ECCTD’01 - European Conference on Circuit Theory and Design, August 28-31, 2001, Espoo, Finland

Asymptotic Equivalent Circuits of Interconnects basedon Complex Frequency Method
A.Ushida ∗ K.Urabe ∗∗ Y.Yamagami∗ and Y.Nishio∗
Abstract— Nowaday, Spice is widely used for circuit designs and simulations of ICs. Therefore, it
is a very important to develop a user friendly simulator with Spice for solving LSI circuits coupledwith interconnects, because LSI circuits are usuallyconnected with PCBs(printed circuit boards) which
are modeled with the transmission lines in high frequency domain. There have been already publishedmany papers for solving the interconnects terminated by nonlinear subnetworks. However, most of
the algorithms are not easy to develop with Spicesimulator. In this paper, we propose a new technique of replacing the interconnect by an asymptotic
equivalent circuits based on a complex frequency
method. We found from many simulation resultsthat we can get the good results even with the loworder approximations. Thus, using the equivalentcircuits, we can easily obtain the transient responsesof LSI circuits coupled with interconnects by Spicesimulator.

In this paper, we propose an asymptotic equivalent circuit technique based on a complex frequency method, which
can be applied to the transient response of LSI circuits coupled with interconnects. Firstly, we calculate the exact poles
and the residues of admittance matrix obtained from the interconnect. Secondly, the equivalent circuit is synthesizedwith the essential poles and the residues. Lastly, we getthe equivalent circuits coupled with the LSI circuits. Thus,
we can calculate the transient responses with Spice. Our
simulator can get the good results, especially for the delay,with the relatively low order approximation. In section 2, weshow how to calculate the exact poles and residues, and ourasymptotic equivalent circuit in section 3. Some interestingillustrative examples are shown in section 4.

2. Calculation of the poles and the residues ofinterconnect
Now, consider a uniformly coupled N conductor interconnect. The telegraph equation is described bydV(x, s)= −(R + sL)I(x, s) dx
dI(x, s)= −(G + sC)V(x, s) dx

1. Introduction
The analysis and design of high speed LSI chips are becoming more and more important, because PCBs connecting LSI chips may cause the signal delays and crosstalks
which sometimes happen the faulty switching operations.
In the last decade, many papers have been published on thetransient analysis of lossy interconnects [1-5]. The recursive
convolution methods using moment-matching technique [13] can be eﬃciently applied to the lossy interconnects terminated by nonlinear elements. However, one of problems inthe moment-matching method is that the poles obtained by
the applications of Maclaurin expansion and Padè approximation may happen to serious errors, when they are located
far from the origin. To overcome the problem, Nakhla etal. propose CFH (complex frequency hopping) [4] which
can calculate the exact poles, and they introduce a recursive convolution technique for Spice simulation. However,it seems that it is not so easy to develop the user friendly
simulator. Another technique based on the inverse Laplacetransformation is already programmed in Spice 3 [6], which
can be applied to the uniform interconnects. It is known thatthe simulator sometimes takes much computational time depending on the parameters. In the reference [7], they propose such a technique that the interconnects are replacedby the discrete π-type and/or T-type models, which can beapplied to the relatively short interconnect, eﬃciently.
∗ Department of Electrical and Electronic Engineering,
Tokushima University, Tokushima, 770-8506 JAPAN∗∗ Dept. Electronic Control Eng., Niihama National TechnicalCollege, Niihama, 792-0805 JAPANIII-29(1)

Thus, we haved2 V(x, s)= (R + sL)(G + sC)V(x, s) dx2
2d I(x, s)= (G + sC)(R + sL)I(x, s)
dx2

(2)

Let us introduce the transfer matrix Pv (s) [8], and get the
eigenvalues as follows:diag[λj (s)2 ] = Pv (s)−1 (R + sL)(G + sC)Pv (s)(3)

We further introduce the transfer matrix Pc (s):Pc (s) = (R + sL)−1Pv (s)Γ(s),Γ(s) = diag[λj (s)](4)

Then, the admittance matrix is described as follows:

I(0, s)I(d, s)=

Y11 (d, s)
Y21 (d, s)

Y12 (d, s)
Y22 (d, s)V(0, s)V(d, s)


(5)

where
Y11 (d, s) = Y22 (d, s) = Pc (s)diag[coth λj (s)d]Pv (s)−1Y12 (d, s) = Y21 (d, s) = −Pc (s)diag[sinh−1 λj (s)d]Pv (s)−1(6)
Observe from (6) that poles of the admittance matrix are
obtained only by solving sinh λj (s) = 0. In this case, wehave the following theorem for calculations of the poles.

