Wavelets trong Electromagnetics và mô hình thiết bị P5 ppsx

CHAPTER FIVE
Sampling Biorthogonal Time
Domain Method (SBTD)
The finite difference time domain (FDTD) method was proposed by K. Yee [1] in
1966. The simplicity of the FDTD method in mathematics has proved to be its great
advantage. The method does not involve any integral equations, Green’s functions,
singularities, nor matrix equations. Neither does it involve functional or variational
principles. In addition the FDTD proves to be versatile when used in complicated
geometries. The computational issues associated with the FDTD are the radiation
boundary conditions or absorption boundary conditions for open structures, numer-
ical dispersion, and stability conditions. Its major drawbacks include its massive
memory consumption and huge computational time.
In these regard wavelets offer significant improvements to the FDTD. It will be
shown that the Yee-based FDTD is identical to the Galerkin method using Haar
wavelets. Since the Haar bases are discontinuous, the slow decay of the frequency
components and the Gibbs phenomena of the Haar basis prevent the use of a coarse
mesh in the FDTD. In contrast, the Daubechies-based sampling functions are contin-
uous basis functions with fast decay in both the spatial and spectral domains. Thus a
more efficient time domain method can be derived: the sampling biorthogonal time
domain (SBTD) algorithm.
5.1 BASIS FDTD FORMULATION
For a lossy medium with a conductivity σ , we begin with Maxwell’s two curl equa-
tions
µ
∂H
∂t
=−∇×E, (5.1.1)

∂E
∂t
+ σ E =∇×H. (5.1.2)

189
Wavelets in Electromagnetics and Device Modeling. George W. Pan
Copyright
¶ 2003 John Wiley & Sons, Inc.
ISBN: 0-471-41901-X
190 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)
We obtain by the leapfrog method [2] a set of finite difference equations
k+(1/2)
H
x

, m +
1
2
, n +
1
2

=
k−(1/2)
H
x

, m +
1
2
, n +
1
2


+
t
µz

k
E
y

, m +
1
2
,(n + 1)

−
k
E
y

, m +
1
2
, n

−
t
µy

k
E
z


, (m + 1), n +
1
2

−
k
E
z

, m, n +
1
2

,
k+(1/2)
H
y

 +
1
2
, m, n +
1
2

=
k−(1/2)
H
y


 +
1
2
, m, n +
1
2

+
t
µx

k
E
z

 + 1, m, n +
1
2

−
k
E
z

, m, n +
1
2

−

t
µz

k
E
x

 +
1
2
, m, n + 1

−
k
E
x

 +
1
2
, m, n

, (5.1.3)
k+(1/2)
H
z

 +
1
2

, m +
1
2
, n

=
k−(1/2)
H
z

 +
1
2
, m +
1
2
, n

+
t
µy

k
E
x

 +
1
2
, m + 1, n


−
k
E
x

 +
1
2
, m, n

−
t
µx

k
E
y

 + 1, m +
1
2
, n

−
k
E
y

, m +

1
2
, n

,
k+1
E
x

 +
1
2
, m, n

=

1 − (σ t /2)
1 + (σ t /2)

k
E
x

 +
1
2
, m, n

+


1
1 + (σ t/2)

t
y

k+(1/2)
H
z

 +
1
2
, m +
1
2
, n

−
k+(1/2)
H
z

 +
1
2
, m −
1
2
, n


BASIS FDTD FORMULATION 191
−
t
z

k+(1/2)
H
y

 +
1
2
, m, n +
1
2

−
k+(1/2)
H
y

 +
1
2
, m, n −
1
2

,

k+1
E
y

, m +
1
2
, n

=

1 −
σ t
2
1 +
σ t
2

k
E
y

, m +
1
2
, n

+

1

1 +
σ t
2


t
z

k+(1/2)
H
x

, m +
1
2
, n +
1
2

−
k+(1/2)
H
x

, m +
1
2
, n −
1
2


−
t
x

k+(1/2)
H
z

 +
1
2
, m +
1
2
, n

−
k+(1/2)
H
z

 −
1
2
, m +
1
2
, n


, (5.1.4)
k+1
E
z

, m, n +
1
2

=

1 −
σ t
2
1 +
σ t
2

k
E
z

, m, n +
1
2

+

1
1 +

σ t
2


t
x

k+(1/2)
H
y

 +
1
2
, m, n +
1
2

−
k+(1/2)
H
y

 −
1
2
, m, n +
1
2


−
t
y

k+(1/2)
H
x

, m +
1
2
, n +
1
2

−
k+(1/2)
H
x

, m −
1
2
, n +
1
2

.
In the equations above, indexes l, m,andn are the node numbers in the x, y,andz
directions, respectively, and the leftscript k denotes the time step. Note that a central

difference scheme has been used in all of the finite difference equations. Figure 5.1
illustrates a unit cell in the FDTD lattice where the electric and magnetic fields are
spaced apart by a half-grid in each dimension. At the interface of two media (e.g.,
at the boundary between a conductor and a dielectric), the average values of  and
192 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)
H
2
3
6
8
7
65
4
8
2
4
1
5
Magnetic cell
Electric cell
3
E
FIGURE 5.1 Standard Yee-FDTD lattice.
σ are used:  =
(

1
+ 
2
)

/2andσ =
(
σ
1
+ σ
2
)
/2. To ensure the stability of the
time-stepping algorithm of (5.1.1) and (5.1.2), a time increment is chosen to satisfy
the inequality
ct ≤
1

1/x
2
+ 1/y
2
+ 1/z
2
, (5.1.5)
where c is the velocity of light in the computational space. Equation (5.1.5) will be
derived in the next section. A Gaussian pulse
E
z
= e
−(t−t
0
)
2
/T

2
is chosen as the excitation pulse and is imposed upon the rectangular region under
the port to be excited.
The finite difference mesh must be truncated because of the finite ability of com-
puters to solve across very large or even infinite 3D volumes. The field components
tangential to the truncation planes cannot be evaluated from the FDTD equations
above since they would require for their evaluation the values of field components
outside the mesh. The tangential electric field components on the truncation planes
must be specified in such a way that outgoing waves are not reflected; this is known
as an absorbing boundary condition (ABC). There are many ABCs, including the
Mur absorbing boundary conditions [3] and the perfectly matched layer absorbing
BASIS FDTD FORMULATION 193
boundary conditions (PML) of Berenger [4], among others. Here we have specified
the boundary values of the fields according to the Engquist-Majda unconditionally
stable, absorbing boundary condition [5]
φ
k+1
0
= φ
k
1
+
ct −x
ct +x
(φ
k+1
1
− φ
k
0

),
where φ
0
and φ
1
are the tangential electric field components at the mesh wall and at
the first node within the wall, respectively.
Figure 5.2 depicts a system consisting of three coupled microstrip lines. In the
direction of signal propagation, the y direction, we have chosen the parameters t ,
x, y,andz such that the wave travels one spatial step in approximately five
temporal steps; this choice in turn requires a priori calculation in order to obtain
the approximate wave velocity in the direction of propagation. At the top and side
boundaries, the local velocity of light at the calculated node is used as the approx-
imate wave velocity. Without a loss of generality, the time domain solution for this
six-port system is obtained by means of the following procedures:
(1) Initialize (at t = kt = 0) all fields to 0.
(2) Impose Gaussian excitation on port 1:
•
H
k+(1/2)
is calculated from the FDTD equations.
•
E
k+1
is calculated from the FDTD equations.
•
The tangential E field is set to 0 on the ground plane and the absorbing
boundary condition is used on the truncation planes.
•
Store port voltages V

(1)
i
(kt) at the reference plane of port i (i = 1, 2, 3, 4,
5, 6), where a port voltage V
i
has been obtained by numerically integrating
the vertical electric field beneath the center of port i.
ref. plane
L
1
ref. plane
x
z
y
1
2
3
4
5
6
=3.5
d
W
W
W
S
S
t
FIGURE 5.2 Coupled three-line system.
194 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)

•
Store port currents I
(1)
i
(kt) at the reference plane of port i (i = 1, 2, 3),
where a port current I
i
has been obtained by numerical integration of the
magnetic field around the strip surface of port i in the reference plane.
•
k → k +1, repeat the previous steps 1 through 5 until the pulse and induced
waves pass through the reference plane of ports (4, 5, 6) completely.
(3) Impose a Gaussian excitation on port 2 and repeat the above six procedures.
Store all of the port voltages V
(2)
i
(i = 1, 2, 3, 4, 5, 6) and port currents
I
(2)
i
(i = 1, 2, 3).
In the previous items the superscripts (1) and (2) represented port 1 excitation and
port 2 excitation, respectively.
The Yee algorithm has been modified and extended into many versions and deriva-
tives, including the nonuniform mesh FDTD, and the finite volume time domain
(FVTD) method [6], nonorthogonal mesh [7], and the like. The transmission line
matrix (TLM) method was proposed by Peter Johns [8] in 1971 independently of
Yee’s work. Nonetheless, it was proven that the TLM is equivalent to FDTD method.
In handling lossy structures, the TLM needs to use artificial “stubs”; this necessity is
inconvenient. Because of its simplicity and popularity, only the standard FDTD will

be discussed in the text.
5.2 STABILITY ANALYSIS FOR THE FDTD
An unstable solution may occur owing to an improper choice of the time step t for
the space intervals x, y,andz. The instability is not due to an accumulation
of errors, but to causality. The analysis is conducted on plane waves, and is quite
general, since any wave may be expressed as a superposition of plane waves. Let us
write FDTD in terms of time–space eigenvalue problems. Space eigenvalues must be
located in stable regions.
The two curl equations in a lossless medium are written in their component forms
∇×H = 
∂E
∂t
,
∂ E
z
∂t
=
1


∂ H
y
∂x
−
∂ H
x
∂y

,
∂ E

y
∂t
=
1


∂ H
x
∂z
−
∂ H
z
∂x

,
∂ E
x
∂t
=
1


∂ H
z
∂y
−
∂ H
y
∂z


,
and
∇×E =−µ
∂H
∂t
,
STABILITY ANALYSIS FOR THE FDTD 195
∂ H
x
∂t
=
1
µ

∂ E
y
∂z
−
∂ E
z
∂y

,
∂ H
y
∂t
=
1
µ


∂ E
z
∂x
−
∂ E
x
∂z

,
∂ H
z
∂t
=
1
µ

∂ E
x
∂y
−
∂ E
y
∂x

.
In the rest of this section, we will only attack 2D problems. In doing so, we will be
able to capture the essence of the algorithms without spending too much time and
effort on tedious details. In 2D problems we will deal with only three rather than six
equations. The extension of the 2D formulation into 3D problems is straightforward,
but time-consuming. Consider a 2D TM

(z)
wave, namely
∂
∂z
= 0, H
z
= 0.
The remaining three equations are
∂ E
z
∂t
=
1


∂ H
y
∂x
−
∂ H
x
∂y

,
∂ H
x
∂t
=−
1
µ

∂ E
z
∂y
,
∂ H
y
∂t
=
1
µ
∂ E
z
∂x
.
Using the center difference Yee scheme and simplified notations, we obtain
n+1
E
z
i, j
−
n
E
z
i, j
t
=
1


n+(1/2)

H
y
i+(1/2), j
−
n+(1/2)
H
y
i−(1/2), j
x
−
n+(1/2)
H
x
i, j +(1/2)
−
n+(1/2)
H
x
i, j −(1/2)
y

,
n+(1/2)
H
x
i, j +(1/2)
−
n−(1/2)
H
x

i, j +(1/2)
t
=−
1
µ
n
E
z
i, j +1
−
n
E
z
i, j
y
,
n+(1/2)
H
y
i+(1/2), j
−
n−(1/2)
H
y
i+(1/2), j
t
=
1
µ
n

