investigation of time and frequency domain based methods for radar cross section calculations
TRANSCRIPT
FOI-R–0149–SEJune 2001
ISSN 1650-1942
Scientific report
Division of Aeronautics, FFASE-172 90 STOCKHOLM
Oswald Fogelklou and Jan Nordstrom
Investigation of time and frequencydomain based methods for radar cross
section calculations
FOI – Swedish Defence Research AgencyDivision of Aeronautics, FFASE-172 90 STOCKHOLM
FOI-R–0149–SEJune 2001
ISSN 1650-1942
Scientific report
Division of Aeronautics, FFASE-172 90 STOCKHOLM
Oswald Fogelklou and Jan Nordstrom
Investigation of time and frequencydomain based methods for radar cross
section calculations
FOI-R–0149–SE
2
FOI-R–0149–SE
Contents1 Introduction 7
2 Maxwell’s equations and the wave equation 9
3 An energy estimate for the continuous problem 11
4 Discretization of the wave equation 13
5 Stability analysis 155.1 Von Neumann analysis . . . . . . . . . . . . . . . . . . 155.2 GKS analysis . . . . . . . . . . . . . . . . . . . . . . . 15
6 The periodic case in time domain 19
7 The pulse in time domain 23
8 The Helmholtz equation 27
9 Efficiency 299.1 The periodic case in time domain . . . . . . . . . . . . 299.2 The pulse in time domain . . . . . . . . . . . . . . . . 299.3 The Helmholtz equation . . . . . . . . . . . . . . . . . 299.4 The efficiency in 2D and 3D . . . . . . . . . . . . . . . 309.5 Multiple frequencies . . . . . . . . . . . . . . . . . . . 31
10 Conclusions 33
References 35
Document information 37
Dokument information 39
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AbstractMost of the radar cross section (RCS) calculations are done in frequencydomain. In this report we investigate if the more general time domainsolvers can be used with reasonable efficiency for RCS calculations. Thewave equation is solved in a 1D domain. At one boundary the in-comingwave is known and at the other, the wave is reflected.
We investigate three different cases. In the first case a periodic waveis sent from one boundary and reflected at the other, in the second casea pulse is sent and in the third case, the Helmholtz equation, correspond-ing to the first case is solved. The orders of accuracy are calculated andthe efficiencies are estimated in all the three cases. More efficient algo-rithms, which take advantage of the precise location of the signal, are alsoconstructed.
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1 IntroductionRoughly speaking, the radar cross section is the size of the particular objectseen from the radar. The radar cross section (RCS) is defined as [1]
σ = limr→∞
4πr2
∣
∣
∣
∣
Ps
Pi
∣
∣
∣
∣
,
wherePs is the power scattered from the object per unit area andPi is thepower from the sender per unit area. SinceP is proportional toE2 or H2,the radar cross section can also be expressed as
σ = limr→∞
4πr2
∣
∣
∣
∣
Es
Ei
∣
∣
∣
∣
2
= limr→∞
4πr2
∣
∣
∣
∣
Hs
Hi
∣
∣
∣
∣
2
. (1)
σ can be calculated with several methods. In MOM (method of moments)the scattered field is calculated with the integral
Es =
∮
A
f(K)dA,
where the integral is taken over the area of the scattering object andf is aknown function of the surface current,K. K is calculated by writing it asa linear combination of surface currents with the coefficients determinedfrom Ei.
Other methods are geometrical optics (GO) and physical optics (PO),which are high frequency approximations. In GO a bundle of parallel raysare studied. The power is always constant but the power perm2 varies,since the cross section area of the ray bundle varies. In PO the scatteredfield is calculated with the integral
Es =
∮
A
fA(E,H)dA.
E andH are calculated with G0 and shadowed regions of the surface areneglected in the integral.
There are advantages and disadvantages related to solving equations intime and in frequency domain respectively. In frequency domain the num-ber of arithmetic operations per frequency is relatively small. On the otherhand one equation per frequency has to be solved. In time domain moregeneral problems can be solved but on the other hand, more arithmetic op-erations per frequency are necessary. For the reasons mentioned above, itis interesting to know how efficiently one can compute periodic solutionsin time domain compared to frequency domain. In this thesis we will tryto answer that question.
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2 Maxwell’s equations and the wave equationMaxwell’s equations are
∇ • E =ρ
ε0, ∇ • B = 0,
∇× E = −∂B
∂tand ∇×B = µ0J + ε0µ0
∂E
∂t.
With no variation in y and z, a linear, isotropic and homogeneous medium,
ρ = 0 andJ = 0 we have{
∂Ez
∂x= µ
∂Hy
∂t∂Hy
∂x = ε∂Ez
∂t
{
∂Ey
∂x= −µ∂Hz
∂t
−∂Hz
∂x = ε∂Ey
∂t
andEx andHx are constant. The componentsEy, Ez, Hy andHz satisfythe wave equation. This can be seen by differentiating once more andobserving thatc = 1√
µ0ε0.
In this paper we will consider
∂2E
∂t2= c2 ∂2E
∂x20 ≤ x ≤ 1 t ≥ 0 (2)
E(x, 0) = f(x)∂E
∂t(x, 0) = h(x) (3)
∂E
∂t(0, t)− c
∂E
∂x(0, t) = g(t) (4)
E(1, t) = 0, (5)
whereE can beEy or Ez. The characteristic boundary condition (4) saysthat the in-going wave is known atx = 0. The boundary condition (5) saysthat there is a perfectly electric conductor (P.E.C) atx = 1. The problem(2)-(5) is a very simple model of a radar reflection scenario. The in-goingwave is sent fromx = 0 and is perfectly reflected atx = 1. It follows thatσ = Ps
Pi= 1 at x = 1. At the other boundaryx = 0 however, nothing is
reflected, which means thatσ = 0.
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3 An energy estimate for the continuous problemMultiplication of (2) by2Et and integration leads to
∫ 1
0
2EtEttdx =
∫ 1
0
2Etc2Exxdx.
Integration by parts leads to
d
dt
(
||Et||2 + c2||Ex||2)
= 2c2[EtEx]10, (6)
where
||u||2 =
∫ 1
0
u2dx.
SinceEt(1, t) = E(1, t) = 0 we only get a boundary contribution from thelower limit. After inserting (4) in (6) we get
d
dt
(
||Et||2 + c2||Ex||2)
= −2c2(
cEx(0, t)2 − g(t)Ex(0, t)
)
= −2c2
(
(√cEx(0, t)−
g(t)
2√
c
)2
− g(t)2
4c
)
≤ cg(t)2
2.
So(||Et||2 + c2||Ex||2)t is bounded by the boundary data and we have anenergy estimate.
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4 Discretization of the wave equationA second order accurate discretization of (2) is
En+1j − 2En
j + En−1j
(∆t)2= c2
Enj+1 − 2En
j + Enj−1
(∆x)2. (7)
A second order accurate approximation requires second order boundaryconditions. One such discretization could be derived from the equations
En+10 − En−1
0
2∆t− c
En1 − En
−1
2∆x= g(tn) (8)
and
En+10 − 2En
0 + En−10
(∆t)2= c2En
1 − 2En0 + En
−1
(∆x)2, (9)
where (8) is the leap frog approximation at the boundary. In (9) we haveassumed that the wave equation is valid in a neighborhood of the boundary.En
−1 is the electric field just outside the boundary. After eliminatingEn−1
from (8) and (9) we get
En+10 =
2(1− λ2)En0 + 2λ2En
1 + (λ − 1)En−10 + 2(∆x)λ2g(tn)
c
λ + 1, (10)
where we have introducedλ = c∆t∆x
.A second order accurate time discretization of (3) is obtained by the
use of the equations
E1j − E−1
j
2∆t= h(xj)
and
E1j − 2E0
j + E−1j
(∆t)2= c2
E0j+1 − 2E0
j + E0j−1
(∆x)2.
The first equation corresponds to the second initial condition and the sec-ond equation is the wave equation att = 0. Together withE0
j = f(xj) weget by eliminatingE−1
j
E1j = h(xj)∆t + f(xj) +
λ2(f(xj+1)− 2f(xj) + f(xj−1))
2, (11)
which is valid forj = 0, . . . , J − 1, i.e. every point in the domain exceptthe boundary pointx = 1. Note that (11) requires knowledge of the initialdata atx−1. At the boundaryx = 1 we haveE = 0, i.e. no informationfrom outside the boundary is required.
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5 Stability analysis
5.1 Von Neumann analysis
The von Neumann ansatzEnj = Enei
ωxjc = Enei
jω∆xc in (7) yields
En+1 − 2
(
1− 2 sin2
(
ω∆x
2c
)
λ2
)
En + En−1 = 0, (12)
whereλ = c∆t/∆x. The two-step method (12) is described by the system
(
En+1
En
)
=
(
2(
1− 2 sin2(
ω∆x2c
)
λ2)
−11 0
) (
En
En−1
)
. (13)
The system (13) is stable if and only if the spectral radius of the systemmatrix is at most one.
The characteristic polynomial is
β2 − 2
(
1− 2λ2 sin2
(
ω∆x
2c
))
β + 1 = 0, (14)
which means that
β1,2 = α ±√
α2 − 1, where α = 1− 2λ2 sin2
(
ω∆x
2c
)
.
If α2 = 1, (14) has the double rootβ1,2 = 1 or the double rootβ1,2 = −1.If α2 < 1, (14) has the solutionβ1,2 = α ± i
√1− α2. It follows that
|β1,2|2 = α2 + (1− α2) = 1. HenceE is bounded in both these cases. Forα2 > 1, we getα < −1 or α > 1. The only possible solution is
α < −1 ⇒ |β2| = |α −√
α2 − 1| = −α +√
α2 − 1 > −α > 1.
Obviously a stability condition is
1− 2λ2 sin2
(
ω∆x
2c
)
= α ≥ −1, or λ ≤ 1∣
∣sin(
ω∆x2c
)∣
∣
.
This inequality should be satisfied by any frequency. In particular it shouldbe satisfied by a frequency such thatsin(ω∆x
2c) = ±1, soλ ≤ 1.
5.2 GKS analysisNext we consider the boundary conditions (10) andEn
J = 0. The ansatzEnj = znEj in (7) leads to
Ej(z − 1)2 = λ2z(Ej+1 − 2Ej + Ej−1). (15)
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This is the resolvent equation, which has the general solution
Ej = σ1κj1 + σ2κ
j2. (16)
The constantsσ1 andσ2 are determined by the boundary conditions. TheansatzEj = κj in (15) leads to
κ(z − 1)2 = λ2z(κ − 1)2, (17)
which is called the characteristic equation.κ1 andκ2 are the solutions ofit and |κ1| < 1 and|κ2| > 1 for |z| > 1. This follows from the followingreason. Ifκ would beei
jωh
c , we would get the von Neumann analysis case.But there are no solutions which satisfy|z| > 1, because such solutionsare unstable according to the von Neumann analysis.
The characteristic equation (17) can be rewritten as
κ2 −(
2 +(z − 1)2
λ2z
)
κ + 1 = 0. (18)
Therefore the roots satisfyκ1κ2 = 1. Since|κ| �= 1, we have|κ1| < 1 and|κ2| > 1. The solution of (18) is
κ = 1 +(z − 1)2
2λ2z±
√
(
1 +(z − 1)2
2λ2z
)2
− 1. (19)
We have a double root if
1 +(z − 1)2
2λ2z= ±1, (20)
which means thatz = 1 or z = 1−2λ2±√
(1− 2λ2)2 − 1. In the first caseκ = 1 and in the secondκ = −1. If there is a double root, the resolventequation (15) has the general solution
Ej = (σ1 + σ2j)κj (21)
instead of the solution in (16).Let us now show stability on0 ≤ x ≤ 1 by showing stability onx ≥ 0
andx ≤ 1. We have onx ≥ 0
limj→∞
|Ej | < ∞ (22)
as an extra boundary condition. It follows thatσ2 = 0 because|κ2| > 1.Enj = σ1z
nκj1 in the homogeneous version of the boundary condition (10)becomes
σ1((z − 1)2 + λ(z2 − 1)− 2λ2z(κ − 1)) = 0. (23)
The relation (23) is called the determinant condition onx ≥ 0. We havean instability if there exist solutions of (17) and the determinant condition
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with σ1 �= 0 with |κ| < 1 and |z| > 1. From the determinant condition(23) we get after division byσ1
κ = 1 +(z − 1)2 + λ(z2 − 1)
2λ2z. (24)
If we insert (24) in the characteristic equation (17) we get
(λ2 − 1)(z − 1)2 = 0. (25)
It follows for λ < 1 that there are no other roots thanz = 1 Suppose wehave solutions close toz = 1 andκ = 1. Setz = 1 + reiθ andκ = 1 + εin (23) and divide byσ1. Then we get, see Fig. 1,
r2ei2θ + λ(r2ei2θ + 2reiθ)− 2λ2(1 + reiθ)ε = 0.
If we neglect higher order terms we getε = reiθ/λ. This means that|z| > 1 ⇔ |κ| > 1 so the boundary condition (10) withg ≡ 0 does nothave a generalized eigenvalue close toκ = 1.
Consider the caseκ = −1. Equation (22) implies thatσ2 = 0 in (21).Enj = σ1(−1)j in the boundary condition (10) withg ≡ 0 becomes
σ1zn+1 = σ1
(
2(1− λ2)zn − 2λ2zn + (λ − 1)zn−1
λ + 1
)
. (26)
If we divide (26) byσ1zn−1 we get
z =1− 2λ2
λ + 1±
√
(
1− 2λ2
λ + 1
)2
+λ − 1
λ + 1
The solutions above coincide with the solutionsz = 1−2λ2±√
(1− 2λ2)2 − 1only in the caseλ = 0. Summing up, we can say that the right half-planeproblem with the boundary condition (10) is stable.
We have onx ≤ 1
limj→−∞
|Ej | < ∞
as an extra boundary condition. It follows thatσ1 = 0 because|κ1| < 1.Enj = σ2z
nκj2 in the boundary conditionEJ = 0 becomesσ2κJ2 zn = 0,
which is the determinant condition onx ≤ 1. Here we have instabilityif there exist solutions of (17) and the determinant condition withσ2 �= 0with |κ| > 1 and|z| > 1, which is impossible. With a double root ansatzwe get the same expression. Hence we have stability for the boundaryconditionEJ = 0 and therefore for the whole problem.
Finally let us verify that the double root solutions satisfy (15) forκ =−1, z = 1−2λ2 ±
√
(1− 2λ2)2 − 1 andκ = 1, z = 1. After inserting theansatzEj = (σ1 + σ2j)κ
j into (15) we get
κ(z2(σ1 + σ2j)− 2z(σ1 + σ2j) + (σ1 + σ2j))
= λ2z(κ2(σ1 + σ2(j + 1))− 2(σ1 + σ2j)κ + (σ1 + σ2(j − 1)))
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FOI-R–0149–SE
Figure 1. z = 1 + δ = 1 + reiθ
where 0 < r ≪ 1. For such val-ues of r |z| > 1 ⇔ −π
2< θ <
π2
.
−1 −0.5 0 0.5 1 1.5−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
r
1
The coefficients in front ofσ1 andσ2 become zero if{
κ(z − 1)2 = λ2z(κ − 1)2
jκ(z − 1)2 = jλ2z(κ − 1)2 + λ2z(κ2 − 1).
If we setκ = ±1 the equations become equivalent and equal to (17), whichimplies thatEj = (σ1 + σ2j)κ
j really satisfies (15).
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Figure 2. E(x, t) is plotted ver-sus x at time t = 0.237. Thesupport of E(x, t) is now ap-proximately [0,474]. c = 2,ω0 = 40, ∆t = 0.001 and ∆x =
0.005.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.2
0
0.2
0.4
0.6
0.8
10.237
6 The periodic case in time domainConsider (2)-(5) withg(t) = ω0 sin(ω0t), E(x, 0) = 0 and ∂E
∂t(x, 0) = 0.
The homogeneous initial conditions imply that solution for the wave equa-tion can be computed efficiently in time domain. At timet < 1
c, E(x, t)
has only support on(0, ct), so it is not necessary to calculateE(x, t) on(ct, 1), which is shown in Fig. 2. After the timet = τ ≥ 2
call transients
have left the domain and E is periodic in both time and space. The Fourierseries can then be calculated by evaluating the integral
Em(x) =1
T
∫ τ+T
τ
E(x, t)e−imωtdt.
using Simpson’s rule.The order of accuracy p in time domain is calculated point-wise with
the formula
E1 − E
E2 − E= 2p. (27)
In (27),E is the exact solution,E1 is the numerical solution with a partic-ular resolution andE2 is the numerical solution with twice that resolutionin time and space.2p is calculated numerically and the result is shown inFig. 3 and 4. In the ordinary method, the algorithm does not exploit thefact thatE(x, t) has support on(0, ct) only. In the efficient method, thealgorithm exploits that fact. The exact solution is (after the transients haveleft the domain)
E(x, t) =cos(ω0(t +
x−2c
))− cos(ω0(t − xc))
2.
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FOI-R–0149–SE
Figure 3. The ratio r = 2p
for the periodic case in time do-main at time t = 3. ω0 =
5π, c = 1, ∆x = 0.01 and∆t = 0.001. Discontinuities ap-pear where the error almost van-ishes. The ordinary method isused.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90
1
2
3
4
5
6
7
8
x
ratio
time=3
Fig. 3 and Fig. 4 show that the order of accuracy is approximately 2.To calculate the order of accuracy for the Fourier series coefficients
Em, we use thel2 norm.
||error|| ≡ ||Em,∆x − Em|| =
√
√
√
√
J∑
j=0
N∑
n=0
|Em,∆x − Em|2∆x,
where the exact Fourier series components are
Em(x) =
{
(eimω0(x−2)
c − e−imω0x
c )/4 m = −1, 10 else.
The procedure is repeated for other values of∆x but with the same∆t and||error|| is plotted versus∆x. The plots are shown in Fig. 5 and Fig. 6.The order of accuracy forE1 (Fig. 5) is 2.0103 and the order of accuracyfor E2 (Fig. 6) is 2.2062.
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Figure 4. The ratio r = 2p
for the periodic case in time do-main at time t = 3. ω0 =
5π, c = 1, ∆x = 0.01 and∆t = 0.001. Discontinuities ap-pear where the error almost van-ishes. The efficient method isused.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90
1
2
3
4
5
6
7
8
x
ratio
time=3
Figure 5. The error in l2 normfor E1 (the first component of theFourier series of E). ∆t=0.0001,∆x=0.02, 0.01, 0.005, 0.0025,0.00125, ω0 = 10π and c = 1.||error|| is on the y-axis and ∆x
is on the x-axis.
10−3
10−2
10−1
10−4
10−3
10−2
10−1
100
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Figure 6. The error in l2
norm for E2 (the second com-ponent of the Fourier series ofE). ∆t=0.0001, ∆x=0.04, 0.02,0.01, 0.005, 0.0025, 0.00125,ω0 = 10π and c = 1. ||error||is on the y-axis and ∆x is on thex-axis.
10−3
10−2
10−1
10−5
10−4
10−3
10−2
10−1
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Figure 7. A comparison be-tween the ordinary and the ef-ficient way to solve the waveequation for a pulse. c=1,h=0.005, k=0.001, τ = 0.1 andt = 1.564. x is on the x-axisand E(x, t) is on the y-axis. Thedotted curve shows the efficientmethod. The support of E(x, t)
is now approximately 0.436 ≤x ≤ 0.636.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
0.51.564
7 The pulse in time domainNext we consider (2)-(5) with
g(t) =
{
6(t−τ)((t−τ)2−τ2)2
τ6 0 ≤ t ≤ 2τ0 else
andE(x, 0) = ∂E∂t
(x, 0) = 0. Note thatgǫC1. The support ofE(x, t)moves with speed c to the right, is reflected atx = 1 and then moveswith speed c to the left. If we use this information and update the solutiononly whereE(x, t) is non-zero, an efficient algorithm (called the efficientmethod) is obtained. If we update the solution on the whole interval0 ≤x ≤ 1, we call the method, the ordinary method.
The efficient and the ordinary method are compared in Fig. 7. To verifythe order of accuracy, we calculate the error in thel2 norm
||error|| ≡ ||E(x, t)∆x,∆t − E(x, t)∆x2,∆t
2||
=
√
√
√
√
J∑
j=0
N∑
n=0
(E(xj , tn)∆x,∆t − E(xj , tn)∆x2,∆t
2)2∆x∆t,
for particular values of∆x and∆t. The procedure is repeated for2n frac-tions of∆x and∆t and the result is shown in Fig. 8. The order of accuracyis 1.9555 for the ordinary method and 1.7595 for the efficient method.
To estimate the order of accuracy in frequency domain, we calculatethe Fourier transform as
E(x, ω) =
∫ +∞
−∞e−iωtE(x, t)dt ≈
∫ 2c+2τ
0
e−iωtE(x, t)dt.
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Figure 8. ∆x = ∆t =0.02,0.01, 0.005, 0.0025, 0.00125,0.000625 at the x-axis and||E(x, t)∆x,∆t−E(x, t)∆x
2, ∆t
2
||at the y-axis. c = 0.5 and τ =
0.2. The dotted curve shows theefficient method.
10−4
10−3
10−2
10−1
10−5
10−4
10−3
10−2
10−1
Note that the pulse enters the region atx = 0 whent = 0 is reflected atx =1 and has almost vanished in the whole region aftert = 2
c+2τ . The integral
is evaluated with Simpson’s rule directly. This procedure is referred to asthe ordinary method. Since the support ofE(x, t) is moving (see Fig. 7)the Fourier transform can also be efficiently calculated ifE(x, t) is onlyintegrated where it is non-zero. The trapezoidal rule is used when thereis an odd number of subintervals, otherwise Simpson’s rule is used. Thisprocedure is called the efficient method.
To get the exact solution we Fourier transform (2) and get
−ω2E(x, ω) = c2 ∂2E
∂x2(x, ω).
The general solution is
E(x, ω) = A(ω)eiωxc + B(ω)e−i
ωxc . (28)
To determineA andB we take the Fourier transforms of (4) and (5) andget
iωE(0, ω)− c∂E
∂x(0, ω) = g(ω), E(1, ω) = 0. (29)
If we insert (28) in (29) we get
E(x, ω) =g(ω)
ωeiωc
sin
(
ω(1− x)
c
)
. (30)
SoE is on the form
E(x, ω) = f(x, ω)g(ω) = |f(x, ω)|ei arg(f(x,ω))g(ω),
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Figure 9. A comparisonbetween the pulse and theperiodic case. The graphsfrom above are E(x, t) =
ω0|f(x, ω0)| sin(ω0t +
arg(f(x, ω0)), E(x, t) wheng(t) = ω0 sin(ω0t) and thedifference between them attime 1.68. E(x, t) is on they-axes. ∆x = 0.01, ∆t = 0.005,τ = 0.1 and ω0 = 10π.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1
−0.5
0
0.5
11.68
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1
−0.5
0
0.5
11.68
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1
−0.5
0
0.51.68
wheref(x, ω) is the transfer function. Supposeg(t) = ω0 sin(ω0t). Then
g(ω) = iω0π(δ(ω + ω0)− δ(ω − ω0))
and consequently
E(x, ω) = iω0π(δ(ω + ω0)− δ(ω − ω0))f(x, ω).
The inverse Fourier transform ofE(x, ω) is
E(x, t) = iω0πf(x,−ω0)e
−iω0t − f(x, ω0)eiω0t
2. (31)
Now f(x, ω0) = f(x,−ω0) implies arg(f(x,−ω0)) = − arg(f(x, ω0))andf(x, ω0)| = |f(x,−ω0)|. If we use this fact in (31) we get
E(x, t) = ω0|f(x, ω0)| sin(ω0t + arg(f(x, ω0))). (32)
It is therefore possible to calculateE(x, t) for a periodicg(t) from E(x, t)obtained wheng(t) is a pulse. One solves the wave equation for the pulse,take the Fourier transform of the solution and take the Fourier transformof g(t). Thenf(x, ω) = E(x, ω)/g(ω) is calculated. To get the solutionof the wave equation in the periodic case just insertf(x, ω) in (32). Thepulse and the periodic case are compared in Fig. 9.
The Fourier transform of the pulse is
g(ω) =96ie−iωt0
(ωt0)6(−(ωt0)
3 cos(ωt0) + 6(ωt0)2 sin(ωt0)
+ 15ωt0 cos(ωt0)− 15 sin(ωt0)). (33)
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FOI-R–0149–SE
Figure 10. r = 2p = E1−E
E2−E
is calculated numerically for thepulse in frequency domain. c =
1, ∆x = 0.01, ∆t = 0.001,ω = 10π and τ = 0.1. Theefficient method is used for thewave equation as well as for theFourier transform.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90
2
4
6
8angular frequency=31.4159
Re(
ratio
)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9−4
−2
0
2
4
x
Im(r
atio
)
The order of accuracy p is calculated point-wise using the formula
r = 2p =E1 − E
E2 − E,
whereE is the exact solution, obtained from inserting (33) in (30),E1 isthe numerical solution with a particular resolution andE2 is the numericalsolution with twice that resolution in space. The time step is chosen, sothat the time error is negligible.r is calculated numerically in Fig. 10.
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FOI-R–0149–SE
8 The Helmholtz equationConsider the problem (2)-(5) withg(t) = ω0 sin(ω0t) in (4) but disregardthe initial conditions. The solution ast → ∞ becomes periodic. Theansatz
E(x, t) =∞
∑
n=−∞En(x)e
inω0t
inserted in (2), (4) and (5) leads to
E ′′n + k2
nEn = 0, wherekn =nω0
c, (34)
{
inω0En(0)− cE ′n(0) = − inω0
2if n=1 orn=-1
En(0) = 0 else(35)
and
En(1) = 0. (36)
ObviouslyEn(x) ≡ 0 unlessn = 1 or n = −1, so let us assume thatn = 1or n = −1 from now on. Letu be the discretization ofEn.
The discretization of (34) and (36) are
uj+1 − 2uj + uj−1
(∆x)2+
ω20
c2uj = 0 jǫ{1, . . . , J − 1}, uJ = 0. (37)
To obtain a discretization of (35) we use the equations
inω0u0 − cu1 − u−1
2∆x= −iω0
n
2, (38)
u1 − 2u0 + u−1
(∆x)2+
ω20
c2u0 = 0, (39)
where (38) is a central discretization of (35) and (39) is (37) withj = 0.After eliminatingu−1 from (38) and (39) we get together with (37).
A(u)n=1 = b andA(u)n=−1 = b, (40)
where
A =
r2− iω0∆x
c1 0 . . . . . . 0
1 r 1. . .
...
0. . . . . . . . . . . .
......
. . . . . . . . . . . . 0...
. . . 1 r 10 . . . . . . 0 1 r
, b =
iω0∆x2c
0...0
and
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FOI-R–0149–SE
Figure 11. The real and imagi-nary part of the ratio r = E1−E
E2−E.
If the order of accuracy is 2, r =
4, which you see in the figure.c = 1, ∆x = 0.01 and ω0 =
10π.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90
2
4
6
8angular frequency=31.4159
Re(
ratio
)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9−4
−2
0
2
4
x
Im(r
atio
)
r =(ω0∆x)2
c2− 2.
It follows from (40) thatun=−1 = (un=1) so it suffices to solve the firstequation in (40). Let us therefore only consider the casen = 1. SinceA is symmetric and tridiagonal, a direct Gauss’ elimination method (LU-decomposition) becomes very easy.
The order of accuracy p is calculated point-wise with the formula
r = 2p =u1 − E
u2 − E,
whereE is the exact solution,u1 is the numerical solution with a particularresolution andu2 is the numerical solution with twice that resolution. Theresult is shown in Fig. 11.
After solving (40) with LU-decomposition we get the final solution as
uj(t) = eiω0tuj + e−iω0tuj = 2Re(eiω0tuj).
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FOI-R–0149–SE
9 EfficiencyIn this section we will consider the efficiency of the previously describedmethods. In the efficiency test below we usec = 1, τ = 0.1, ∆x = ∆t =0.01, ω = ω0 = 10π andT = 2π
ω0= 0.2.
9.1 The periodic case in time domainIn the periodic case we solve the wave equation untilt = 2
c+ T − ∆t =
2.19. The result 286680 flops for the ordinary method and 225642 flopsfor the efficient method is obtained. For the transformation to frequencydomain we can choose the trapezoidal rule or Simpson’s rule. We get44903 flops for Simpson’s rule and 42738 flops for the trapezoidal rule.However the transformation to frequency domain is unnecessary, since (1)is independent of the phase and frequency ofEi andEs. Therefore onlyabout 225000 flops are required.
9.2 The pulse in time domainFor the pulse we solve the wave equation untilt = 2
c+ 2τ = 2.2. This
requires 287817 flops for the ordinary method and 56462 flops for the effi-cient. For the Fourier transforms we need 449512 flops and 447609 flops ifwe integrate over the whole domain (use the ordinary method) with Simp-son’s rule and the trapezoidal rule respectively. The efficient Fourier trans-forms are calculated with the trapezoidal rule and a method which usesthe Simpson’s rule if possible and otherwise the trapezoidal rule. In thefirst case we have 92983 flops and in the second case we have 94797 flops.However, we only have to transform the part of the domain where the re-flected signal is received i.e. atx = 0. The number of flops for the Fouriertransforms is therefore about 1/100 of the number of flops above. So in themost efficient case we have totally about 57000 flops.
9.3 The Helmholtz equationFor the Helmholtz equation we have a tridiagonal symmetric system. If wecalculate it with a general tridiagonal symmetric system algorithm, we get14425 flops. If we use all possible information, (basicly that only the firstand last row differ from the other rows), we get 2632 flops. The most natu-ral choice is here144425 ≈ 15000 flops, because we normally solve moregeneral problems and are not able to use all information for this particularproblem. The relationship between the flops for the Helmholtz equationand the flops in the periodic case is≈ 15000
225000≈ 7% and the relationship
between the flops for the Helmholtz equation and the flops for the pulse is≈ 14000
56000= 25%.
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FOI-R–0149–SE
9.4 The efficiency in 2D and 3DA discretization of the Helmholtz equation in 2D is
El+1,m − 2El,m + El−1,m
(∆x)2+
El,m+1 − 2El,m + El,m−1
(∆y)2= −ω2
c2El,m.
If we assumel = 1, . . . , L andm = 1, . . . , M we solve the systems ofequationsAE = b. If we neglect boundary conditions the band widthbecomesd = 2min(L, M). The number of arithmetic operations with adirect method is approximatelyMLd2/2 [2], which is 2L4 if we have aquadratic grid. In this caseL = 100 so the number of flops is2 · 108.
What about iterative methods? A typical row in the system matrixlooks like
[1 0 . . . 0 1 − 4 + (ω∆/c)2 1 0 . . . 0 1].
It follows that the absolute value of the diagonal element is always lessthan the sum of the off diagonal elements. Therefore it is not clear thatwe can use an iterative method like Gauss-Seidel or Jacobi. The conju-gate gradient method cannot be used directly either, since the matrix is notpositive definite.
If we discretize the wave equation in 2D we get
En+1j,k − 2En
j,k + En−1j,k
(c∆t)2
=Enj+1,k − 2En
j,k + Enj−1,k
(∆x)2+
Enj,k+1 − 2En
j,k + Enj,k+1
(∆y)2.
To solve this equation for a quadratic grid we need 2L=200 times as manyflops as for the 1D periodic case and the 1D pulse. The Fourier transfor-mation must only be done at the boundary points, therefore the number offlops for the Fourier transform can be neglected. The total number of flopsis approximately4.5 · 107 for the periodic case,1.1 · 107 for the pulse and2 · 108 for the Helmholtz equation.
A discretization of the Helmholtz equation in 3D is
El+1,m,n − 2El,m,n + El−1,m,n
(∆x)2+
El,m+1,n − 2El,m,n + El,m−1,n
(∆y)2+
El,m,n+1 − 2El,m,n + El,m,n−1
(∆z)2= −ω2
c2El,m,n.
If we assumel = 1, . . . , L, m = 1, . . . , M andn = 1, . . . , N , we solvethe systems of equationsAE = b. If we neglect boundary conditions, theband width becomesd = 2min(LM, NL, MN). The number of arith-metic operations with a direct method is approximatelyMLNd2/2 [2],which is2L7 = 2 · 1014 for a cubic grid.
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FOI-R–0149–SE
The wave equation in 3D requires3L2=30000 times as many flops asfor the 1D periodic case and 1D pulse for a cubic grid. The total numberof flops is approximately6.8 · 109 in the periodic case and1.7 · 109 for thepulse.
In reality L=100 is normally a too low resolution. A typical radar hasa frequency of 10 GHz which corresponds to a wavelength of 3 cm. Ifwe suppose that an aircraft is 9 m and a minimum of 10 grid points perwavelength is required, we getL=3000 instead. An interested reader caninsert this value in the efficiency calculations.