Theorem 1: The locations of poles satisfying relations (6)are found by solving the following equation: nπ 2 I = 0, n = 0, 1, 2, . . .
det (R + sL)(G + sC) +d(7)
Proof: From (6), the poles satisfy the following relation:
det|Pc(s)diag[sinh−1 λj (s)d]Pv (s)−1 | = 0(8)

Diﬀerentiating (16) by s, we have∂λj (s)∂{Z(s)Y(s)} − 2λj (s)
I Uj
∂s
∂s
∂Uj
=0
(17)
∂s
Since Z(s)Y(s) is the symmetric matrix, the eigenvector isdescribed by a normalized orthogonal vector. Thus, we have+(Z(s)Y(s) − λj (s)2 I)

UTj Uj = 1 ⇒ UTj

Note that since the transfer matrices Pv (s) and Pc (s) are
nonsingular, we have

∂Uj
=0
∂s

Combining (17) with (18), we havesinh λj (s)d = 0, j = 1, 2, . . . , N(9)Z(s)Y(s) − λj (s)2 I
UTj

The poles are found by solving (9). The relation can be
rewritten as follows:
λj (s)d = jnπ, j = 1, 2, . . . , N n = 0, 1, 2, . . .

(10)ing (3). Hence, shifting the eigenvalue of (3) −(nπ/d) , thecharacteristic equation is written as follows: 2 
nπI = 0det (R + sL)(G + sC) − λ −
d

(11)

j=1


 2 nπ
λj = det (R + sL)(G + sC) +I = 0d

(12)Thus, if we can ﬁnd s satisfying (12) for each n, they arepoles of the admittance matrix given by (6).
Q.E.D.
Observe that (12) is an algebraic equation of s, so thatthey are numerically calculated for each n.
Now, let us evaluate the residues of admittance matrices (6).
Theorem 2: The residues of Y12 (d, s) and Y21 (d, s) in (6)for the pole sj is given by




P
(s)
k12,j = − Pc (s)diagv∂λj (s)
cosh(λj (s)d) ∂s
d
s=s
1




k11,j = Pc (s)diag ∂λ (s)Pv (s)
j

d
∂s
s=s



∂λj (s)
∂s



×

−

−2λj (s)Uj0∂{Z(s)Y(s)}Uj
∂s
0

−1

Z(s) = R + sL, Y(s) = G + sC

(15)

(16)

III-30(21)

Consider the special case of a single interconnect. The admittance equation is given byI(0, s)I(d, s)=1
Z0 (s)

coth λ(s)d− sinh−1 λ(s)d×V (0, s)V (d, s)− sinh−1 λ(s)dcoth λ(s)d




(22)

(R + sL)/(G + sC), λ(s) =(R + sL)(G + sC)

Thus, the poles are solution of sinh λ(s)d = 0. Namely, wehave(23.1)
s0 = −R/L
and1





n = 0, 1, 2, . . .

(23.2)nπ 2) ) + (RC + LG)2
d
2CL
(24)
Now, let us compare the poles obtained by the moment
matching method with our exact method.
sn =and Uj is the eigenvector for λj (sj ).
Proof: We have(Z(s)Y(s) − λj (s) I)Uj = 0

k1,ij
k2,ij
k3,ij
+
+
+···
s − p1s − p2s − p3Thus, we have

where

2

Yij (s) =(R + sL)(G + sC) = jnπ/d,

(14)

(20)
j

Then, using these poles and the residues, the element ofadmittance matrix are described by the partial fractions asfollows:

Z0 (s) =is obtained as follows:∂Ujλj (s)2 I ∂s  = Z(s)Y(s) −∂λj (s)
UTj
∂s

1

where

(13)
where

(19)

∂λj (s)for sj and λj (s), and the residue is∂scalculated by (13).
Q.E.D.
Corollary: The residue of Y11 (d, s) is given by

−1

j

∂Uj
 ∂s 
∂λj (s)∂sThus, we have

Therefore, if we can ﬁnd the solutions satisfying λ = 0 givenby (11), they are equal to the eigenvalues satisfying (10).
Hence, from the relation between the roots and coeﬃcientsin the characteristic equation, we have
N





∂
(Z(s)Y(s))Uj
∂s
0

=−

Note that λ2j (s) from (10) is equal to the eigenvalue satisfy2−2λj (s)Uj
0(18)−(RC + LG)±j
2CL−4CL(RG + (1 The root s = −G/C is excluded, because
lim Z0 (s) sinh λ(s)d = 0.

s→−G/C 2 {ki }s + 2 {ki pi }k0+
s + p0(s + ui )2 + vi2
2M

Table 1: Comparison between our method and the moment
matching method [2]2 . The length of line is d = 5[mm],and the circuit parameters are
R = 0.5[Ω/mm],
C = 4[pF/mm]
No.
p0
p1
p2
p3
p4
p5
p6
p7

Y12 (d, s) = Y21 (d, s)

L = 10[nH/mm]
G = 0.5[mS/mm]

Exact poles
with our method-0.500E-01-0.875E-01+ j0.314E+01-0.875E-01+ j0.628E+01-0.875E-01+ j0.942E+01-0.875E-01+ j0.125E+02-0.875E-01+ j0.157E+02-0.875E-01+ j0.188E+02-0.875E-01+ j0.219E+02

 2 × (−1)i−1 ( {ki }s + {ki pi })k0+
s + p0(s + ui )2 + vi2
2M−

Poles from the moment
matching method-0.500E-01-0.875E-01+j0.314E+01-0.875E-01+j0.628E+01-0.875E-01+j0.942E+01-0.867E-01+j0.121E+02-0.196E+01+j0.128E+02
0.178E+01+j0.120E+02-0.470E+01+j0.133E+02

For simplicity, we set here k0 = k0,11 = −k0,12 , ki = ki,11 =|ki,12 |, and 2M + 1 means the order of approximation. Set
each term of (28) as follows:b0
bi1 s + bi0, Yi (s) = 2, i = 1, 2, . . . , 2M
s + a0s + ai1 s + ai0(29)Then, the circuits given by (29) are realized by Fig.1.

Y0 (s) =

Ci
R0

Ri

L0

Li

(a)

3. Asymptotic equivalent circuit

Gi

(b)

Figure 1: Equivalent circuits of eq.(29.1) and (29.2).

Note that, for the lossy interconnects, the poles far fromthe imaginary axis in s-plane don’t give large eﬀect in thetransient analysis, and the residues usually become thesmaller as the poles leave from the origin [2]. Therefore,
the lossy interconnect can be well-modeled with relatively
few terms of (21). In this section, we consider the modelingof the admittance matrix described by the partial fractionsin the form of (21). For simplicity, we consider an uncoupled
uniform interconnect.
Put the poles as follows:
pi = −ui ± jvi ,(28.2)

i=1

Observe that the moment matching method has the serious error after no.5 pole. It seems that the error happens
from the Padè approximation, and the Maclaurin expansion
which can be applied only around the origin.

p0 = −u0 ,(28.1)

i=1

i = 1, 2, . . .

Y2
I (0,s)I (d,s)Y1

V (0,s)Y1
I (d,s)r

Near end

V (d,s)I r (0,s)

Far end

(a)

Y2
R0

(25)

Y1

R2

R 2N

R1

R3

R 2N-1

L2

L 2N

L1

L3

L 2N-1

L0
C2

where

LG + RCui =2LC

u0 = R/L,


vi =



2
1

, ki,11 =i = 1, 2, . . .
k0,11 =Ld
Z0 (s)D{λ(s)}d s=pi(26.1)

L
∂λ(s)
=
∂s
2

C1

G1 C 3

G3

C1

G2N-1

I (d,s)r

Y1

Y1

V(0,s)

C 2N-1

R1

R3

R 2N-1

R 2N-1

R3

R1

L1

L3

L 2N-1

L 2N-1L3

L1

G1 C 3G3

C 2N-1

G2N-1

G2N-1

2N-1 C

G3

3C

G1

V(d,s)
C1

(e)

(d)

Figure 2: Equivalent circuits of an interconnect.

i = 1, 2, . . .
s=pi(26.2)
where k0,22 = k0,11 , ki,22 = ki,11 , k0,21 = k0,12 , ki,21 = ki,12andD{λ(s)} =

G 2N

(c)

I r (0,s)2LC
Then, the corresponding residues are given by



C 2N

(b)

4LC(RG + ( nπ)2 ) − (LG + RC)2d(−1)i−12k0,12 = − , ki,12 =Ld
Z0 (s)D{λ(s)}d 

G2

CG + sC+R + sL2



R + sL
G + sC(27)

Since the poles and residues are composed of the complex
conjugate ones, the admittance matrices can be described asfollows:
Y11 (d, s) = Y22 (d, s)
2 We

used a double precision computer for the moment
matching.

III-31

where the parameters are given by

Li =

L0 =

1
,
b0

R0 =

a0
b0

1
,
bi1

Ri =

ai1 bi1 − bi0
b2i1(30.1)








b3i1
a0i b2i1 + (bi0 − ai1 bi1 )bi0 
b2i1 bi0Gi =2
a0i bi1 + (bi0 − ai1 bi1 )bi0
Ci =(30.2)

Observe that the admittance of (Y11 (s), Y22 (s)) and(Y12 (s), Y21 (s)) are composed of the same terms except of
their signs. Hence, put the admittance composed of thesame sign to Y1 (s), and that of the opposite sign to Y2 (s) as
follows:

 2 {k2i−1 }s + 2 {k2i−1 p2i−1 }

[v]
3.0
VS

2.0

N

Y1 (s) =1.0(31.1)

2(s + u2i−1 )2 + v2i−1

i=1

[v]
3.0

 2 {k2i }s + 2 {k2i p2i }k0+2s + p0(s + u2i )2 + v2i

VCN

Y2 (s) =0.0

2.0(31.2)

i=1

1.0
0.0

Then, we can realize the interconnect by the equivalent circuit as shown in Fig.2. Note that the current controlledcurrent sources Ir (d, s) in the ﬁgure shows the reﬂection at
the far end, and Ir (0, s) the reﬂection at the near end.

