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A Discrete Hartley Transform based on Simpson’s rule
89
A Discrete Hartley Transform based on Simpson’s rule Ashai Ramsunder December 2015
Transcript of A Discrete Hartley Transform based on Simpson’s rule
A Discrete Hartley Transform based on
Simpson’s rule
Ashai Ramsunder
December 2015
A Discrete Hartley Transform based on
Simpson’s rule
by
Ashai Ramsunder
Submitted in fulfilment of the
academic requirements for the degree of
Master of Science
in the
School of
Mathematics, Statistics and Computer Science
University of KwaZulu-Natal
Durban
December 2015
As the candidate’s supervisors, we have approved this thesis for submission.
Signed:............................ Name:...Dr P.Singh.............. Date:
Signed:............................ Name:...Dr V.Singh.............. Date:
Preface
The work described in this dissertation was carried out in the School of Mathematical
Sciences, University of KwaZulu-Natal, Durban, from February 2015 to December 2015,
under the supervision of Dr P. Singh and co-supervised by Dr V. Singh. This study
represents original work by the author and has not otherwise been submitted in any
form for any degree or diploma to any other tertiary institution. Where use has been
made of the work of others, it is duly acknowledged in the text.
Signed:............................
Mr Ashai Ramsunder
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FACULTY OF SCIENCE AND AGRICULTURE
DECLARATION 1 - PLAGIARISM
I, Ashai Ramsunder, declare that
1. The research reported in this thesis represents my own original research, except
where due reference is made.
2. This thesis has not been submitted in the past for any degree at any other
institution (academic or otherwise).
3. This thesis does not contain other persons’ data, pictures, graphs or other infor-
mation, unless specifically acknowledged as being sourced from the said persons.
4. This thesis does not contain other persons’ writing, unless specifically acknowl-
edged as being sourced from the said persons. Where other written sources have
been quoted, then:
a. Their words have been re-written but the general information attributed to
them has been referenced.
b. Where their exact words have been used, their writing has then been placed
in italics, within quotation marks, and referenced.
5. This thesis does not contain text, graphics or tables copied and pasted from the
Internet, unless specifically acknowledged, and the source being detailed in the
thesis and in the Bibliography.
Signed:...............................................
Mr Ashai Ramsunder
iv
Dedication
This thesis is dedicated to my father Dave Ramsunder, my mother Saira Ramsunder,
my brother Mihir Ramsunder and all the family members who have been lost over the
years.
v
Acknowledgement
I would firstly like to thank God and my immediate family which include my parents
Dave and Saira Ramsunder as well as my brother Mihir Ramsunder for making me
the person I am today. My utmost gratitude and appreciation is expressed to my
supervisor Dr Pravin Singh and co-supervisor Dr Virath Singh. Dr P.Singh and Dr
V.Singh have provided me with a vast set of new skills in both life and mathematics.
They have ensured that the work presented is of the highest quality and I have always
challenged myself to make an impression and meet their high standards. Although
the work was demanding, I have not once been afraid or stressed to ask either of my
supervisors for assistance.
I would like to also thank the National Research Foundation (NRF) for providing me
with funding required to pursue my studies further. A special thank you to Mr Soren
Greenwood for the computer support as well as insightful conversations.
Finally I would like to thank all my friends, both inside as well as out of UKZN and
the staff at the department of mathematics, who made my life easier.
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Notation
A,AN Matrix of order N
C(k), C(k, l) Approximation to Hartley coefficients
f Signal or image matrix
fe Even part of signal f
fo Odd part of signal f
fR Signal f reversed
f0 Even indexed components of a signal
f1 Odd indexed components of a signal
H Hartley transformation matrix
H Simpson Hartley transformation matrix
I, IN Identity matrix of order N
P,PN Permutation matrix of order N
⊗ Convolution
? Correlation
.? Element-wise multiplication
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Abstract
The Discrete Hartley Transform and Discrete Fourier Transform are classical transfor-
mations designed for efficient computations in the frequency domain. We introduce
a relatively new transformation based on the existing Discrete Hartley Transform by
applying Simpson’s quadrature for N = 4m + 2 quadrature nodes. The majority of
our investigation involves exploring the mathematical properties satisfied by our newly
derived transformation. We formulate the convolution and cross correlation properties
both in the real and frequency domain. An intensive spectral analysis is performed
to ascertain the multiplicities of the eigenvalues corresponding to the transformation
matrix.
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Contents
Introduction 1
1 Discrete Hartley Transform 3
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Derivation of the DHT . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Simpson Discrete Hartley Transform 7
3 Properties of the SDHT 11
3.1 Transform of an even signal . . . . . . . . . . . . . . . . . . . . . . 11
3.2 Transform of an odd signal . . . . . . . . . . . . . . . . . . . . . . . 12
3.3 Time reversal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.4 Frequency shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.5 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.6 Convolution in the real domain . . . . . . . . . . . . . . . . . . . . 19
3.7 Convolution in the frequency domain . . . . . . . . . . . . . . . . . 25
3.8 Cross correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
4 Spectrum of H 43
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4.1 Minimal polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.2 Traces of powers of H . . . . . . . . . . . . . . . . . . . . . . . . . 48
4.3 Multiplicity of eigenvalues . . . . . . . . . . . . . . . . . . . . . . . 53
5 2-Dimensional SDHT 57
5.1 Derivation of the 2-D SDHT . . . . . . . . . . . . . . . . . . . . . . . . 58
6 Applications of the 2-D SDHT 68
6.1 Watermarking in the frequency domain . . . . . . . . . . . . . . . . . . 69
6.2 Noise cleaning in the frequency domain . . . . . . . . . . . . . . . . . . 72
Conclusion 76
Bibliography 77
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List of Figures
5.1 A 6× 10 mesh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
6.1 Cameraman . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
6.2 Transform of cameraman . . . . . . . . . . . . . . . . . . . . . . . . . . 69
6.3 fftshift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
6.4 Peppers and it’s transform . . . . . . . . . . . . . . . . . . . . . . . . . 71
6.5 Watermarking in the frequency domain in an area of high amplitude . . 71
6.6 Watermarking in an intermediate region in the frequency domain . . . 72
6.7 House contaminated with noise . . . . . . . . . . . . . . . . . . . . . . 73
6.8 Effect of noise in the frequency domain . . . . . . . . . . . . . . . . . . 73
6.9 Block filter applied in the frequency domain . . . . . . . . . . . . . . . 74
6.10 House after applying filter . . . . . . . . . . . . . . . . . . . . . . . . . 74
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Introduction
In chapter 1, we provide a brief literature survey of the Discrete Hartley Transform
(DHT) and show how it arises from trapezoidal quadrature.
In chapter 2, we derive the Simpson Discrete Hartley Transform (SDHT) from Simp-
son’s quadrature. We provide an expression for the transformation matrix as well as an
expression for the inverse. We investigate the effect of the transform on a real signal.
In chapter 3, we provide detailed proofs for several properties satisfied by the Simpson
Discrete Hartley Transform. We prove the convolution property for both the real and
frequency domains as well as the cross correlation property. We also prove various
other properties such as the frequency shift property and the duality property.
In chapter 4, we provide a detailed spectral analysis of the transformation matrix. We
use the duality property to obtain an expression for the minimal polynomial. From the
minimal polynomial we obtain four distinct eigenvalues for the transformation matrix.
We employ powers of the traces of the matrix and the theory of Gauss sums to obtain
the multiplicities of the eigenvalues.
In chapter 5, we derive a two-dimensional Simpson Discrete Hartley Transform.
In chapter 6, we apply this transform to the useful area of image processing namely
1
watermarking and cleaning sinusoidal noise in the frequency domain.
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Chapter 1
Discrete Hartley Transform
1.1 Introduction
Ralph Hartley was an innovative scientist responsible for mathematical triumphs such
as the Hartley transform and Hartley oscillator [9]. The Hartley transform is closely re-
lated to the Fourier transform. These are both integral transforms, which however have
one main difference being that the former is real valued whilst the latter is generally
complex valued [9].
While investigating the relationship of Fourier analysis to images in 1983, Ronald
Bracewell, who was in the research field of electronic engineering discovered a new
factorization of the discrete Fourier transformation matrix which resulted in a fast
algorithm for spectral analysis [4]. This discovery led to the proposal of the Discrete
Hartley Transform (DHT) and the first Fast Hartley Transform (FHT) algorithm by
Ronald Bracewell in 1983 [5]. The DHT is closely related to the Discrete Fourier
transform (DFT) whereby both consist of discrete periodic data. Ronald Bracewell
3
also proposed that the FHT algorithm is more computationally efficient than the FFT
(Fast DFT) when the data is real, however this was disproved by tweaking the FFT
for real inputs [7].
The properties of the DHT enables one to design fast algorithms to compute it [14] and
thus allows applications in various areas of signal processing such as power engineering
and digital filtering [10]. The DHT is also used extensively in biomedical imaging for
edge enhancement and medical resonance imaging and in addition can also be useful
in image compression and reconstruction.
1.2 Derivation of the DHT
We will now show how the DHT arises by applying trapezoidal quadrature. Recall that
the coefficients in a Hartley series for a continuous, 2π-periodic function f(x) have the
form
c(k) =1
2π
∫ 2π
0
f(x) cas(kx)dx, (1.1)
where cas(x) = cos(x) + sin(x). Divide the interval [0, 2π] into N equally spaced
subintervals of length ω = 2πN
with nodes xj = ωj, j = 0, 1, . . . , N . The approximation
C(k) to c(k) using the trapezoidal rule gives
C(k) =1
2N
[f(x0) cas(kx0) + f(xN) cas(kxN) + 2
N−1∑j=1
f(xj) cas(kxj)]
=1
2N
[2f(x0) cas(kx0) + 2
N−1∑j=1
f(xj) cas(kxj)]
=1
N
N−1∑j=0
f(xj) cas(kxj).
(1.2)
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We used the fact that f(x0) = f(xN) due to the periodicity of f . We adopt the notation
f(xj) = f(j) and rewrite equation (1.2) as
C(k) =1
N
N−1∑j=0
f(j) cas(kjω). (1.3)
The derivation above enables us to define the DHT of a N point real sequence f(j),
j = 0, 1, . . . , N − 1 by [6]
F (k) = Hf(k) =N−1∑j=0
cas(kωj)f(j), (1.4)
k = 0, 1, . . . , N − 1 along with transformation matrix H defined by components hkj =
cas(kωj), j = 0, 1, . . . , N − 1, k = 0, 1, . . . , N − 1. To be consistent we have chosen to
index the components from zero.
In order to recover the signal f = [f(0), f(1), . . . , f(N−1)]T multiply (1.4) by cas(kωp),
p ∈ {0, 1, . . . , N − 1} and sum over k to obtain
N−1∑k=0
cas(kωp)F (k) =N−1∑j=0
[N−1∑k=0
cas(kωj) cas(kωp)]f(j). (1.5)
We will now prove the following lemma.
Lemma 1.1.
N−1∑k=0
cas(kωj) cas(kωp) =
N when j = p,
0 when j 6= p,
(1.6)
where p ∈ {0, 1, . . . , N − 1}.
Proof. Using the identity
cas(A) cas(B) = cos(A−B) + sin(A+B), (1.7)
enables us to expand the left hand side of (1.6) which results in
N−1∑k=0
cos(kω(j − p)
)+
N−1∑k=0
sin(kω(j + p)
). (1.8)
5
Let m = j − p then m ∈ {−N + 1,−N + 2, . . . , N − 1} and it suffices to show that
N−1∑k=0
cos(kωm) =
N when m = 0,
0 when m 6= 0.
(1.9)
When m = 0 the result is trivial. Consider the case when m 6= 0. Let
Sm =N−1∑k=0
eikωm = 1 + eiωm + e2iωm + . . .+ eiωm(N−1)
= 1 +N−1∑k=1
eikωm. (1.10)
Multiply equation (1.10) by eiωm to get
Sm(eiωm) =N−1∑k=0
ei(k+1)ωm
=N∑k=1
eikωm
=N−1∑k=1
eikωm + eiωNm. (1.11)
Subtracting (1.11) from (1.10) yields
Sm =1− eiωNm
1− eiωm. (1.12)
We observe that Sm = 0 since the numerator is zero and the denominator is non zero.
Taking the real part of Sm proves (1.9). Furthermore it may be shown in a similar
manner thatN−1∑k=0
sin(kω(j + p)) = 0 (1.13)
Applying lemma 1.1 to the right hand side of (1.5) yields Nf(p). Finally from (1.5),
by replacing the index p by j, we get the inverse DHT
f(j) =1
N
N−1∑k=0
cas(kωj)F (k). (1.14)
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Chapter 2
Simpson Discrete Hartley
Transform
Applying Simpson quadrature to equation (1.1) yields
C(k) =1
3N
[f(x0) cas(kx0) + f(xN) cas(kxN) + 2
N2−1∑
j=1
f(x2j) cas(kx2j)
+ 4
N2−1∑
j=0
f(x2j+1) cas(kx2j+1)]
=1
N
[2
3
N2−1∑
j=0
cas(2kωj)f(2j) +4
3
N2−1∑
j=0
cas(kω(2j + 1))f(2j + 1)]. (2.1)
Hence the Simpson Discrete Hartley Transform (SDHT) is defined component-wise by
Hf(k) =2
3
N2−1∑
j=0
cas(2kωj)f(2j) +4
3
N2−1∑
j=0
cas(kω(2j + 1))f(2j + 1). (2.2)
Define
F0(k) =2
3
N2−1∑
j=0
cas(2kωj)f(2j) (2.3)
and
F1(k) =4
3
N2−1∑
j=0
cas(kω(2j + 1))f(2j + 1), (2.4)
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for k = 0, 1, 2, . . . , N − 1. Using the fact that the cas function is anti-periodic in π it
follows from (2.4) that
F1(k + N2
) =4
3
N2−1∑
j=0
cas((k + N2
)ω(2j + 1))f(2j + 1)
=4
3
N2−1∑
j=0
cas(kω(2j + 1) + (2j + 1)π)f(2j + 1)
= −4
3
N2−1∑
j=0
cas(kω(2j + 1))f(2j + 1)
= −F1(k). (2.5)
Define F1 = [F1(0), F1(1), . . . , F1(N2− 1)]T then the vector F1 is anti-periodic in N
2.