E
z
i+1, j
−
n
E
z
i, j
x
. (5.2.1)
C
ASE 1. TIME EIGENVALUE PROBLEM Separating the time derivatives in the pre-
ceding equations, we arrive at
n+1
E
z
i, j
−
n
E
z
i, j
t
= λ
n+(1/2)
E
z
i, j
, (5.2.2)
196 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)

n+(1/2)
H
x
i, j +(1/2)
−
n−(1/2)
H
x
i, j +(1/2)
t
= λ
n
H
x
i, j +(1/2)
, (5.2.3)
n+(1/2)
H
y
i+(1/2), j
−
n−(1/2)
H
y
i+(1/2), j
t
= λ
n
H
y

i+(1/2), j
. (5.2.4)
The general form of (5.2.2) through (5.2.4) is
n+(1/2)
V
i
−
n−(1/2)
V
i
t
= λ
n
V
i
. (5.2.5)
Let us define a factor
q
i
=
n+(1/2)
V
i
n
V
i
. (5.2.6)
In order to have a stable solution of (5.2.2) through (5.2.4), we must meet the condi-
tion
|q

i
|≤1.
Substituting (5.2.6) into (5.2.5), we have
n+(1/2)
V
i
n
V
i
−
n−(1/2)
V
i
n
V
i
= λt,
or equivalently
q
i
−
1
q
i
= λt.
Thus we obtain
q
2
i
− λtq

i
− 1 = 0,
q
i
=
λt
2
±

1 +

λt
2

2
. (5.2.7)
In order to have |q
i
|≤1, we need

Re{λ}=0
−
2
t
≤ Im{λ}≤
2
t
.
Letting λ = µ + jν, (5.2.7) gives
q

i
= j
νt
2
±

1 −
(νt)
2
4
.
C
ASE 2. SPACE EIGENVALUE PROBLEM The right-hand side of (5.2.1) provides
the following eigenvalue equations
STABILITY ANALYSIS FOR THE FDTD 197
H
y
i+(1/2), j
− H
y
i−(1/2), j
x
−
H
x
i, j +(1/2)
− H
x
i, j −(1/2)
y

= λ E
z
i, j
, (5.2.8)
E
z
i, j +1
− E
z
i, j
y
=−λµH
x
i, j +(1/2)
, (5.2.9)
E
z
i+1, j
− E
z
i, j
x
= λµH
y
i+(1/2), j
. (5.2.10)
Again, a nonplane wave can be expanded into a superposition of plane waves. Thus
we may work with the following plane waves:
E
z

I,J
= E
z
e
j (k
x
I  x +k
y
J y)
,
H
x
I,J
= H
x
e
j (k
x
I  x +k
y
J y)
,
H
y
I,J
= H
y
e
j (k
x

I  x +k
y
J y)
. (5.2.11)
Substitution of (5.2.11) into (5.2.8–5.2.10) leads to
E
z
= j
2
λ

H
y
x
sin
k
x
x
2
−
H
x
y
sin
k
y
y
2

,

H
x
=−j
2E
z
λµ y
sin
k
y
y
2
,
H
y
= j
2E
z
λµ x
sin
k
x
x
2
,
and
λ
2
=−
4
µ



sin k
x
x/2
x

2
+

sin k
y
y/2
y

2

.
Note that |sin(·)|≤1. Hence for ∀ k
x
, k
y
,
Re{λ}=0
Im{λ}≤2v


1
x


2
+

1
y

2

1/2
.
C
ASE 3. NUMERICAL STABILITY Relating the time eigenvalue problem to the
space eigenvalue problem, we have the 2D stability condition
2v


1
x

2
+

1
y

2

1/2
≤
2

t
,
198 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)
namely
t ≤
1
v

(
1/ x
)
2
+
(
1/y
)
2
.
For 3D cases, the stability condition is
t ≤
1
v

(
1/ x
)
2
+
(
1/y

)
2
+
(
1/z
)
2
=

v
√
3
,
if x = y = z = l. For 1D, this condition reduces to
t ≤
x
v
.
5.3 FDTD AS MAXWELL’S EQUATIONS WITH HAAR EXPANSION
The finite difference time domain (FDTD) formulas of (5.1.3) to (5.1.4) are derived
from the two Maxwell curl equations, using the finite difference to approximate the
differential operators. In this section we will see that the FDTD can be derived as a
special case of wavelet expansion using the Haar system.
To simplify our mathematical notation without losing generality, we consider the
one-dimensional case, namely the telegraphers’ equations in the frequency domain.
The telegraphers’ equations are








−
dI
dx
= jωCV
−
dV
dx
= jωLI.
(5.3.1)
When the finite difference method is applied, the expected result is







I
n+1
− I
n
x
=−j ωCV
n+(1/2)
V
n+(1/2)
− V

n−(1/2)
x
=−j ωLI
n
.
(5.3.2)
Let us expand the unknown current and voltage in terms of Haar scalets, that is, pulse
functions

I =

m
I
m
P
m
(x)
V =

m
V
m+(1/2)
P
m+(1/2)
(x).
(5.3.3)
FDTD AS MAXWELL’S EQUATIONS WITH HAAR EXPANSION 199
Notice that the voltage node and current node are offset by a half unit in space. The
pulse function can be written in the form
P

k
(x) = P

x
x
− k

,
where
P(x ) =







1if|x| <
1
2
1
2
if |x|=
1
2
0if|x| >
1
2
.
It can easily be seen that


∞
−∞
P
k
(x)P
l
(x) dx = (x)δ
k,l
, (5.3.4)
which is analogous to the orthogonality for wavelets

∞
−∞
ϕ
j,k
(x)ϕ
j,l
(x) dx = δ
k,l
.
Show.
(1) If k = l, P
k
(x ) and P
l
(x ) have no overlaps; hence

