0
n
C
k
(
i
2
π
y
)
k
)
F
(
f
)
=
F
(
g
)
.
If ∑
k
=
0
n
‍
C
k
(
i
2
π
y
)
k
≠
0 for arbitrary y, then F
(
f
)
=
(
1
/
(
∑
k
=
0
n
‍
C
k
(
i
2
π
y
)
k
)
)
F
(
g
).Therefore,f
=
F
-
1
(
(
1
/
(
∑
k
=
0
n
‍
C
k
(
i
2
π
y
)
k
)
)
F
(
g
)
), formally.For example, we consider the ordinary differential equation(45)
L
f
(
x
)
+
2
π
f
(
x
)
=
δ
(
x
)
,
where δ
(
x
) is the delta function. Since F
(
δ
)
(
y
) is equal to exp
⁡
(
i
π
y
2
), the solution f is just F
-
1
(
(
1
/
(
2
π
i
y
+
2
π
)
)
exp
⁡
(
i
π
y
2
)
), that is, (
(
1
/
2
π
i
)
∫
-
∞
∞
‍
(
(
exp
⁡
(
i
2
π
x
y
)
)
/
(
y
-
i
)
)
d
y
)
exp
⁡
(
-
i
π
x
2
). The integral (
(
1
/
2
π
i
)
∫
-
∞
∞
‍
(
(
exp
⁡
⁡
(
i
2
π
x
y
)
)
/
(
y
-
i
)
)
d
y
) is exp
⁡
(
-
2
π
x
), 1
/
2, 0, in case x
>
0, x
=
0
,
x
<
0, respectively. Hence, f
(
x
)
=
exp
⁡
(
-
π
(
i
x
2
+
2
x
)
)
,
1
/
2
,
0 for x
>
0
,
x
=
0
,
x
<
0, respectively.


## 6. Application for Partial Recursion Formulas and Differential Equations

In this section, we consider the application for partial recursion formulas and differential equations.(i) Letf
(
x
,
t
) be two variable function on {
0,1
,
…
,
N
-
1
}
×
R, and let
(46)
∂
f
∂
t
=
∑
k
=
0
m
‍
C
k
exp
⁡
(
i
2
π
(
x
+
k
)
k
N
)
T
k
f
(
x
)
.
Then
(47)
F
(
∂
f
∂
t
)
=
F
(
∑
k
=
0
m
C
k
exp
⁡
⁡
(
i
2
π
(
x
+
k
)
k
N
)
T
k
f
(
x
)
)
,
∂
(
F
f
)
∂
t
=
∑
k
=
0
m
C
k
exp
⁡
(
i
π
(
2
y
+
k
)
k
N
)
·
F
f
,
log
⁡
(
F
f
)
=
(
∑
k
=
0
m
C
k
exp
⁡
(
i
π
(
2
y
+
k
)
k
N
)
)
t
+
C
,
F
f
=
C
exp
⁡
(
t
∑
k
=
0
m
C
k
exp
⁡
(
i
π
(
2
y
+
k
)
k
N
)
)
,
f
=
C
F
-
1
(
exp
⁡
(
t
∑
k
=
0
m
C
k
exp
⁡
(
i
π
(
2
y
+
k
)
k
N
)
)
)
.(ii) Letf
(
x
,
t
) be two variables function on R
×
R. We write L
=
(
∂
/
∂
x
)
+
i
2
π
x.Then, it is already shown as(48)
F
(
L
f
)
(
y
)
=
i
2
π
y
F
(
f
)
(
y
)
.LetC
k  (
0
≤
k
≤
n
) be complex coefficients. Then, we consider the following partial differential equation:
(49)
∂
∂
t
f
=
∑
k
=
0
n
C
k
·
L
k
f
.
We put the chirp transform. Consider
(50)
F
∂
∂
t
f
=
F
(
∑
k
=
0
n
C
k
·
L
k
f
)
,
∂
∂
t
(
F
f
)
=
(
∑
k
=
0
n
C
k
(
i
2
π
y
)
k
)
F
(
f
)
.
We divide the equation by F
f,
(51)
1
F
f
∂
∂
t
(
F
f
)
=
∑
k
=
0
n
C
k
(
i
2
π
y
)
k
.
We put ∫
d
t,
(52)
l
n
(
F
f
)
=
(
∑
k
=
0
n
C
k
(
i
2
π
y
)
k
)
t
+
C
.
We write e
c
=
C,
(53)
F
f
=
C
exp
⁡
(
(
∑
k
=
0
n
C
k
(
i
2
π
y
)
k
)
t
)
.
Hence,
(54)
f
=
C
F
-
1
exp
⁡
(
(
∑
k
=
0
n
C
k
(
i
2
π
y
)
k
)
t
)
.


## 7. Conclusions

In this paper, we define the chirp signal transform using the chirp signal. Furthermore, we develop the theory of the chirp signal transform and apply it to the recursion problem, the ordinary differential equation, and the partial differential equation.



---

*Source: 161989-2014-10-20.xml*