Similarly it may be shown that the vector F0 = [F0(0), F0(1), . . . , F0(N2−1)]T is periodic
in N2
.
Equation (2.3) leads us to define a (N2× N
2) matrix A by
A =
1 1 1 · · · 1
1 cas(2ω) cas(4ω) · · · cas(N1ω)
1 cas(4ω) cas(8ω) · · · cas(2N1ω)
......
......
...
1 cas(N1ω) cas(2N1ω) · · · cas(N1
2N1ω)
. (2.6)
Hence (2.3) may be written as
F0 =2
3Af0,
where f0 = [f(0), f(2), . . . , f(N − 2)]T . Equation (2.4) leads us to define a (N2× N
2)
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matrix B by
B =
1 1 1 · · · 1
cas(ω) cas(3ω) cas(5ω) · · · cas((N1 + 1)ω)
cas(2ω) cas(6ω) cas(10ω) · · · cas(2(N1 + 1)ω)
......
......
...
cas(N1
2ω) cas(N1
23ω) cas(N1
25ω) · · · cas(N1
2(N1 + 1)ω)
, (2.7)
where N1 = N − 2. Hence (2.4) may be written as
F1 =4
3Bf1,
where f1 = [f(1), f(3), . . . , f(N − 1)]T .
Considering (2.2) for k = 0, 1, . . . , N2−1 the right hand side can be written in the form
2
3Af0 +
4
3Bf1.
Considering (2.2) for k = N2, N
2+ 1, . . . , N − 1 the right hand side can be written in
the form
2
3Af0 −
4
3Bf1,
where we have used the periodicity and anti-periodicity of F0 and F1 respectively.
Finally (2.2) in block matrix vector form is represented by
Hf =2
3
A 2B
A −2B
f0
f1
. (2.8)
Since [f0 f1]T = Pf where the (N ×N) permutation matrix P is defined by Pj,2j = 1
and PN2+j,2j+1 = 1 for j = 0, 1, . . . , N
2− 1. With this substitution in (2.8) we recover
the transformation matrix
H =2
3
A 2B
A −2B
P. (2.9)
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Taking the inner product of the jth and pth columns of A results in
N2−1∑
i=0
cas(2ijω) cas(2ipω) =
N2
when j = p,
0 when j 6= p,
(2.10)
where we have used a variation of lemma 1.1. Hence we conclude that√
2N
A is or-
thogonal. Since A is symmetric it follows that√
2N
A is involutory. Similarly it may
be shown that√
2N
B is orthogonal.
Using these properties it may be shown by multiplication that the inverse of H is given
by
H−1 =3
2NPT
A A
12BT −1
2BT
. (2.11)
From (2.11) the inverse transform may be represented component-wise by
f(2j) =3
2N
N2−1∑
k=0
cas(2kωj)Hf(k), (2.12)
f(2j + 1) =3
4N
N2−1∑
k=0
cas(kω(2j + 1))Hf(k). (2.13)
10
Chapter 3
Properties of the SDHT
In this chapter we state and prove various properties of the SDHT.
An even signal f is defined as one for which f(−j) = f(j) whilst an odd signal is
defined as one for which f(−j) = −f(j).
3.1 Transform of an even signal
Given an even signal f then Hf(−k) = Hf(k).
Proof. Replace k by −k in equation (2.2) to obtain
Hf(−k) =2
3
N2−1∑
j=0
cas(−2kωj)f(2j) +4
3
N2−1∑
j=0
cas(−kω(2j + 1))f(2j + 1) (3.1)
=2
3
0∑j=−N
2+1
cas(2kωj)f(−2j) +4
3
0∑j=−N
2+1
cas(kω(2j − 1))f(−2j + 1) (3.2)
=2
3
0∑j=−N
2+1
cas(2kωj)f(−2j) +4
3
−1∑j=−N
2
cas(kω(2j + 1))f(−2j − 1) (3.3)
=2
3
N2−1∑
j=0
cas(2kωj)f(2j) +4
3
N2−1∑
j=0
cas(kω(2j + 1))f(2j + 1) (3.4)
= Hf(k),
11
where we replaced j by −j from (3.1) to (3.2). From (3.2) to (3.3) we have replaced j
by j + 1 in the second summation. We notice that the arguments of both summations
in (3.3) are periodic in j of periodicity N2
, hence we may sum over any period of length
N2
[2] resulting in (3.4).
3.2 Transform of an odd signal
Given an odd signal f then Hf(−k) = −Hf(k).
Proof. Replace k by −k in equation (2.2) to obtain
Hf(−k) =2
3
N2−1∑
j=0
cas(−2kωj)f(2j) +4
3
N2−1∑
j=0
cas(−kω(2j + 1))f(2j + 1) (3.5)
=2
3
0∑j=−N
2+1
cas(2kωj)f(−2j) +4
3
0∑j=−N
2+1
cas(kω(2j − 1))f(−2j + 1) (3.6)
=2
3
0∑j=−N
2+1
cas(2kωj)f(−2j) +4
3
−1∑j=−N
2
cas(kω(2j + 1))f(−2j − 1) (3.7)
= −2
3
N2−1∑
j=0
cas(2kωj)f(2j)− 4
3
N2−1∑
j=0
cas(kω(2j + 1))f(2j + 1) (3.8)
= −Hf(k).
From (3.7) to (3.8) we have used the fact that f is an odd signal.
3.3 Time reversal
By this property it is meant that the reversal of the transformed signal is equal to the
transform of the reversed signal, that is Hf(−k) = HfR(k) where
fR = [f(0), f(N − 1), . . . , f(1)]T .
12
Proof. Replace k by −k in equation (2.2) to get
Hf(−k) =2
3
N2−1∑
j=0
cas(−2kωj)f(2j) +4
3
N2−1∑
j=0
cas(−kω(2j + 1))f(2j + 1) (3.9)
=2
3
N2∑j=1
cas(−2kω(−j + N2
))f(2(−j + N2
)) (3.10)
+4
3
N2∑j=1
cas(−kω(2(−j + N2
) + 1))f(2(−j + N2
) + 1),
where we have replaced j by −j + N2
.
Equation (3.10) simplifies to
Hf(−k) =2
3
N2∑j=1
cas(2kωj − kωN)f(N − 2j)
+4
3
N2∑j=1
cas(kω(2j −N − 1))f(N − 2j + 1) (3.11)
=2
3
N2∑j=1
cas(2kωj − 2kπ)f(N − 2j)
+4
3
N2∑j=1
cas(kω(2j − 1)− 2kπ)f(N − 2j + 1) (3.12)
=2
3
N2∑j=1
cas(2kωj)f(−2j)
+4
3
N2∑j=1
cas(kω(2j − 1))f(−2j + 1), (3.13)
=2
3
N2−1∑
j=0
cas(2kωj)f(−2j)
+4
3
N2−1∑
j=0
cas(kω(2j + 1))f(−2j − 1) (3.14)
= HfR(k).
From (3.12) to (3.13) we have used the periodicity of both f and the cas function. From
(3.13) to (3.14) we replaced j by j + 1 in the second summation.
13
In order to proceed we require the following lemma.
Lemma 3.1. The double angle formula of the cas function is given by
cas(A+B) = 12
[cas(A) cas(B)+cas(A) cas(−B)+cas(−A) cas(B)−cas(−A) cas(−B)
].
Proof.
cas(A+B)
= cos(A+B) + sin(A+B)
= cos(A) cos(B) + cos(A) sin(B) + sin(A) cos(B)− sin(A) sin(B)
= cos(A) cas(B) + sin(A) cas(−B)
= 12[cas(A) + cas(−A)] cas(B) + 1
2[cas(A)− cas(−A)] cas(−B)
= 12
[cas(A) cas(B) + cas(A) cas(−B) + cas(−A) cas(B)
− cas(−A) cas(−B)],
where we have used cos(A) = 12[cas(A) + cas(−A)] and sin(A) = 1
2[cas(A)− cas(−A)].
Given a sequence f define the even part fe by fe(j) = f(j)+f(−j)2
and the odd part
fo by fo(j) = f(j)−f(−j)2
. In addition, define the vector v by v(j) = cas(k0ωj), j =
0, 1, . . . , N − 1, where k0 is an integer.
3.4 Frequency shift
A shift by k0 of the transformed signal f is equivalent to the transformed signal v. ?
fe +vR.? fo where vR is the reversal of v and .? represents element-wise multiplication.
14
Proof. From (2.2) replacing k by k − k0 gives
Hf(k−k0) =2
3
N2−1∑
j=0
cas((k−k0)ω2j)f(2j)+4
3
N2−1∑
j=0
cas((k−k0)ω(2j+1))f(2j+1). (3.15)
Applying lemma 3.1 to equation (3.15) yields
Hf(k − k0)
=1
2
[2
3
N2−1∑
j=0
(cas(2kωj) cas(−2k0ωj) + cas(2kωj) cas(2k0ωj)
+ cas(−2kωj) cas(−2k0ωj)− cas(−2kωj) cas(2k0ωj))]f(2j)
+1
2
[4
3
N2−1∑
j=0
(cas(kω(2j + 1)) cas(−k0ω(2j + 1))
+ cas(kω(2j + 1)) cas(k0ω(2j + 1))
+ cas(−kω(2j + 1)) cas(−k0ω(2j + 1))
− cas(−kω(2j + 1)) cas(k0ω(2j + 1)))]f(2j + 1). (3.16)
Rearranging yields
Hf(k − k0)
=1
2
[(2
3
N2−1∑
j=0
cas(2kωj) cas(−2k0ωj)f(2j)
+4
3
N2−1∑
j=0
cas(kω(2j + 1)) cas(−k0ω(2j + 1))f(2j + 1))
(3.17)
+(2
3
N2−1∑
j=0
cas(2kωj) cas(2k0ωj)f(2j)
+4
3
N2−1∑
j=0
cas(kω(2j + 1)) cas(k0ω(2j + 1))f(2j + 1))
(3.18)
15
+(2
3
N2−1∑
j=0
cas(−2kωj) cas(−2k0ωj)f(2j)
+4
3
N2−1∑
j=0
cas(−kω(2j + 1)) cas(−k0ω(2j + 1))f(2j + 1))
(3.19)
−(2
3
N2−1∑
j=0
cas(−2kωj) cas(2k0ωj)f(2j)
+4
3
N2−1∑
j=0
cas(−kω(2j + 1)) cas(k0ω(2j + 1))f(2j + 1))]. (3.20)
Statement (3.19) can be simplified by replacing j by −j in the first term and j by
−j − 1 in the second term to give
2
3
N2−1∑
j=0
cas(2kωj) cas(2k0ωj)f(−2j)+4
3
N2−1∑
j=0
cas(kω(2j+1)) cas(k0ω(2j+1))f(−2j−1).
(3.21)
Similarly statement (3.20) simplifies to
− 2
3
N2−1∑
j=0
cas(2kωj) cas(−2k0ωj)f(−2j)
− 4
3
N2−1∑
j=0
cas(kω(2j + 1)) cas(−k0ω(2j + 1))f(−2j − 1). (3.22)
We can combine (3.18) and (3.21) to yield
2
3
N2−1∑
j=0
cas(2kωj) cas(2k0ωj)[f(2j) + f(−2j)
2
]
+4
3
N2−1∑
j=0
cas(kω(2j + 1)) cas(k0ω(2j + 1))[f(2j + 1) + f(−2j − 1)
2
]. (3.23)
In a similar manner we can combine (3.17) and (3.22) to get
2
3
N2−1∑
j=0
cas(2kωj) cas(−2k0ωj)[f(2j)− f(−2j)
2
]
+4
3
N2−1∑
j=0
cas(kω(2j + 1)) cas(−k0ω(2j + 1))[f(2j + 1)− f(−2j − 1)
2
]. (3.24)
16
With this simplification (3.15) reduces to
Hf(k − k0) = H[v. ? fe] + H[vR. ? fo]
= H[v. ? fe + vR. ? fo]. (3.25)
3.5 Duality
By this we mean the effect of H2 on f . From (2.2) we obtain the even components
H2f(2k)
=2
3
N2−1∑
j=0
cas(4kωj)Hf(2j) +4
3
N2−1∑
j=0
cas(2kω(2j + 1))Hf(2j + 1) (3.26)
=2
3
N2−1∑
j=0
cas(4kωj)[2
3
N2−1∑
p=0
cas(4jωp)f(2p) +4
3
N2−1∑
p=0
cas(2jω(2p+ 1))f(2p+ 1)]
+4
3
N2−1∑
j=0
cas(2kω(2j + 1))[2
3
N2−1∑
p=0
cas(2ωp(2j + 1))f(2p)
+4
3
N2−1∑
p=0
cas(ω(2j + 1)f(2p+ 1))]
(3.27)
=4
9
N2−1∑
p=0
N2−1∑
j=0
cas(4kωj) cas(4jωp)f(2p) (3.28)
+8
9
N2−1∑
p=0
N2−1∑
j=0
cas(4kωj) cas(2ωj(2p+ 1))f(2p+ 1) (3.29)
+8
9
N2−1∑
p=0
N2−1∑
j=0
cas(2kω(2j + 1)) cas(2pω(2j + 1))f(2p) (3.30)
+16
9
N2−1∑
p=0
N2−1∑
j=0
cas(2kω(2j + 1)) cas(ω(2p+ 1)(2j + 1))f(2p+ 1) (3.31)
Statement (3.28) simplifies to
4
9
N2−1∑
p=0
N2−1∑
j=0
cos(4ωj(k − p))f(2p) +4
9
N2−1∑
p=0
N2−1∑
j=0
sin(4ωp(k + p))f(2p). (3.32)
17
It may be proved as in (1.13) that the second term in (3.32) is zero. There is con-
tribution from the first term in (3.32) only when p = k. Hence (3.32) simplifies to
29Nf(2k).