∞
−∞

P
k
(x )P
l
(x ) dx = 0. (5.3.5)
(2) If k = l,

∞
−∞
P
k
(x )P
l
(x ) dx =

P
2

x
x
− k


 
u
dx
= x

P
2


x
x
− k

d

x
x

= x

1/2
−1/2
P
2
(u) du
= x

1/2
−1/2
1 du = x. (5.3.6)
Combining (5.3.5) and (5.3.6), we arrive at (5.3.4).
Next we will show that

P
m
(x)
∂
∂x

P
m

+(1/2)
(x) dx = δ
m,m

− δ
m,m

+1
. (5.3.7)
200 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)
In fact
∂ P(x)
∂x
= δ

x +
1
2

− δ

x −
1
2

,
where the Dirac delta function δ(x −τ) has been used. More rigorously, we can use

the Heaviside step function H(x − τ)
H ( x − τ) =
1
2πi

e
(x −τ)s
s
ds,
where we integrate along the imaginary axis. The Dirac delta function is defined by
following integral

b
a
δ(x − τ)dx =

0ifb <τor a >τ
1ifa <τ<b.
It is possible to write an integral representation that is similar to the step function
δ(x − τ) =
1
2πi

e
(x −τ)s
ds.
The Dirac delta function and the step function are related by the expression
H

(x −τ) = δ(x −τ). (5.3.8)

The pulse function can be written as
P(x ) = H

x +
1
2

− H

x −
1
2

. (5.3.9)
Equations (5.3.9) and (5.3.8) lead to (5.3.7) as follows: using the previous results,
we obtain

P
m
(x)
∂
∂x
P
m

+(1/2)
(x) dx
=

P


x
x
− m

∂
∂x
P

x
x
−

m

+
1
2

dx
=

P

x
x
− m

∂
∂x


H

x
x
−

m

+
1
2

+
1
2

−H

x
x
−

m

+
1
2

−

1
2

dx
=

P

x
x
− m

1
x

δ

x
x
− m


− δ

x
x
− (m

+ 1)


dx
= δ
m,m

− δ
m,m

+1
.
FDTD WITH BATTLE–LEMARIE WAVELETS 201
Thus far we have sufficient knowledge to derive the finite difference equation (5.3.2).
By substituting (5.3.3) into (5.3.1), we obtain
−

m
V
m+(1/2)
d
dx
P
m+(1/2)
(x) = jωL

m
I
m
P
m
(x).
Multiplying both sides by P

n
(x) and integrating, we arrive at
RHS = jωL

m
I
m

P
m
(x)P
n
(x) dx
= jωLI
n
x,
where the orthogonality of P
m
(x) and P
n
(x) has been employed in order to simplify
the summation. In the meantime
LHS =−

m
V
m+(1/2)

P
n

(x)
d
dx
P
m+(1/2)
(x) dx
=−

m
V
m+(1/2)
[δ
n,m
− δ
n,m+1
]
=−[V
n+(1/2)
− V
n−(1/2)
].
Equating the two sides, we finally have
V
n+(1/2)
− V
n−(1/2)
x
=−j ωLI
n
,

which is exactly the centralized finite difference expression of (5.3.2). Notice that
the derivation is totally new and never makes use of the finite difference concept.
5.4 FDTD WITH BATTLE–LEMARIE WAVELETS
Battle–Lemarie wavelets possess better regularity than Haar wavelets. The Battle–
Lemarie based time domain method, referred to as the multiresolution time domain
(MRTD), improves numerical dispersion of the FDTD significantly [9]. However, the
MRTD is not widespread in the field computation because of its high computational
cost, complexity of its algorithm, CPU time required, and the difficulties in incorpo-
rating boundary conditions. For reasons of historical development and completeness,
we will briefly discuss the scalet-based MRTD.
In the MRTD the time dependencies of the field quantities are still treated as
pulse functions while the space dependencies are expanded in terms of the Battle–
Lemarie (B-L) scalets instead of Haar scalets. The six components field equations
are
202 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)
E
x
(r, t) =

k,l,m,n
k
E
ϕx
l+(1/2),m,n
P
k
(t)ϕ
l+(1/2)
(x)ϕ
m

(y)ϕ
n
(z),
E
y
(r, t) =

k,l,m,n
k
E
ϕy
l,m+(1/2),n
P
k
(t)ϕ
l
(x)ϕ
m+(1/2)
(y)ϕ
n
(z),
E
z
(r, t) =

k,l,m,n
k
E
ϕz
l,m,n+(1/2)

P
k
(t)ϕ
l
(x)ϕ
m
(y)ϕ
n+(1/2)
(z), (5.4.1)
H
x
(r, t) =

k,l,m,n
k+(1/2)
H
ϕx
l,m+(1/2),n+(1/2)
P
k+(1/2)
(t)ϕ
l
(x)ϕ
m+(1/2)
(y)ϕ
n+(1/2)
(z),
H
y
(r, t) =


k,l,m,n
k+(1/2)
H
ϕy
l+(1/2),m,n+(1/2)
P
k+(1/2)
(x)ϕ
l+(1/2)
(x)ϕ
m
(y)ϕ
n+(1/2)
(z),
H
z
(r, t) =

k,l,m,n
k+(1/2)
H
ϕz
l+(1/2),m+(1/2),n
P
k+(1/2)
(x)ϕ
l+(1/2)
(x)ϕ
m+(1/2)

(y)ϕ
n
(z).
The Fourier transform pair of the cubic spline Battle–Lemarie scalet is
ˆϕ(ω) =

∞
−∞
ϕ(x)e
−iωx
dx
and
ϕ(x) =
1
2π

∞
−∞
ˆϕ(ω)e
iωx
dω.
It can be verified that the Fourier transform of the cubic Battle–Lemarie scalet
is
ˆϕ(ω) =

sin ω/2
ω/2

4
1


1 −(4/3) sin
2
(ω/2) +(2/5) sin
4
(ω/2)) −(4/315) sin
6
(ω/2)
.
(5.4.2)
Using properties of the Fourier integral, it is possible to write

∞
−∞
ϕ
m
(x)
∂
∂x
ϕ
m

+(1/2)
(x) dx
=

∞
−∞
dx


1
2π

dω ˆϕ(ω)e
−iωm+iωx

1
2π

dω

∂
∂x
ˆϕ(ω

)e
−iω

[m

+(1/2)]
e
iω

x

=

dω dω



∞
−∞
dx

1
2π
e
ix(ω+ω

)

iω

2π
ˆϕ(ω

) ˆϕ(ω)e
−i(ωm)−iω

[m

+(1/2)]
=
1
2π

∞
−∞
dω ˆϕ(ω)e

−iωm

dω

δ(ω + ω

)(iω

) ˆϕ(ω

)e
−iω

[m

+(1/2)]
FDTD WITH BATTLE–LEMARIE WAVELETS 203
=
1
2π

∞
−∞
dω ˆϕ(ω)e
−i(ωm)
(−iω) ˆϕ(−ω)e
iω[m

+(1/2)]
=

1
π

∞
0
ω|ˆϕ(ω)|
2
sin

ω

m

− m +
1
2

dω.
This integral can be evaluated numerically. We can rewrite the expression above
as

∞
−∞
ϕ
m
(x)
∂ϕ
m

+(1/2)

(x)
∂x
dx =
∞

−∞
a
i
δ
m+i,m

. (5.4.3)
It can be seen that
a
0
=
1
π

∞
0
|ˆϕ(ω)|
2
ω sin
ω
2
dw
a
1
=

1
π

∞
0
|ˆϕ(ω)|
2
ω sin
3
2
ω dω
a
2
=
1
π

∞
0
|ˆϕ(ω)|
2
ω sin
5
2
ω dω
.
.
.
The Battle–Lemarie scalets decay rapidly, and the coefficients a
i

are negligible for
i > 8andi < −9. Thus the summation can be truncated as

∞
−∞
ϕ
m
(x)
∂ϕ
m

+(1/2)
(x)
∂x
dx =
8

−9
a
i
δ
m+i,m

. (5.4.4)
For negative indexes
a
−1−i
=−a
i
, i = 0, 1, ,8.

Table 5.1 provides the values of a
i
, i = 0, 1, ,8. Consider the x-component of
Ampere’slaw
∂ H
z
∂y
−
∂ H
y
∂z
= 
∂ E
x
∂t
. (5.4.5)
We approximate the right-hand side as
∂ E
x
∂t
=

k

,l

,m

,n


k

E
ϕx
l

+(1/2),m

,n

ϕ
l

+(1/2)
(x)ϕ
m

(y)ϕ
n

(z)
∂
∂t
P
k

(t).
After sampling the right-hand side of (5.4.5), we obtain ∂ E
x
/∂t, in space and

time,

dx dydz dtϕ
l+(1/2)
(x)ϕ
m
(y)ϕ
n
(z)P
k+(1/2)
(t)
∂ E
x
∂t
204 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)
TABLE 5.1. Coefficients
a
i
ia
i
0 1.2918462
1 −0.1560761
2 0.0596391
3 −0.0293099
4 0.0153716
5 −0.0081892
6 0.0043788
7 −0.0023433
8 0.0012542
=


k

,l

,m

,n

k

E
ϕx
l

+(1/2),m

,n


ϕ
l+(1/2)
(x)ϕ
l

+(1/2)
(x) dx

ϕ
m

(y)ϕ
m

(y) dy

ϕ
n
(z)ϕ
n

(z)dz

P
k+(1/2)
(t)
∂
∂t
P
k

(t)
=

k

,l

,m

,n

k

E
ϕx
l

+(1/2),m

,n

x δ
l,l

y δ
m,m

z δ
n,n

(δ
k+1,k

− δ
k,k

)
=

k+1
E

ϕx
l+(1/2),m,n
−
k
E
ϕx
l+(1/2),m,n

x y z.
Then, sampling the first term of the left-hand side, ∂ H
z
/∂y, and using the same
testing functions as for the RHS, we have

dt dx dy dz

k

,l

,m

,n

k

+(1/2)
H
ϕz
l


+(1/2),m

+(1/2),n

P
k

+(1/2)
(t)ϕ
l

+(1/2)
(x)
∂
∂y
ϕ
m

+(1/2)
(y)ϕ
n

(z)P
k+(1/2)
(t)ϕ
l+(1/2)
(x)ϕ
m
(y)ϕ

n
(z)
=

k

,l

,m

,n

k

+(1/2)
H
ϕz
l

+(1/2),m

+(1/2),n

t δ
k,k

x δ
l,l

z δ

n,n


dyϕ
m
(y)
∂
∂y
ϕ
m

+(1/2)
(y)
≈

8

i=−9
a
i
k+(1/2)
H
ϕz
l+(1/2),m+(1/2)+i,n

t x z.
Applying the same procedure to the term ∂ H
y
/∂z,wefinally obtain a difference
equation

POSITIVE SAMPLING AND BIORTHOGONAL TESTING FUNCTIONS 205

t

k+1
E
ϕx
l+(1/2),m,n
−
k
E
ϕx
l+(1/2),m,n

=
1
y
8

i=−9
a
i
k+(1/2)
H
ϕz
l+(1/2),m+(1/2)+i,n
−
1
z
8


i=−9
a
i
k+(1/2)
H
ϕy
l+(1/2),m,n+(1/2)+i
.
(5.4.6)
Note that the space differential operator is approximated by an 18-term summation
in the MRTD versus a 2-term summation in the traditional FDTD. The other five
equations can be derived in the same manner.
5.5 POSITIVE SAMPLING AND BIORTHOGONAL TESTING FUNCTIONS
Recall that in communication theory Shannon’s sampling theorem [10] is given by
x(t) =
∞

k=−∞
x(kT)
sin σ(t −kT)
σ(t −kT)
, T =
π
σ
for σ-band limited signals. For these signals x(t ) ∈ L
2
(R), and the Fourier transform
F (x(t)) has finite support [−σ, σ]. Often we use the notation sinc as
ϕ(t) = sinc (t) :=

sin πt
πt
.
As studied in Chapter 3 that the Shannon sinc function is a scalet satisfying the
sampling property
ϕ(n) = δ
0,n
.
In addition the sinc forms an orthogonal system

