A Frequency Domain Analysis of Electromagnetic Fields
of Printed Circuit Boards
Akiko Kusaka†, Takaaki Kinouchi†, Yoshihiro Yamagami†, Yoshifumi Nishio†, Akio Ushida‡†Department of Electrical and Electronic Engineering,
Tokushima University, Tokushima, 770-8506 JAPAN
Email: {kusaka, t-kino, yamagami, nishio}@ee.tokushima-u.ac.jp‡Department of Mechanical and Electronic Engineering,
Tokushima Bunri University, Kagawa, 769-2193 JAPAN
Email: ushida@fe.bunri-u.ac.jp

Abstract
For designing PCBs (printed circuit boards), it is very
important to find out the locations gradating strong EMF(electromagnetic fields), where the voltages will have large
peak value. Although they may be found by AC analysis of
Spice, it is diﬃcult to find the resonant points and values if
the quality factor Q is too large. Because we may pass overthem when we choose a large step size. We need to find out
both the exact peak voltages and locations on the PCB.
In this study, we propose an eﬃcient Spice-oriented frequency domain peak detection algorithm, which is basedon the HB (harmonic balance) method using Sine-Cosine
circuit, where the frequency response curve is traced by
the variable step size transient analysis of Spice.

L
r
j(t)

C

r
g

L

Figure 1: LRCG plane model of PCB.

1. Introduction
Nowadays, sizes of LSIs and PCBs become smaller andsmaller, and the operating frequency higher and higher. In
these cases, PCBs are modeled by linear LRCG large scaleplane circuits as shown in Fig. 1. They can be solved bytime-domain and/or frequency domain [1][2] and we concentrate on the latter technique in this study. In this case,the frequency response curve may have many sharp resonant points for high Q circuits. It is diﬃcult to find theexact peak points.
The resonant and anti-resonant points appear at poles
and zeros of driving and transfer functions, respectively.
We can not find out them with analytical method when thecircuit size becomes larger. They are also found by tracing
the frequency response curve |V(ω)|, where the resonant
and anti-resonant points correspond to the highest and lowest values. In this case, when we use AC analysis of Spicesimulator, we may pass over to find out all the resonant
points especially for the high Q circuits. Furthermore, it is
impossible to find out the exact resonant voltages. Therefore, we need to propose a new algorithm based on the HB
method where the determining equation is formulated by
the coupled Sine-Cosine circuits [3]-[5]. Both the resonant
and anti-resonant points correspond to the solutions satisfying f 0 (ω) = 0 on the frequency response curve, so that

we use a diﬀerentiator and a nonlinear limiter to find thef 0 (ω) = 0 points. The circuit is solved by the variable step
size transient analysis of Spice1 . Thus, we can develop an
eﬃcient simulator to trace the frequency response curves,and to find the exact peak voltages.

2. Reactance circuit
Now, we review the circuit properties of a general linear
LRC circuit as shown Fig. 1. The circuit equation is given
by

 


 Y11 Y12 ... Y1n   V1   J 

 

 Y
 21 Y22 ... Y2n   V2   0 
(1)
 ..
..
..   ..  =  .. .
.   .   .  .
0Vn
Yn1 Yn2 ... Ynn
where Vs are nodal voltages for the input J 2 , and the mutual
admittance is given by
Yi j = jωCi j + Gi j +1jωLi j(2)

1 Transient analysis of Spice is performed using the variable step size

algorithm such that the step size becomes very small around rapidlychanging regions.
2 Although the forced input can be chosen at any nodes, we choose thenode 1 without loss of generality.

- 114 -

Then, we haveThen, the current iC is given by

∆1k J
Vk =
∆

where
∆=

Y11
Y21
..
.

Y12
Y22
..
.

Yn1

Yn2

...
...
...

(3)

iC = CThus, the coeﬃcients of sin ωt, cos ωt are described by

Y1n
Y2n
..
.

ILS = −ωCVLC , ILC = ωCVLS

(4)

Ynn

Now, we consider the case r = g = 0 in Fig. 1 which corresponds to the reactance circuit. It is known that the voltage(3) can be written byn a0k ∑
ja2i,k ω+Vk (ω) = + ja∞,k ω J(5)22jω i=1 ω2i − ωObserve that the resonant points correspond to the poles ofboth the driving and the transfer functions. They arise at
the same frequencies. They also satisfy
0 < ω1 < ω2 < ω3 < . . . < ∞Node voltage
Sine-Cosine
Circuitω

Differentiator Nonlinear limitter

Figure 2: Peak detector of frequency response curve.
Since ωs at the peak voltages satisfy

3. Peak points tracing algorithm

ω = Kt(7)

Now, let us discuss the Sine-Cosine circuit [3][4] corresponding to the determining equation of the HB method.
Let the current through an inductance L be
(8.1)

Then, the voltage vL is given by
vL = LdiL= − ωLILC sin ωt + ωLILS cos ωt
dt

(8.2)

Thus, the coeﬃcients of sin ωt, cos ωt are described by
VLS = −ωLILC , VLC = ωLILS(8, 3)

Namely, the inductance is replaced by coupled currentcontrolled voltage sources in the Sine-Cosine transformation of the HB method. In the same way, let the voltage
across a capacitor C be
vC = VCS sin ωt + VCC cos ωtE(6)

Although, for relative low Q circuits, the AC analysis of
Spice and a cubic spline combining Newton method [8] canbe usefully applied to find out the peak points, it may passover the points for high Q circuits. Thus, we propose a newalgorithm based on the HB method such that the determining equation is solved by transient analysis of Spice, whereω is a function of time t as follows:(9.1)(9, 3)

Namely, the capacitor is replaced by the coupled voltagecontrolled current sources in the Sine-Cosine transformation. The circuit corresponding to the determining equationof the HB method is driven by a constant voltage E = J andω is given by (7).

Note that r and g in the high frequency domain are small
compared to the reactance values, so that the frequency response behaves like as that of reactance circuits, and it has
sharp resonant points.

iL = ILS sin ωt + ILC cos ωtdvC= − ωCVLC sin ωt + ωCVLS cos ωt (9.2)
dt

d|Vk (ω)|= 0, k = 1, 2, . . . n,
dω

(10)

on the response curve, we need to find the zero points satisfying (10)3 . Hence, |Vk (ω)| need to be firstly diﬀerentiatedby a diﬀerentiator. In order to detect the exact peak points,
the output is limited and expanded with a nonlinear limiteras shown in Fig. 2, which consists of a limiter and nonlinear
diodes. The output of the limiter is given by−Vmax : for vin < −VLkv: for − VL ≥ vin ≥ VL
(11.1)
vL = 
in V: for v > −V
max

in

L

where
kVL = Vmax

(11.2)

In our example of section 4, we set VL = 2[µV], Vmax =100[V], and k = 0.5 × 108 . The outputs of diodes are{
I s exp(λvo ):for vo > 0
io =(11.3)−I s exp(−λvo ) :for vo < 0for I s = 10−12 and λ = 40.
The characteristic of the nonlinear limiter is shown in
Fig. 3. This means that the regions around d|Vk (ω)|/dω = 0
are largely expanded. Furthermore, the characteristic haslarge nonlinearity around the zeros. Thus, the transient
analysis around the zero points is executed with a verysmall step size, and we can find out precise peak points.
3 Note that although the peak voltages are diﬀerent, the resonant frequencies arise at the same points. So, the peak detector should be attached
to arbitrary selected one of the nodes in Fig. 1.

- 115 -

4
[kA]

where

Nonlinear limitter

un = −

2

Lg + rC,
2LC√
4LC(gr + (nπ/d)2 ) − (Lg + rC)2
.
vn =
2LC

0

-2

Observe that the transmission line has an infinite number ofpoles, where the frequency response curve has peak points.

-4
-4

-2

0
Input

2

[µV] 4

Figure 3: Input and output characteristic of nonlinear limiter.

4.2 Discrete model of transmission line
Transmission line is usually modeled by discrete RLCG
ladder circuit as shown in Fig.4, where we neglected g.

R

L

R

L

L

R

L

R

4. Illustrative example
4.1 Theoretical analysis of transmission line
We consider a single transmission line of the length d[mm],whose parameters are

E
ω

C

C

C

Figure 4: LRC ladder circuit C = 1[nF], L = 10[mH],e(t) = E cos ωt and E = 1[µV].

r[Ω/mm], L[H/mm], g[S /mm], C[F/mm].
The circuit equation with a complex frequency s at x[mm]from the near end is given bydV(x, s)= −(r + sL)I(x, s)dx(12.1)

dI(x, s)= −(g + sC)V(x, s)dx(12.2)In order to obtain the frequency response curve, the circuit is transformed by HB method into the Sine-Cosine circuit. The peak detector given in section 3 is attached asshown in Fig.5. Note that the detector can be attached to anarbitrary node.

Thus, we haved2 V(x, s)= (r + sL)(g + sC)V(x, s)
dx2

CSine-circuit

R I1S- ωI 1CV R I 2S- ωI 2C
1S(13.1)E

V2S R

I 3S- ωI3C

V3S R

-ω C V2C-ω C V1C

I 4S- ωI4C

V4S

Differentiator
Cd-ω C V4C-ω C V 3C

2

d I(x, s)= (r + sL)(g + sC)I(x, s)(13.2)
dx2
Set the boundary conditions V(0, s), I(0, s) at the near end
and V(d, s), I(d, s) at the far end. Then, we have
[

]
[
]
1  coth γ(s)d − sinh 1γ(s)d  V(0, s)
I(0, s)= V(d, s)I(d, s)Z0 (s) − sinh 1γ(s)d coth γ(s)d
(14)
where
√
Z0 (s) = (r + sL)/(g + sC)√
γ(s) = (r + sL)(g + sC)
Thus, the poles are located at the frequencies satisfying
sinh γ(s)d = 0, namely(r + sL)(g + sC) = ( jnπ/d)2 , n = 1, 2, . . .

(15)rp0 = − , pn = un ± jvn , n = 1, 2, . . .
L(16)Hence,

Rd

Cosine-circuit

R I1C ωI

1S V1CR I 2C ωI

2S V2C R I 3C ωI 3SV

ω CV1Sω CV2S

3C R

I 4C ωI 4S

V4C
2

ω CV3sω CV4S2SQRT(V kS+ VkC)

Nonlinear limitter

Figure 5: Sine-Cosine circuit combining with peak detector. Cd = 1[pF] and Rd = 1[Ω].

Now, we show the simulation results using the transient
analysis of Spice. The circuit has 4 resonant points asshown by Figs. 6(a), (b), and (c) which show the voltages at
the far end of the circuit in Fig. 4, where we set the peaks
1,2,3,4 from the left hand side. The sharpnesses dependon the quality factors so that we have changed them by Rs.
We found that although the resonant frequencies are almost
same for all Rs as shown in Table 1, the peak values arelargely diﬀerent.

- 116 -

case, we need to trace the resonant curve around the peak 4
with a selection of much smaller initial step size.

0.014

0.012

0.01

5. Conclusions and remarks

0.008

0.006

V[V]

In this study, we have proposed an algorithm to find thepeaks of the frequency response curve by combining theSine-Cosine circuit based on the HB method. The circuit
is traced by the transient analysis of Spice. Since the peak
points correspond to the gradient being to zeros, we diﬀerentiate the curve and find its zero points. In order to findthe exact zero points, we have developed the peak detector using nonlinear limiter such that the regions are largely
expanded. Thus, the step size around zero values becomes
small when we use variable step size transient analysis.

0.004

0.002

0

-0.002-0.00402e-0104e-0106e-010Frequency[rad/sec]

8e-0101e-009(a)0.00140.00120.001

0.0008Future works

V[V]0.0006

1. Although we use a diode circuit for developing thenonlinear limiter, we need to find much more eﬃcient
circuit.

0.00040.00020-0.0002

2. We need to apply our method to larger plane circuit asshown in Fig. 1.

-0.000402e-0104e-0106e-010Frecuency[rad/sec]

8e-0101e-009(b)0.00014

References

0.00012[1] T.Tang and M.S.Nakhla, “Analysis of high-speed
VLSI interconnect using the asymptotic waveform
evaluation technique,” IEEE Trans. Computer-Aided
Design, Vol.14, pp.341-352, 1992 (Japanese).

0.0001V[V]8e-0056e-0054e-0052e-005[2] Y.Tanji, Y.Nishio and A.Ushida, “Analysis of nonuniform and nonlinear transmission lines via frequency
domain techniques,” IEICE Trans. on Fundamentals,
Vol.E79-A, pp.1486-1494.

0-2e-00502e-0104e-0106e-010Frecuency[rad/sec]

8e-0101e-009(c)Figure 6: Frequency characteristics of the circuit in Fig. 5.
(a) R=1[Ω]. (b) R=10[Ω]. (c) R=100[Ω].

[3] A.Ushida, Y.Yamagami and Y.Nishio, “Frequency responses of nonlinear networks using curve tracing algorithm,”Proc. of ISCAS’02, vol.I, pp.641-644, 2002.
[4] A.Ushida and M.Tanaka, Simulations of Electronic
Circuits, Corona Publishing Co., Ltd, 2002.

Table 1 Calculated peak values.
R=1[Ω]
ω[prad/sec]V[mV]
R=10[Ω]
ω[prad/sec]
V[µV]
R=100[Ω]ω[prad/sec]
V[µV]

peak 1
175.1
13.6
peak 1
175.5
1256
peak 1
175.4
136.2

peak 2503.7
10.0
peak 2504.2
989.5
peak 2504.1
104.6

peak 3771.6
5.8
peak 3772.1
572.5
peak 3772.2
57.9[5] T.Kinouchi, Y.Yamagami, Y.Nishio and A.Ushida,“Spice-oriented harmonic balance Volterra series
methods,” Proc. of NOLTA’07, pp.513-516, 2007.

peak 4946.7
0.5
peak 4947
163.7
peak 4
944.8
16.5[6] Y.Nakano, Electronic Circuit Theory, Hirokawa publishing Co. 1976 (Japanese).
[7] A.S.Sedra and K.C.Smth, Microelectronic Circuits,
Oxford Univ. Press, 2004.
[8] H.Hasegawa, et al., Numerical Computation Methods, Ohmsha publishing Co. 1998 (Japanese).

It seems that the calculated peak 4 for R = 1 and R = 10
is a little bit larger than the exact value. It is due to passover the point because the peak is too narrow and small todetect even when the nonlinear limiter is introduced. In this- 117 -