Similarly (3.29) reduces to
8
9
N2−1∑
p=0
N2−1∑
j=0
cos(2ωj(2k− 2p− 1))f(2p+ 1) +8
9
N2−1∑
p=0
N2−1∑
j=0
sin(2ωj(2k + 2p+ 1))f(2p+ 1).
(3.33)
Once again the second summation in (3.33) can be shown to be zero. It must be
observed that 2k − 2p − 1 is non zero. There is contribution from the first term only
when 2k − 2p − 1 = N2
. Hence (3.33) simplifies to 49Nf(2k + N
2). Similarly (3.30)
simplifies to 49Nf(2k) and (3.31) simplifies to −8
9Nf(2k + N
2).
In summary we combine all four terms to get the duality property for the even com-
ponents, namely
H2f(2k) =2
9Nf(2k) +
4
9Nf(2k + N
2) +
4
9Nf(2k)− 8
9Nf(2k + N
2)
=2
3Nf(2k)− 4
9Nf(2k + N
2). (3.34)
The odd components for the duality property is done in a similar manner to yield
H2f(2k + 1)
=4
9
N2−1∑
p=0
N2−1∑
j=0
cas(2jω(2k + 1)) cas(4jωp)f(2p) (3.35)
+8
9
N2−1∑
p=0
N2−1∑
j=0
cas(2ωj(2k + 1)) cas(2ωj(2p+ 1))f(2p+ 1) (3.36)
+8
9
N2−1∑
p=0
N2−1∑
j=0
cas(ω(2j + 1)(2k + 1)) cas(2ωp(2j + 1))f(2p) (3.37)
18
+16
9
N2−1∑
p=0
N2−1∑
j=0
cas(ω(2j + 1)(2k + 1)) cas(ω(2j + 1)(2p+ 1))
× f(2p+ 1). (3.38)
Statements (3.35)-(3.38) evaluate to 29Nf(2k+1+ N
2), 4
9Nf(2k+1), −4
9Nf(2k+1+ N
2)
and 89Nf(2k + 1) respectively.
In summary we combine all four terms to get the duality property for the odd compo-
nents, namely
H2f(2k + 1)
=2
9Nf(2k + 1 + N
2) +
4
9Nf(2k + 1)− 4
9Nf(2k + 1 + N
2) +
8
9Nf(2k + 1)
=4
3Nf(2k + 1)− 2
9Nf(2k + 1 + N
2). (3.39)
Let f and g be vectors of period N . The circular convolution z = f ⊗ g is defined
component-wise by
z(j) =N−1∑m=0
f(m)g(j −m), (3.40)
for j = 0, 1, . . . , N − 1.
3.6 Convolution in the real domain
We derive an expression for the transform of the convolution f ⊗ g.
Hz(k) =2
3
N2−1∑
j=0
cas(2kωj)z(2j) +4
3
N2−1∑
j=0
cas(kω(2j + 1))z(2j + 1), (3.41)
19
where
z(2j) =
N2−1∑
m=0
f(2m)g(2j − 2m) +
N2−1∑
m=0
f(2m+ 1)g(2j − 2m− 1), (3.42)
z(2j + 1) =
N2−1∑
m=0
f(2m)g(2j + 1− 2m) +
N2−1∑
m=0
f(2m+ 1)g((2j + 1)− (2m+ 1)).
(3.43)
Upon substitution of (3.42) and (3.43) into (3.41), we get
Hz(k)
=2
3
N2−1∑
j=0
cas(2kωj)z(2j) +4
3
N2−1∑
j=0
cas(kω(2j + 1))z(2j + 1)
=2
3
N2−1∑
j=0
cas(2kωj)[N2 −1∑m=0
f(2m)g(2j − 2m) +
N2−1∑
m=0
f(2m+ 1)g(2j − 2m− 1)]
+4
3
N2−1∑
j=0
cas(kω(2j + 1))[N2 −1∑m=0
f(2m)g(2j + 1− 2m)
+
N2−1∑
m=0
f(2m+ 1)g((2j + 1)− (2m+ 1))]
(3.44)
=2
3
[N2 −1∑j=0
N2−1∑
m=0
cas(2kωj)f(2m)g(2(j −m)) (3.45)
+
N2−1∑
j=0
N2−1∑
m=0
cas(2kωj)f(2m+ 1)g(2(j −m)− 1)]
(3.46)
+4
3
[N2 −1∑j=0
N2−1∑
m=0
cas(kω(2j + 1))f(2m)g(2(j −m) + 1) (3.47)
+
N2−1∑
j=0
N2−1∑
m=0
cas(kω(2j + 1))f(2m+ 1)g(2(j −m))]. (3.48)
20
Statement (3.45) simplifies to
2
3
N2−1−m∑l=−m
N2−1∑
m=0
cas(2kω(l +m))f(2m)g(2l)
=1
3
N2−1∑
l=0
N2−1∑
m=0
[cas(2kωl) cas(2kωm) + cas(2kωl) cas(−2kωm) + cas(−2kωl) cas(2kωm)
− cas(−2kωl) cas(−2kωm)]f(2m)g(2l). (3.49)
The reindexing of the argument of g allows the double summation to be written as
products of single summations giving
1
3
[N2 −1∑m=0
cas(2kωm)f(2m)
N2−1∑
l=0
cas(2kωl)g(2l) (3.50)
+
N2−1∑
m=0
cas(−2kωm)f(2m)
N2−1∑
l=0
cas(2kωl)g(2l) (3.51)
+
N2−1∑
m=0
cas(2kωm)f(2m)
N2−1∑
l=0
cas(−2kωl)g(2l) (3.52)
−N2−1∑
m=0
cas(−2kωm)f(2m)
N2−1∑
l=0
cas(−2kωl)g(2l)]. (3.53)
Statement (3.46) simplifies to
2
3
N2−1−m−1∑l=−m−1
N2−1∑
m=0
cas(2kω(l +m+ 1))f(2m+ 1)g(2l + 1)
=
N2−1∑
l=0
N2−1∑
m=0
cas(kω(2l + 1) + kω(2m+ 1))f(2m+ 1)g(2l + 1)
=1
3
N2−1∑
l=0
N2−1∑
m=0
[cas(kω(2m+ 1)) cas(kω(2l + 1)) + cas(kω(2m+ 1)) cas(−kω(2l + 1))
+ cas(−kω(2m+ 1)) cas(kω(2l + 1))− cas(−kω(2m+ 1)) cas(kω(2l + 1))]
× f(2m+ 1)g(2l + 1) (3.54)
21
The reindexing of the argument of g once again allows the double summation to be
written as products of single summations giving
1
3
[N2 −1∑m=0
cas(kω(2m+ 1))f(2m+ 1)
N2−1∑
l=0
cas(kω(2l + 1))g(2l + 1) (3.55)
+
N2−1∑
m=0
cas(kω(2m+ 1))f(2m+ 1)
N2−1∑
l=0
cas(−kω(2l + 1))g(2l + 1) (3.56)
+
N2−1∑
m=0
cas(−kω(2m+ 1))f(2m+ 1)
N2−1∑
l=0
cas(kω(2l + 1))g(2l + 1) (3.57)
−N2−1∑
m=0
cas(−kω(2m+ 1))f(2m+ 1)
N2−1∑
l=0
cas(−kω(2l + 1))g(2l + 1)]. (3.58)
It should be noted that (3.47) and (3.48) can be simplified in a similar manner to
obtain
2
3
[N2 −1∑m=0
cas(2kωm)f(2m)
N2−1∑
l=0
cas(kω(2l + 1))g(2l + 1) (3.59)
+
N2−1∑
m=0
cas(2kωm)f(2m)
N2−1∑
l=0
cas(−kω(2l + 1))g(2l + 1) (3.60)
+
N2−1∑
m=0
cas(−2kωm)f(2m)
N2−1∑
l=0
cas(kω(2l + 1))g(2l + 1) (3.61)
−N2−1∑
m=0
cas(−2kωm)f(2m)
N2−1∑
l=0
cas(−kω(2l + 1))g(2l + 1)]
(3.62)
and
2
3
[N2 −1∑m=0
cas(kω(2m+ 1))f(2m+ 1)
N2−1∑
l=0
cas(2kωl)g(2l) (3.63)
+
N2−1∑
m=0
cas(kω(2m+ 1))f(2m+ 1)
N2−1∑
l=0
cas(−2kωl)g(2l) (3.64)
+
N2−1∑
m=0
cas(−kω(2m+ 1))f(2m+ 1)
N2−1∑
l=0
cas(2kωl)g(2l) (3.65)
−N2−1∑
m=0
cas(−kω(2m+ 1))f(2m+ 1)
N2−1∑
l=0
cas(−2kωl)g(2l)]. (3.66)
22
For any signal x we make the following observation
Hx(k) =2
3
N2−1∑
j=0
cas(2kωj)x(2j) +4
3
N2−1∑
j=0
cas(kω(2j + 1))x(2j + 1), (3.67)
Hx(k + N2
) =2
3
N2−1∑
j=0
cas(2kωj)x(2j)− 4
3
N2−1∑
j=0
cas(kω(2j + 1))x(2j + 1). (3.68)
By adding (3.67) to (3.68) we obtain the result
N2−1∑
j=0
cas(2kωj)x(2j) =3
4Hx(k) +
3
4Hx(k + N
2). (3.69)
Similarly it may be shown that
N2−1∑
j=0
cas(kω(2j + 1))x(2j + 1) =3
8Hx(k)− 3
8Hx(k + N
2), (3.70)
N2−1∑
j=0
cas(−2kωj)x(2j) =3
4Hx(−k) +
3
4Hx(−k − N
2), (3.71)
N2−1∑
j=0
cas(−kω(2j + 1))x(2j + 1) =3
8Hx(−k)− 3
8Hx(−k − N
2). (3.72)
By using the results in (3.69)-(3.72) in (3.50)-(3.53), (3.55)-(3.58), (3.59)-(3.62) and
(3.63)-(3.66) we arrive at
Hz(k)
=39
64
[Hf(k)Hg(k) + Hf(k)Hg(−k) + Hf(−k)Hg(k)−Hf(−k)Hg(−k)
]− 9
64
[Hf(k + N
2)Hg(k + N
2) + Hf(k + N
2)Hg(−k − N
2) + Hf(−k − N
2)Hg(k + N
2)
−Hf(−k − N2
)Hg(−k − N2
)]
+9
64
[Hf(k)Hg(k + N
2) + Hf(k)Hg(−k − N
2)
+ Hf(−k)Hg(k + N2
)−Hf(−k)Hg(−k − N2
) + Hf(k + N2
)Hg(k)
+ Hf(k + N2
)Hg(−k) + Hf(−k − N2
)Hg(k)−Hf(−k − N2
)Hg(−k)]. (3.73)
23
Define vectors S and T as follows
S = Hf . ?Hg + Hf . ?HgR + HfR. ?Hg −HfR. ?HgR, (3.74)
T = Hf . ?HgN2
+ Hf . ?HgN2R
+ HfR. ?HgN2
−HfR. ?HgN2R, (3.75)
where HgR represents the reversal of Hg, HgN2
represents Hg shifted by N2
and HgN2R
represents the reversal of HgN2
. A similar explanation applies to f . Finally we can
express equation (3.73) in the form
H(f ⊗ g) =39
64S− 9
64S(N/2) +
9
64
(T + T(N/2)
). (3.76)
We now prove the following lemma.
Lemma 3.2.
cas(C) cas(A−B) =cas(A)
2
[cas(C −B) + cas(C +B)
]+
cas(−A)
2
[cas(−C −B)− cas(−C +B)
].
Proof.
cas(C) cas(A−B) = cas(C)[sin(A−B) + cos(A−B)
]= cas(C)
[sin(A)
(sin(B) + cos(B)
)+ cos(A)
(cos(B)− sin(B)
)]= cas(C)
[sin(A) cas(B) + cos(A) cas(−B)
]=
1
2cas(C) cas(B)
[cas(A)− cas(−A)
]+
1
2cas(C) cas(−B)
[cas(A) + cas(−A)
]=
1
2cas(C) cas(B) cas(A)− 1
2cas(C) cas(B) cas(−A)
+1
2cas(C) cas(−B) cas(A) +
1
2cas(C) cas(−B) cas(−A)
=cas(A)
2
[cas(C) cas(B) + cas(C) cas(−B)
]+
cas(−A)
2
[cas(C) cas(−B)− cas(C) cas(B)
]. (3.77)
24
In order to simplify the terms in (3.77) we use the following identities.
cas(C +B) =1
2
[cas(C) cas(B) + cas(C) cas(−B) + cas(−C) cas(B)
− cas(−C) cas(−B)]
(3.78)
cas(C −B) =1
2
[cas(C) cas(−B) + cas(C) cas(B) + cas(−C) cas(−B)
− cas(−C) cas(B)]
(3.79)
cas(−C +B) =1
2
[cas(−C) cas(B) + cas(−C) cas(−B) + cas(C) cas(B)
− cas(C) cas(−B)]
(3.80)
cas(−C −B) =1
2
[cas(−C) cas(−B) + cas(−C) cas(B) + cas(C) cas(−B)
− cas(C) cas(B)]. (3.81)
Adding (3.78) and (3.79) gives
cas(C) cas(B) + cas(C) cas(−B) = cas(C +B) + cas(C −B). (3.82)
Subtracting (3.80) from (3.81) gives
cas(C) cas(−B)− cas(C) cas(B) = cas(−C −B)− cas(−C +B). (3.83)
With this substitution in (3.77) we complete the proof.
3.7 Convolution in the frequency domain
We derive an expression for Hf ⊗Hg. This is also known as the product theorem.
(Hf ⊗Hg)(j) =N−1∑m=0
Hf(m)Hg(j −m)
=
N2−1∑
m=0
Hf(2m)Hg(j − 2m) +
N2−1∑
m=0
Hf(2m+ 1)Hg(j − 2m− 1). (3.84)
25
We now simplify the first summation in (3.84). This reads
N2−1∑
m=0
Hf(2m)Hg(j − 2m)
=
N2−1∑
m=0
([2
3
N2−1∑
p=0
cas(4mωp)f(2p) +4
3
N2−1∑
p=0
cas(2mω(2p+ 1))f(2p+ 1)]×
[2
3
N2−1∑
l=0
cas(2lω(j − 2m))g(2l) +4
3
N2−1∑
l=0
cas(ω(j − 2m)(2l + 1))g(2l + 1)])
=4
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(4mωp) cas(2lω(j − 2m))f(2p)g(2l)+ (3.85)
8
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(4mωp) cas(ω(j − 2m)(2l + 1))f(2p)g(2l + 1)+ (3.86)
8
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(2mω(2p+ 1)) cas(2lω(j − 2m))f(2p+ 1)g(2l)+ (3.87)
16
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(2mω(2p+ 1)) cas(ω(j − 2m)(2l + 1))f(2p+ 1)g(2l + 1). (3.88)
Using lemma 3.2, (3.85) simplifies to
4
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
(cas(2jωl)
2
[cas(4mωp+ 4mωl) + cas(4mωp− 4mωl)
]+
cas(−2jωl)
2
[cas(−4mωp− 4mωl)− cas(−4mωp+ 4mωl)
])f(2p)g(2l) (3.89)
=2
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(2jωl) cas(4mω(p+ l))f(2p)g(2l) (3.90)
+2
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(2jωl) cas(4mω(p− l))f(2p)g(2l) (3.91)
+2
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(−2jωl) cas(−4mω(p+ l))f(2p)g(2l) (3.92)
− 2
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(−2jωl) cas(−4mω(l − p))f(2p)g(2l). (3.93)
26
The summations in (3.90) may be further simplified as follows.
2
9
N2−1∑
l=0
cas(2jωl)
N2−1∑
p=0
N2−1∑
m=0
cas(4mω(p+ l))f(2p)g(2l) (3.94)
=2
9
N2−1∑
p=0
cas((N − 2p)ωj)f(2p)g(N − 2p)
N2−1∑
m=0
cas(4πm) (3.95)
=N
9
N2−1∑
p=0
cas(−2pωj)f(2p)g(−2p). (3.96)
We have used the fact that the only contribution to the summation in (3.94) occurs
for p + l = N2
, the periodicity N of g and the periodicity 2π of the cas function. The
summations in (3.91) may be further simplified as follows.
2
9
N2−1∑
l=0
cas(2jωl)
N2−1∑
p=0
N2−1∑
m=0
cas(4mω(p− l))f(2p)g(2l) (3.97)
=2
9
N2−1∑
p=0
cas(2jωp)f(2p)g(2p)
N2−1∑
m=0
1 (3.98)
=N
9
N2−1∑
p=0
cas(2jωp)f(2p)g(2p). (3.99)
We have used the fact that the only contribution to the summation in (3.97) occurs
for p = l. The summations in (3.92) may be further simplified as follows.
2
9
N2−1∑
l=0
cas(−2jωl)
N2−1∑
p=0
N2−1∑
m=0
cas(−4mω(p+ l))f(2p)g(2l) (3.100)
=2
9
N2−1∑
p=0
cas((2p−N)ωj)f(2p)g(N − 2p)
N2−1∑
m=0
cas(−4πm) (3.101)
=N
9
N2−1∑
p=0
cas(2pωj)f(2p)g(−2p). (3.102)
27
We have used the fact that the only contribution occurs for p+l = N2
. The summations
in (3.93) may be further simplified as follows.
2
9
N2−1∑
l=0
cas(−2jωl)
N2−1∑
p=0
N2−1∑
m=0
cas(−4mω(l − p))f(2p)g(2l) (3.103)
=2
9
N2−1∑
p=0
cas(−2jωp)f(2p)g(2p)
N2−1∑
m=0
1 (3.104)
=N
9
N2−1∑
p=0
cas(−2jωp)f(2p)g(2p). (3.105)
We have used the fact that the only contribution to the summation occurs for p = l.
Using lemma 3.2, (3.86) simplifies to
8
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
(cas(ωj(2l + 1))
2
[cas(4mωp+ 2ωm(2l + 1))
+ cas(4mωp− 2ωm(2l + 1))]
+cas(−ωj(2l + 1))
2×[
cas(−4mωp− 2ωm(2l + 1))− cas(−4mωp+ 2ωm(2l + 1))])f(2p)g(2l + 1) (3.106)
=4
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(ωj(2l + 1)) cas(4mωp+ 2ωm(2l + 1))f(2p)g(2l + 1) (3.107)
+4
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(ωj(2l + 1)) cas(4mωp− 2ωm(2l + 1))f(2p)g(2l + 1) (3.108)
+4
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(−ωj(2l + 1)) cas(−4mωp− 2ωm(2l + 1))f(2p)
× g(2l + 1) (3.109)
− 4
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(−ωj(2l + 1)) cas(−4mωp+ 2ωm(2l + 1))f(2p)
× g(2l + 1). (3.110)
28
The summations in (3.107) may be further simplified as follows.
4
9
N2−1∑
l=0
cas(ωj(2l + 1))
N2−1∑
p=0
N2−1∑
m=0
cas(2ωm(2p+ 2l + 1))f(2p)g(2l + 1) (3.111)
=4
9
N2−1∑
p=0
cas((N2− 2p)ωj)f(2p)g(N
2− 2p)
N2−1∑
m=0
cas(2πm) (3.112)
=2N
9
N2−1∑
p=0
(−1)j cas(−2pωj)f(2p)g(N2− 2p). (3.113)
We have used the fact that the only contribution to the summation occurs for 2p+2l+
1 ∈ {N2, 3N
2}. Now 3N
2is equivalent to N
2due to periodicity N . Therefore it suffices
to consider only 2p + 2l + 1 = N2
. Hence cas((N2− 2p)ωj) = (−1)j cas(−2pωj). The
summations in (3.108) may be further simplified as follows.
4
9
N2−1∑
l=0
cas(ωj(2l + 1))
N2−1∑
p=0
N2−1∑
m=0
cas(2ωm(2p− 2l − 1))f(2p)g(2l + 1) (3.114)
=4
9
N2−1∑
p=0
cas((2p+ N2
)ωj)f(2p)g(2p+ N2
)
N2−1∑
m=0
cas(2πm) (3.115)
=2N
9
N2−1∑
p=0
(−1)j cas(2pωj)f(2p)g(2p+ N2
). (3.116)
We used the fact that the only contribution to the summation occurs for 2p−2l−1 = N2
and the fact that cas((2p + N2
)ωj) = (−1)j cas(2pωj). The summations in statement
(3.109) may be further simplified as follows.
4
9
N2−1∑
l=0
cas(−ωj(2l + 1))
N2−1∑
p=0
N2−1∑
m=0
cas(−2mω(2p+ 2l + 1))f(2p)g(2l + 1) (3.117)
=4
9
N2−1∑
p=0
cas(−(N2− 2p)ωj)f(2p)g(N
2− 2p)
N2−1∑
m=0
cas(−2πm) (3.118)
=2N
9
N2−1∑
p=0
(−1)j cas(2pωj)f(2p)g(N2− 2p). (3.119)
We used the fact that the only contribution in the summation occurs for 2p+ 2l+ 1 ∈
{N2, 3N
2} which is equivalent to 2p+ 2l+ 1 = N
2. The summations in statement (3.110)
29
may be further simplified as follows.
4
9
N2−1∑
l=0
cas(−ωj(2l + 1))
N2−1∑
p=0
N2−1∑
m=0
cas(2ωm(−2p+ 2l + 1))f(2p)g(2l + 1) (3.120)
=
N2−1∑
l=0
cas((N2
+ 2p)ωj)f(2p)g(2p+ N2
)
N2−1∑
m=0
cas(2πm) (3.121)
=2N
9
N2−1∑
p=0
(−1)j cas(−2pωj)f(2p)g(2p+ N2
). (3.122)
We used the fact that the only contribution in the summation occurs for −2p+2l+1 =
N2
.
Using lemma 3.2, (3.87) simplifies to
8
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
(cas(2jωl)
2
[cas(2mω(2p+ 1) + 4lωm)
+ cas(2mω(2p+ 1)− 4lωm)]
+cas(−2jωl)
2
[cas(−2mω(2p+ 1)− 4lωm)
− cas(−2mω(2p+ 1) + 4lωm)])f(2p+ 1)g(2l) (3.123)
=4
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(2jωl) cas(2mω(2p+ 1) + 4lωm)f(2p+ 1)g(2l) (3.124)
+4
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(2jωl) cas(2mω(2p+ 1)− 4lωm)f(2p+ 1)g(2l) (3.125)
+4
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(−2jωl) cas(−2mω(2p+ 1)− 4lωm)f(2p+ 1)g(2l) (3.126)
− 4
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(−2jωl) cas(−2mω(2p+ 1) + 4lωm)f(2p+ 1)g(2l). (3.127)
The summations in (3.124) may be further simplified as follows.
4
9
N2−1∑
l=0
cas(2jωl)
N2−1∑
p=0
N2−1∑
m=0
cas(2mω(2p+ 2l + 1))f(2p+ 1)g(2l) (3.128)
=4
9
N2−1∑
p=0
cas((N2− 2p− 1)ωj)f(2p+ 1)g(N
2− 2p− 1)
N2−1∑
m=0
cas(2πm) (3.129)
30
=2N
9
N2−1∑
p=0
(−1)j cas(−ωj(2p+ 1))f(2p+ 1)g(N2− 2p− 1). (3.130)
We have used the fact that the only contribution in the summation essentially occurs
when 2p+2l+1 = N2
. The summations in (3.125) may be further simplified as follows.
4
9
N2−1∑
l=0
cas(2jωl)
N2−1∑
p=0
N2−1∑
m=0
cas(2mω(2p− 2l + 1))f(2p+ 1)g(2l) (3.131)
=4
9
N2−1∑
p=0
cas((N2
+ 2p+ 1)ωj)f(2p+ 1)g(N2
+ 2p+ 1)
N2−1∑
m=0
cas(2πm) (3.132)
=2N
9
N2−1∑
p=0
(−1)j cas(jω(2p+ 1))f(2p+ 1)g(N2
+ 2p+ 1). (3.133)
We have used the fact that the only contribution in the summation occurs when 2p−
2l+ 1 = N2
and periodicity 2π of the cas function. The summations in (3.126) may be
further simplified as follows.
4
9
N2−1∑
l=0
cas(−2jωl)
N2−1∑
p=0
N2−1∑
m=0
cas(−2mω(2p+ 2l + 1))f(2p+ 1)g(2l) (3.134)
=4
9
N2−1∑
p=0
cas((N2
+ 2p+ 1)ωj)f(2p+ 1)g(N2− 2p− 1)
N2−1∑
m=0
cas(−2πm) (3.135)
=2N
9
N2−1∑
p=0
(−1)j cas(jω(2p+ 1))f(2p+ 1)g(N2− 2p− 1). (3.136)
We have used the fact that the only contribution in the summation essentially occurs
when 2p + 2l + 1 = N2
and periodicity 2π of the cas function. The summations in
(3.127) may be further simplified as follows.
4
9
N2−1∑
l=0
cas(−2jωl)
N2−1∑
p=0
N2−1∑
m=0
cas(2mω(−2p+ 2l − 1))f(2p+ 1)g(2l) (3.137)
=4
9
N2−1∑
p=0
cas((N2− 2p− 1)ωj)f(2p+ 1)g(N
2+ 2p+ 1)
N2−1∑
m=0
cas(2πm) (3.138)
=2N
9
N2−1∑
p=0
(−1)j cas(−jω(2p+ 1))f(2p+ 1)g(N2
+ 2p+ 1). (3.139)
31
We have used the fact that the only contribution in the summation occurs when −2p+
2l − 1 = N2
.
Using lemma 3.2, (3.88) simplifies to
16
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
(cas(ωj(2l + 1))
2
[cas(2mω(2p+ 1) + 2mω(2l + 1))
+ cas(2mω(2p+ 1)− 2mω(2l + 1))]
+cas(−ωj(2l + 1))
2×[
cas(−2mω(2p+ 1)− 2mω(2l + 1))− cas(−2mω(2p+ 1) + 2mω(2l + 1))])
× f(2p+ 1)g(2l + 1) (3.140)
=8
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(ωj(2l + 1)) cas(2mω(2p+ 1) + 2mω(2l + 1))f(2p+ 1)
× g(2l + 1) (3.141)
+8
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(ωj(2l + 1)) cas(2mω(2p+ 1)− 2mω(2l + 1))f(2p+ 1)
× g(2l + 1) (3.142)
+8
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(−ωj(2l + 1)) cas(−2mω(2p+ 1)− 2mω(2l + 1))f(2p+ 1)
× g(2l + 1) (3.143)
− 8
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(−ωj(2l + 1)) cas(−2mω(2p+ 1) + 2mω(2l + 1))f(2p+ 1)
× g(2l + 1). (3.144)
32
The summations in statement (3.141) may be further simplified as follows.
8
9
N2−1∑
l=0
cas(ωj(2l + 1))
N2−1∑
p=0
N2−1∑
m=0
cas(4mω(p+ l + 1))f(2p+ 1)g(2l + 1) (3.145)
=8
9
N2−1∑
p=0
cas((N − 2p− 1)ωj)f(2p+ 1)g(−2p− 1)
N2−1∑
m=0
cas(4πm) (3.146)
=4N
9
N2−1∑
p=0
cas(−ωj(2p+ 1))f(2p+ 1)g(−2p− 1). (3.147)
We used the fact that the only contribution to the summation occurs for p+ l+1 = N2
.