∞
−∞
ϕ(t)ϕ(t − n) dt = δ
0,n
.
Now we will construct sampling functions using the Daubechies scalets. Letting ϕ(x)
be the Daubechies scalet of N = 2, we can write a positive sampling function
S(x) =
2ν
ν − 1
∞

k=0

1 + ν
1 − ν

k
ϕ(x − k + 1), (5.5.1)
where ν =−1/

√
3. S(x) was used to eliminate the Gibbs phenomenon [11]. We will
demonstrate that S(x) has a sampling property similar to the sinc function. By the
factor ϕ(1) = (ν −1)/2ν, ϕ(2) = (ν + 1)/2ν (3.8.2), (5.5.1) may be rewritten in a
206 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)
more specificformas
S(x) =
1
ϕ(1)
∞

k=0

|ϕ(2)|
ϕ(1)

k
ϕ(x − k + 1).
Introducing the notation
S
m
(x) := S(x − m) =
1
ϕ(1)
∞

k=0

|ϕ(2)|
ϕ(1)


k
ϕ(x − m − k + 1), (5.5.2)
we will show the sampling property
S
m
(n) = δ
m,n
. (5.5.3)
Show.
From the definition of S
m
(x ),wehave
S
m
(n) =
1
ϕ(1)
∞

k=0

|ϕ(2)|
ϕ(1)

k
ϕ(n − m − k + 1). (5.5.4)
Notice that for the Daubechies scalet of N = 2, only two terms on the RHS of (5.5.4) are
nonzero because supp{ϕ}=[0, 3]. Therefore
n − m − k +1 =


1
2
or
k =

n − m
n − m −1.
Hence we may write (5.5.4) explicitly as
S
m
(n) =
1
ϕ(1)


|ϕ(2)|
ϕ(1)

n−m−1
ϕ(2) +

|ϕ(2)|
ϕ(1)

n−m
ϕ(1)

. (5.5.5)
From (5.5.5) we immediately see that S

m
(n) = δ
m,n
. In fact we can verify this property:
(1) When n = m, k = n − m − 1 =−1. However, the summation in (5.5.4) begins with
k = 0. Thus the first term in (5.5.5) must be dropped, yielding
S
m
(m) =
1
ϕ(1)


|ϕ(2)|
ϕ(1)

0
ϕ(1)

= 1.
(2) When n = m, we can use the fact that ϕ(2) is negative and obtain from (5.5.5),
S
m
(n) =
1
ϕ(1)

−
|ϕ(2)|
n−m

ϕ(1)
n−m−1
+
|ϕ(2)|
n−m
ϕ(1)
n−m−1

= 0.
As x →∞, the D
2
(Daubechies scalet of N = 2) based sampling function S
m
(x)
decays much faster than the sinc function and is compactly supported on the left
POSITIVE SAMPLING AND BIORTHOGONAL TESTING FUNCTIONS 207
endpoint of −1. As a matter of fact, supp{S(x)}≈[−1, 3] or [−1, 4]. Unfortunately,
S
m
(x) is not orthogonal to its shifted versions, namely

∞
−∞
S
m
(x)S
n
(x) dx = δ
m,n
.

The biorthogonal testing functions Q
n
(x) were introduced by Walter [12] as the
reproducing kernel
Q
n
(x) =

p∈Z
ϕ(n − p)ϕ(x − p).
It has been shown that {Q
n
(x)} forms a Riesz basis [11]. We will now demonstrate
that {Q
n
(x)} is biorthogonal to {S
m
(x)}, namely

∞
−∞
S
m
(x)Q
n
(x) dx = δ
m,n
. (5.5.6)
Owing to the finite support of D
2

, the testing functions have a closed-form expression
Q
n
(x) = ϕ(1)ϕ(x −n + 1) + ϕ(2)ϕ(x −n + 2). (5.5.7)
From the previous equation and the support of D
2
,wefind immediately that
supp{Q
n
(x)}=[n − 2, n + 2]. (5.5.8)
Let us show the biorthogonality of (5.5.6).
Show.

∞
−∞
S
m
(x )Q
n
(x ) dx =
1
ϕ(1)
+∞

k=0

|ϕ(2)|
ϕ(1)

k


∞
−∞
[ϕ(1)ϕ(x − n + 1) + ϕ(2)ϕ(x − n + 2)]ϕ(x − m − k + 1) dx
=
1
ϕ(1)


|ϕ(2)|
ϕ(1)

n−m
ϕ(1) +

|ϕ(2)|
ϕ(1)

n−m−1
ϕ(2)

, (5.5.9)
where we have used the orthogonality of the Daubechies scalets
ϕ(x −k), ϕ(x − )=δ
k,
.
Note that the right-hand side of (5.5.9) is identical to the expression (5.5.5), which is equal to
δ
m,n
. Therefore


∞
−∞
S
m
(x )Q
n
(x ) dx = δ
m,n
.
208 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)
The sampling function S(x) and testing function Q(x) are plotted in Figs. 5.3
and 5.4.
In the Battle–Lemarie based MRTD, one must compute ahead of time the coeffi-
cients
a
i
=

∞
−∞
ϕ
−i
(x)
dϕ
1/2
(x)
dx
dx.
In a similar fashion we need to evaluate the coefficients

c
i
=

∞
−∞
Q
−i
(x)
dS
1/2
(x)
dx
dx. (5.5.10)
2 1 0 1 2 3 4 5 6 7 8
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
S(x)
S
0

(x)
S
1
(x)
S
2
(x)
FIGURE 5.3 Daubechies-based positive sampling functions S(x).
−3 −2 −1 0 1 2 3
−0.5
0
0.5
1
1.5
2
x
Q(x)
Q
−1
(x)
Q
0
(x)
Q1(x)
FIGURE 5.4 Reproducing kernel Q(x) as the biorthogonal dual of S(x).
POSITIVE SAMPLING AND BIORTHOGONAL TESTING FUNCTIONS 209
TABLE 5.2. Coefficients
c
i
ic

i
0 1.22916661202745
1 −0.09374997764764
2 0.01041666418309
(1) For i ≥ 3, c
i
= 0 exactly, due to the finite support of D
2
.
(2) For i ≤−4, the values of c
i
evaluated according to (5.5.10) are identically
zero, as remains to be shown later.
(3) For −3 ≤ i ≤ 2, it can be proved analytically that
c
i
:=

∞
−∞
Q
−i
(x)
dS
1/2
(x)
dx
dx
=


∞
−∞
ϕ
−i
(x)
dϕ
1/2
(x)
dx
dx (5.5.11)
Thus we can compute the inner product
Q

(x),
d
dx
S


+(1/2)
(x)=
2

i=−3
c
i
δ
+i,

,

where the values of c
i
have been evaluated numerically, similar to (5.4.4), as
c
i
=

ϕ
−i
,
d
dx
ϕ
1/2
(x)

=
1
π

∞
−∞
ω|ˆϕ(ω)|
2
sin

ω

i +
1

2

dω. (5.5.12)
The values of c
i
are tabulated in Table 5.2.
For negative indexes, the symmetry holds, that is,
c
−1−i
=−c
i
, i = 0, 1, 2,
although ϕ(x) is not symmetric. Before presenting an elegant proof in the transform
domain, let us examine (5.5.11) in the spatial domain for i = 1. Other cases will
follow the same procedure. The following proofs are lengthy, but they provide some
physical insight.
Show.
Using (5.5.7) for Q
n
(x ) and (5.5.2) for S
m
(x ),wehave
c
i
|
i=1
=

∞
−∞

Q
−1
(x )
d
dx
S
1/2
(x ) dx
=
1
ϕ(1)
+∞

k=0

|ϕ(2)|
ϕ(1)

k

∞
−∞
dx[ϕ(1)ϕ(x +2)
210 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)
+ ϕ(2)ϕ(x + 3)]
dϕ

x +
1
2

− k

dx
=
1
ϕ(1)
[I
1
+ I
2
]. (5.5.13)
In the evaluation of I
1
,andI
2
of the equation above, we need to utilize the fact that for the
Daubechies scalet of N = 2, supp{ϕ(x)}=[0, 3]; only k = 0andk = 1 remain in the infinite
summation. For k ≥ 2 the integrand of (5.5.13) is zero because the two scalets ϕ(x + 2) or
ϕ(x +3) and ϕ(x +
1
2
− k) do not overlap. Hence
I
1
= ϕ(1)



|ϕ(2)|
ϕ(1)


0

∞
−∞
ϕ(x +2)
dϕ

x +
1
2

dx
dx
+

|ϕ(2)|
ϕ(1)

1

∞
−∞
ϕ(x +2)
dϕ

x −
1
2


dx
dx


= ϕ(1)

∞
−∞
ϕ(x +2)
dϕ

x +
1
2

dx
dx
+|ϕ(2)|

∞
−∞
ϕ(x +2)
dϕ

x −
1
2

dx
dx (5.5.14)

and
I
2
= ϕ(2)



|ϕ(2)|
ϕ(1)

0

∞
−∞
ϕ(x +3)
dϕ

x +
1
2

dx
dx


= ϕ(2)



∞

−∞
ϕ(y +2)
dϕ

y −
1
2

dy
dy


=−|ϕ(2)|

∞
−∞
ϕ(x +2)
dϕ

x −
1
2

dx
dx, (5.5.15)
where we have used the substitution y = x + 1 and the fact that ϕ(2) =−|ϕ(2)| because
ϕ(2)<0.
Combining (5.5.14) and (5.5.15), we obtain from (5.5.13) that
c
1

=
1
ϕ(1)
[I
1
+ I
2
]
=

∞
−∞
ϕ(x +2)
dϕ

x +
1
2

dx
dx
=

∞
−∞
ϕ(x +1)
dϕ

x −
1

2

dx
dx
=

∞
−∞
ϕ
−1
(x )
dϕ
1/2
(x )
dx
dx. (5.5.16)
The last integral in (5.5.16) is exactly equal to c
i
(i = 1) given by (5.5.11).
POSITIVE SAMPLING AND BIORTHOGONAL TESTING FUNCTIONS 211
The coefficients c
i
are identically zero for i > 2ori < −3, in contrast to the
MRTD where a
i
are approximately zero for i > 8ori < −9. The verification of
c
−4
= 0 is provided below.
Show.

We begin with
c
−4
=

∞
−∞
Q
4
(x )
dS
1/2
(x )
dx
dx.
Notice that
supp{S
1/2
(x )}=

−
1
2
, +∞

and that from (5.5.8),
supp{Q
4
(x )}=[2, 6].
Following the procedure in (5.5.13), we have

c
−4
=
1
ϕ(1)
+∞

k=0

|ϕ(2)|
ϕ(1)

k

∞
−∞
[ϕ(1)ϕ(x − 3) +ϕ(2)ϕ(x − 2)]
dϕ

x +
1
2
− k

dx
dx
=
1
ϕ(1)
[I

1
+ I
2
].
In the equation above the nonzero terms are k = 1, 2, ,7forI
1
and k = 0, 1, ,6forI
2
.
We s ee that
I
1
= ϕ(1)



|ϕ(2)
ϕ(1)

1

∞
−∞
ϕ(x −3)
dϕ

x −
1
2


dx
dx
+

|ϕ(2)
ϕ(1)