9.5 Multiple frequenciesConsider the case when we are interested in more than one frequency. Inthe periodic case in time and frequency domain the number of flops areproportional to the number of frequencies. However, for the pulse thenumber of flops for the Fourier transform is proportional to the numberof frequencies but we only need to solve the wave equation once for allfrequencies.
For 10 frequencies in 1D, we need1.5 · 105 flops for Helmholtz equa-tion, 2.25·106 for the periodic case but only 66000 flops for the pulse. In2D and 3D we can neglect the number of flops for the Fourier transforms.It follows that number of flops for the periodic case and Helmholtz equa-tion is multiplied by a factor of 10 but for the pulse the number of flops isalmost equal to the number of flops for one frequency.
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FOI-R–0149–SE
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FOI-R–0149–SE
10 ConclusionsAs expected, the Helmholtz equation requires less work for a single fre-quency in 1D than the other methods if a direct method is used.
On the other hand the Helmholtz equation (using a direct solver) re-quires much more work in two or three dimensions, especially in a com-parison with the pulse if many frequencies are of interest. With a sufficientamount of frequencies, the pulse method is the most efficient even in 1D.
The higher the dimension, the more efficient iterative methods are com-pared to direct methods. But it is a non-trivial problem to find an iterationmatrix for the Helmholtz equation, see for example [2], [3]. If an effi-cient iteration matrix cannot be found, time domain algorithms are morepreferable than frequency domain methods.
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FOI-R–0149–SE
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FOI-R–0149–SE
References[1] Fagerstrom, JanBerakningsmetoder for radarmalarean. Enoversikt,
Forsvarets forskningsanstalt, Sweden, 1997
[2] Johnson, ClaesNumerical solution of partial differential equation bythe finite element method, Studentlitteratur, Sweden, 1987
[3] Otto, Kurt and Larsson, ElisabethIterative solution of the Helmholtzequation by a second order method, Department for Scientific Com-puting, Uppsala University, Sweden, 1999
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FOI-R–0149–SE
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FOI – Swedish Defence Research AgencyDivision of Aeronautics, FFASE-172 90 STOCKHOLM
FOI-R–0149–SE Scientific report
June 2001 E840343
3. Aeronautical Research
6. Electric Warfare
62. Stealth Technology
Oswald Fogelklou and Jan Nordstrom Jan Nordstrom
Torsten BerglindHead, Computational Aerodynamics Department
Jan Nordstrom
Investigation of time and frequency domain based methods for radar cross section calculations
Most of the radar cross section (RCS) calculations are done in frequency domain. In this reportwe investigate if the more general time domain solvers can be used with reasonable efficiency forRCS calculations. The wave equation is solved in a 1D domain. At one boundary the in-comingwave is known and at the other, the wave is reflected.We investigate three different cases. In the first case a periodic wave is sent from one boundaryand reflected at the other, in the second case a pulse is sent and in the third case, the Helmholtzequation, corresponding to the first case is solved. The orders of accuracy are calculated and theefficiencies are estimated in all the three cases. More efficient algorithms, which take advantageof the precise location of the signal, are also constructed.
Finite differences, characteristic boundary conditions, efficiency, accuracy, Fourier transform, Helmholtzequation
ISSN 1650-1942 39 English
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Totalforsvarets Forskningsinstitut – FOIAvdelningen for Flygteknik, FFASE-172 90 STOCKHOLM
FOI-R–0149–SE Vetenskaplig rapport
Juni 2001 E840343
3. Flygteknisk forskning
6. Telekrig
62. Signaturanpassning
Oswald Fogelklou och Jan Nordstrom Jan Nordstrom
Torsten BerglindChef, Institutionen for Berakningsaerodynamik
Jan Nordstrom
Undersokning av tidsdomans- och frekvensdomansbaserade metoder for radarmalareaberakningar
De flesta radarmalareaberakningar gors i frekvensdoman. I den har rapporten under-soker vi om de mer generella tidsdomanlosarna kan anvandas med rimlig effektivitet forradarmalareaberakningar. Vagekvationen loses i en endimensionell doman. Vid en rand arden inkommande vagen kand och vid den andra reflekteras den.Vi undersoker tre olika fall. I det forsta fallet sands en periodisk vag fran en rand ochreflekteras vid den andra, i det andra fallet sands en puls och i tredje fallet loses denHelmholtzekvation som motsvarar det forsta fallet. Noggrannhetsordningarna beraknas ocheffektiviteterna uppskattas i alla tre fallen. Effektivare metoder, som utnyttjar signalenslokalisering, konstrueras ocksa.
Finita differenser, karaktaristiska randvillkor, effektivitet, noggrannhet, Fouriertransform, Helmholtz ek-vation
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