The summations in statement (3.142) may be further simplified as follows.
8
9
N2−1∑
l=0
cas(ωj(2l + 1))
N2−1∑
p=0
N2−1∑
m=0
cas(4mω(p− l))f(2p+ 1)g(2l + 1) (3.148)
=8
9
N2−1∑
p=0
cas((2p+ 1)ωj)f(2p+ 1)g(2l + 1)
N2−1∑
m=0
1 (3.149)
=4N
9
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(2p+ 1). (3.150)
We used the fact that the only contribution to the summation occurs for p = l. The
summations in statement (3.143) may be further simplified as follows.
8
9
N2−1∑
l=0
cas(−ωj(2l + 1))
N2−1∑
p=0
N2−1∑
m=0
cas(−4mω(p+ l + 1))f(2p+ 1)g(2l + 1) (3.151)
=8
9
N2−1∑
p=0
cas((N + 2p+ 1)ωj)f(2p+ 1)g(−2p− 1)
N2−1∑
m=0
cas(−4πm) (3.152)
=4N
9
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(−2p− 1). (3.153)
We used the fact that the only contribution to the summation occurs for p+ l+1 = N2
.
33
The summations in statement (3.144) may be further simplified as follows.
8
9
N2−1∑
l=0
cas(−ωj(2l + 1))
N2−1∑
p=0
N2−1∑
m=0
cas(4mω(l − p))f(2p+ 1)g(2l + 1) (3.154)
=8
9
N2−1∑
p=0
cas(−ωj(2p+ 1))f(2p+ 1)g(2p+ 1)
N2−1∑
m=0
1 (3.155)
=4N
9
N2−1∑
p=0
cas(−ωj(2p+ 1))f(2p+ 1)g(2p+ 1). (3.156)
We used the fact that the only contribution to the summation occurs for p = l.
Combining (3.96), (3.99), (3.102) and (3.105) statement (3.85) simplifies to
N
9
[N2 −1∑p=0
cas(2pωj)f(2p)g(2p) +
N2−1∑
p=0
cas(2pωj)f(2p)g(−2p)
+
N2−1∑
p=0
cas(−2pωj)f(2p)g(−2p)−N2−1∑
p=0
cas(−2pωj)f(2p)g(2p)]. (3.157)
Combining (3.113), (3.116), (3.119) and (3.122), statement (3.86) simplifies to
2N
9
[N2 −1∑p=0
(−1)j cas(−2pωj)f(2p)g(N2− 2p) +
N2−1∑
p=0
(−1)j cas(2pωj)f(2p)g(2p+ N2
)
+
N2−1∑
p=0
(−1)j cas(2pωj)f(2p)g(N2− 2p)−
N2−1∑
p=0
(−1)j cas(−2pωj)f(2p)g(2p+ N2
)].
(3.158)
Combining (3.130), (3.133), (3.136) and (3.139), statement (3.87) simplifies to
2N
9
[N2 −1∑p=0
(−1)j cas(−ωj(2p+ 1))f(2p+ 1)g(N2− 2p− 1)
+
N2−1∑
p=0
(−1)j cas(jω(2p+ 1))f(2p+ 1)g(N2
+ 2p+ 1)
+
N2−1∑
p=0
(−1)j cas(jω(2p+ 1))f(2p+ 1)g(N2− 2p− 1)
−N2−1∑
p=0
(−1)j cas(−jω(2p+ 1))f(2p+ 1)g(N2
+ 2p+ 1)]. (3.159)
34
Combining (3.147), (3.150), (3.153) and (3.156), statement (3.88) simplifies to
4N
9
[N2 −1∑p=0
cas(−ωj(2p+ 1))f(2p+ 1)g(−2p− 1)
+
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(2p+ 1)
+
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(−2p− 1)
−N2−1∑
p=0
cas(−ωj(2p+ 1))f(2p+ 1)g(2p+ 1)]. (3.160)
By a similar analysis the second summation in (3.84) simplifies to the following four
terms:
4
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(2ωp(2m+ 1)) cas(2ωl(j − 2m− 1))f(2p)g(2l)
=N
9
[N2 −1∑p=0
cas(2pωj)f(2p)g(2p) +
N2−1∑
p=0
cas(2pωj)f(2p)g(−2p)
+
N2−1∑
p=0
cas(−2pωj)f(2p)g(−2p)−N2−1∑
p=0
cas(−2pωj)f(2p)g(2p)]. (3.161)
8
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(2pω(2m+ 1)) cas(ω(2l + 1)(j − 2m− 1))f(2p)g(2l + 1)
= −2N
9
[N2 −1∑p=0
(−1)j cas(−2pωj)f(2p)g(N2− 2p) +
N2−1∑
p=0
(−1)j cas(2pωj)f(2p)g(2p+ N2
)
+
N2−1∑
p=0
(−1)j cas(2pωj)f(2p)g(N2− 2p)−
N2−1∑
p=0
(−1)j cas(−2pωj)f(2p)g(2p+ N2
)].
(3.162)
35
8
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(ω(2p+ 1)(2m+ 1)) cas(2ωl(j − 2m− 1))f(2p+ 1)g(2l)
= −2N
9
[N2 −1∑p=0
(−1)j cas(−ωj(2p+ 1))f(2p+ 1)g(N2− 2p− 1)
+
N2−1∑
p=0
(−1)j cas(jω(2p+ 1))f(2p+ 1)g(N2
+ 2p+ 1)
+
N2−1∑
p=0
(−1)j cas(jω(2p+ 1))f(2p+ 1)g(N2− 2p− 1)
−N2−1∑
p=0
(−1)j cas(−jω(2p+ 1))f(2p+ 1)g(N2
+ 2p+ 1)]. (3.163)
16
9
N2−1∑
m=0
N2−1∑
p=0
N2−1∑
l=0
cas(ω(2p+ 1)(2m+ 1)) cas(ω(j − 2m− 1)(2l + 1))f(2p+ 1)g(2l + 1)
=4N
9
[N2 −1∑p=0
cas(−ωj(2p+ 1))f(2p+ 1)g(−2p− 1)
+
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(2p+ 1)
+
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(−2p− 1)
−N2−1∑
p=0
cas(−ωj(2p+ 1))f(2p+ 1)g(2p+ 1)]. (3.164)
An interesting observation is that (3.158) is the negative of (3.162) as well as (3.159) is
the negative of (3.163) whilst (3.157) and (3.160) are duplicates of (3.161) and (3.164)
respectively. Hence equation (3.84) simplifies to
(Hf ⊗Hg)(j) =2N
9
[N2 −1∑p=0
cas(2pωj)f(2p)g(2p) +
N2−1∑
p=0
cas(2pωj)f(2p)g(−2p)
36
+
N2−1∑
p=0
cas(−2pωj)f(2p)g(−2p)−N2−1∑
p=0
cas(−2pωj)f(2p)g(2p)]
+8N
9
[N2 −1∑p=0
cas(ωj(2p+ 1))f(2p+ 1)g(2p+ 1)
+
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(−2p− 1)
+
N2−1∑
p=0
cas(−ωj(2p+ 1))f(2p+ 1)g(−2p− 1)
−N2−1∑
p=0
cas(−ωj(2p+ 1))f(2p+ 1)g(2p+ 1)]. (3.165)
We wish to express equation (3.165) in terms of the transform and this requires some
manipulation to yield
(Hf ⊗Hg)(j)
=N
3
[2
3
N2−1∑
p=0
cas(2pωj)f(2p)g(2p) +4
3
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(2p+ 1)
+2
3
N2−1∑
p=0
cas(2pωj)f(2p)g(−2p) +4
3
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(−2p− 1)
+2
3
N2−1∑
p=0
cas(2pωj)f(−2p)g(2p) +4
3
N2−1∑
p=0
cas(ωj(2p+ 1))f(−2p− 1)g(2p+ 1)
− 2
3
N2−1∑
p=0
cas(2pωj)f(−2p)g(−2p) +4
3
N2−1∑
p=0
cas(ωj(2p+ 1))f(−2p− 1)g(−2p− 1)]
+4N
9
[N2 −1∑p=0
cas(ωj(2p+ 1))f(2p+ 1)g(2p+ 1)
+
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(−2p− 1)
+
N2−1∑
p=0
cas(ωj(2p+ 1))f(−2p− 1)g(2p+ 1)
37
−N2−1∑
p=0
cas(ωj(2p+ 1))f(−2p− 1)g(−2p− 1)]
=N
3
[H(f . ? g)(j) + H(f . ? gR)(j) + H(fR. ? g)(j)−H(fR. ? gR)(j)
]+
4N
9
[N2 −1∑p=0
cas(ωj(2p+ 1))f(2p+ 1)g(2p+ 1)
+
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(−2p− 1)
+
N2−1∑
p=0
cas(ωj(2p+ 1))f(−2p− 1)g(2p+ 1)
−N2−1∑
p=0
cas(ωj(2p+ 1))f(−2p− 1)g(−2p− 1)]. (3.166)
It should be noted that the reason we have arrived at a reversal of signals is simply
due to the fact that we have replaced p by −p in order to introduce the transform.
The remaining summations in (3.166) may be written in terms of the transform by
employing (3.69) and (3.70) to obtain
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(2p+ 1) =3
8
[H(f . ? g)(j)−H(f . ? g)(j + N
2)]. (3.167)
N2−1∑
p=0
cas(ωj(2p+ 1))f(2p+ 1)g(−2p− 1) =3
8
[H(f . ? gR)(j)−H(f . ? gR)(j + N
2)].
(3.168)N2−1∑
p=0
cas(ωj(2p+1))f(−2p−1)g(2p+1) =3
8
[H(fR. ? g)(j)−H(fR. ? g)(j+N
2)]
(3.169)
N2−1∑
p=0
cas(ωj(2p+ 1))f(−2p− 1)g(−2p− 1) =3
8
[H(fR. ? gR)(j)−H(fR. ? gR)(j + N
2)].
(3.170)
We now define a vector S by
S = f . ? g + f . ? gR + fR. ? g − fR. ? gR (3.171)
38
thus allowing us to write the final form of equation (3.84) as
(Hf ⊗Hg) =N
2
(HS− 1
3S(N/2)
). (3.172)
The circular cross correlation r = f ? g is defined component-wise by
r(j) =N−1∑m=0
f(m)g(j +m). (3.173)
The cross correlation property bares striking similarities to that of the convolution
property. As such we will not consider the detailed derivation for this property, rather
we will use an alternative approach to arrive at the final result.
Component-wise the effect of the transpose on a signal can be obtained from (2.9) and
is given by [15]
HT f(2k) =2
3
N−1∑j=0
cas(2kωj)f(j)
=2
3
N2−1∑
j=0
cas(4kωj)f(2j) +2
3
N2−1∑
j=0
cas(2kω(2j + 1))f(2j + 1) (3.174)
and
HT f(2k + 1) =4
3
N−1∑j=0
cas((2k + 1)ωj)f(j)
=4
3
N2−1∑
j=0
cas((2k + 1)ω2j)f(2j) +4
3
N2−1∑
j=0
cas((2k + 1)ω(2j + 1))f(2j + 1).
(3.175)
We see that the index 2k + N2
is odd, hence from (3.175) it follows that
HT f(2k + N2
) =4
3
N2−1∑
j=0
cas(4kωj)f(2j)− 4
3
N2−1∑
j=0
cas(2kω(2j + 1))f(2j + 1). (3.176)
We see that the index 2k + N2
+ 1 is even, hence from (3.174) it follows that
HT f(2k+1+ N2
) =2
3
N2−1∑
j=0
cas((2k+1)ω2j)f(2j)− 2
3
N2−1∑
j=0
cas((2k+1)ω(2j+1))f(2j+1).
(3.177)
39
From (2.2) and (3.174) it follows that
Hf(2k) = HT f(2k) +2
3
N2−1∑
j=0
cas(2kω(2j + 1))f(2j + 1). (3.178)
From (3.174) and (3.176) it can be shown that
2
3
N2−1∑
j=0
cas(2kω(2j + 1))f(2j + 1) =1
2HT f(2k)− 1
4HT f(2k + N
2). (3.179)
Hence (3.178) can be written in the form
Hf(2k) =3
2HT f(2k)− 1
4HT f(2k + N
2). (3.180)
It may be shown similarly from (2.2), (3.175) and (3.177) that
Hf(2k + N2
) =3
4HT f(2k + N
2)− 1
2HT f(2k). (3.181)
From (3.180) and (3.181) we may recover the effect of the transpose on a signal to get
HT f(2k) =3
4Hf(2k) +
1
4Hf(2k + N
2) (3.182)
HT f(2k + N2
) =1
2Hf(2k) +
3
2Hf(2k + N
2). (3.183)
3.8 Cross correlation
It is shown that the even component of the circular cross correlation is given by [15]
Hr(2k) =9
8
[HT f(2k)HTg(2k)−HT f(2k)HTg(−2k)
+ HT f(−2k)HTg(2k) + HT f(−2k)HTg(−2k)]
− 3
32
[HT f(2k + N
2)HTg(2k + N
2)−HT f(2k + N
2)HTg(−2k − N
2)
+ HT f(−2k − N2
)HTg(2k + N2
) + HT f(−2k − N2
)HTg(−2k − N2
)]. (3.184)
40
Using (3.182) and (3.183) in (3.184) we may write the correlation purely in terms of
the transform to get
Hr(2k) =9
8
[(3
4Hf(2k) +
1
4Hf(2k + N
2))(3
4Hg(2k) +
1
4Hg(2k + N
2))
−(3
4Hf(2k) +
1
4Hf(2k + N
2))(3
4Hg(−2k) +
1
4Hg(−2k − N
2))
+(3
4Hf(−2k) +
1
4Hf(−2k − N
2))(3
4Hg(2k) +
1
4Hg(2k + N
2))
+(3
4Hf(−2k) +
1
4Hf(−2k − N
2))(3
4Hg(−2k) +
1
4Hg(−2k − N
2))]
− 3
32
[(1
2Hf(2k) +
3
2Hf(2k + N
2))(1
2Hg(2k) +
3
2Hg(2k + N
2))
−(1
2Hf(2k) +
3
2Hf(2k + N
2))(1
2Hg(−2k) +
3
2Hg(−2k − N
2))
+(1
2Hf(−2k) +
3
2Hf(−2k − N
2))(1
2Hg(2k) +
3
2Hg(2k + N
2))
+(1
2Hf(−2k) +
3
2Hf(−2k − N
2))(1
2Hg(−2k) +
3
2Hg(−2k − N
2))]
(3.185)
=39
64
[Hf(2k)Hg(2k)−Hf(2k)Hg(−2k) + Hf(−2k)Hg(2k) + Hf(−2k)Hg(−2k)
]− 9
64
[Hf(2k + N
2)Hg(2k + N
2)−Hf(2k + N
2)Hg(−2k − N
2)
+ Hf(−2k − N2
)Hg(2k + N2
) + Hf(−2k − N2
)Hg(−2k − N2
)]
+9
64
[Hf(2k)Hg(2k + N
2)−Hf(2k)Hg(−2k − N
2)
+ Hf(−2k)Hg(2k + N2
) + Hf(−2k)Hg(−2k − N2
) + Hf(2k + N2
)Hg(2k)
−Hf(2k + N2
)Hg(−2k) + Hf(−2k − N2
)Hg(2k) + Hf(−2k − N2
)Hg(−2k)]. (3.186)
It may be proved that (3.186) is also valid for the odd components. In order to write
the correlation in a more compact form we define two vectors
R = Hf . ?Hg −Hf . ?HgR + HfR. ?Hg + HfR. ?HgR (3.187)
M = Hf . ?HgN2−Hf . ?HgN
2R + HfR. ?HgN
2+ HfR. ?HgN
2R. (3.188)
41
Hence the correlation can be written as
Hr =39
64R− 9
64R(N/2) +
9
64
(M + M(N/2)
). (3.189)
42
Chapter 4
Spectrum of H
From (3.34) and (3.39) we deduce that the matrix H2 is tridiagonal. The diagonal of
H2 has the form[23N, 4
3N, 2
3N, 4
3N, . . . , 2
3N, 4
3N]
of length N , the super diagonal has the
form[−4
9N,−2
9N,−4
9N,−2
9N, . . . ,−2
9N,−4
9N]
of length N2
located at column N2
in the
0th row and the sub diagonal has the form[−2
9N,−4
9N,−2
9N,−4
9N, . . . ,−4
9N,−2
9N]
of length N2
located at row N2
in the 0th column.
Let X = [X(0), X(1), X(2), . . . , X(N − 1)] be an eigenvector of H2 which corresponds
to the eigenvalue λ, then we can write
H2X = λX. (4.1)
Now let p be the index such that |X(2k)| ≤ |X(2p)| and m be the index such that
|X(2k + 1)| ≤ |X(2m+ 1)| for k = 0, 1, . . . , N2− 1. Using the duality property (3.34),
(3.39) results in
H2X(2p) =2
3NX(2p)− 4
9NX(2p+ N
2) (4.2)
H2X(2p+ N2
) =4
3NX(2p+ N
2)− 2
9NX(2p). (4.3)
43
Since X is an eigenvector of H2 it must be true that
2
3NX(2p)− 4
9NX(2p+ N
2) = λX(2p) (4.4)
4
3NX(2p+ N
2)− 2
9NX(2p) = λX(2p+ N
2). (4.5)
Hence system (4.4)-(4.5) can be written as
(23N − λ)X(2p)− 4
9NX(2p+ N
2) = 0 (4.6)
−29NX(2p) + (4
3N − λ)X(2p+ N
2) = 0. (4.7)
Now if X(2p) = X(2m + 1) = 0 then X cannot be an eigenvector. If X(2p) 6= 0 then
X(2p + N2
) = 0 will contradict (4.7). Hence the linear system (4.6)-(4.7) has a non
trivial solution.
Using the duality property (3.34), (3.39) results in
H2X(2m+ 1) =4
3NX(2m+ 1)− 2
9NX(2m+ 1 + N
2) (4.8)
H2X(2m+ 1 + N2
) = −4
9NX(2m+ 1) +
2
3NX(2m+ 1 + N
2). (4.9)
Since X is an eigenvector of H2 it must be true that
4
3NX(2m+ 1)− 2
9NX(2m+ 1 + N
2) = λX(2m+ 1) (4.10)
−4
9NX(2m+ 1) +
2
3NX(2m+ 1 + N
2) = λX(2m+ 1 + N
2). (4.11)
Hence system (4.10)-(4.11) can be written as
(43N − λ)X(2m+ 1)− 2
9NX(2m+ 1 + N
2) = 0 (4.12)
−49NX(2m+ 1) + (2
3N − λ)X(2m+ 1 + N
2) = 0. (4.13)
If X(2m + 1) 6= 0 then X(2m + 1 + N2
) = 0 will contradict (4.13). Hence the linear
44
system (4.12)-(4.13) has a non trivial solution. Linear systems (4.6)-(4.7) and (4.12)-
(4.13) imply that ∣∣∣∣∣∣∣∣23N − λ −4
9N
−29N 4
3N − λ
∣∣∣∣∣∣∣∣ = 0. (4.14)
This results in (2
3N − λ
)(4
3N − λ
)− 8
81N2 = 0, (4.15)
which can easily be solved to yield the eigenvalues of H2, namely r21 = N + N9
√17 and
r22 = N − N9
√17 where
r1 =
√9 +√
17√N
3(4.16)
r2 =
√9−√
17√N
3. (4.17)
Hence we may deduce that the spectrum σ(H) ⊆ {±r1,±r2}.
4.1 Minimal polynomial
The minimal polynomial Pm of an n×n matrix H over a field F is the monic polynomial
over F of least degree such that Pm(H) = 0. Any other polynomial Q with Q(H) = 0
is a polynomial multiple of Pm.
In order to derive an expression for the minimal polynomial of H we require the effect
of H4 on a signal f . Hence
H4f(2k) = H2[H2f ](2k)
= H2g(2k), (4.18)
45
where g = H2f . By recursively using the duality property (3.34) and (3.39) we have
H2g(2k) =2
3Ng(2k)− 4
9Ng(2k + N
2)
=2
3NH2f(2k)− 4
9NH2f(2k + N
2)
=2
3N[2
3Nf(2k)− 4
9Nf(2k + N
2)]− 4
9N[4
3Nf(2k + N
2)− 2
9Nf(2k)
]=
4
9N2f(2k)− 8
27N2f(2k + N
2)− 16
27N2f(2k + N
2) +
8
81N2f(2k)
=44
81N2f(2k)− 8
9N2f(2k + N
2) (4.19)
= H4f(2k).
We also note that
−(r21 + r22)H2f(2k) = −2N
(2
3Nf(2k)− 4
9Nf(2k + N
2))
= −4
3N2f(2k) +
8
9N2f(2k + N
2) (4.20)
and
r21r22If(2k) =
64
81N2f(2k). (4.21)
Summing equations (4.19), (4.20) and (4.21) we observe the following
[H4 − (r21 + r22)H
2 + r21r22I]f(2k)
=44
81N2f(2k)− 8
9N2f(2k + N
2)− 4
3N2f(2k) +
8
9N2f(2k + N
2) +
64
81N2f(2k)
=108
81N2f(2k)− 4
3N2f(2k)
= 0. (4.22)
In a similar manner it may be shown that
[H4 − (r21 + r22)H
2 + r21r22I]f(2k + 1) = 0.
46
Hence we may conclude that
H4 − (r21 + r22)H2 + r21r
22I = 0. (4.23)
It follows from (4.23) that the minimal polynomial has degree four and is given by
P4(λ) = λ4 − (r21 + r22)λ2 + r21r
22
= (λ2 − r21)(λ2 − r22)
= (λ− r1)(λ+ r1)(λ− r2)(λ+ r2). (4.24)
In our case we can see that H is diagonalizable since the minimal polynomial has linear
elementary divisors [13], furthermore σ(H) = {±r1,±r2}.
We now derive an expression for the effect of H3 on a signal f . Hence
H3f(2k) = H2[Hf ](2k)
= H2g(2k), (4.25)
where g = Hf . Using the duality property again we arrive at
H2g(2k)
=2
3Ng(2k)− 4
9Ng(2k + N
2)
=2
3NHf(2k)− 4
9NHf(2k + N
2)
=2
3N[2
3
N2−1∑
j=0
cas(4kωj)f(2j) +4
3
N2−1∑
j=0
cas(2kω(2j + 1))f(2j + 1)]
− 4
9N[2
3
N2−1∑
j=0
cas(2(2k + N2
)ωj)f(2j) +4
3
N2−1∑
j=0
cas(ω(2k + N2
)(2j + 1))f(2j + 1)].
(4.26)
47
Using the fact that the cas function is periodic in 2π and anti-periodic in π (4.26)
simplifies to
H3f(2k) =4
9N
N2−1∑
j=0
cas(4kωj)f(2j) +8
9N
N2−1∑
j=0
cas(2kω(2j + 1))f(2j + 1)
− 8
27N
N2−1∑
j=0
cas(4kωj)f(2j) +16
27N
N2−1∑
j=0
cas(2kω(2j + 1))f(2j + 1)
=4
27N
N2−1∑
j=0
cas(4kωj)f(2j) +40
27N
N2−1∑
j=0
cas(2kω(2j + 1))f(2j + 1). (4.27)
It can be shown in a similar manner that
H3f(2k+1) =20
27N
N2−1∑
j=0
cas(2ωj(2k+1))f(2j)+56
27N
N2−1∑
j=0
cas(ω(2j+1)(2k+1))f(2j+1).
(4.28)
4.2 Traces of powers of H
For future use we require expressions for the traces of H, H2 and H3. For this we
require the following summary
Hf(2k) =2
3
N2−1∑
j=0
cas(4kωj)f(2j) +4
3
N2−1∑
j=0
cas(ω(2j + 1)(2k + 1))f(2j + 1) (4.29)
Hf(2k + 1)
=2
3
N2−1∑
j=0
cas(2ωj(2k + 1))f(2j) +4
3
N2−1∑
j=0
cas(ω(2j + 1)(2k + 1))f(2j + 1) (4.30)
H2f(2k) =2
3Nf(2k)− 4
9Nf(2k + N
2) (4.31)
H2f(2k + 1) =4
3Nf(2k + 1)− 2
9Nf(2k + 1 + N
2) (4.32)
H3f(2k) =4
27N
N2−1∑
j=0
cas(4kωj)f(2j) +40
27N
N2−1∑
j=0
cas(2kω(2j + 1))f(2j + 1) (4.33)
48
H3f(2k + 1) =20
27N
N2−1∑
j=0
cas(2ωj(2k + 1))f(2j) +56
27N
N2−1∑
j=0
cas(ω(2j + 1)(2k + 1))
× f(2j + 1). (4.34)
By expressing the right hand side of (4.29) and (4.30) in matrix vector form we observe
that the diagonal entries of H consist of alternating terms of the form 23
cas((2p)2ω)
and 43
cas((2p+ 1)2ω) hence
trace(H) =2
3
N2−1∑
p=0
cas((2p)2ω) +4
3
N2−1∑
p=0
cas((2p+ 1)2ω). (4.35)
The diagonal of H2 has already been expressed on page 43 and can be expressed as
trace(H2) =N
2
(2N
3+
4N
3
)= N2. (4.36)
The trace of H3 is deduced (from (4.33),(4.34)) in a similar manner to the trace of H
to give
trace(H3) =4
27N
N2−1∑
p=0
cas((2p)2ω) +56
27N
N2−1∑
p=0
cas((2p+ 1)2ω). (4.37)
In order to further simplify the trace of H we consider the first term of (4.35) which
simplifies as follows
N2−1∑
p=0
cas(4p2ω)
= 1 +
N2−1∑
p=1
cas(4p2ω)
= 1 +
N4− 1
2∑p=1
cas(4p2ω) +
N2−1∑
p=N4+ 1
2
cas(4p2ω)
49
= 1 +
N4− 1
2∑p=1
cas(4p2ω) +
N4− 1
2∑p=1
cas(4(p+ N4− 1
2)2ω)
= 1 +
N4− 1
2∑p=1
cas(4p2ω) +
N4− 1
2∑p=1
cas((2p+ N2− 1)2ω). (4.38)
The argument of the summation of the sum in the third term of (4.38) simplifies as
follows
cas(
(2p− 1 + N2
)2ω)
= cas(
(2p− 1)2ω + (2p− 1)Nω + N2
4ω)
= cas(
(2p− 1)2ω + (2p− 1)2π + N2π)
= cas(
(2p− 1)2ω + N2π)
= cas(
(2p− 1)2ω + (2m+ 1)π)
= − cas(
(2p− 1)2ω). (4.39)
Using (4.39) in (4.38) we obtain
N2−1∑
p=0
cas(4p2ω) = 1 +
N4− 1
2∑p=1
cas(4p2ω)−N4− 1
2∑p=1
cas((2p− 1)2ω). (4.40)
Now let us investigate the second term of (4.35) which simplifies as follows
N2−1∑
p=0
cas((2p+ 1)2ω)
=
N4− 1
2∑p=0
cas((2p+ 1)2ω) +
N2−1∑
p=N4+ 1
2
cas((2p+ 1)2ω)
=
N4− 1
2∑p=0
cas((2p+ 1)2ω) +
N4− 1
2∑p=1
cas((2(p+ N4− 1
2) + 1)2ω)
=
N4+ 1
2∑p=1
cas((2p− 1)2ω) +
N4− 1
2∑p=1
cas((2(p+ N4− 1
2) + 1)2ω)
50
=
N4− 1
2∑p=1
cas((2p− 1)2ω) + cas(
(2(N4
+ 12)− 1)2ω
)+
N4− 1
2∑p=1
cas((2(p+ N4− 1
2) + 1)2ω)
=
N4− 1
2∑p=1
cas((2p− 1)2ω)− 1 +
N4− 1
2∑p=1
cas((2(p+ N4− 1
2) + 1)2ω). (4.41)
The argument of the summation of the sum in the third term of (4.41) simplifies as
follows
cas(
(2(p+ N4− 1
2) + 1)2ω
)= cas
(4p2ω + 2pNω +
N2
4ω)
= cas(
4p2ω + 2p2π + N2π)
= cas(
4p2ω + N2π)
= cas(
4p2ω + (2m+ 1)π)
= − cas(4p2ω). (4.42)
Using (4.42) in (4.41) we obtain
N2−1∑
p=0
cas((2p+ 1)2ω) = −1−N4− 1
2∑p=1
cas(4p2ω) +
N4− 1
2∑p=1
cas((2p− 1)2ω). (4.43)
From (4.40) and (4.43) we deduce that∑N
2−1
p=0 cas(4p2ω) = −∑N
2−1
p=0 cas((2p + 1)2ω).
With this substitution in (4.35) we obtain
trace(H) = −2
3
N2−1∑
p=0
cas(4p2ω). (4.44)
In order to explicitly evaluate the sum in (4.44) we employ the result of generalized
quadratic Gauss sums [3] stated in the theorem below.
Theorem 4.1. The reciprocity formula for generalized quadratic Gauss sums can be
written as|c|−1∑p=0
eiπ(ap2+bp)
c =∣∣ ca
∣∣12 e iπ(|ac|−b2)4ac
|a|−1∑p=0
e−iπ(cp2+bp)
a (4.45)
51
where a, b and c are integers with ac 6= 0 and ac+ b being even.
Choosing parameters c = N2
, a = 4 and b = 0 in theorem 4.1 gives
N2−1∑
p=0
eπi(
4p2
N2
)=
N2−1∑
p=0
ei4p2ω
=
√N
8eπ4i
3∑p=0
e−πi( N
2p2
4
)
=
√N
8eπ4i
3∑p=0
e−πiN8p2
=1
2
√N
2eπ4i[1 + e−i
π8N + e−i
4π8N + e−i
9π8N]
=1
2
√N
2eπ4i[1 + e−i
π8N + e−i
π8N + e−i
π2N]
=1
2
√N
2eπ4i[1 + 2e−i
π8N − 1
]=
√N
2ei(
π4−π
8N). (4.46)
Thus if we take the real and imaginary parts respectively of equation (4.46) we arrive
at
N2−1∑
p=0
cos(4p2ω) =
√N
2cos(π
4− π
8N)
(4.47)
N2−1∑
p=0
sin(4p2ω) =
√N
2sin(π
4− π
8N). (4.48)
We then add equations (4.47) and (4.48) to get
N2−1∑
p=0
cas(4p2ω) =
√N
2cas(π
4− π
8N). (4.49)
Hence
trace(H) = −2
3
√N
2cas(π
4− π
8N). (4.50)
A similar calculation yields
trace(H3) = −52
27N
√N
2cas(π
4− π
8N). (4.51)
52
4.3 Multiplicity of eigenvalues
Let m10,m11,m20,m21 represent the multiplicities of the eigenvalues r1,−r1, r2,−r2
respectively. Using the fact that the trace of a matrix is the sum of all it’s eigenvalues
we form the following system of linear equations in order to ascertain the multiplicities
of the eigenvalues of H.
m10 +m11 +m20 +m21 = N (4.52)
r1m10 − r1m11 + r2m20 − r2m21 = trace(H) (4.53)
r21m10 + r21m11 + r22m20 + r22m21 = trace(H2) (4.54)
r31m10 − r31m11 + r32m20 − r23m21 = trace(H3). (4.55)
In order to solve the system (4.52)-(4.55) we let
A = m10 +m11 (4.56)
B = m20 +m21 (4.57)
C = m10 −m11 (4.58)
D = m20 −m21. (4.59)
Hence (4.52) and (4.54) can be written as
A+B = N (4.60)
Ar21 +Br22 = N2. (4.61)
53
The simplified system (4.60)-(4.61) can be solved easily by Cramer’s rule. This leads
us to the solution
A =Nr22 −N2
r22 − r21(4.62)
B =N2 −Nr21r22 − r21
. (4.63)
In a similar manner we simplify equations (4.53) and (4.55) which leads us to
r1C + r2D = −2
3
√N
2cas(
π
4− π
8N) (4.64)
r31C + r32D = −52
27N
√N
2cas(
π
4− π
8N). (4.65)
System (4.64)-(4.65) yields the solution
C =
√N2
cas(π4− π
8N)(52
27Nr2 − 2
3r32)
r1r32 − r2r31(4.66)
D =
√N2
cas(π4− π
8N)(2
3r31 − 52
27Nr1)
r1r32 − r2r31. (4.67)
Since r22 − r21 = −29
√17N , this substitution in (4.62) yields
A =N(r22 −N)
r22 − r21
=N2(9−√17
9− 1)
−29
√17N
=N
2. (4.68)
54
Similarly B = N2
. We now attempt to simplify C from (4.66) as follows.
C =r2
√N2
cas(π4− π
8N)
(5227N − 2
3r22)
r1r2(r22 − r21)
=
√N2
cas(π4− π
8N)(
5227N − 2
27N(9−
√17))
13
√9 +√
17√N(−2
9
√17N)
=
√N2
cas(π4− π
8N)
(3427N + 2
√17
27N)
− 227
√9 +√
17√NN√
17
=
227N√N 1√
2cas(π4− π
8N)
(17 +√
17)
− 227
√9 +√
17√NN√
17
=
1√2
cas(π4− π
8N)
(17 +√
17)
−√
17(9 +√
17)
=− cas
(π4− π
8N)
(17 +√
17)√34(9 +
√17)
= − cas(π
4− π
8N). (4.69)
The result in (4.69) is due to the fact that S = 17+√17√
34(9+√17)
= 1 as shown below:
S2 =(17)2 + 34
√17 + 17
34(9 +√
17)
=306 + 34
√17
306 + 34√
17
= 1.
Similarly D = cas(π4− π
8N)
. From equations (4.56)-(4.59) we obtain
A+ C = 2m10 (4.70)
C − A = −2m11 (4.71)
B +D = 2m20 (4.72)
D −B = −2m21, (4.73)
55
from which the multiplicities can easily be recovered and expressed as
m10 =1
2
[N2− cas
(−mπ2
)](4.74)
m11 =1
2
[N2
+ cas(−mπ
2
)](4.75)
m20 =1
2
[N2
+ cas(−mπ
2
)](4.76)
m21 =1
2
[N2− cas
(−mπ2
)], (4.77)
by using the fact that cas(π4− π
8N)
= cas(−mπ2
).
56
Chapter 5
2-Dimensional SDHT
The focus of this chapter is going to be the two-dimensional (2-D) SDHT. The digital
world has evolved drastically over the past decade with the ability to communicate vast
amounts of data in the form of audio or visual signals. This leads us to the fact that
digital images are of vital importance to various aspects of life including but not limited
to medical imaging, astronomy, image processing, geophysics and magnetic resonance
imaging [11]. It may be apparent that images could be contaminated by sinusoidal
noise. This can be easily cleaned up in the frequency domain using specific types of
filters whose purpose is to eliminate the contaminating noise. As a two-dimensional
(2-D) Discrete Hartley Transform is obtained by an application of the trapezoidal
rule in 2-D, here we similarly derive a 2-D Simpson Discrete Hartley Transform using
Simpson’s rule. A 2-D transform is necessary for application to image processing as
images are represented digitally in the form of a 2-D array on the gray scale or a
collection of 2-D arrays on the colour scale.
57
5.1 Derivation of the 2-D SDHT
In order to proceed we adopt the notation HN where N denotes the order of the SDHT
matrix H defined in equation (2.9). This implies that HN is given component-wise by
HNf(k) =2
3
N2−1∑
j=0
cas(2kωNj)f(2j) +4
3
N2−1∑
j=0
cas(kωN(2j + 1))f(2j + 1), (5.1)
where ωN = 2πN
and k = 0, 1, . . . , N −1. The transformation matrix can be represented
by
HN =2
3
AN 2BN
AN −2BN
PN, (5.2)
where the (N2× N
2) matrices AN and BN are defined respectively by
AN =
1 1 1 · · · 1
cas(ωN) cas(3ωN) cas(5ωN) · · · cas((N1 + 1)ωN)
cas(2ωN) cas(6ωN) cas(10ωN) · · · cas(2(N1 + 1)ωN)
......
......
...
cas(N1
2ωN) cas(N1
23ωN) cas(N1
25ωN) · · · cas(N1
2(N1 + 1)ωN)
, (5.3)
BN =
1 1 1 · · · 1
cas(ωN) cas(3ωN) cas(5ωN) · · · cas(N1ωN)
cas(2ωN) cas(6ωN) cas(10ωN) · · · cas(2N1ωN)
......
... · · · ...
cas(N1ωN) cas(3N1ωN) cas(5N1ωN) · · · cas(N1
2(N1 + 1)ωN)
(5.4)
and the N × N permutation matrix PN defined by components PN(j,2j)= 1 and
PN(N2 +j,2j+1)
= 1 for j = 0, 1, . . . , N2− 1 and N1 = N − 2.
58
The two-dimensional Discrete Hartley Transform in a separable form is defined as [16]
f(x, y) =∞∑
l=−∞
∞∑k=−∞
C(k, l) cas(2πNxk) cas(2π
Myl), (5.5)
with f(x, y) representing a periodic function of 2π in the variables x and y. The Hartley
coefficients are represented by
c(k, l) =1
4π2
∫ 2π
0
∫ 2π
0
f(x, y) cas(2πNxk) cas(2π
Myl)dxdy. (5.6)
In order to proceed with the derivation we need to approximate the double integral
(5.6). To do this we employ Simpson’s quadrature with the aid of a grid in the x
direction defined by the nodes xn = 2πnN
, n = 0, 1, . . . , N with the step length defined
by ωN = 2πN
. With this discretization equation (5.6) can be written as
1
4π2
∫ 2π
0
ωN3
[f(x0, y) cas(2π
Nx0k) + f(xN , y) cas(2π
NxNk)
+ 2
N2−1∑
n=1
f(x2n, y) cas(2πNx2nk) + 4
N2−1∑
n=0
f(x2n+1, y) cas(2πNx2n+1k)
]cas(2π
Myl)dy
=ωN
12π2
[∫ 2π
0
f(x0, y) cas(2πNx0k) cas(2π
Myl)dy +
∫ 2π
0
f(xN , y) cas(2πNxNk) cas(2π
Myl)dy
+ 2
N2−1∑
n=1
∫ 2π
0
f(x2n, y) cas(2πNx2nk) cas(2π
Myl)dy
+ 4
N2−1∑
n=0
∫ 2π
0
f(x2n+1, y) cas(2πNx2n+1k) cas(2π
Myl)dy
]. (5.7)
Using the fact that f(x0, y) = f(xN , y), (5.7) simplifies to
ωN12π2
[2
N2−1∑
n=0
∫ 2π
0
f(x2n, y) cas(2πNx2nk) cas(2π
Myl)dy
+ 4
N2−1∑
n=0
∫ 2π
0
f(x2n+1, y) cas(2πNx2n+1k) cas(2π
Myl)dy
]
59
=ωN6π2
[N2 −1∑n=0
∫ 2π
0
f(x2n, y) cas(2πNx2nk) cas(2π
Myl)dy
+ 2
N2−1∑
n=0
∫ 2π
0
f(x2n+1, y) cas(2πNx2n+1k) cas(2π
Myl)dy
]. (5.8)
We now apply Simpson quadrature in the y direction with nodes ym = 2πmM
where
m = 0, 1, . . . ,M . We now define G(x2n, y) = f(x2n, y) cas(2πNx2nk) cas(2π
Myl) and obtain
the following approximation
∫ 2π
0
G(x2n, y)dy ≈ ωM3
[G(x2n, y0) +G(x2n, yM)
+ 2
M2−1∑
m=1
G(x2n, y2m) + 4
M2−1∑
m=0
G(x2n, y2m+1)]
=ωM3
[2
M2−1∑
m=0
G(x2n, y2m) + 4
M2−1∑
m=0
G(x2n, y2m+1)]
=2ωM
3
M2−1∑
m=0
G(x2n, y2m) +4ωM
3
M2−1∑
m=0
G(x2n, y2m+1), (5.9)
where ωM = 2πM
is the step length in the y direction and m = 0, 1, . . . ,M . We now
substitute (5.9) into the first part of (5.8) to get
ωN6π2
N2−1∑
n=0
∫ 2π
0
G(x2n, y)dy
=ωN6π2
N2−1∑
n=0
[2ωM3
M2−1∑
m=0
G(x2n, y2m) +4ωM
3
M2−1∑
m=0
G(x2n, y2m+1)]
=ωNωM
9π2
N2−1∑
n=0
M2−1∑
m=0
G(x2n, y2m) +2ωNωM
9π2
N2−1∑
n=0
M2−1∑
m=0
G(x2n, y2m+1)
=ωNωM
9π2
N2−1∑
n=0
M2−1∑
m=0
f(x2n, y2m) cas(2πNx2nk) cas(2π
My2ml)
+2ωNωM
9π2
N2−1∑
n=0
M2−1∑
m=0
f(x2n, y2m+1) cas(2πNx2nk) cas(2π
My2m+1l). (5.10)
Adopting the notation f(xn, ym) = f(n,m), (5.10) can be written in the simplified
60
form
4
9NM
N2−1∑
n=0
M2−1∑
m=0
f(2n, 2m) cas(2nωNk) cas(2ωMml)
+8
9NM
N2−1∑
n=0
M2−1∑
m=0
f(2n, 2m+ 1) cas(2nωNk) cas(ωM(2m+ 1)l). (5.11)
Similarly the second part of (5.8) can be shown to simplify to
8
9NM
N2−1∑
n=0
M2−1∑
m=0
f(2n+ 1, 2m) cas(ωN(2n+ 1)k) cas(2mωM l)
+16
9NM
N2−1∑
n=0
M2−1∑
m=0
f(2n+ 1, 2m+ 1) cas(ωN(2n+ 1)k) cas(ωM(2m+ 1)l). (5.12)
We now define the following terms
T1(k, l) =4
9
N2−1∑
n=0
M2−1∑
m=0
f(2n, 2m) cas(2nωNk) cas(2ωMml) (5.13)
T2(k, l) =8
9
N2−1∑
n=0
M2−1∑
m=0
f(2n, 2m+ 1) cas(2nωNk) cas(ωM(2m+ 1)l) (5.14)
T3(k, l) =8
9
N2−1∑
n=0
M2−1∑
m=0
f(2n+ 1, 2m) cas(ωN(2n+ 1)k) cas(2ωMml) (5.15)
T4(k, l) =16
9
N2−1∑
n=0
M2−1∑
m=0
f(2n+ 1, 2m+ 1) cas(ωN(2n+ 1)k) cas(ωM(2m+ 1)l). (5.16)
We now let
F (k, l) = T1(k, l) + T2(k, l) + T3(k, l) + T4(k, l). (5.17)
Thus from (5.8) the approximation C(k, l) to c(k, l) is given by
C(k, l) =F (k, l)
NM, (5.18)
where F (k, l) is defined as the Simpson Discrete Hartley Transform in 2-D. Summing
61
(5.13) and (5.15) we get
T1(k, l) + T3(k, l)
=2
3
M2−1∑
m=0
[2
3
N2−1∑
n=0
f(2n, 2m) cas(2nωNk) cas(2ωMml)
+4
3
N2−1∑
n=0
f(2n+ 1, 2m) cas(ωN(2n+ 1)k) cas(2ωMml)]
=2
3
M2−1∑
m=0
[2
3
N2−1∑
n=0
f(2n, 2m) cas(2nωNk)
+4
3
N2−1∑
n=0
f(2n+ 1, 2m) cas(ωN(2n+ 1)k)]
cas(2ωMml)
=2
3
M2−1∑
m=0
F̂ (k, 2m) cas(2ωMml), (5.19)
where
F̂ (k, 2m) =2
3
N2−1∑
n=0
f(2n, 2m) cas(2nωNk) +4
3
N2−1∑
n=0
f(2n+ 1, 2m) cas(ωN(2n+ 1)k).
(5.20)
In a similar manner by summing (5.14) and (5.16) we get
T2(k, l) + T4(k, l)
=4
3
M2−1∑
m=0
[2
3
N2−1∑
n=0
f(2n, 2m+ 1) cas(2nωNk) cas(ωM(2m+ 1)l)
+4
3
N2−1∑
n=0
f(2n+ 1, 2m+ 1) cas(ωN(2n+ 1)k) cas(ωM(2m+ 1)l)]
=4
3
M2−1∑
m=0
[2
3
N2−1∑
n=0
f(2n, 2m+ 1) cas(2nωNk)
+4
3
N2−1∑
n=0
f(2n+ 1, 2m+ 1) cas(ωN(2n+ 1)k)]
cas(ωM(2m+ 1)l)
=4
3
M2−1∑
m=0
F̂ (k, 2m+ 1) cas(ωM(2m+ 1)l), (5.21)
62
where
F̂ (k, 2m+ 1) =2
3
N2−1∑
n=0
f(2n, 2m+ 1) cas(2nωNk)
+4
3
N2−1∑
n=0
f(2n+ 1, 2m+ 1) cas(ωN(2n+ 1)k). (5.22)
Using (5.19) and (5.21) allows us to write
F (k, l) =2
3
M2−1∑
m=0
F̂ (k, 2m) cas(2ωMml) +4
3
M2−1∑
m=0
F̂ (k, 2m+ 1) cas(ωM(2m+ 1)l). (5.23)
It can be seen that equations (5.20) and (5.22) are equivalent to the 1-D SDHT on
the columns of the N ×M matrix f which consists of elements f(n,m). This can be
represented as F̂ = HNf . An interesting observation is that (5.23) represents the 1-D
SDHT on the rows of F̂. Let us consider transforming the rows of F̂. In order to
achieve this, it would be appropriate to transform the columns of F̂T. Performing the
transformation yields
FT = HMF̂T
= HMfTHTN. (5.24)
From (5.24) we obtain
F = HNfHTM. (5.25)
Equation (5.25) is the representation of the 2-D SDHT on matrix f . Equation (5.17)
essentially represents the Simpson’s quadrature for a periodic function in two variables.
This is characterized by the mesh (xn, ym) for n = 0, 1, . . . , N −1, m = 0, 1, . . . ,M −1.
This is illustrated for an image f of size 6×10 in figure 5.1 with the appropriate weights.
63
Figure 5.1: A 6× 10 mesh
Let us now investigate some properties of (5.13)-(5.16).
Theorem 5.1. Let T1 be the matrix of size N2× M
2consisting of elements T1(k, l).
Then T1 is periodic with periodicity of N2
in the first variable and M2
in the second
variable for k = 0, 1, . . . , N2− 1 and l = 0, 1, . . . , M
2− 1.
Proof. From equation (5.13) we have
T1(k, l) =4
9
N2−1∑
n=0
M2−1∑
m=0
f(2n, 2m) cas(2nωNk) cas(2ωMml).
64
Thus we see that
T1(k + N2, l + M
2)
=4
9
N2−1∑
n=0
M2−1∑
m=0
f(2n, 2m) cas(2nωN(k + N2
)) cas(2ωMm(l + M2
))
=4
9
N2−1∑
n=0
M2−1∑
m=0
f(2n, 2m) cas(2nωNk + 2πn) cas(2ωMml + 2πm)
=4
9
N2−1∑
n=0
M2−1∑
m=0
f(2n, 2m) cas(2nωNk) cas(2ωMml) (5.26)
= T1(k, l),
due to the fact that the cas function is periodic in 2π. This completes the proof.
Theorem 5.2. Let T3 be the matrix of size N2× M
2consisting of elements T3(k, l) for
k = 0, 1, . . . , N2−1 and l = 0, 1, . . . , M
2−1. Then T3 is anti-periodic in the first variable
and periodic in the second variable.
Proof. To proceed we first will show that T3 is anti-periodic in the first variable. Recall
from equation (5.15) that
T3(k, l) =8
9
N2−1∑
n=0
M2−1∑
m=0
f(2n+ 1, 2m) cas(ωN(2n+ 1)k) cas(2ωMml).
Thus we see that
T3(k + N2, l)
=8
9
N2−1∑
n=0
M2−1∑
m=0
f(2n+ 1, 2m) cas(ωN(2n+ 1)(k + N2
)) cas(2ωMml)
=8
9
N2−1∑
n=0
M2−1∑
m=0
f(2n+ 1, 2m)(−1)2n+1 cas(ωN(2n+ 1)k) cas(2ωMml)
65
= −8
9
N2−1∑
n=0
M2−1∑
m=0
f(2n+ 1, 2m) cas(ωN(2n+ 1)k) cas(2ωMml) (5.27)
= −T3(k, l),
due to the fact that the cas function is anti-periodic in π. This completes the first part
of the proof. To prove the second part of theorem 5.2 we are required to calculate the
following:
T3(k, l + M2
)
=8
9
N2−1∑
n=0
M2−1∑
m=0
f(2n+ 1, 2m) cas(ωN(2n+ 1)k) cas(2ωMm(l + M2
))
=8
9
N2−1∑
n=0
M2−1∑
m=0
f(2n+ 1, 2m) cas(ωN(2n+ 1)k) cas(2ωMml + 2πm)
=8
9
N2−1∑
n=0
M2−1∑
m=0
f(2n+ 1, 2m) cas(ωN(2n+ 1)k) cas(2ωMml) (5.28)
= T3(k, l).
We have again employed the fact that the cas function is periodic in 2π. This completes
the proof of theorem 5.2.
Let T4 be the N2× M
2matrix consisting of elements T4(k, l), then similarly it can be
shown that T4(k + N2, l) = −T4(k, l) and T4(k, l + M
2) = −T4(k, l), k = 0, 1, . . . , N
2− 1,
l = 0, 1, . . . , M2− 1. It can be also shown that if we let T2 be the N
2× M
2matrix
consisting of elements T2(k, l) then T2(k+ N2, l) = T2(k, l) and T2(k, l+
M2
) = −T2(k, l),
k = 0, 1, . . . , N2− 1, l = 0, 1, . . . , M
2− 1.
Thus to summarize we see that T1(k, l) is periodic in both the first and second variables,
T2(k, l) is periodic in the first variable and anti-periodic in the second variable, T3(k, l)
is anti-periodic in the first variable and periodic in the second and finally T4(k, l) is
66
anti-periodic in both the first and second variables. Using the properties listed above
we are permitted to write (5.17) in block matrix form as
F =
T1 + T2 + T3 + T4 T1 −T2 + T3 −T4
T1 + T2 −T3 −T4 T1 −T2 −T3 + T4
. (5.29)
67
Chapter 6
Applications of the 2-D SDHT
In the preceding chapter we have constructed the theory of the 2-D SDHT and we
now illustrate it’s usefulness to two applications in the frequency domain. Firstly we
examine the amplitude spectrum in the frequency domain of the Hartley Transform
and SDHT applied to a typical image illustrated in figure 6.1.
Figure 6.1: Cameraman
68
(a) Frequency domain (Hartley Transform) (b) Frequency domain (SDHT)
Figure 6.2: Transform of cameraman
We observe four distinct bright regions (corresponding to large amplitudes) in figure
6.2 (a), whilst in figure 6.2 (b) we observe nine such bright regions. The extra regions
could be exploited in applications such as watermarking.
6.1 Watermarking in the frequency domain
Digital watermarks are of vital importance in this digital age due to the digital medium
in which data is transmitted. A digital watermark can be regarded as a covertly
embedded distortion in a noise tolerant signal such as audio, image or video data [8].
These watermarks are used to typically verify authenticity or ownership of the signal.
Watermarking is the process where the information is hidden within the signal via
some sort of algorithmic method [12]. To demonstrate the process of watermarking we
will now introduce a watermark into an image. The watermark will be inserted into
the frequency domain by using the 2-D SDHT transformation derived in chapter 5. By
69
implementing equation (5.25) to an image, we successfully insert a watermark into an
image of peppers using a distortion algorithm.
In order to do this we first use a function fftshift in MATLAB. This essentially moves
the low frequency regions to the center and the high frequency into the periphery as
illustrated in the figures 6.3 (a) and (b).
(a) Unshifted transform (b) Shifted transform
Figure 6.3: fftshift
Figure 6.4 (a) is an illustration of the original image while it’s corresponding shifted
frequency domain is illustrated in figure 6.4 (b).
70
(a) Peppers (b) fftshifted transform
Figure 6.4: Peppers and it’s transform
(a) Watermark in area of high amplitude. (b) Recovered peppers from figure 6.5 (a)
Figure 6.5: Watermarking in the frequency domain in an area of high amplitude
Figure 6.5 (a) represents a triangle watermark placed in the region of high amplitude
and low frequency components. We see that a very small distortion in the central region
of the frequency domain results in visible blurriness on the image when transformed
back to the real domain. This visible distortion is illustrated in figure 6.5 (b) which is
71
why it is imperative to place a watermark in an intermediate to high frequency region.
Figure 6.6 (a) shown above represents the frequency domain in which we have inserted
a watermark in the high frequency region. This results in an unnoticeable difference of
the watermarked image when transformed back into the real domain using the inverse
transform.
(a) Watermark in area of low amplitude (b) Recovered peppers from figure 6.6 (a)
Figure 6.6: Watermarking in an intermediate region in the frequency domain
6.2 Noise cleaning in the frequency domain
There are various algorithms which are responsible for the removal of unwanted noise in
a signal. The cleaning algorithms are also dependent on the type of noise being cleaned
up, and the process of removing noise from a signal is known as filtering. This filtering
is usually done in the real domain. It is of vital importance that these filter algorithms
are implemented in the correct manner as they occur in life altering instruments such
as an electrocardiogram machine. The filter for the electrocardiogram machine has to
be designed such that only the electric signals from the heart are shown and all other
72
neurological impulses are suppressed.
The experiment which we are going to conduct comprises the removal of sinusoidal
noise from an image. This filtering is best accomplished in the frequency domain.
Figure 6.7: House contaminated with noise
Figure 6.7 is contaminated by unwanted sinusoidal noise. Figure 6.8 illustrates the
frequency domain corresponding to figure 6.7. It is observed that the sinusoidal noise
appears as sixteen bright crosses.
Figure 6.8: Effect of noise in the frequency domain
73
We remove the noise in the frequency domain and this is achieved by applying a
simple block filter algorithm. This is certainly not an optimal filter. This filtration is
illustrated in figure 6.9 below.
Figure 6.9: Block filter applied in the frequency domain
Figure 6.10: House after applying filter
By inverting the filtered transform we obtain a reasonably cleaned image. Figure 6.10
represents figure 6.7 after the application of a filter. There is clearly visual difference
in quality of the image. The lack of perfect clarity in the image is due to the non
74
optimal filter implemented. We can see that a simple filter algorithm has demonstrated
reasonable cleaning ability when transformed into the frequency domain using the 2-D
SDHT.
75
Conclusion
In this thesis we have introduced a relatively new transform which was derived by
application of Simpson’s numerical quadrature. We have derived in detail the math-
ematical properties satisfied by this transform, thus we have placed the SDHT on a
firm mathematical foundation. We have not exploited the full potential of the SDHT,
but rather restricted it to two applications. Further applicability is left to the field
of engineering. We have also not compared the DHT with the SDHT as this was not
our intention. A similar investigation of this nature is left as a future task for the
dimension N ≡ 0 (mod 4).
76
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