2

∞
−∞
ϕ(x −3)
dϕ

x −
3
2

dx
dx
+

|ϕ(2)
ϕ(1)

3

∞
−∞
ϕ(x −3)

dϕ

x −
5
2

dx
dx
+

|ϕ(2)
ϕ(1)

4

∞
−∞
ϕ(x −3)
dϕ

x −
7
2

dx
dx
+

|ϕ(2)
ϕ(1)


5

∞
−∞
ϕ(x −3)
dϕ

x −
9
2

dx
dx
+

|ϕ(2)
ϕ(1)

6

∞
−∞
ϕ(x −3)
dϕ(x −
11
2
)
dx
dx

+

|ϕ(2)
ϕ(1)

7

∞
−∞
ϕ(x −3)
dϕ

x −
13
2

dx
dx]


and
212 SAMPLING BIORTHOGONAL TIME DOMAIN METHOD (SBTD)
I
2
= ϕ(2)



|ϕ(2)
ϕ(1)


0

∞
−∞
ϕ(x −2)
dϕ

x +
1
2

dx
dx
+

|ϕ(2)
ϕ(1)

1

∞
−∞
ϕ(x −2)
dϕ

x −
1
2


dx
dx
+

|ϕ(2)
ϕ(1)

2

∞
−∞
ϕ(x −2)
dϕ

x −
3
2

dx
dx
+

|ϕ(2)
ϕ(1)

3

∞
−∞
ϕ(x −2)

dϕ

x −
5
2

dx
dx
+

|ϕ(2)
ϕ(1)

4

∞
−∞
ϕ(x −2)
dϕ

x −
7
2

dx
dx
+

|ϕ(2)
ϕ(1)


5

∞
−∞
ϕ(x −2)
dϕ

x −
9
2

dx
dx
+

|ϕ(2)
ϕ(1)

6

∞
−∞
ϕ(x −2)
dϕ

x −
11
2


dx
dx


.
Further simplification yields
I
1
=|ϕ(2)|

∞
−∞
ϕ(x −3)
dϕ

x −
1
2

dx
dx
+
|ϕ(2)|
2
ϕ(1)

∞
−∞
ϕ(x −3)
dϕ


x −
3
2

dx
+
|ϕ(2)|
3
ϕ(1)
2

∞
−∞
ϕ(x −3)
dϕ

x −
5
2

dx
+
|ϕ(2)|
4
ϕ(1)
3

∞
−∞

ϕ(x −3)
dϕ

x −
7
2

dx
+
|ϕ(2)|
5
ϕ(1)
4

∞
−∞
ϕ(x −3)
dϕ

x −
9
2

dx
+
|ϕ(2)|
6
ϕ(1)
5


∞
−∞
ϕ(x −3)
dϕ

x −
11
2

dx
I
2
=−|ϕ(2)|

∞
−∞
ϕ(x −2)
d
dx
ϕ

x +
1
2

dx
−
|ϕ(2)|
2
ϕ(1)


∞
−∞
ϕ(x −2)
d
dx
ϕ

x −
1
2

dx
POSITIVE SAMPLING AND BIORTHOGONAL TESTING FUNCTIONS 213
−
|ϕ(2)|
3
ϕ(1)
2

∞
−∞
ϕ(x −2)
d
dx
ϕ

x −
3
2


dx
−
|ϕ(2)|
4
ϕ(1)
3

∞
−∞
ϕ(x −2)
d
dx
ϕ

x −
5
2

dx
−
|ϕ(2)|
5
ϕ(1)
4

∞
−∞
ϕ(x −2)
d

dx
ϕ

x −
7
2

dx
−
|ϕ(2)|
6
ϕ(1)
5

∞
−∞
ϕ(x −2)
d
dx
ϕ

x −
9
2

dx.
I
1
and I
2

cancel each other out exactly item by item due to the fact that

∞
−∞
ϕ(x −3)
dϕ

x −
1
2

dx
dx =

∞
−∞
ϕ(x −2)
dϕ

x +
1
2

dx
dx

∞
−∞
ϕ(x −3)
dϕ


x −
3
2

dx
dx =

∞
−∞
ϕ(x −2)
dϕ

x −
1
2

dx
dx.
···=···
Proof. To begin with, we rewrite (5.5.12) below

∞
−∞
ϕ
−l
(x )
d
dx
ϕ

1/2
(x ) dx =
1
2π

∞
−∞
dω ˆϕ(ω) ˆϕ(−ω)e
iω(l+(1/2))
(−iω)
=
1
π

∞
0
ω|ˆϕ(ω)|
2
sin

ω

l +
1
2

dω.
On the other hand,

∞

−∞
Q
−l
(x )
d
dx
S
1/2
(x ) dx
=

∞
−∞
dx

1
2π

∞
−∞
dω
ˆ
Q(ω)e
(iωl+iωx)
1
2π

∞
−∞
dω


ˆ
S(ω

)e
−iω

(1/2)+iω

x

=

dω dω

1
2π

∞
−∞
dxe
ix(ω

+iω

)
iω

2π
ˆ

Q(ω)
ˆ
S(ω

)e
iωl−iω

(1/2)
=
1
2

∞
−∞
dω
ˆ
Q(ω)e
iωl

(dω

)δ(ω +ω

)(iω

)
ˆ
S(ω

)e

iω

(1/2)
=
1
2

∞
−∞
dω
ˆ
Q(ω)
ˆ
S(−ω)e
iω(l+(1/2))
(−iω) (5.5.17)
where
ˆ
Q(ω) is the Fourier transform of Q
0
(x ),and
ˆ
S(ω) is the Fourier transform of S(x)
given by (5.5.1). Since Q
0
(x ) and
ˆ
S(ω) are real, their Fourier transform
ˆ
Q(ω) and

ˆ
S(ω),are
conjugate symmetric.
According to (5.5.2)
S(x) =
1
ϕ(1)
∞

k=0
r
k
ϕ(x −k +1)