of 10
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.

Design of circularly symmetric two-dimensional linear-phase low-pass FIR filters using closed-form expressions


Recruiting & HR

Publish on:

Views: 6 | Pages: 10

Extension: PDF | Download: 0

Abstract A design technique is presented for two-dimensional linear phase low-pass FIR filters based on a constrained minimization formulation. The minimization criterion is a convex combination of two criteria representing measures of the error
  Design of Circularly Symmetric Two-Dimensional Linear-PhaseLowpass FIR Filters using Closed-Form Expressions By Magdy T. Hanna Department of Electrical EngineeringUniversity of BahrainP.O. Box 32038Isa TownBahrainTel : (973) 688308Fax : (973) 686300  1 ABSTRACT A design technique is presented for two-dimensional linear phase lowpass FIR filters basedon a constrained minimization formulation. The minimization criterion is a convexcombination of two criteria representing measures of the error between the actual and idealfrequency responses in the pass and stop bands. The constraints can for example representflatness conditions at the srcin. The result is applied to the case of a circularly symmetriclowpass filter and closed-form expressions are derived for the elements of the relevant matriceswhich appear in the minimization criterion so that numerical integration can be avoided. I. INTRODUCTION Recently 2-D FIR eigenfilters were designed by minimizing a quadratic error criteriondefined by the integral of the square of the difference between the frequency response of thedesigned filter and a scaled version of the ideal frequency response [1]. The scaling factor wasintroduced in order to get a quadratic form as an essential requirement of the eigenfilterformulation. The technique proposed here is distinct from that of [1] in two respects. First, amore meaningful minimization criterion is used where no scaling of the ideal frequencyresponse is employed. Second, instead of the single quadratic normalization constraint requiredby the eigenfilter formulation of [1], a set of linear constraints will be imposed which can forexample represent flatness conditions of the frequency response at the srcin of the 2Dfrequency plane. In the case of a circularly symmetric lowpass filter, closed-form expressionswill be derived for the elements of the pertinent matrices so that numerical integration can beavoided.The frequency response of a 2-D FIR filter having a first quadrant causality and a realimpulse response which is symmetric about the center of its support can be expressed as : ( )( ) ( )( ) [ ] { } )2,1(125.02115.01exp)2,1( ω ω ω ω  ω ω  a H  N  N  j  je je H  −+−−= (1)where Ha( ω 1, ω 2) is given below for the case of odd N1and N2being treated in this paper : ( ) ( ) ( ) ∑ −= ∑ −== )11(5.0 01)12(5.0 0222cos11cos2,1)2,1(  N n N nnnnna a H  ω ω ω ω  . (2)The amplitude function Ha( ω 1, ω 2) can be compactly expressed as 1 : ( ) )2()1(2,1 ω ω ω ω  BvT ua H  = (3)where ( )( ) T  M u 111cos...) 12cos()1cos(1)1( ω ω ω ω  −= (4)and v( ω 2) is analogously defined with ω 1and M1respectively replaced by ω 2and M2where : 1 The superscript T denotes the transpose.  2 15.0 += i N i M  , i = 1 , 2 . (5)In Eq (3) the elements of the square matrix B are related to the coefficients a(n1,n2) of Eq (2)which in turn are related to the impulse response of the filter.In order to be able to express the amplitude function Ha( ω 1, ω 2) as an inner product of twovectors; one of them consisting of the filter coefficients, a vector x will be defined as aconcatenation of the rows of matrix B and a vector s( ω 1, ω 2) will be defined as the Kroneckerproduct of the vectors u( ω 1) and v( ω 2) :)2()1(2,1 ω ω ω ω  vus ⊗= (6)Therefore Eq (3) can be rewritten as : ( ) xT sa H  )2,1(2,1 ω ω ω ω  = . (7) II. The Constrained Minimization Problem The total mean squared error criterion is defined as [1] : 1 0 )1( ≤≤−+= α α α   p E s E  E  (8)where Esand Epare respectively mean squared error measures in the stop and pass domains Ω sand Ω pdefined by : ∫∫ Ω= sd d eW  s E  21)2,1(2)2,1( ω ω ω ω ω ω  (9)and an analogous expression holds for Epwith the subscript s replaced by p. In the aboveequation W  (,) ω ω  12is a positive weighting function and e( ω 1, ω 2) is the error between theactual frequency response Ha( ω 1, ω 2) and the desired one Hd( ω 1, ω 2) , i.e., e H a H d  (,)(,)(,) ω ω ω ω ω ω  121212 = −   . (10)Let Hd( ω 1, ω 2) be unity in the passband and zero in the stopband; it can be shown that the errorcriterion (8) can be evaluated as :  E xT Qx xT  pbpc = − − + − 211()() α α  (11)where Q PsPa = + − α α  ()1 , (12) ∫∫ Ω= sd d T ssW sP 21)2,1()2,1()2,1( ω ω ω ω ω ω ω ω  , (13) ∫∫ Ω=  pd d T ssW aP 21)2,1()2,1()2,1( ω ω ω ω ω ω ω ω  , (14) ∫∫ Ω=  pd d sW  b p 21)2,1()2,1( ω ω ω ω ω ω  , (15) ∫∫ Ω=  pd d W c p 21)2,1( ω ω ω ω  , (16)  3          ⊗= T vvT uuT ss . (17)Let us impose a set of constraints - for example flatness constraints at the srcin - on thefrequency response of (7). This results in the following set of linear constraints : Cx K  = (18)where C is a rectangular matrix and K is a vector.The vector x of filter coefficients appearing in (7) will be evaluated by solving theconstrained optimization problem :Minimize E of (11) subject to Cx = K .Using the Lagrange multipliers technique which has been applied to a related 1-D filterdesign problem [2], one gets the unique solution :  −−−−−−+−−= b pCQK T C CQT C Qb pQ x 1)1(1)1(11)1( α α  . (19) III. CIRCULARLY SYMMETRIC LOWPASS FILTER Let ω pand ω sbe respectively the edges of the pass and stop bands of Hd(,) ω ω 12of acircularly symmetric lowpass filter. In order to evaluate vector x of (19), one should evaluatethe matrices Psand Paand vector pbdefined by (13)-(15). Taking the weight function to beunity everywhere, using (17) and (6), and integrating only over the first quadrant because of thecircular symmetry, one gets : PsPsPs = + 12 , (20) ∫ =⊗ ∫ −==  sd T vvsd T uusP ω ω ω ω ω π ω ω ω ω ω ω  022)2()2(22211)1()1(1 , (21) ∫ ⊗ ∫ = π ω ω ω ω π ω ω ω  sd T vvd T uusP 2)2()2(01)1()1(2 , (22) ∫ =⊗ ∫ −==   pd T vv pd T uuaP ω ω ω ω ω ω ω ω ω ω ω  022)2()2(222011)1()1( , (23) ∫ =⊗ ∫ −==   pd v pd ub p ω ω ω ω ω ω ω ω ω  022)2(222011)1( . (24)Using (4), one can derive the following expressions for the inner integrals in (21), (23) and(24) and the single integrals in (22) :1M,...,1=r,  )1())1sin(( 0)( −−= ∫ r ar r ad u ω ω  (25) 11,...,=,,  )(2))sin(()2(2 ))2sin(( ,0)()( M cr cr acr cr acr cr ad T uu −−+−+−+= ∫  ω ω ω  (26)  4  ≠−−−−+−+−−−−−−= ∫ 11,...,=,,  )(2))sin(()2(2 ))2sin(( 1,...,2==  )1(4 ))1(2sin( 221==  ,)()(  M cr cr  cr acr cr acr  M cr r ar acr acr ad T uu π π π ω ω ω  (27) { } ∫ = π π π π ω ω ω  05.0,,5.0,)()( L  Diagd T uu . (28)Substituting (25) in (24), the r th element of vector  pb reduces to : ( ) ∫ −−      −−=  pd r r  pr r b p ω ω ω ω ω  02)2)12cos(()11(222)11(sin(29)where the index r is related to the indices r1and r2by :rrMrrMrM = − + = = (),...,,...,1122111212, , . (30)In order to evaluate the integral in (29), the definition of the following function isintroduced : ∫     −= λ ω ω ω λ λ  0)cos(22sin1),,( d baabaF  (31)In the appendix , it will be proved that the integral in (31) has the closed-form expression :      ++= 221222),,( ba J babaF  λ λ π λ  (32)where J1(x) is Bessel function of the first kind of order one. In the same Appendix it will beproved that (32) is also valid in the limiting cases of a = 0 or b = 0 , i.e. ∫ −≡ λ ω ω ω λ λ  0)cos(22),0,( d bbF  (33-a) F bb J b (,,)() λ π λ λ  021 = (33-b) ∫     −≡ λ ω ω λ λ  022sin1)0,,( d aaaF  (34-a) F aa J a (,,)() λ π λ λ  021 = (34-b)Defining the function G( λ ,a) as : G a F a (,)(,,) λ λ  ≡ 0 (35-a)and using (33-b) and (34-b), one gets : G a F a (,)(,,) λ λ  = 0 . (35-b)Moreover in the extreme case of a = b = 0 , (31), (33) - (35) reduce to :24022)0,()0,0,( λ π λ ω ω λ λ λ  = ∫ −≡≡ d GF  . (36)After the above digression, Eq (29) can be expressed using definition (31) as :)12,11,( −−= r r  pF r b p ω  . (37)
Similar documents
View more...
Search Related
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks