Miniaturization and control of split ring structures from an analytic solution of their resonance
11 Miniaturization and control of split ring structures from ananalytic solution of their resonance
S. T. Chui , Y. Zhang and L. Zhou Bartol Research Institute and Department of Physics and Astronomy, University ofDelaware, Newark, DE 19716, Surface Physics Laboratory (State Key Laboratory) and Physics Department, FudanUniversity, Shanghai 200433, China
AbstractWe derived simple polynomial equations to determine the entire resonance spectra ofsplit ring structures. For double stacking split rings made with flat wires, we showedthat the resonance frequency depends linearly on the ring-ring separation. In particular,we found that the wavelength of the lowest resonance mode can be made as large as the geometrical size of the ring for realistic experimental conditions, whereas forcurrent systems this ratio is of the order of 10. Finite-difference-time-domainsimulations on realistic structures verified the analytic predictions. I. Introduction
Stimulated by interests in negative reflecting materials [1-2], there has been muchinterest in split ring resonators (SRR) [3-7]. A theme in current research is thedevelopment of resonators with geometrical sizes much smaller than the wavelengthsat resonance. Miniaturization of resonators will open the door to differenttechnological applications such as the subwavelength ultra-compact dipole antennas [8]and filters [9] that have been proposed. Recently, we showed that for two rings of radii R placed on the same plane made by wires possessing a circular cross-section withradius a , the resonance frequency exhibits a lower bound [ ] ln( / ) R a − ∝ [7,10]. Toproduce a significant change to the slowly changing log function requires a big reduction of a, which increases the resistance of the wire and hence the damping of thestructure.In this paper we show how to reduce the resonance frequency of structures madewith rings. Our insight is obtained through our recent efforts to understand theresonance properties of metallic ring systems more rigorously and analytically [7]. Incontrast, most previous calculations on metallic rings are numerical in nature. In thispaper, we show further that the entire spectra of the SRR systems (both single- anddouble-ring) can be determined by simple polynomial equations (Eqs. 4 and 11). Whenthe rings are made with flat wires of a film-like rectangular cross-section and placed ondifferent planes [5], their separation d needs only be larger than the thickness t of therings which can be much smaller than width 2 a of the wires that make up the ring. Theresonance frequency is controlled by the difference between the inverses of the self andthe mutual capacitances, which can be very small. For example, it is experimentallypossible to make bilayer systems separated by a distance of less than 10 nm. For ringsof radii 10 mm, the resonance wavelength can be made the radius of the ring.This huge ratio is previously inaccessible (For current systems, this ratio is 10).Such a double-stacking SRR was first proposed in Ref. [5] and studied by anempirical method [5], but the attention mainly focused on reducing the bianisotropy ofthe SRR systems. Here, we employed a more rigorous theory [7] to study the samesystem, our attention mainly focused on how to lower the resonance frequency withsuch a system. The extreme limit when the ring radii are much larger than theirseparation involves the cancellation of two quantities close in values. This requires careful control of the numerical accuracies of the calculation complemented byanalytic considerations, which is feasible in the present approach. We now describe ourresults in detail. II. Single Ring
We first recapitulate our results for the single ring case [7]. Take the coordinatesystem so that the z axis is perpendicular to the ring and assume a time dependence i t e ω . Driven by an external field, the current along the ring is determinedby ( ') /( ) ( ) m m m m extm m m I iI L m C E m ρ ω ω (cid:1) (cid:2) − + − = (cid:3) (cid:4) (cid:1) , where ( ), , ( ) m ext m I E m ρ arethe Fourier components of the resistivity function ( ) ρφ of the ring, the current ( ) I φ flowing on the ring, and the external field ( ) ext E φ on the ring [7]. Here, , m m L C are theself inductance and capacitance [7], and φ is the azimuthal angle. The resonancefrequencies are obtained by solving the above circuit equation setting ( ) 0 ext E φ = [7].For a single-ring SRR with a gap at φ= , the resistivity function is ,0 ( ) m c m r r ρ δ≈ + in the limit of gap width approaching zero; here r and c r are theresistances per unit length of the insulating gap and the conductor [7]. We take thelimit of r → ∞ in the end. When the gap is located at φ π= , weget ,0 ( ) ( 1) m m c m r r ρ δ≈ − + . In matrix form, the circuit equation to determine theresonance frequencies is HI = 0 , (1)with H = H + X , (2)where r = H M , M is a matrix with all elements equal to 1, ie. , i j = M . X is adiagonal matrix with [ /( )] m c m m X r i L m C ω ω= + − . For our problem, m m
X X − = .Because of the m , - m symmetry of X , there are two classes of solutions,corresponding to even and odd symmetries under the transformation from m to - m .Those with odd symmetries (i.e., m m I I − = − ) are not coupled to each other and theresonance frequencies are given by the conditions that 0 m X = . In the limit of / 0, a R → the resonance frequencies (obtained by solving 0 m X = with 0 c r = ) are / , 1, 2,... m m m u m L C m m ω ω= → = (3)with / u c R ω = ( c is the speed of light) being the frequency unit of the present problem[7]. Equation (3) accurately accounts for all the even-numbered resonance modes of asingle-ring SRR obtained numerically (see Fig. 2 of [7]).We next consider the solutions with even symmetry (i.e., m m I I − = ). Now0 = (0)0 H I so long as (0) mm I = (cid:1) . This current distribution is such that its magnitudeis zero at the gap. (0) I is in general not a solution of the circuit equation because theinternal emf inside the ring is not zero: int ( ) exp( ) 0 mm E E im φ φ= ≠ (cid:1) , where (0) m m m
E X I = . The only way the circuit equation can be satisfied is if int ( ) ( ) E φ δ φ∝ ; theinternal emf is zero inside the ring except at the gap where it is counterbalanced by thegap resistance. For this to be true, it is necessary that all m E ’s be the same. We thuswrite (0) ' m m m E X I E = = and (0) '/ m m I E X = . Since (0) mm I = (cid:1) , we arrive at theeigenvalue equation 1/ 0 mm X = (cid:1) . Substituting in the expression for m X , we get thefollowing equation for the entire spectrum of the even modes:
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2( )1 0/ cm m m c
L irL m C ir ω ωω ω ∞= −+ =− − (cid:1) . (4)For the low lying modes, contributions from terms with large m are not importantbecause of the m coefficient multiplying1/ m C . The eigenmode only involves a fewFourier components, which agrees well with the eigenvector calculated numerically(See Fig. 3 of [7]).We now describe a formal approach to solve the circuit equation, which can beeasily extended to more complicated situations. This approach explicitly displays theeigenvectors, which can be used to calculate the responses of the structures to externalfields. Because r is very big, we solve the matrix problem (1) by a standardperturbation method. For the unperturbed matrix problem, (0)0 H I = 0 , any vector (0) I satisfying (0) mm I = (cid:1) (6)is a solution. Now consider the full matrix problem Eq. (1). Assume the solution to Eq.(1) can be written as (0) (1) I = I + I , where (1) I is of the order of r − (0) I . Put I into Eq. (1), since (0)0 H I = 0 , we get ( ) ( ) ( ) 0 r r o r − + = (0) (1) (1) (0) HI = M + X I + I = MI + XI .Substituting in the elements of X and M, we get (0) '/ m m I E X = where (1)'' ' mm E r I = − (cid:1) is a constant independent of m . Employing the constrain (6), wearrive at the equation
1/ 0 mm X = (cid:1) , which again leads to Eq. (4). Choosing anappropriate normalization constant, in the limit
1/ 0 r → , we find the eigenvector atresonance to be [ ] ..., / , / ,1, / , / ,... T X X X X X X X X = = (0)
I I . (7)We emphasize that Eqs. (4) and (7) are the exact solutions of the matrix problem (1),since the perturbation theory becomes exact in the limit of
1/ 0 r → . III. Double Ring
Realizing the difficulties of miniaturizing the resonance structure of a single-ringSRR, we now consider the double-ring system, in which one ring has a gap at φ = andthe other one, φ π = . In what follows, we develop analytical formulas suitable for thefollowing two cases that were widely studied in literature: (1) the two rings are ofslightly different size and located on the same plane [3-6,10]; (2) the two rings are ofthe same size and located on different planes separated by a distance d with centersboth on the z axis [5]. For both cases, the circuit equation to determine the resonancefrequencies is given by ( ) = HI = H + Y I . The unperturbed and the perturbationmatrices are r r (cid:1) (cid:2)(cid:3) (cid:4)(cid:5) (cid:6) a0 b
M 0H = 0 M , (cid:1) (cid:2)(cid:3) (cid:4)(cid:5) (cid:6) X X'Y = X' X (8)where the matrix a M (with elements , ' m m = a M ) is for ring 1 and the matrix b M ( with elements ', ' ( 1) m mm m − = − b M ) is for ring 2 . X, X' are diagonal matriceswith m X same as before and ' ' 2 ' [ /( )] m m m X i L m C ω ω= − , in which ' m L and ' m C are themutual inductances and capacitances between two rings. Here, we have neglected thesize differences of the two rings in writing the self interaction terms. Again, there areeven and odd modes under the transformation m to – m . Here we only focus on the evenmodes. Following the perturbation theory, we first consider the unperturbed matrix problem H w = 0 . The solutions can be written as (cid:1) (cid:2)(cid:3) (cid:4)(cid:5) (cid:6) ab ww = w which satisfy0, ( 1) 0 mm mm m = − = (cid:1) (cid:1) a b w w . (9)Now assume the full solution as I = w + ε where ε is of the order of r , we get ( ) 0 o r − + = HI = Yw + H ε . Putting Eqs. (8) into the above matrix and doing thealgebra, we find ' 2 '2' 2 '2 ( 1) /( 1) / a m bm m m m mm b am m m m m X C X C X XX C X C X X (cid:1) (cid:2) (cid:3) (cid:2) (cid:3) = − − − (cid:4) (cid:5) (cid:6) (cid:5) (cid:6)(cid:7) (cid:2) (cid:3) (cid:2) (cid:3) = − − − (cid:4) (cid:5) (cid:6) (cid:5) (cid:6)(cid:8) ab ww (10)where '' '' ' , ( 1) a b mm mm m C r C r = − = − − (cid:1) (cid:1) a b ε ε are two constants independent ofindex m . Applying the constrains (9) to (10), we finally arrive at the equation ' mm m m X X =− (cid:1) (cid:1) to determine the resonance frequencies of the double-ringsystem. We note that this equation recovers the single-ring results (i.e., Eq. (4)) in theabsence of mutual-interaction terms (i.e., ' m X = ). In explicit form, this equationbecomes cmm c m m m m L L iri r L m C L m C ω ωω ω ω ∞= ± − = (cid:1) (cid:2) − + ± − − + (cid:3) (cid:4) (cid:1) (11)Again, we emphasize that Eqs. (11) is the exact solution of the double-ring problem(recalling r → ∞ ).Under the three mode approximation [retaining only the m = term in Eq. (11)], weobtain, for 0 c r = , the resonance condition for the lowest (magnetic) mode (plus sign), )''22/()'/1/1( LLLLCC +−+−=ω . (12)The frequency squared is reduced because it is proportional to the difference betweenthe inverse self capacitance C and the inverse mutual capacitance '1 C . Physicallythis comes about because the charge distributions on the two rings are proportional to sin φ and sin φ − and nearly cancel each other. For rings made with wires of circularcross sections, such a difference is proportional to ( ) ln / d a [10], and the resonancefrequency is reduced when two rings approach each other ( d decreases). Unfortunately,since the wires are of circular cross sections, the minimum value for d is 2 a , so that the resonance frequency never approaches zero but possesses a natural lower bound [10].This problem can be remedied if the rings are made with flat wires, and are placed ondifferent planes [see inset to Fig. 1(a)]. We describe this next. IV. Non coplanar rings
For rings made with flat wires not on the same plane, the separation d betweenrings need only be larger than the thickness t of each ring. In general t can be muchsmaller than width 2 a of the flat wires, so that the difference of capacitances, and inturn, the resonance frequency, can be much smaller.Our conclusion remains valid when higher modes are included. To illustrate, whenthe m = ± modes are included, the resonance frequency is given by, for 0 c r = , '2 ' ' '0 1 1 0 0 1 1 (1 )(1/ 1/ ) / 2( ) (1 )( ) C C L L L L ω δ δ (cid:1) (cid:2) = − − + + − − (cid:3) (cid:4) where )'/1/1/()'(2
CCLL ++= ωδ . Thus as '1 1
1/ 1/
C C − is made small, the resonancefrequency becomes small as well.We now demonstrate the above argument by numerically evaluating the resonancefrequencies of double stacking rings made with flat wires. As shown in the inset to Fig.1(a), the planes of the rings are defined by an angle ( ) tan / 2 d R β − = . Extending ourprevious theory for wires with circular cross sections [7,10] to the present case withrectangular cross sections [11], we are able to calculate all the circuit parameters ' ' , , , m m m m L L C C in terms of the geometrical parameters d , R , a , and t . For example, theself- and mutual- capacitance parameters are found to be (cid:1) (cid:1)(cid:2) +− +− +><∞= +−= aR aR aR aR llmlmlm ddPmlmlamC /'/|)0(|)!/()!()8/(/1 ρρρρρε ,
1/ ' /(8 ) ( )!/( )! ( sin ) (sin ) / ' /
R a R am m l lm l ll m R a R a
C m a l m l m P P d d r r ε β β ρ ρ ρ + +∞ +< >= − − = − + − (cid:1) (cid:2) (cid:2) , where ( / 2) , ' ' ( / 2) r d r d ρ ρ = + = + , ( ) r r > < and ( ) ρ ρ > < take the larger(smaller) values of , ' r r and , ' ρ ρ , respectively. Similar expressions are foundfor ' , m m L L . It is important to note that parameter t does not enter the expressions of thecircuit parameters in the thin-wire limit ( t → ) [11].We have numerically evaluated these circuit parameters, and found that in the limitof / 0 a R → , both m L and 1/ m C still exhibit the ln( / ) R a dependences, similar to the circular cross-section case [7,10]. Thus the analytical results obtained for circularcross-section case, e.g., Eqs. (3,5), remain valid for the single-ring SRR made with flatwires. We next consider the double-ring case.When both a and R are fixed, the normalized inverse-capacitancedifferences, ' m m C C − , approach zero as d → , as shown in Fig. 1(a) calculated for / 0.025 a R = . As d → , sin 0, , ' ' r r β ρ ρ → → → ; thus '
1/ 1/ m m
C C → . As a result,the lowest resonance frequency ( ω ) of the double-ring SRR is significantly reducedas d → , as shown in Fig. 1 (b) by the solid line, obtained by numerically solving Eq.(11) setting / 0.025 a R = with the circuit parameters described above and similarexpressions for m L and ' m L [11].We performed finite-difference-time-domain (FDTD) simulations [12] on a seriesof realistic double-ring SRR’s made by flat wires with different separation d . TheFDTD calculated ω are shown in Fig. 1(b) as solid stars, which agree quite well withthe analytic solutions. The small discrepancies between the FDTD and the analyticresults can be attributed to the approximations adopted in calculating the circuitparameters [11].However, FDTD simulations are difficult to perform for the very small d cases,since in such cases the basic mesh discretizing the structure becomes too fine.Fortunately, analytic formulas are available when d → . Expanding, r r < > and (sin ) ml P β as a power series in / d R and keeping the lowest order terms, wefind that ( ) C C F d R − ≈ ⋅ , where F is a dimensionless coefficient depending onlyweakly on / a R . Considering the 3-mode expression as shown in Eq. (12), we furtherfind that / ( / ) u F d R ω ω ≈ ⋅ (cid:1) with F (cid:1) being another dimensionless parameter. Shown inFig. 2 are '1 1 C C − and / u ω ω as functions of / d R , calculated with the full theorysetting / 0.025 a R = . These numerical results accurately confirmed the above twoformulas, and suggested that F = (cid:1) and F = for the present structure.This analytical formula enables us to estimate the lowest possible value of / u ω ω in practical situations. Experimentally, it is possible to make bilayer systems separatedby a distance of d of the order of 10nm. For rings of radius R = mm and a = mm, we find from the formula that u ω ω − ≈ × , indicating that the longest resonance wavelength can be made of the order of the radius of the ring if c r canbe ignored. This huge ratio is previously inaccessible. Experimentally bothsuperconducting [13] and ordinary SRR’s have been studied. For nonsuperconductingrings, a detailed examination of Eq. (11) under the three-mode-approximation showsthat our estimate is valid (i.e., ω taking a non-zero real part) onlywhen ( ) (4 / 3) / 1 / c r L C C C < − . This implies that the total resistance of the ring,2 c r R π⋅ , has to be approximately less than ( ) Z g C C − , where ( / ) 377 Z µ ε = = Ω is the vacuum impedance, and g L C RZ π= is adimensionless constant, depending only weakly on / a R (for / 0.025 a R = , we get g = ). V. Conclusions
In short, through consistently analytical, numerical and brute-force simulationstudies, we have demonstrated how to create resonance structures whose sizes aremuch smaller than the resonance wavelength. The calculation described in this papercan be generalized to other structures, say, a four-ring system, which can further lowerthe resonance frequency, and can be extended to calculate analytically the responses ofresonant structures to external field. STC is partly supported by a grant from the DOE.LZ thanks supports from the National Basic Research Program of China, the NSFC,Fok Ying Tung Education Foundation, and PCSIRT.
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168 (2005).[7] L. Zhou and S. T. Chui, Phys. Rev. B /( ) mC m m E m I i C ω= − , slightly different from the originaldefinition (Eq. (8)) in Ref. 7.[8] R. W. Ziolkowski and A. D. Kipple, IEEE Trans. Antennas Propag. 51, 2626(2003).[9] Joan Garcia-garcia et al, IEEE Transactions on Microwave Theory and Techniques,53, 1997 (2005).[10] L. Zhou and S. T. Chui, Appl. Phys. Lett. E r (cid:1) (cid:1) and the currentdensity ( ) j r (cid:1) (cid:1) is generally non-local inside the cross section area, making it difficult todefine the inductance/capacitance parameters as in Refs. [7,10]. To overcome thesedifficulties, we have assumed that the current distributes uniformly over the rectanglecross section area [shaded areas in the inset to Fig. 2(a)], and averaged both ( )
E r (cid:1) (cid:1) and( ) j r (cid:1) (cid:1) over the cross section area to get the averaged field ( ) E φ and the totalcurrent ( ) I φ . By doing so, we can define and calculate the inductance/capacitanceparameters of such structures unambiguously.[12] Simulations were performed by using the package CONCERTO 4.0, Vector FieldsLimited, England, 2005. In our simulations, the geometrical parameters were taken as a = mm, R = mm, t = mm, and gap width = degrees. We tookperfect metal boundary conditions on the metal surfaces, and discretized the wholestructure with a basic mesh sized d × × , and adopted finer mesheswhenever necessary. Refer to Refs. [7,10] for more simulation details.[13] M. Ricci, N. Orloff and S. Anlage, Appl. Phys. Lett., , 034102 (2005). Figures