4. Illustrative examples
4.1 Interconnect terminated by linear resistors
To investigate the accuracy of our asymptotic method,
consider an interconnect terminated by 10[Ω] at near and
far ends, respectively.

0

10

20

30

40

50

[n sec] 60(c) Transient response of the fulladder coupled withinterconnects with 37th order.
Figure 4: R = 500[Ω/mm], L = 10[µ/mm], C = 0.01[pF/mm],
G = 0.5[µS/mm], d = 5[mm]In this example, we replaced the interconnects by the asymptotic equivalent circuits, and get the results with transient
analysis of Spice. We found from these results that it haslarge time delay. Our simulator is easily applied to any kindof circuits.

5. Conclusions and remarks
5.0

5.0

Near end

Near end
2.5

2.5

Far endFar end0.0

0.0

T=1.2ns

T=1.2ns
-2.5
0

2T

4T

6T

8T

10T

-2.5

0

2T

4T

6T

8T

10T

(a) Solution by the numerical inverse Laplace transformation.
(b) Solution by our asymptotic method with our 15th order.
Figure 3: Comparison our asymptotic method to the numericalinverse Laplace transformation.
L = 1[nH/mm],

R = 0.5[Ω/mm],

G = 0.5[mS/mm],

C = 4[pF/mm],

Referencesd = 5[mm]From these results, we found that our method can get thegood results with 15th order of the approximation. Furthermore, the time delay is exactly estimated even with our low
order model.

4.2 A LSI circuit coupled with interconnects
Half Adder 2T3
v1[4] E.Chiprout and M.S.Nakhla, “ Analysis of interconnect
networks using complex frequency hopping(CHF),” IEEE
Trans. on Computer Aided Design, vol.CAD-14, pp.186200,1995.
[5] R.Achar, P.K.Gunupudi, M.S.Nakhla andE.Chiprout,“Passive interconnect reduction algorithm for
distributed/measured networks,” IEEE Trans. Circuits andSystems-II, vol.CAS-47, pp.287-301, 2000.
[6] J.S.Roychowdhury, A.R.Newton and D.O.Pederson, “Algorithms for the transient simulation of lossy interconnects, IEEE Trans. on Computer Aided Design, vol.CAD13, pp.96-104, 1994
[7] T.Dhaene and D.D.Zutter, “Selection of lumped element
models for coupled lossy transmission lines,” IEEE Trans.
on Computer Aided Design, vol.CAD-11, pp.805-815, 1992.

vs
v2

T2
vc

v3

T4

T1

[1] L.T.Pillage and R.A.Rohrer, “Asymptotic waveform evaluation for timing analysis,” IEEE Trans. on Computer Aided
Design, vol.CAD-9, pp.352-355, 1990.
[2] E.Chiprout and M.S.Nakhla, Asymptotic Waveform Evaluation and Moment Matching for Interconnect Analysis,
Kluwer Academic Pub., 1994.
[3] S.Lin and E.S.Kuh,“Transient simulation of lossy interconnects based on the recursive convolution formulation,”IEEETrans. Circuits and Systems-I, vol.CAS-39,pp.879-892,
1992.

Consider a fulladder circuit coupled with interconnects.
Half Adder 1

In this paper, we have proposed an asymptotic equivalentcircuit for interconnets. Thus, replacing the interconnect
by the circuit, we can easily calculate the transient responsewith Spice. At ﬁrst step, we calculate the exact poles and theresidues of admittance matrix, and describe it in the form ofpartial fractions. Secondly, we realize the admittance matrix
by the equivalent circuit. We found that we can get goodresult even with our low order approximation.
As the future problem, we need to extend our algorithm tononuniform interconnects.

(a) Fulladder circuit consisted of nand circuits.
[v]
3.0
VS

2.0
1.0
0.0
[v]
3.0
VC

2.0
1.0
0.0
0

10

20

30

40

50

[n sec] 60(b) Transient response of the fulladder without interconnects.

III-32[8] Y.Tanji, L.Jiang and A.Ushida, “Analysis of pulse responseof multi-conductor transmission lines by a partitioning technique,” IEICE Trans. on Fundamentals, vol.E77-A,no.12,
pp.2017-2027, 1994.